Download Algebra - University at Albany

Algebra
Mark Steinberger
The University at Albany
State University of New York
August 31, 2006
Preface
The intent of this book is to introduce the student to algebra from a point of view that
stresses examples and classification. Thus, whenever possible, the main theorems are
treated as tools that may be used to construct and analyze specific types of groups,
rings, fields, modules, etc. Sample constructions and classifications are given in both
text and exercises.
It is also important to note that many beginning graduate students have not taken
a deep, senior-level undergraduate algebra course. For this reason, I have not assumed
a great deal of sophistication on the part of the reader in the introductory portions of
the book. Indeed, it is hoped that the reader may acquire a sophistication in algebraic
techniques by reading the presentations and working the exercises. In this spirit, I have
attempted to avoid any semblance of proofs by intimidation. The intent is that the
exercises should provide sufficient opportunity for the student to rise to mathematical
challenges.
The first chapter gives a summary of the basic set theory we shall use throughout
the text. Other prerequisites for the main body of the text are trigonometry and the
differential calculus of one variable. We also presume that the student has seen matrix
multiplication at least once, though the requisite definitions are provided.
Chapter 2 introduces groups and homomorphisms, and gives some examples that will
be used as illustrations throughout the material on group theory.
Chapter 3 develops symmetric groups and the theory of G-sets, giving some useful
counting arguments for classifying low order groups. We emphasize actions of one group
on another via automorphisms, with conjugation as the initial example.
Chapter 4 studies consequences of normality, beginning with the Noether Isomorphism Theorems. We study simple groups and composition series, and then classify
all finite abelian groups via the Fundamental Theorem of Finite Abelian Groups. The
automorphism group of a cyclic group is calculated. Then, semidirect products are introduced, using the calculations of automorphism groups to construct numerous examples
that will be essential in classifying low order groups. We then study extensions of groups,
developing some useful classification tools.
In Chapter 5, we develop some additional tools, such as the Sylow Theorems, and
apply the theory we’ve developed to the classification of groups of many of the orders
≤ 63, in both text and exercises. The methods developed are sufficient to classify the
rest of the orders in that range. We also study solvable and nilpotent groups.
Chapter 6 is an introduction to basic category theoretic notions, with examples drawn
from the earlier material. The concepts developed here are useful in understanding rings
and modules, though they are used sparingly in the text on that material. Pushouts of
groups are constructed and the technique of generators and relations is given.
Chapter 7 provides a general introduction to the theory of rings and modules. Exam p , the cyclotomic
ples are introduced, including the quaternions, H, the p-adic integers, Z
ix
Preface
x
integers and rational numbers, Z[ζn ] and Q[ζn ], polynomials, group rings, and more. Free
modules and chain conditions are studied, and the elementary theory of vector spaces
and matrices is developed. The chapter closes with the study of rings and modules of
fractions, which we shall apply to the study of P.I.D.s and fields.
Chapter 8 develops the theory of P.I.D.s and covers applications to field theory. The
exercises treat prime factorization in some particular Euclidean number rings. The basic
theory of algebraic and transcendental field extensions is given, including many of the
basic tools to be used in Galois theory. The cyclotomic polynomials over Q are calculated,
and the chapter closes with a presentation of the Fundamental Theorem of Finitely
Generated Modules over a P.I.D.
Chapter 9 gives some of the basic tools of ring and module theory: Nakayama’s
Lemma, primary decomposition, tensor products, extension of rings, projectives, and
the exactness properties of tensor and hom. In Section 9.3, Hilbert’s Nullstellensatz is
proven, using many of the tools so far developed, and is used to define algebraic varieties.
In Section 9.5, extension of rings is used to extend some of the standard results about
injections and surjections between finite dimensional vector spaces to the study of maps
between finitely generated free modules over more general rings. Also, algebraic K-theory
is introduced with the study of K0 .
Chapter 10 gives a formal development of linear algebra, culminating in the classification of matrices via canonical forms. This is followed by some more general material
about general linear groups, followed by an introduction to K1 .
Chapter 11 is devoted to Galois theory. The usual applications are given, e.g. the
Primitive Element Theorem, the Fundamental Theorem of Algebra, and the classification
of finite fields. Sample calculations of Galois groups are given in text and exercises,
particularly for splitting fields of polynomials of the form X n − a. The insolvability of
polynomials of degree ≥ 5 is treated.
Chapter 12 gives the theory of hereditary and semisimple rings with an emphasis on
Dedekind domains and group algebras. Stable and unstable classifications of projective
modules are given for Dedekind domains and semisimple rings.
The major dependencies between chapters are as follows, where A → B means that
B depends significantly on material in A.
1−4
/ 7
5 ?
8
??
??
??
??
? 6
11
/ 9
/ 10
/ 12
Additionally there are some minor dependencies. For instance, Chapter 9 assumes an
understanding of the basic definitions of categories and functors, as given in the first two
sections in Chapter 6. Some categorical notions would also be useful in Chapter 7, but
are optional there. Also, the last section in Chapter 5 assumes an understanding of some
of the material on linear algebra from Chapter 10. Some material on linear algebra is also
used in the closing section of Chapter 11. And optional references are made in Chapter
10 to the material on exterior algebras from Chapter 9. Other than the references to
linear algebra in Chapter 5, the chapters may be read in order.
Preface
xi
A solid box symbol ( ) is used to indicate either the end of a proof or that the proof
is left to the reader.
Those exercises that are mentioned explicitly in the body of the text are labeled
with a dagger symbol (†) in the margin. This does not necessarily indicate that the
reference is essential to the main flow of discussion. Similarly, a double dagger symbol
(‡) indicates that the exercise is explicitly mentioned in another exercise. Of course,
many more exercises will be used implicitly in other exercises.
Acknowledgments
Let me first acknowledge a debt to Dock Rim, who first introduced me to much of this
material. His love for the subject was obvious and catching. I also owe thanks to Hara
Charalambous, SUNY at Albany, and David Webb, Dartmouth College, who have used
preliminary versions of the book in the classroom and made some valuable suggestions.
I am also indebted to Bill Hammond, SUNY at Albany, for his insights. I’ve used this
material in the classroom myself, and would like to thank the students for many useful
suggestions regarding the style of presentation and for their assistance in catching typos.
Mark Steinberger
Ballston Lake, NY
Contents
1 A Little Set Theory
1.1 Properties of Functions . . . . . . .
1.2 Factorizations of Functions . . . . .
1.3 Relations . . . . . . . . . . . . . . .
1.4 Equivalence Relations . . . . . . . .
1.5 Generating an Equivalence Relation
1.6 Cartesian Products . . . . . . . . . .
1.7 Formalities about Functions . . . . .
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1
1
3
6
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11
13
2 Groups: Basic Definitions and Examples
2.1 Groups and Monoids . . . . . . . . . . . .
2.2 Subgroups . . . . . . . . . . . . . . . . . .
2.3 The Subgroups of the Integers . . . . . . .
2.4 Finite Cyclic Groups: Modular Arithmetic
2.5 Homomorphisms and Isomorphisms . . . .
2.6 The Classification Problem . . . . . . . .
2.7 The Group of Rotations of the Plane . . .
2.8 The Dihedral Groups . . . . . . . . . . . .
2.9 Quaternions . . . . . . . . . . . . . . . . .
2.10 Direct Products . . . . . . . . . . . . . . .
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15
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19
24
27
29
35
37
38
40
42
3 G-sets and Counting
3.1 Symmetric Groups: Cayley’s Theorem .
3.2 Cosets and Index: Lagrange’s Theorem
3.3 G-sets and Orbits . . . . . . . . . . . . .
3.4 Supports of Permutations . . . . . . . .
3.5 Cycle Structure . . . . . . . . . . . . . .
3.6 Conjugation and Other Automorphisms
3.7 Conjugating Subgroups: Normality . . .
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46
47
51
55
64
66
71
77
4 Normality and Factor Groups
4.1 The Noether Isomorphism Theorems . . . .
4.2 Simple Groups . . . . . . . . . . . . . . . .
4.3 The Jordan–Hölder Theorem . . . . . . . .
4.4 Abelian Groups: the Fundamental Theorem
4.5 The Automorphisms of a Cyclic Group . . .
4.6 Semidirect Products . . . . . . . . . . . . .
4.7 Extensions . . . . . . . . . . . . . . . . . . .
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82
. 83
. 90
. 96
. 98
. 105
. 111
. 119
xii
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CONTENTS
xiii
5 Sylow Theory, Solvability, and Classification
5.1 Cauchy’s Theorem . . . . . . . . . . . . . . .
5.2 p-Groups . . . . . . . . . . . . . . . . . . . .
5.3 Sylow Subgroups . . . . . . . . . . . . . . . .
5.4 Commutator Subgroups and Abelianization .
5.5 Solvable Groups . . . . . . . . . . . . . . . .
5.6 Hall’s Theorem . . . . . . . . . . . . . . . . .
5.7 Nilpotent Groups . . . . . . . . . . . . . . . .
5.8 Matrix Groups . . . . . . . . . . . . . . . . .
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134
136
137
141
149
150
153
156
158
6 Categories in Group Theory
6.1 Categories . . . . . . . . . . . . . . . . .
6.2 Functors . . . . . . . . . . . . . . . . . .
6.3 Universal Mapping Properties: Products
6.4 Pushouts and Pullbacks . . . . . . . . .
6.5 Infinite Products and Coproducts . . . .
6.6 Free Functors . . . . . . . . . . . . . . .
6.7 Generators and Relations . . . . . . . .
6.8 Direct and Inverse Limits . . . . . . . .
6.9 Natural Transformations and Adjoints .
6.10 General Limits and Colimits . . . . . . .
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and Coproducts
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163
164
167
171
176
183
186
189
191
195
198
7 Rings and Modules
7.1 Rings . . . . . . . . . . . . . .
7.2 Ideals . . . . . . . . . . . . . .
7.3 Polynomials . . . . . . . . . . .
7.4 Symmetry of Polynomials . . .
7.5 Group Rings and Monoid Rings
7.6 Ideals in Commutative Rings .
7.7 Modules . . . . . . . . . . . . .
7.8 Chain Conditions . . . . . . . .
7.9 Vector Spaces . . . . . . . . . .
7.10 Matrices and Transformations .
7.11 Rings of Fractions . . . . . . .
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201
202
215
222
234
239
245
250
268
273
277
284
8 P.I.D.s and Field Extensions
8.1 Euclidean Rings, P.I.D.s, and U.F.D.s
8.2 Algebraic Extensions . . . . . . . . . .
8.3 Transcendence Degree . . . . . . . . .
8.4 Algebraic Closures . . . . . . . . . . .
8.5 Criteria for Irreducibility . . . . . . . .
8.6 The Frobenius . . . . . . . . . . . . .
8.7 Repeated Roots . . . . . . . . . . . . .
8.8 Cyclotomic Polynomials . . . . . . . .
8.9 Modules over P.I.D.s . . . . . . . . . .
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295
296
308
314
317
320
324
325
327
332
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CONTENTS
xiv
9 Radicals, Tensor Products, and Exactness
9.1 Radicals . . . . . . . . . . . . . . . . . . . . .
9.2 Primary Decomposition . . . . . . . . . . . .
9.3 The Nullstellensatz and the Prime Spectrum
9.4 Tensor Products . . . . . . . . . . . . . . . .
9.5 Tensor Products and Exactness . . . . . . . .
9.6 Tensor Products of Algebras . . . . . . . . . .
9.7 The Hom Functors . . . . . . . . . . . . . . .
9.8 Projective Modules . . . . . . . . . . . . . . .
9.9 The Grothendieck Construction: K0 . . . . .
9.10 Tensor Algebras and Their Relatives . . . . .
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341
342
346
351
365
377
383
386
392
398
406
10 Linear algebra
10.1 Traces . . . . . . . . . . . . .
10.2 Multilinear alternating forms
10.3 Properties of determinants . .
10.4 The characteristic polynomial
10.5 Eigenvalues and eigenvectors
10.6 The classification of matrices
10.7 Jordan canonical form . . . .
10.8 Generators for matrix groups
10.9 K1 . . . . . . . . . . . . . . .
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416
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447
11 Galois Theory
11.1 Embeddings of Fields . . . .
11.2 Normal Extensions . . . . . .
11.3 Finite Fields . . . . . . . . .
11.4 Separable Extensions . . . . .
11.5 Galois Theory . . . . . . . . .
11.6 The Fundamental Theorem of
11.7 Cyclotomic Extensions . . . .
11.8 n-th Roots . . . . . . . . . .
11.9 Cyclic Extensions . . . . . . .
11.10Kummer Theory . . . . . . .
11.11Solvable Extensions . . . . . .
11.12The General Equation . . . .
11.13Normal Bases . . . . . . . . .
11.14Norms and Traces . . . . . .
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Algebra
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450
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12 Hereditary and Semisimple Rings
12.1 Maschke’s Theorem and Projectives
12.2 Semisimple Rings . . . . . . . . . . .
12.3 Jacobson Semisimplicity . . . . . . .
12.4 Homological Dimension . . . . . . .
12.5 Hereditary Rings . . . . . . . . . . .
12.6 Dedekind Domains . . . . . . . . . .
12.7 Integral Dependence . . . . . . . . .
Bibliography
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550
Chapter 1
A Little Set Theory
Here, we discuss the basic properties of functions, relations and cartesian products.
The first section discusses injective and surjective functions and their behavior under
composition. This is followed in Section 1.2 by a discussion of commutative diagrams
and factorization. Here, duality is introduced, and injective and surjective maps are
characterized in terms of their properties with respect to factorization. In the latter
case, the role of the Axiom of Choice is discussed. Some introductory remarks about the
axiom are given, to be continued in Chapter 6, where infinite products are introduced.
In the next three sections, we discuss relations, including equivalence relations and
partial orderings. Particular attention is given to equivalence relations. For an equivalence relation ∼, the set of equivalence classes, X/∼, is studied, along with the canonical
map π : X → X/∼. In Section 1.5, we analyze the process by which an arbitrary relation may be used to generate an equivalence relation. We shall make use of this in
constructing free products of groups.
We then describe the basic properties of cartesian products, showing that they satisfy the universal mapping property for the categorical product. We close with some
applications to the theory of functions and to the basic structure of the category of sets.
With the exception of the discussion of the Axiom of Choice, we shall not delve into
axiomatics here. We shall assume a familiarity with basic naive set theory, and shall dig
deeper only occasionally, to elucidate particular points.
1.1
Properties of Functions
We use the notation f : X → Y to denote that f is a function from X to Y , and say that
X is its domain, and Y its codomain. We shall assume an intuitive understanding of
functions at this point in the discussion, deferring a more formal discussion to Section 1.7.
We shall use the word “map” as a synonym for function.
Definitions 1.1.1. Let X be a set. The identity map 1X : X → X is the function
defined by 1X (x) = x for all x ∈ X.
If Y is a subset of X, then the inclusion map i : Y → X is obtained by setting
i(y) = y for all y ∈ Y .
We give names for some important properties of functions:
Definitions 1.1.2. A function f : X → Y is one-to-one, or injective, if f (x) = f (x ) for
x = x . An injective function is called an injection.
1
CHAPTER 1. A LITTLE SET THEORY
2
A function f : X → Y is onto, or surjective, if for each y ∈ Y , there is at least one
x ∈ X with f (x) = y. A surjective function is called a surjection.1
A function is bijective if it is both one-to-one and onto. Such a function may also be
called a one-to-one correspondence, or a bijection.
Composition of functions is useful in studying the relationships between sets. The
reader should verify the following properties of composition.
Lemma 1.1.3. Suppose given functions f : X → Y , g : Y → Z, and h : Z → W . Then
1. (h ◦ g) ◦ f = h ◦ (g ◦ f ).
2. f ◦ 1X = f .
3. 1Y ◦ f = f .
Here are some more interesting facts to verify regarding composition.
Lemma 1.1.4. Suppose given functions f : X → Y and g : Y → Z. Then the following
relationships hold.
1. If f and g are injective, so is g ◦ f .
2. If f and g are surjective, so is g ◦ f .
3. If g ◦ f is injective, so is f .
4. If g ◦ f is surjective, so is g.
Two sets which are in one-to-one correspondence may be thought of as being identical
as far as set theory is concerned. One way of seeing this is in terms of inverse functions.
Definition 1.1.5. Let f : X → Y . We say that a function g : Y → X is an inverse
function for f if the composite f ◦ g is the identity map of Y and the composite g ◦ f is
the identity map of X.
Proposition 1.1.6. A function f : X → Y has an inverse if and only if it is bijective.
If f does have an inverse, the inverse is unique, and is defined by setting f −1 (y) to be
the unique element x ∈ X such that f (x) = y.
Proof If f is bijective, then for each y ∈ Y , there is a unique x ∈ X such that f (x) = y.
Thus, there is a function f −1 : Y → X defined by setting f −1 (y) to be equal to this
unique x. The reader may now easily verify that f −1 is an inverse function for f .
Conversely, suppose that f has an inverse function g. Note that identity maps are
always bijections. Thus, f ◦g = 1Y is surjective, and hence f is surjective by Lemma 1.1.4.
Also, g ◦ f = 1X is injective, so that f is injective as well.
Finally, since f (g(y)) = y for all y, we see that g(y) is indeed the unique x ∈ X such
that f (x) = y.
Definition 1.1.7. Let f : X → Y be a function. Then the image of f , written im f , is
the following subset of Y : im f = {y ∈ Y | y = f (x) for some x ∈ X}.
1 “One-to-one” and “onto” are the terms native to the development of mathematical language in
English, while “injection” and “surjection” come from French mathematics.
CHAPTER 1. A LITTLE SET THEORY
3
Thus, the image of f is the set that classically would be called the range of f .
Since the values f (x) of the function f : X → Y all lie in im f , f defines a function,
which we shall call by the same name, f : X → im f . Notice that by the definition of
the image, f : X → im f is surjective. Notice also that f factors as the composite
f
X
/ im f ⊂ Y,
where the second map is the inclusion of im f in Y . Since inclusion maps are always
injections, the next lemma is immediate.
Lemma 1.1.8. Every function may be written as a composite g ◦ f , where g is injective
and f is surjective.
Exercises 1.1.9.
1. Give an example of a pair of functions f : X → Y and g : Y → X, where X and Y
are finite sets, such that g ◦ f = 1X , but f is not surjective, and g is not injective.
2. Show that if X and Y are finite sets with the same number of elements, then any
injection f : X → Y is a bijection. Show also that any surjection f : X → Y is a
bijection.
3. Give an example of a function f : X → X which is injective but not bijective.
4. Give an example of a function f : X → X which is surjective but not bijective.
1.2
Factorizations of Functions
Commutative diagrams are a useful way to summarize the relationships between different
sets and functions.
Definition 1.2.1. Suppose given a diagram of sets and functions.
X@
@@
@@
f @@
h
Y
/ Z.
>
}
}}
}
}g
}}
We say that the diagram commutes if h = g ◦ f . Similarly, given a diagram
X
f
g
g
Z
/ Y
f
/ W
we say the diagram commutes if g ◦ f = f ◦ g.
More generally, given any diagram of maps, we say the diagram commutes if, when
we follow two different paths of arrows which start and end at the same points (following
the directions of the arrows, of course) and compose the maps along these paths, then
the compositions from the two paths agree.
Commutative diagrams are frequently used to model the factorization of functions.
CHAPTER 1. A LITTLE SET THEORY
4
Definitions 1.2.2. Let f : X → Y be a function. We say that a function g : X → Z
factors through f if there is a function g : Y → Z such that the following diagram
commutes.
X@
@@
@@
f @@
g
Y
/Z
?
~
~
~
~~ ~~ g
We say that g factors uniquely through f if there is exactly one function g : Y → Z that
makes the diagram commute.
The language of factorization can be confusing, as there is a totally different meaning
to the words “factors through,” which is also used frequently:
Definitions 1.2.3. Let f : X → Y be a function. We say that a function g : Z → Y
factors through f if there is a function g : Z → X such that the following diagram
commutes.
X?
??f
?

??


/ Y
Z
g
g
We say that g factors uniquely through f if there is exactly one function g : Z → X
that makes the diagram commute.
Notice that despite the fact that there are two possible meanings to the statement
that a function factors through f : X → Y , the meaning in any given context is almost
always unique. Thus, if we say that a map g : X → Z factors through f : X → Y , then
the only possible interpretation is that g = h ◦ f for some h : Y → Z. Similarly, if we say
that a map g : Z → Y factors through f : X → Y , then the only possible interpretation
is that g = f ◦ h for some h : Z → X. Indeed, the only way that ambiguity can enter is
if we ask whether a function g : X → X factors through another function f : X → X.
In this last case, one must be precise as to which type of factorization is being discussed.
Factorization properties may be used to characterize injective and surjective maps.
Proposition 1.2.4. Let f : X → Y . Then the following two statements are equivalent.
1. The map f is injective.
2. Any function g : Z → Y with the property that im g ⊂ im f factors uniquely through
f.
Proof Suppose that f is injective. Let g : Z → Y , and suppose that im g ⊂ im f . Then
for each z ∈ Z there is an x ∈ X such that g(z) = f (x). Since f is injective, this x is
unique, and hence there is a function g : Z → X defined by setting g (z) equal to the
unique x ∈ X such that g(z) = f (x). But then f (g (z)) = g(z), and hence f ◦ g = g, so
that g factors through f .
The factorization is unique, since if f ◦ h = g, we have f (h(z)) = g(z), and hence
f (h(z)) = f (g (z)) for all z ∈ Z. Since f is injective, this says that h(z) = g (z) for all
z ∈ Z, and hence h = g .
CHAPTER 1. A LITTLE SET THEORY
5
Conversely, suppose that any function g : Z → Y with im g ⊂ im f factors uniquely
through f . Let Z be the set with one element: Z = {z}. Then if f (x) = f (x ), define
g : Z → Y by g(z) = f (x). Then if we define h, h : Z → X by h(z) = x and h (z) = x ,
respectively, we have f ◦ h = f ◦ h = g. By uniqueness of factorization, h = h , and
hence x = x . Thus, f is injective.
The uniqueness of the factorization is the key in the above result. There is an entirely different characterization of injective functions in terms of factorizations, where
the factorization is not unique.
Proposition 1.2.5. Let f : X → Y with X = ∅. Then the following two statements are
equivalent.
1. The map f is injective.
2. Any function g : X → Z factors through f .
Proof Suppose that f is injective and that g : X → Z. Choose any z ∈ Z,2 and define
g : Y → Z by
g(x) if y = f (x)
g (y) =
z
if y ∈ im f .
Since f is injective, g is a well defined function. By construction, g ◦ f = g, and hence
g provides the desired factorization of g. (Note that if Z has more than one element,
then g is not unique. For instance, we could choose a different element z ∈ Z.)
Conversely, suppose that every map g : X → Z factors through f . We apply this by
setting g equal to the identity map of X. Thus, there is a function g : Y → X such that
g ◦ f = 1X . But then f is injective by Lemma 1.1.4.
There is a notion in mathematics called dualization. The dual of a statement is the
statement obtained by turning around all of the arrows in it. It is often interesting to
see what happens when you dualize a statement, but one should bear in mind that the
dual of a true statement is not always true.
The dual of Proposition 1.2.4 is true, but its proof is best deferred to our discussion
of equivalence relations below. The dual of Proposition 1.2.5 is also true. In fact, it turns
out to be equivalent to the famous Axiom of Choice.
Axiom of Choice (first form) Let f : X → Y be a surjective function. Then there is
a function s : Y → X such that f ◦ s = 1Y .
In this form, the Axiom of Choice seems to be obviously true. In order to define s,
all we have to do is choose, for each y ∈ Y , an element x ∈ X with f (x) = y. Since such
an x is known to exist for each y ∈ Y (because f is surjective), intuition says we should
be able to make these choices.
Sometimes the desired function s : Y → X may be constructed by techniques that
depend only on the other axioms of set theory. One example of this is when f is a
bijection. Here, s is the inverse function of f , and is constructed in the preceding section.
Alternatively, if Y is finite, we may construct s by induction on the number of elements.
Other examples exist where we may construct s by a particular formula.
2 Since
X = ∅ and g : X → Z, we must have Z = ∅ as well. (See Proposition 1.7.2.)
CHAPTER 1. A LITTLE SET THEORY
6
Nevertheless, intuition aside, for a generic surjection f : X → Y , we cannot construct
the desired function s via the other axioms, and, if we wish to be able to construct s
in all cases, we must simply accept the Axiom of Choice as part of the foundations of
mathematics. Most mathematicians are happy to do this without question, but some
people prefer models of set theory in which the Axiom of Choice does not hold. Thus,
while we shall use it freely, we shall give some discussion of the ways in which it enters
our arguments.3
We now give the dual of Proposition 1.2.5.
Proposition 1.2.6. Let f : X → Y . Then the following two statements are equivalent.
1. The map f is surjective.
2. Any function g : Z → Y factors through f .
Proof Suppose that f : X → Y is surjective. Then the Axiom of Choice provides a
function s : Y → X with f ◦ s = 1Y . But then if g : Z → Y is any map, s ◦ g : Z → X
gives the desired factorization of g through f .
Conversely, suppose that any function g : Z → Y factors through f . Applying this
for g = 1Y , we obtain a function s : Y → X with f ◦ s = 1Y . But then f is surjective by
Lemma 1.1.4.
1.3
Relations
We shall make extensive use of both equivalence relations and order relations in our work.
We first consider the general notion of a relation on a set X.
Definitions 1.3.1. A relation R on a set X is a subset R ⊂ X × X of the cartesian
product of X with itself. We shall use x R y as a shorthand for the statement that the
pair (x, y) lies in the relation R.
We shall often discuss a relation entirely in terms of the expressions x R y, and dispense with discussion of the associated subset of the product. This is an example of
what’s known as “abuse of notation”: we can think and talk about a relation as a property x R y which holds for certain pairs (x, y) of elements of X. We can then, if we choose,
define an associated subset R ⊂ X × X by letting R equal the set of all pairs such that
x R y.
For instance, there is a relation on the real numbers R, called ≤, which is defined in
the usual way: we say that x ≤ y if y − x is a non-negative number. We can discuss all
the properties of the relation ≤ without once referring to the subset of R × R which it
defines.
Not all relations on a set X have a whole lot of value to us. We shall enumerate
some properties which could hold in a relation that might make it interesting for certain
purposes.
3 The desired functions s : Y → X often exist for reasons that do not require the assumption of the
choice axiom. In the study of the particular examples of groups, rings, fields, topological spaces, etc.,
that mathematicians tend to be most interested in, the theorems that we know and love tend to be true
without assuming the axiom. However, the proofs of these theorems are often less appealing without
the choice axiom. Moreover, if we assume the axiom, we can prove very general theorems, which then
may be shown to hold without the axiom, for different reasons in different special cases, in the cases
of greatest interest. Thus, the Axiom of Choice permits us to look at mathematics from a broader and
more conceptual viewpoint than we could without it. And indeed, in the opinion of the author, a world
without the Axiom of Choice is a poorer one, with less functions and narrower horizons.
CHAPTER 1. A LITTLE SET THEORY
7
Definitions 1.3.2. Let R be a relation on a set X. If x R x for each x ∈ X, then we say
that R is reflexive.
Let R be a relation on a set X. Suppose that y R x if and only if x R y. Then we say
that R is symmetric.
Alternatively, we say that R is antisymmetric if x R y and y R x implies that x = y.
Let R be a relation on a set X. Then we say that R is transitive if x R y and y R z
together imply that x R z.
These definitions can be used to define some useful kinds of relations.
Definitions 1.3.3. A relation R on X is an equivalence relation if it is reflexive, symmetric, and transitive.
A relation R on X is a partial ordering if it is reflexive, antisymmetric, and transitive.
A set X together with a partial ordering R on X is called a partially ordered set.
A relation R on X is a total ordering if it is a partial ordering with the additional
property that for all x, y ∈ X, either x R y or y R x. A set X together with a total
ordering R on X is called a totally ordered set.
Examples 1.3.4.
1. The usual ordering ≤ on R is easily seen to be a total ordering.
2. There is a partial ordering, ≤, on R × R obtained by setting (x, y) ≤ (z, w) if both
x ≤ z and y ≤ w. Note that this example differs from the last one in that it is not
a total ordering.
3. There is a different partial ordering, called the lexicographic ordering, on R × R
obtained by setting (x, y) ≤ (z, w) if either x < z, or both x ≤ z and y ≤ w.
(Thus, both (1, 5) and (2, 3) are ≤ (2, 4).) Note that in this case, we do get a total
ordering.
4. There is an equivalence relation, ≡, on the integers, Z, obtained by setting m ≡ n
if m − n is even.
5. For any set X, there is an equivalence relation, ∼, on X defined by setting x ∼ y
for all x, y ∈ X (i.e., the subset of X × X in question is all of X × X).
6. For any set X, there is exactly one relation which is both a partial ordering and
an equivalence relation: here x R y if and only if x = y.
It is customary to use symbols such as ≤ and for a partial ordering, and to use
symbols such as ∼, , and ≡ for an equivalence relation.
1.4
Equivalence Relations
We now look at equivalence relations, as defined in the preceding section, in greater
depth.
Definition 1.4.1. Let ∼ be an equivalence relation on X and let x ∈ X. The equivalence
class of x in X (under the relation ∼), denoted E(x), is defined by
E(x) = {y ∈ X | x ∼ y}.
In words, E(x) is the set of all y ∈ X which are equivalent to x under the relation ∼.
CHAPTER 1. A LITTLE SET THEORY
8
Lemma 1.4.2. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then y is in
the equivalence class E(x) of x if and only if E(x) = E(y).
Proof Suppose y ∈ E(x). This just says x ∼ y. But if z ∈ E(y), then y ∼ z, and hence
x ∼ z by the transitivity of ∼. Therefore, E(y) ⊂ E(x).
But equivalence relations are also symmetric. Thus, x ∼ y if and only if y ∼ x, and
hence y ∈ E(x) if and only if x ∈ E(y). In particular, the preceding argument then
shows that x ∼ y also implies that E(x) ⊂ E(y), and hence E(x) = E(y).
Conversely, if E(x) = E(y), then reflexivity shows that y ∈ E(y) = E(x).
In consequence, we see that equivalence classes partition the set X into distinct
subsets:
Corollary 1.4.3. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then the
equivalence classes of x and y have a common element if and only if they are equal. In
symbols,
E(x) ∩ E(y) = ∅
if and only if
E(x) = E(y).
Proof If z ∈ E(x) ∩ E(y), then both E(x) and E(y) are equal to E(z) by Lemma 1.4.2.
The global picture is now as follows.
Corollary 1.4.4. Let ∼ be an equivalence relation on X. Then every element of X
belongs to exactly one equivalence class under ∼.
Proof By Corollary 1.4.3, every element belongs to at most one equivalence class. But
reflexivity shows that x ∈ E(x) for each x ∈ X, and the result follows.
Given a set with an equivalence relation, the preceding observation allows us to define
a new set from the equivalence classes.
Definitions 1.4.5. Let ∼ be an equivalence relation on a set X. By the set of equivalence classes under ∼, written X/∼, we mean the set whose elements are the equivalence
classes under ∼ in X. In particular, each element of X/∼ is a subset of X.
We define π : X → X/∼ by π(x) = E(x). Thus, π takes each element of X to its
equivalence class. We call π the canonical map from X to X/∼.
Example 1.4.6. Let X = {1, 2, 3, 4} and let ∼ be the equivalence relation on X in
which 1 ∼ 3, 2 ∼ 4, and the other related pairs are given by reflexivity and symmetry.
Then there are two elements in X/∼: {1, 3} and {2, 4}. We have π(1) = π(3) = {1, 3}
and π(2) = π(4) = {2, 4}.
We shall show that the canonical map π characterizes the equivalence relation completely.
Lemma 1.4.7. Let ∼ be an equivalence relation on X and let π : X → X/∼ be the
canonical map. Then π is surjective, and satisfies the property that π(x) = π(y) if and
only if x ∼ y.
Proof Surjectivity of π follows from the fact that every equivalence class has the form
E(x) for some x ∈ X, and E(x) = π(x). Now π(x) = π(y) if and only if E(x) = E(y).
By Lemma 1.4.2, the latter condition is equivalent to the statement that y ∈ E(x), which
is defined to mean that x ∼ y.
CHAPTER 1. A LITTLE SET THEORY
9
Intuitively, this says that X/∼ is the set obtained from X by identifying two elements
if they lie in the same equivalence class.
The canonical map has an important universal property.4
Proposition 1.4.8. Let ∼ be an equivalence relation on X. Then a function g : X → Y
factors through the canonical map π : X → X/∼ if and only if x ∼ x implies that
g(x) = g(x ) for all x, x ∈ X. Moreover, if g factors through π, then the factorization is
unique.
Proof Suppose that g = g ◦ π for some g : X/∼ → Y . If x ∼ x ∈ X, then
π(x) = π(x ), and hence g (π(x)) = g (π(x )). Since g = g ◦ π, this gives g(x) = g(x ),
as desired. Also, since π is surjective, every element of X/∼ has the form π(x) for some
x ∈ X. Since g (π(x)) = g(x), the effect of g on the elements of X/∼ is determined by
g. In other words, the factorization g , if it exists, is unique.
Conversely, suppose given a function g : X → Y with the property that x ∼ x implies
that g(x) = g(x ). Now define g : X/∼ → Y by setting g (π(x)) = g(x) for all x ∈ X.
Since π(x) = π(x ) if and only if x ∼ x , this is well defined by the property assumed
for the function g. By the construction of g , g ◦ π = g, and hence g gives the desired
factorization of g through π.
We have seen that an equivalence relation determines a surjective function: the canonical map. We shall give a converse to this, and show that specifying an equivalence
relation on X is essentially equivalent to specifying a surjective function out of X.
Definition 1.4.9. Let f : X → Y be any function. By the equivalence relation determined by f , we mean the relation ∼ on X defined by setting x ∼ x if f (x) = f (x ).
The reader may easily verify that the equivalence relation determined by f is indeed
an equivalence relation.
We now show that if ∼ is the equivalence relation determined by f : X → Y , then
there is an important relationship between the canonical map π : X → X/∼ and f .
Proposition 1.4.10. Let f : X → Y be any function and let ∼ be the equivalence
relation determined by f . Then f factors uniquely through the canonical map π : X →
X/∼, via a commutative diagram
f
/ Y.
XD
DD
z=
z
DD
zz
D
zzf π DD
z
z
!
X/∼
Moreover, the map f : X/∼ → Y gives a bijection from X/∼ to the image of f . In
particular, if f is surjective, then f gives a bijection from X/∼ to Y .
Proof Recall that the definition of ∼ was that x ∼ x if and only if f (x) = f (x ). Thus,
the universal property of the canonical map (i.e., Proposition 1.4.8) provides the unique
factorization of f through π: just set f (π(x)) = f (x) for all x ∈ X.
Since every element of X/∼ has the form π(x), the image of f is the same as the image
of f . Thus, it suffices to show that f is injective. Suppose that f (π(x)) = f (π(x )).
Then f (x) = f (x ), so by the definition of ∼, x ∼ x . But then π(x) = π(x ), and hence
f is injective as claimed.
4 The expression “universal property” has a technical meaning which we shall explore in greater depth
in Chapter 6 on category theory.
CHAPTER 1. A LITTLE SET THEORY
10
In particular, if f : X → Y is surjective, and if ∼ is the equivalence relation determined by f , then we may identify Y with X/∼ and identify f with π. In particular, the
dual of Proposition 1.2.4 is now almost immediate from the universal property of the
canonical map π.
Proposition 1.4.11. Let f : X → Y . Then the following two statements are equivalent.
1. The map f is surjective.
2. Let g : X → Z be such that f (x) = f (x ) implies that g(x) = g(x ) for all x, x ∈ X.
Then g factors uniquely through f .
Proof Suppose that f is surjective, and let ∼ be the equivalence relation determined
by f . Then we may identify f with the canonical map π : X → X/∼. But under this
identification, the second statement is precisely the universal property of π.
Conversely, suppose that the second condition holds. Then the hypothesis says that
there is a unique map g : Y → Y such that the following diagram commutes.
X@
@@
@@
f @@
f
Y
/Y
~?
~
~
~~ ~~ g
Of course, g = 1Y will do, but the key is the uniqueness: if f were not surjective, we
could alter g on those points of Y which are not in im f without changing the fact that
the diagram commutes, contradicting the uniqueness of the factorization.
Exercises 1.4.12.
† 1. Let n be a positive integer. Define a relation ≡ on the integers, Z, by setting i ≡ j
if j − i is a multiple of n. Show that ≡ is an equivalence relation. (We call it
congruence modulo n.)
2. Let ∼ be any equivalence relation. Show that the equivalence relation determined
by the canonical map π : X → X/∼ is precisely ∼.
1.5
Generating an Equivalence Relation
There are many situations in mathematics where we are given a relation R which is
not an equivalence relation, and asked to form the “smallest” equivalence relation, ∼,
with the property that x R y implies that x ∼ y. Here, “smallest” means that if ≡ is
another equivalence relation with the property that x R y implies that x ≡ y, then x ∼ y
also implies that x ≡ y. In other words, the equivalence classes of ∼ are subsets of the
equivalence classes of ≡.
This smallest equivalence relation is called the equivalence relation generated by R,
and its canonical map π : X → X/∼ will be seen to have a useful universal property with
respect to R. Of course, it is not immediately obvious that such an equivalence relation
exists, so we shall have to construct it.
Definition 1.5.1. Let R be any relation on X. The equivalence relation, ∼, generated
by R is the relation obtained by setting x ∼ x if there is a sequence x1 , . . . , xn ∈ X such
that x1 = x, xn = x , and for each i with 1 ≤ i < n, either xi = xi+1 , or xi R xi+1 , or
xi+1 R xi .
CHAPTER 1. A LITTLE SET THEORY
11
The intuition is that we take the relation R to be a collection of “basic equivalences,”
and declare that x ∼ x if they may be connected by a chain of basic equivalences, some
in one direction and some in the other.
Example 1.5.2. Let n be a positive integer, and define a relation R on the integers, Z,
by setting i R j if j = i + n. Note then that if ∼ is the equivalence relation generated by
R and if i ∼ j, then j = i + kn for some k ∈ Z. But the converse is also clear: in fact,
i ∼ j if and only if j = i + kn for k ∈ Z. Thus, ∼ is the equivalence relation known as
“congruence modulo n,” which was given in Problem 1 of Exercises 1.4.12.
We now verify that the equivalence relation generated by R has the desired properties.
Proposition 1.5.3. Let R be any relation and let ∼ be the equivalence relation generated
by R. Then ∼ is, in fact, an equivalence relation, and has the property that x R x implies
that x ∼ x . Moreover, if ≡ is another equivalence relation with the property that x R x
implies that x ≡ x , then x ∼ x also implies that x ≡ x .
Proof If x R x , then the sequence x1 = x, x2 = x shows that x ∼ x , so that x R x
implies that x ∼ x , as desired. Similarly, we see that ∼ is reflexive. Symmetry of ∼ may
be obtained by reversing the sequence x1 , . . . , xn that’s used to show that x ∼ x , while
transitivity is obtained by concatenating two such sequences.
Now let ≡ be another equivalence relation with the property that x R x implies that
x ≡ x , and suppose that x ∼ x . Thus, we are given a sequence x1 , . . . , xn ∈ X such
that x1 = x, x2 = x , and for each i with 1 ≤ i < n, either xi = xi+1 , or xi R xi+1 ,
or xi+1 R xi . But in any of the three situations, this says that xi ≡ xi+1 , so that
x = x1 ≡ x2 ≡ · · · ≡ xn = x . Since ≡ is transitive, this gives x ≡ x , as desired.
The fact that ∼ is the smallest equivalence relation containing R translates almost
immediately into a statement about the canonical map π : X → X/∼.
Corollary 1.5.4. Let R be a relation on X and let ∼ be the equivalence relation generated by R. Then a map f : X → Y factors through the canonical map π : X → X/∼
if and only if x R x implies that f (x) = f (x ) for all x, x ∈ X. The factorization, if it
exists, is unique.
Proof Suppose that f : X → Y has the property that x R x implies that f (x) = f (x )
for all x, x ∈ X. Let be the equivalence relation induced by f (i.e., x x if and only
if f (x) = f (x )). Then x R x implies that x x .
Thus, by Proposition 1.5.3, x ∼ x implies that x x . By the definition of , this
just says that x ∼ x implies that f (x) = f (x ). But now the universal property of the
canonical map π : X → X/∼ (Proposition 1.4.8) shows that f factors uniquely through
π.
For the converse, if f factors through π, then Proposition 1.4.8 shows that x ∼ x
implies that f (x) = f (x ). But x R x implies that x ∼ x , so the result follows.
1.6
Cartesian Products
The next definition should be familiar.
Definition 1.6.1. Let X and Y be sets. The cartesian product X × Y is the set of all
ordered pairs (x, y) with x ∈ X and y ∈ Y .
CHAPTER 1. A LITTLE SET THEORY
12
When describing a function f : X × Y → Z, it is cumbersome to use multiple
parentheses, as in f ((x, y)). Instead, we shall write f (x, y) for the image of the ordered
pair (x, y) under f .
Definition 1.6.2. There are projection maps π1 : X × Y → X and π2 : X × Y → Y
defined by π1 (x, y) = x, and π2 (x, y) = y, respectively.
An understanding of products with the null set is useful for the theory of functions.
Lemma 1.6.3. Let X be any set and let ∅ be the null set. Then X × ∅ = ∅ × X = ∅.
Proof If one of the factors of a product is empty, then there are no ordered pairs: one
cannot choose an element from the empty factor.
We shall make extensive use of products in studying groups and modules. The next
proposition is useful for studying finite groups. Here, if X is finite, we write |X| for the
number of elements in X.
Proposition 1.6.4. Let X and Y be finite sets. Then X × Y is finite, and |X × Y | =
|X| · |Y |, i.e., the number of elements in X × Y is the product of the number of elements
in X and the number of elements in Y .
Proof If either X or Y is empty, the result follows from Lemma 1.6.3. Thus, we assume
that both X and Y are nonempty.
Consider the projection map π1 : X ×Y → X. For each x0 ∈ X, write π1−1 (x0 ) for the
set of ordered pairs that are mapped onto x0 by π1 . Thus, π1−1 (x0 ) = {(x0 , y) | y ∈ Y }.5
For each x ∈ X, there is an obvious bijection between π1−1 (x) and Y , so that
each
π1−1 (x) has |Y | elements. Now π1−1 (x) ∩ π1−1 (x ) = ∅ for x = x , and X × Y =
−1
x∈X π1 (x). Thus, X × Y is the union of |X| disjoint subsets, each of which has |Y |
elements, and hence |X × Y | = |X| · |Y | as claimed.
Products have an important universal property. Let f : Z → X × Y be a map, and
write f (z) = (f1 (z), f2 (z)) for z ∈ Z. In the usual parlance, f1 : Z → X and f2 : Z → Y
are the component functions of f . Note that if π1 : X × Y → X and π2 : X × Y → Y
are the projection maps, then fi = πi ◦ f , for i = 1, 2.
Since a pair (x, y) ∈ X × Y is determined by its coordinates, we see that a function
f : Z → X×Y is determined by its component functions. In other words, if g : Z → X×Y
is a map, and if g1 : Z → X and g2 : Z → Y are its component functions, then f = g if
and only if f1 = g1 and f2 = g2 .
Finally, if we are given a pair of functions f1 : Z → X and f2 : Z → Y , then we can
define a function f : Z → X × Y by setting f (z) = (f1 (z), f2 (z)) for z ∈ Z. In other
words, we have shown the following.
Proposition 1.6.5. For each pair of functions f1 : Z → X and f2 : Z → Y , there is a
unique function f : Z → X × Y such that fi = πi ◦ f for i = 1, 2. Here, π1 : X × Y → X
and π2 : X × Y → Y are the projection maps.
that the subsets π1−1 (x), as x ranges over the elements of X, are precisely the equivalence
classes for the equivalence relation determined by π1 .
5 Note
CHAPTER 1. A LITTLE SET THEORY
1.7
13
Formalities about Functions
Intuitively, a function f : X → Y is a correspondence which assigns to each element
x ∈ X an element f (x) ∈ Y . The formal definition of function is generally given in terms
of its graph.
Definitions 1.7.1. A subset f ⊂ X × Y is the graph of a function from X to Y if for
each x ∈ X there is exactly one ordered pair in f whose first coordinate is x.
If f ⊂ X × Y is the graph of a function and if (x, y) ∈ f , we write y = f (x). Thus,
f = {(x, f (x)) | x ∈ X}. By the function determined by f we mean the correspondence
f : X → Y that assigns to each x ∈ X the element f (x) ∈ Y .
Two functions from X to Y are equal if and only if their graphs are equal.
Thus, every function is determined uniquely by its graph.
We obtain some immediate consequences.
Proposition 1.7.2. Let Y be any set. Then there is one and only one function from
the null set, ∅, to Y .
On the other hand, if X is a set with X = ∅, then there are no functions from X to
∅.
Proof For a set Y , we have ∅ × Y = ∅. Thus, ∅ × Y has exactly one subset, and hence
there is only one candidate for a function. The subset in question is the graph of a
function, because, since the null set has no elements, the condition that must be satisfied
is automatically true.
On the other hand, if X = ∅, then the unique subset of X × ∅ is not a function: for
any given x ∈ X, there is no ordered pair in this subset whose first coordinate is x.
The next proposition is important in the foundations of category theory.
Proposition 1.7.3. Let X and Y be sets. Then the collection of all functions from X
to Y forms a set.
Proof The collection of all functions from X to Y is in one-to-one correspondence with
the collection of their graphs. But the latter is a subcollection of the collection of all
subsets of X × Y . The fact that the collection of all subsets of a set, W , is a set (called
the power set of W ) is one of the standard axioms of set theory.
An important special case of this is when X is finite. For simplicity of discussion, let
us establish a notation.
Notation 1.7.4. Let X and Y be sets. We write F (X, Y ) for the set of functions from
X to Y .
Proposition 1.7.5. Let X1 and X2 be subsets of X such that X = X1 ∪ X2 and X1 ∩
X2 = ∅. Write ι1 : X1 ⊂ X and ι2 : X2 ⊂ X for the inclusions of these two subsets in
X. Let Y be any set, and define
α : F (X, Y ) → F (X1 , Y ) × F (X2 , Y )
by α(f ) = (f ◦ ι1 , f ◦ ι2 ). Then α is a bijection.6
6 In the language of category theory, this says that the disjoint union of sets is the coproduct in the
category of sets.
CHAPTER 1. A LITTLE SET THEORY
14
Proof Suppose that α(f ) = α(g). Then f ◦ ι1 = g ◦ ι1 , and hence f and g have the
same effect on the elements of X1 . Similarly, f ◦ ι2 = g ◦ ι2 , and hence f and g have the
same effect on the elements of X2 . Since a function is determined by its effect on the
elements of its domain, f = g, and hence α is injective.
Now suppose given an element (f1 , f2 ) ∈ F (X1 , Y ) × F (X2 , Y ). Thus, f1 : X1 → Y
and f2 : X2 → Y are functions. Define f : X → Y by
f1 (x) if x ∈ X1
f (x) =
f2 (x) if x ∈ X2 .
Then f ◦ ιi = fi for i = 1, 2, and hence α(f ) = (f1 , f2 ). Thus, α is surjective.
As a result, we can count the number of distinct functions between two finite sets.
Corollary 1.7.6. Let X and Y be finite. Then the set of functions from X to Y is
finite. Specifically, if X has n elements and Y has |Y | elements, then the set F (X, Y ) of
n
functions from X to Y has |Y | elements. (Here, we take 00 to be equal to 1.)
Proof If Y = ∅, the result follows from Proposition 1.7.2. Thus, we assume that Y = ∅,
and argue by induction on n. If n = 0, then the result is given by Proposition 1.7.2. For
n = 1, say X = {x}, there is a one-to-one correspondence ν : F (X, Y ) → Y given by
ν(f ) = f (x). Thus, we assume n > 1.
Write X = X ∪ {x}, with x ∈ X . Then we have a bijection α : F (X, Y ) →
n−1
elements, while F ({x}, Y ) has
F (X , Y ) × F ({x}, Y ). By induction, F (X , Y ) has |Y |
|Y | elements by the case n = 1, above. Thus, the result follows from Proposition 1.6.4.
Chapter 2
Groups: Basic Definitions and
Examples
In the first four sections, we define groups, monoids, and subgroups, and develop the
most basic properties of the integers and the cyclic groups Zn .
We then discuss homomorphisms and classification. The homomorphisms out of
cyclic groups are determined, and the abstract cyclic groups classified. Classification is
emphasized strongly in this volume, and we formalize the notion as soon as possible. We
then discuss isomorphism classes and show that there are only finitely many isomorphism
classes of groups of any given finite order.
We then give some examples of groups, in particular the dihedral and quaternionic
groups, which we shall use throughout the book as illustrations of various concepts. We
close with a section on cartesian products, giving a way to create new groups out of old
ones.
2.1
Groups and Monoids
A binary operation on a set X is a function
μ : X × X → X.
That is, μ assigns to each ordered pair (x, y) of elements of X an element μ(x, y) ∈ X.
We think of these as multiplications, and generally write x · y, or just xy (or sometimes
x + y), instead of μ(x, y). Note that the order of x and y is important. Generally x · y
and y · x are different elements of X.
Binary operations without additional properties, while they do arise in various situations, are too general to study in any depth. The condition we shall find most valuable
is the associative property: our binary operation is associative if
(x · y) · z = x · (y · z)
for every x, y, and z in X.
Another important property is the existence of an identity element. An identity
element for the operation is an element e ∈ X such that
e·x=x=x·e
for every x in X. The next result is basic.
15
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
16
Lemma 2.1.1. A binary operation can have at most one identity element.
Proof Suppose that e and e are both identity elements for the binary operation ·.
Since e is an identity element, we have e · e = e . But the fact that e is also an identity
element shows that e · e = e. Thus, e = e .
Definition 2.1.2. A monoid is a set with an associative binary operation that has an
identity element.
Examples 2.1.3.
1. Let Z+ be the set of positive integers: Z+ = {1, 2, . . . }. Then ordinary multiplication gives Z+ the structure of a monoid, with identity element 1.
2. Let N ⊂ Z be the set of non-negative integers: N = {0, 1, 2, . . . }. Then ordinary
addition gives N the structure of a monoid, with identity element 0.
Monoids occur frequently in mathematics. They are sufficiently general that their
structure theory is extremely difficult.
Definition 2.1.4. Let X be a monoid. We say that an element x ∈ X is invertible if
there is an element y ∈ X with
x · y = e = y · x,
where e is the identity. We call y the inverse for x and write y = x−1 .
The next result shows that the above definition uniquely defines y.
Lemma 2.1.5. Let X be a monoid. Then any element x ∈ X which is invertible has a
unique inverse.
Proof Suppose that y and z are both inverses for x. Then x · y = e = y · x, and
x · z = e = z · x. Thus,
y = y · e = y · (x · z) = (y · x) · z = e · z = z.
Note that the statement that x · y = e = y · x, used in defining inverses, is symmetric
in x and y, establishing the following lemma.
Lemma 2.1.6. Let X be a monoid, and let x ∈ X be invertible, with inverse y. Then y
is also invertible, and y −1 = x.
Finally, we come to the object of our study.
Definition 2.1.7. A group is a monoid in which every element is invertible.
Notice that the monoid, Z+ , of positive integers under multiplication is not a group,
as, in fact, no element other than 1 is invertible: if n ∈ Z+ is not equal to 1, then n > 1.
Thus, for any m ∈ Z+ , n · m > 1 · m, and hence n · m > 1.
The monoid, N, of the non-negative integers under addition also fails to be a group,
as the sum of two positive integers is always positive.
We shall generally write G for a group, rather than X. We do have numerous examples
of groups which come from numbers.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
17
Examples 2.1.8. The simplest example of a group is the group with one element, e.
There is only one binary operation possible: e · e = e. We call this group the trivial
group and denote the group itself by e.
Let Z be the integers: Z = {0, ±1, ±2, . . . }. Then addition of numbers gives Z the
structure of a group: the identity element is 0, and the inverse of n ∈ Z is −n. Similarly,
the rational numbers, Q, and the real numbers, R, form groups under addition.
We also know examples of groups obtained from multiplication of numbers. Write
Q× and R× for the nonzero numbers in Q and R, respectively. Since the product of two
nonzero real numbers is nonzero, multiplication provides Q× and R× with an associative
operation with identity element 1. And if x ∈ R× , the reciprocal, 1/x, provides an inverse
to x in R× . Moreover, if x = m/n ∈ Q× , with m, n ∈ Z, then 1/x = n/m ∈ Q× , and
hence x has an inverse in Q× .
×
Now write Q×
+ and R+ for the positive rational numbers and the positive real numbers, respectively. Since the product of two positive numbers is positive and the reciprocal
×
of a positive number is positive, the above arguments show that Q×
+ and R+ are groups
under multiplication.
The examples above are more than just groups. They also satisfy the commutative
law : x · y = y · x for all x and y in G. In fact, the ubiquity of the commutative law in
our background can exert a pernicious influence on our understanding of groups, as most
groups don’t satisfy it. The ones that do merit a special name:
Definitions 2.1.9. A group or monoid is abelian if it satisfies the commutative law. A
group or monoid in which the commutative law is known to fail is said to be nonabelian.
We will often use + for the operation in an abelian group. In fact, we shall restrict
the use of + to these groups.
Notice that the groups in the examples above all have infinitely many elements. Our
main focus here will be on finite groups: those with finitely many elements.
Definition 2.1.10. If G is a finite group, then the order of G, written |G|, is the number
of elements in G. If G is infinite, we say |G| = ∞.
Let X be a monoid. Write Inv(X) ⊂ X for the collection of invertible elements in X.
Lemma 2.1.11. Inv(X) is a group under the multiplication inherited from that of X.
Proof Let x and y be elements of Inv(X). Thus, x has an inverse, x−1 , and y has an
inverse y −1 . Now the element y −1 x−1 may easily be seen to be an inverse for xy, and
hence xy ∈ Inv(X), also.
Since the operation on Inv(X) is inherited from that on X, it must be associative.
Since e−1 = e, e is an element of Inv(X), and acts as an identity element there. And for
x ∈ Inv(X), x−1 is invertible, with inverse x, and hence x−1 ∈ Inv(X), and is the inverse
of x in Inv(X). In particular, every element of Inv(X) is invertible, so that Inv(X) is a
group.
A primary source for nonabelian groups is from matrices. We assume familiarity
with the basic properties of matrix multiplication shown in a beginning course in linear
algebra. We shall state the properties we are to use, and leave them as exercises to the
reader.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
18
Definitions 2.1.12. We write Mn (R) for the set of n×n matrices over the real numbers,
R. We shall make use of the binary operation of matrix multiplication, which is defined
as follows. If
⎛
⎞
⎛
⎞
a11 . . . a1n
b11 . . . b1n
⎜
⎟
⎜
⎟
..
..
A=⎝
and
B=⎝
⎠
⎠,
.
.
an1
...
ann
bn1
...
bnn
then the product AB is the matrix whose ij-th coordinate is
cij =
n
aik bkj = ai1 b1j + · · · + ain bnj .
k=1
We write I = In for the n × n identity matrix: the matrix whose diagonal entries are
all equal to 1 and whose off-diagonal entries are all equal to 0.
1 0 0
Thus, I2 = 10 01 , I3 = 0 1 0 , etc.
001
The properties we need here regarding matrix multiplication are few. We shall discuss
matrices in greater detail in Section 7.10 and Chapter 10.
Lemma 2.1.13. Multiplication of n × n matrices gives an associative binary operation
on Mn (R). Moreover, In is an identity element for this operation. Thus, Mn (R) is a
monoid under matrix multiplication.
The invertible elements in this monoid structure on Mn (R) are precisely the invertible
matrices in the usual sense. As such, they merit a name.
Definition 2.1.14. We write Gln (R) for Inv(Mn (R)), the group of invertible elements
of Mn (R). We call it the n-th general linear group of R.
Later in this chapter, we shall construct two infinite families of finite nonabelian
groups, the dihedral groups and the quaternionic groups, as explicit subgroups of Gln (R)
for n = 2 and 4, respectively. We shall show in Chapter 10 that every finite group is a
subgroup of Gln (R) for some value of n.
It is sometimes useful to study partial inverses in a monoid.
Definitions 2.1.15. Let X be a monoid with identity element e. If x, y ∈ X with
xy = e, then we say that x is a left inverse for y and that y is a right inverse for x.
Exercises 2.1.16.
1. Let M be the set of all nonzero integers. Then M is a monoid under multiplication.
What is Inv(M )?
† 2. Let G be a group and let x, y ∈ G. Show that x and y commute if and only if
x2 y 2 = (xy)2 .
† 3. Let G be a group such that x2 = e for all x ∈ G. Show that G is abelian.
−1
4. Let G be a group and let x1 , . . . , xk ∈ G. Show that (x1 . . . xk )−1 = x−1
k . . . x1 .
5. Let G be a group and let x, y ∈ G. Show that x and y commute if and only if
(xy)−1 = x−1 y −1 .
6. Let G be a group and let x, y ∈ G. Suppose there are three consecutive integers n
such that xn y n = (xy)n . Show that x and y commute.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
19
7. Let G be an abelian group and let x, y ∈ G. Show that xn y n = (xy)n for all n ∈ Z.
8. Verify that multiplication of n × n matrices is associative.
9. Show that Gl1 (R) is isomorphic to R× , the group of non-zero real numbers under
multiplication.
10. Show that Gln (R) is nonabelian for n ≥ 2.
11. Let
G=
a b
0 c
b ∈ R, a, c ∈ Q and ac = 0 .
Show that G is a group under matrix multiplication. Is G abelian?
12. Let X be a monoid. Suppose that x ∈ X has a left inverse, y, and a right inverse,
z. Show that y = z, and that x is invertible with inverse y.
† 13. Let X be a monoid with the property that every element of X has a left inverse.
Show that X is a group.
14. Let X be a finite monoid with the left cancellation property: if xy = xz, then
y = z. Show that X is a group.
15. Let X be a finite set with an associative binary operation. Suppose this operation
has both the left and the right cancellation properties. Show that X is a group.
2.2
Subgroups
One can tell quite a bit about about a group by knowing its subgroups.
Definitions 2.2.1. A subset S of a group G is said to be closed under multiplication if
for x and y in S, the product xy is also in S.
A subset H of G is said to be a subgroup if it is nonempty and closed under multiplication, and if for each x ∈ H, the inverse element x−1 is also in H.
Examples 2.2.2.
1. The groups Z and Q are subgroups of R.
2. The inclusions below are all inclusions of subgroups.
Q×
+
∩
R×
+
⊂ Q×
∩
⊂ R×
3. Z+ ⊂ R×
+ is closed under multiplication, but is not a subgroup, because the inverses
in R×
+ of the non-identity elements of Z+ do not lie in Z+ .
4. Any group G is a subgroup of itself.
5. For any group G, consider the subset {e} ⊂ G, consisting of the identity element
alone. Because e · e = e and e−1 = e, {e} is a subgroup of G, called the trivial
subgroup, or identity subgroup. By abuse of notation (i.e., for convenience), we
shall generally write e in place of {e} for this subgroup. (When the identity element
is called 1 or 0, we shall write 1 or 0 for the trivial subgroup as well.)
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
20
The next lemma is immediate from the definitions.
Lemma 2.2.3. Let H ⊂ G be a subgroup. Then H is a group under the operation
inherited from that of G.
We also have what is sometimes called Gertrude Stein’s Theorem.1 We leave the
proof as an exercise to the reader.
Lemma 2.2.4. A subgroup of a subgroup is a subgroup. (I.e., if H is a subgroup of G
and K is a subgroup of H, then K is a subgroup of G.)
Identifying the subgroups of a given group is a deep and interesting activity, and the
result says a lot about the group. As we shall see presently, the easiest subgroups to find
are the smallest ones.
Definition 2.2.5. Let G be a group and let g ∈ G. For any integer n, we define the
n-th power of g, g n , as follows. For n = 0, 1, g n is given by e and g, respectively. For
n > 1, we define g n by induction on n: g n = g n−1 · g. For negative exponents, we require
that g −1 be the inverse of g, and if n is positive, then g −n = (g −1 )n .
If G is abelian and written in additive notation, we write n · g, or ng for the n-th
power of G in the above sense.
The power laws for a group are fundamental:
Lemma 2.2.6. Let G be a group and let g ∈ G. Then for any integers m and n,
gm · gn
m n
(g )
= g m+n ,
= g
mn
and
.
If the group operation in G is written additively, this translates to the following:
mg + ng = (m + n)g,
n(mg) = (nm)g.
and
Proof We use the multiplicative notation. Consider the first statement first. For n = 0,
it reduces to g m · e = g m which is certainly true.
For n = 1, the statement depends on the value of m: if m > 0 it just gives the
definition of g m+1 , and hence is true. For m = 0, it says e · g = g. For m < 0, we have
m = −k, say, and g m = (g −1 )k = (g −1 )k−1 · g −1 . Thus,
g m · g = (g −1 )k−1 (g −1 · g) = (g −1 )k−1 = g −(k−1) = g m+1 ,
because −k = m.
For n > 1, we argue by induction on n. We have g n = g n−1 · g, so that g m · g n =
m
(g · g n−1 ) · g. By induction, this is g m+n−1 · g, which is g m+n by the case n = 1 above.
The case of n < 0 follows from that of n > 0 by exchanging the roles of g and g −1 .
We also use induction for the second statement. As with the first, we may assume
n ≥ 0. For n = 0, 1 it says nothing more than e = e, or g m = g m , respectively. For
n > 1, we have (g m )n = (g m )n−1 · g m . By induction, this is g mn−m · g m , which is g mn
by an application of the first statement.
1 This
usage was coined by Arunas Liulevicius.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
21
Notice that there’s a potential conflict between the additive notation for powers and
the notation for ordinary multiplication if G = Z. This is resolved by the following
lemma.
Lemma 2.2.7. Let m and n be integers. Then mn (as defined by ordinary multiplication
of integers) is the m-th power of n in the group Z.
Proof First note the definition of multiplication in Z: if m and n are both positive,
then mn is the sum of m copies of n and (−1)k m · (−1)l n = (−1)k+l mn.
The sum of m copies of n is by definition the m-th power of n. The remaining cases
follow from Lemma 2.2.6.
Lemma 2.2.6 gives us a convenient source of subgroups.
Definition 2.2.8. Let G be a group and let g ∈ G. Define g ⊂ G to be the set of all
integer powers of g:
g = {g n | n ∈ Z}.
Our next result will show that g is a subgroup of G. We call it the cyclic subgroup
generated by g.
Note that if the group operation in G is written additively, then g is defined by
g = {ng | n ∈ Z}.
In particular, in the case G = Z, for n ∈ Z, n = {nk | k ∈ Z} is the set of all integer
multiples of n. In this case, there are a couple of alternate notations that come out of
ring theory: for n ∈ Z we may write either (n) or nZ for n.
Lemma 2.2.9. Let G be a group and let g ∈ G. Then g is a subgroup of G. Moreover,
g has the property that it is the smallest subgroup of G which contains g. In other
words, if H ⊂ G is a subgroup, and if g ∈ H, then g ⊂ H.
Proof Since g m · g n = g m+n , g is closed under multiplication in G. Also, the inverse
of g m is g −m ∈ g, so that g is a subgroup of G.
If g ∈ H ⊂ G, with H a subgroup of G, then g n must lie in H for all n ∈ Z by the
closure properties of a subgroup. Thus, g ⊂ H. Indeed, viewing g as an element of H,
g is the subgroup of H generated by g.
We shall see in the next section that the cyclic subgroups n ⊂ Z give all of the
subgroups of Z.
Cyclic subgroups have a very special kind of group structure. They are cyclic groups:
Definition 2.2.10. A group G is cyclic if there is an element g ∈ G such that G = g.
Cyclic groups are simplest kind of group there is. For this reason, we can answer
some very complicated questions about them, questions we could not answer for more
complicated groups. Nevertheless, there are many deep and interesting questions about
cyclic groups (some of them connected with algebraic number theory) which have yet to
be answered.
We can also talk about the subgroup generated by a family of elements.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
22
Definitions 2.2.11. Let G be a group. For any subset S ⊂ G, the subgroup, S,
generated by S is the set of all products (of arbitrary length) of powers of elements of S:
S = {g1n1 . . . gknk | k ≥ 1, gi ∈ S
and ni ∈ Z for 1 ≤ i ≤ k}.
Warning: In the products g1n1 . . . gknk , we do not, and cannot, assume that the g1 , . . . , gk
are distinct. Indeed, if g, h ∈ S do not commute, then, for instance, g 2 h2 and ghgh are
distinct elements of S, by Problem 2 of Exercises 2.1.16.
If S is finite, say S = {x1 , . . . , xn }, then by abuse of notation, we shall write
x1 , . . . , xn for S.
If S = G, we say that G is generated by S. If G is generated by a finite set, we say
that G is finitely generated.
The following lemma is straightforward.
Lemma 2.2.12. S is the smallest subgroup of G which contains S.
We now give a method for detecting finite subgroups.
Proposition 2.2.13. Let G be a group and let H be a nonempty finite subset of G which
is closed under multiplication. Then H is a subgroup of G.
Proof We must show H is closed under inverses. Let x ∈ H and let S = {xk | 1 ≤ k ∈
Z}. Then S is contained in H, and hence S must be finite. Thus, the elements xk with
k ≥ 1 cannot all be distinct. We must have xk = xl for some k = l. Say l > k. But
then multiplying both sides by x−k , we get e = xl−k = xl−k−1 · x. Thus, xl−k−1 = x−1 .
But l − k − 1 ≥ 0 since l > k. If l − k − 1 = 0, we have x = e = x−1 ∈ H. Otherwise,
xl−k−1 ∈ S ⊂ H, and hence H is closed under inverses.
We can give one useful operation on subgroups at this point. The proof is left to the
reader.
Lemma 2.2.14. Let H and K be subgroups of G. Then H ∩ K is also a subgroup of G.
Warning: The union of two subgroups is almost never a subgroup.
The collection of all subgroups of a group G is easily seen to form a partially ordered
set as defined in Definition 1.3.3. Here the order relation is given by inclusion as subsets
of G; explicitly, H is less than or equal to K if H ⊂ K.
Definition 2.2.15. Let G be a group. The partially ordered set of subgroups of G is
known as the lattice of subgroups of G.
It is often useful to diagram the lattice of subgroups of a group G. Right now we
have neither enough examples of groups nor enough theory about subgroups to be able
to give a good example and verify its accuracy. However, we shall give one anyway, for
a group we shall encounter in this chapter. The reader should reconsider the following
example as the theory develops.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
23
Example 2.2.16. We write D6 for the dihedral group of order 6. As will become clear
later, the following is the lattice of subgroups of D6 .
D6
??
oo
o
o

??
oo 
o
??
o

o
o

?

oo
o

o
o
b O
a
ab
ab2 OOO
?
?

OOO

OOO ???

OOO ??

OOO? 
e
Here, b has order 3, while a, ab, and ab2 have order 2. The upward-slanted lines
represent the inclusions of subgroups.
Because the elements of monoids don’t necessarily have inverses, the definition of a
submonoid will have to be different from that of a subgroup.
Definition 2.2.17. A submonoid of a monoid M is a subset which is closed under
multiplication and contains the identity element.
Exercises 2.2.18.
1. Show that in the real numbers R, the cyclic subgroup generated by 1 is the integers.
In particular, Z is cyclic.
2. In Z, show that n = Z if and only if n = ±1.
3. Consider the group, Q×
+ , of positive rational numbers under multiplication. What
are the elements of 2 ⊂ Q×
+?
4. Consider the group Q× of nonzero rational numbers under multiplication. What
are the elements of −1 ⊂ Q× ? What are the elements of −2?
† 5. Let M be a monoid. We can define the positive powers of the elements in M in
exactly the same way that positive powers in a group are defined. We have m1 = m
for all m ∈ M , and the higher powers are defined by induction: mk = mk−1 m.
Show that for m ∈ M and for i, j ≥ 1, we have
(a) mi · mj = mi+j , and
(b) (mi )j = mij .
6. In Z, show that 2, 3 = Z.
7. In Z, show that 3n, 5n = n for any n ∈ Z.
8. In the group, Q×
+ , of positive rational numbers under multiplication, show that
2, 3 is not a cyclic subgroup. In other words, there is no rational number q such
that 2, 3 = q.
9. Show that Q×
+ is generated by the set of all prime numbers.
10. Show that the group Q× of nonzero rational numbers is generated by the set
consisting of −1 and all of the prime numbers.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
24
11. Let G be a group and let a, b ∈ G such that aba−1 ∈ b. Show that H =
{ai bj | i, j ∈ Z} is a subgroup of G. Deduce that H = a, b.
12. In Z, show that 2 ∩ 3 = 6.
13. In Z, show that m ∩ n = k, where k is the least common multiple of m and n.
14. In the group Q×
+ of positive rational numbers under multiplication, show that
2 ∩ 3 = 1. Here, 1 is the trivial subgroup of Q×
+.
15. Let M be a monoid. Show that the group of invertible elements Inv(M ) is a
submonoid of M .
16. Show that not every submonoid of a group is a group.
17. Show that every submonoid of a finite group is a group.
2.3
The Subgroups of the Integers
One of the simplest yet most powerful results in mathematics is the Euclidean Algorithm.
We shall use it here to identify all subgroups of Z and to derive the properties of prime
decomposition in Z.
Theorem 2.3.1. (The Euclidean Algorithm2 ) Let m and n be integers, with n > 0.
Then there are integers q and r, with 0 ≤ r < n, such that m = qn + r.
Proof First, we assume that m ≥ 0, and argue by induction on m. If m < n, we may
take q = 0 and r = m. If m = n, we take q = 1 and r = 0. Thus, we may assume that
m > n and that the result holds for all non-negative integers less than m.
In particular, the induction hypothesis gives
m − 1 = q n + r ,
for integers q and r with 0 ≤ r < n. If r < n − 1, we may take q = q and r = r + 1
for the desired result. Otherwise, m = (q + 1)n, and the proof for m ≥ 0 is complete.
If m < 0, then −m is positive, and hence −m = q n + r with 0 ≤ r < n. If r = 0,
this gives m = −q n. Otherwise, we have m = −q n − r . Subtracting and adding a copy
of n on the right of the equation gives m = (−q − 1)n + (n − r ), and since 0 < r < n,
we have 0 < n − r < n as well.
We shall use this to characterize the subgroups of Z. The subgroups we already know
are the cyclic ones:
n = {qn | q ∈ Z}.
We shall next show that these are all the subgroups of Z. First, note that if H is a nonzero
subgroup of Z, then there must be a nonzero element in it, and hence, by closure under
inverses, a positive element. Since the set of positive integers less than or equal to a
given one is finite, there is a unique smallest positive element in it.
2 The terminology that we’ve chosen here is not universal. There are some mathematicians who refer
to Theorem 2.3.1 as the Division Algorithm, and use the term Euclidean Algorithm for the procedure
for calculating greatest common divisors, outlined in Problem 3 of Exercises 2.3.18.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
25
Proposition 2.3.2. Let H ⊂ Z be a subgroup, and suppose that H = 0. Let n be the
smallest positive element of H. Then H = n.
Proof Since n ∈ H, n ⊂ H. Thus, it suffices to show that H ⊂ n.
For m ∈ H, write m = qn + r with q and r integers and 0 ≤ r < n. Now m is in
H, as is qn. Thus, m − qn = r must be in H. But r is non-negative and is less than n.
Since n is the smallest positive integer in H, we must have r = 0. Thus, m = qn ∈ n.
The language of divisibility is useful for discussing subgroups of Z:
Definition 2.3.3. Let m and n be integers. We say n divides m (or that m is divisible
by n) if there is an integer q such that m = nq.
The next lemma is now immediate from the fact that n is the set of all multiples
of n.
Lemma 2.3.4. Let m and n be integers. Then m ∈ n if and only if n divides m.
By Lemma 2.2.9, Lemma 2.3.4 may be restated this way.
Lemma 2.3.5. Let m and n be integers. Then n divides m if and only if m ⊂ n.
We shall also need a uniqueness statement about generators of cyclic subgroups.
Proposition 2.3.6. Let m and n be integers such that m = n. Then m = ±n.
Proof Clearly, if either of m and n is 0, so is the other. By passage to negatives, if
necessary, we may assume that m and n are both positive. By Lemma 2.3.5, they must
divide each other. Say m = qn and n = rm. But then q and r are positive. If q > 1,
then m > n, and hence n = rm > rn ≥ n, i.e., n > n, which is impossible. Thus, q = 1,
and hence m = n.
Corollary 2.3.7. Let H be a subgroup of Z. Then there is a unique non-negative integer
which generates H.
Definition 2.3.8. Let m and n be integers. We write
m + n = {rm + sn | r, s ∈ Z}.
Lemma 2.3.9. m + n is a subgroup of Z.
Proof The inverse of rn + sm is (−r)m + (−s)n. The result follows since (rm + sn) +
(r m + s n) = (r + r )m + (s + s )n ∈ m + n.
In fact, it is easy to see that m + n is the subgroup of Z generated by the set
{m, n}.
Definition 2.3.10. Let m, n ∈ Z. We write (m, n) for the unique non-negative integer
which generates the subgroup m + n of Z:
m + n = (m, n).
We shall refer to (m, n) as the greatest common divisor, or g.c.d., of m and n.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
26
We shall now derive the familiar properties of the g.c.d. and of prime decomposition
from the above results.
Proposition 2.3.11. Let m and n be integers. Then an integer k divides both m and n
if and only if it divides (m, n). In particular, if m and n are not both 0, then (m, n) is
the largest integer which divides both m and n.
In addition, there exist integers r and s with rm + sn = (m, n).
Proof The last statement follows because
(m, n) ∈ m + n = {rm + sn | r, s ∈ Z}.
Now let k divide both m and n. Then it must divide any number of the form rm+sn,
including (m, n). Conversely, to show that the divisors of (m, n) must divide both m and
n, it suffices to show that (m, n) divides both m and n. But m = 1 · m + 0 · n and
n = 0 · m + 1 · n both lie in (m, n), so the result follows from Lemma 2.3.4.
Definition 2.3.12. We say that the integers m and n are relatively prime if (m, n) = 1.
Note that if 1 ∈ m + n, then m + n = Z. The following lemma is immediate.
Lemma 2.3.13. The integers m and n are relatively prime if and only if there exist
integers r and s with rm + sn = 1.
Definition 2.3.14. We say that an integer p > 1 is prime if its only divisors are ±1 and
±p. By Lemma 2.3.5, this says that p ⊂ k if and only if either k = p or k = Z.
Lemma 2.3.15. Let p and a be integers, with p prime. Then p and a are relatively
prime if and only if p does not divide a.
Proof The greatest common divisor of p and a must be either p or 1. In the former
case, p divides a. In the latter, it does not.
Proposition 2.3.16. Let p be a prime number. Then p divides a product ab if and only
if it divides at least one of a and b.
Proof Clearly, if p divides a or b, it must divide their product. Conversely, suppose p
divides ab, but does not divide a. We shall show that it must divide b.
Since p does not divide a, we have (p, a) = 1, so that qp + ra = 1 for some integers q
and r. But then qpb + rab = b. Since p divides both qpb and rab, it must divide b.
We can now give the fundamental theorem regarding prime factorization.
Theorem 2.3.17. Every integer m > 1 has a unique prime decomposition: there is a
factorization
m = pr11 . . . prkk ,
where pi is prime and ri > 0 for 1 ≤ i ≤ k, and pi < pj for i < j. Uniqueness means
that if m = q1s1 . . . qlsl with qi prime and si > 0 for 1 ≤ i ≤ l and qi < qj for i < j, then
we must have k = l and have pi = qi and ri = si for 1 ≤ i ≤ k.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
27
Proof The existence of a prime decomposition amounts to showing that any integer
m > 1 is a product of primes. Clearly, this is true for m = 2. We argue by induction on
m, assuming that any integer k with 1 < k < m has a prime decomposition.
If m is prime, we are done. Otherwise, m = kl, where both k and l are greater than
1 and less than m. But then k and l are products of primes by induction, and hence so
is m.
For uniqueness, suppose given decompositions pr11 . . . prkk and q1s1 . . . qlsl of m as above.
We argue by induction on t = r1 + · · · + rk .
If t = 1, then m = p1 , a prime. But then p1 is divisible by q1 , and since both are prime,
they must be equal. We then have p1 (1 − q1s1 −1 . . . qlsl ) = 0, so that (1 − q1s1 −1 . . . qlsl ) = 0,
and hence q1s1 −1 . . . qlsl = 1. Since the qi are prime and since no positive number divides
1 other than 1 itself, we must have l = 1 and s1 = 1.
Suppose t > 1. Since p1 divides m, Proposition 2.3.16, together with an induction
on s1 + · · · + sl , shows that p1 must divide qi for some i. But since p1 and qi are prime,
we must have p1 = qi . A similar argument shows that q1 = pj for some j, and our
order assumption then shows that i = j = 1, so that p1 = q1 . Now p1 (p1r1 −1 . . . prkk −
q1s1 −1 . . . qlsl ) = 0, so that pr11 −1 . . . prkk = q1s1 −1 . . . qlsl . But our inductive hypothesis
applies here, and the resulting equality is exactly what we wanted.
Exercises 2.3.18.
1. Let p1 , . . . , pk be distinct primes and let ri ≥ 0 for 1 ≤ i ≤ k. Show that an integer
n divides pr11 . . . prkk if and only if n = ±ps11 . . . pskk with 0 ≤ si ≤ ri for 1 ≤ i ≤ k.
2. Show that there are infinitely many prime numbers.
† 3. Let n = qm + r with 0 ≤ r < m. Show that (m, n) = (r, m). Use this to obtain an
iterative procedure to calculate (m, n).
4. Implement the iterative procedure above by a computer program to calculate greatest common divisors.
† 5. A subgroup G ⊂ R is called discrete if for each g ∈ G there is an open interval
(a, b) ⊂ R such that (a, b) ∩ G = {g}. Show that every discrete subgroup of R is
cyclic.
2.4
Finite Cyclic Groups: Modular Arithmetic
Here, we develop Zn , the group of integers modulo n.
Definition 2.4.1. Let n > 0 be a positive integer. We say that integers k and l are
congruent modulo n, written k ≡ l mod n, or sometimes k ≡ l (n), if k − l is divisible
by n (i.e., k − l ∈ n).
Lemma 2.4.2. Congruence modulo n is an equivalence relation.
Proof Clearly, k − k ∈ n for any n, so congruence modulo n is reflexive. It is also
symmetric, since l − k = −(k − l) is contained in any subgroup containing k − l. If
k ≡ l mod n and l ≡ m mod n, then both k − l and l − m are in n, and hence so is
k − m = (k − l) + (l − m). Thus, the relation is transitive as well.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
28
Definition 2.4.3. We write Zn for the set of equivalence classes of integers modulo n,
and write m ∈ Zn for the equivalence class containing m ∈ Z. 3 We shall also use 0 to
denote 0.
Zn is called the group of integers modulo n.
Lemma 2.4.4. There are exactly n elements in Zn , represented by the elements 0 =
0, 1, . . . , n − 1.
Proof The elements listed are precisely the classes r for 0 ≤ r < n. For m ∈ Z, the
Euclidean algorithm provides an equation m = qn + r, with 0 ≤ r < n. But then m = r,
and hence m is in the stated list of elements.
To see that these classes are all distinct, suppose that 0 ≤ k < l < n. Then 0 <
l − k < n, so that l − k is not divisible by n, and hence k = l in Zn .
There are two important binary operations on Zn .
Definitions 2.4.5. The operations of addition and multiplication in Zn are defined by
setting k + l = k + l and k · l = kl, respectively, for all k, l ∈ Z.
We must show that these operations are well defined.
Lemma 2.4.6. The operations of addition and multiplication are well defined binary
operations on Zn .
Proof Suppose that k = q and l = r in Zn . In other words, k ≡ q mod n and l ≡ r
mod n. We must show that k + l ≡ q + r mod n and kl ≡ qr mod n.
Addition is easy: k − q ∈ n and l − r ∈ n. Now, (k + l) − (q + r) = (k − q) + (l − r).
This lies in n, as n closed under addition.
For multiplication, we have kl − qr = kl − kr + kr − qr = k(l − r) + r(k − q). This
last is divisible by n because l − r and k − q are.
The chief properties of addition and multiplication are inherited from those in the
integers:
Proposition 2.4.7. Addition gives Zn the structure of an abelian group with identity
0. Multiplication makes Zn into an abelian monoid with identity 1. The two operations
satisfy the distributive law: a(b + c) = ab + a c.
Proof
(a + b) + c = a + b + c = a + b + c.
Since a + (b + c) has the same expansion, addition is associative. The other assertions
follow by similar arguments.
In the language of abstract algebra, Proposition 2.4.7 says, simply, that the operations
of addition and multiplication give Zn the structure of a commutative ring.
When we refer to Zn as a group, we, of course, mean with respect to addition. By
Lemma 2.4.4, Zn has order n.
Corollary 2.4.8. Let n be a positive integer. Then there are groups of order n.
3 In
the notation of Section 1.4, Zn = Z/ ≡, where ≡ denotes equivalence modulo n.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
29
Definition 2.4.9. The canonical map π : Z → Zn is defined by setting π(k) = k for all
k ∈ Z.
Given that Zn is the set of equivalence classes of integers with respect to the equivalence relation of congruence modulo n, and that π takes each integer to its equivalence
class, we see that this coincides with the canonical map defined in Section 1.4 for an
arbitrary equivalence relation.
We shall see in the next section that the canonical map π : Z → Zn is what’s known
as a homomorphism of groups.
Exercises 2.4.10.
1. Show that Zn is generated as a group by 1. Thus, Zn is a cyclic group.
2. Show that 2 does not generate Z4 . Deduce that there are finite cyclic groups that
possess nontrivial proper subgroups.
3. Let p be a prime. Show that the nonzero elements of Zp form a group under
multiplication modulo p.
4. If n is not prime, show that the nonzero elements of Zn do not form a group under
multiplication modulo n.
2.5
Homomorphisms and Isomorphisms
There are many ways to construct a finite set, but the thing that describes the essence
of the structure of such a set is the number of elements in it. Two sets with the same
number of elements may be placed in one-to-one correspondence, at which point their
properties as sets are indistinguishable.
In the case of groups, the structure is too complicated to be captured by the number
of elements alone. We shall see examples of finite groups with the same order whose
group properties are quite different. For instance, one may be abelian and the other not.
But the idea of a one-to-one correspondence makes a good start at deciding when two
groups may be identified, provided that the correspondence respects the multiplications
of the groups.
Definition 2.5.1. An isomorphism between two groups is a function f : G → G which
is one-to-one and onto, such that f (x · y) = f (x) · f (y) for all x, y ∈ G. Here, · is used
for the multiplications of both G and G .
We say that the groups G and G are isomorphic if there exists an isomorphism from
G to G , and write G ∼
= G .
This turns out to be enough to capture the full structure of the groups in question.
A glance at the definition shows that there is a weaker notion which is part of it:
Definition 2.5.2. Let G and G be groups. A homomorphism from G to G is a function
f : G → G such that f (x · y) = f (x) · f (y) for all x, y ∈ G.
Thus, an isomorphism is a homomorphism which is one-to-one and onto.
Recall that one-to-one functions are sometimes called injective functions, or injections,
and that onto functions are known as surjective functions, or surjections. Also, a function
which is both one-to-one and onto is called bijective.
Clearly, an understanding of homomorphisms will contribute to our understanding
of isomorphisms, but in fact, homomorphisms will turn out to say a lot more about the
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
30
structure of a group than may be apparent at this point. Before we dive into theory, let’s
consider some examples.
Examples 2.5.3.
1. For any two groups, G and G , there is a homomorphism f : G → G obtained by
setting f (g) = e for all g ∈ G. We call this the trivial homomorphism from G to
G .
2. Let H be a subgroup of the group G. Then the inclusion i : H ⊂ G is a homomorphism, because the multiplication in H is inherited from that of G.
3. Recall that the canonical map π : Z → Zn is defined by π(k) = k for all k ∈ Z.
Thus, π(k + l) = k + l = k + l = π(k) + π(l), and hence π is a homomorphism.
For some pairs G, G of groups, the trivial homomorphism is the only homomorphism
from G to G .
The relationship between a homomorphism and the power laws in a group is an
important one:
Lemma 2.5.4. Let f : G → G be a homomorphism of groups and let g ∈ G. Then
f (g n ) = f (g)n for all n ∈ Z. In particular, f (e) = e (recall that e = g 0 ) and f (g −1 ) =
f (g)−1 .
Proof For n > 0, this is a quick induction from the definition of powers. For n = 0,
we have f (e) = f (e · e) = f (e) · f (e). Multiplying both sides of the resulting equation
f (e) = f (e) · f (e) by f (e)−1 , we get e = f (e). (Any group element which is its own
square must be the identity element of the group.) For n < 0, we have e = f (e) =
f (g n g −n ) = f (g n )f (g −n ) = f (g n )f (g)−n by the two cases above, so that f (g n ) is the
inverse of f (g)−n .
Since the canonical map π : Z → Zn is a homomorphism, Lemma 2.2.7 gives us the
next corollary.
Corollary 2.5.5. Let m ∈ Zn and let k be any integer. Then the k-th power of m with
respect to the group operation in Zn is km.
Another consequence of Lemma 2.5.4 is a determination of all of the homomorphisms
from Z to an arbitrary group G.
Proposition 2.5.6. Let G be a group and let g ∈ G. Then there is a unique homomorphism fg : Z → G with fg (1) = g.
Proof Define fg by fg (n) = g n for all n ∈ Z. This is a homomorphism by the power
laws of Lemma 2.2.6.
For uniqueness, suppose that f : Z → G is an arbitrary homomorphism with f (1) = g.
We’d like to show that f (n) = g n . Since n is the n-th power of 1 with respect to the
group operation in Z, this follows from Lemma 2.5.4.
The most basic properties of homomorphisms are the following. We leave the proof
to the reader.
Lemma 2.5.7. Let f : G → G and g : G → G be homomorphisms. Then so is the
composite g ◦ f : G → G . If f and g are both isomorphisms, then so is g ◦ f . If f is an
isomorphism, then so is the inverse map f −1 : G → G.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
31
There are two subgroups associated with a homomorphism that can be useful in
detecting its deviation from being an isomorphism. One of them is the kernel:
Definition 2.5.8. Let f : G → G be a homomorphism. Then the kernel of f , written
ker f , is given by
ker f = {x ∈ G | f (x) = e}.
Recall that the image of a function f : X → Y is the subset im f ⊂ Y given by
im f = {y ∈ Y | y = f (x) for some x ∈ X}.
An easy check, using Lemma 2.5.4, shows that the kernel and image of a homomorphism f : G → G are subgroups of G and G , respectively.
Lemma 2.5.9. Let f : G → G be a homomorphism. Then ker f detects the deviation of
f from being injective: we have f (x) = f (x ) if and only if x−1 x ∈ ker f . In particular,
f is injective if and only if the kernel of f is the trivial subgroup, e, of G.
The image detects whether f is onto: it is onto if and only if im f = G .
Proof Suppose that f (x) = f (x ). Then f (x−1 x ) = f (x)−1 f (x ) = e, and hence
x−1 x ∈ ker f . Conversely, if x−1 x ∈ ker f , then f (x ) = f (xx−1 x ) = f (x)f (x−1 x ) =
f (x) · e, and hence f (x ) = f (x).
Suppose that f is injective. Since f (e) = e, no element of G other than e may be
carried to e by f , and hence ker f = e. Conversely, if ker f = e, then x−1 x ∈ ker f
implies that x−1 x = e, so that x = x . But then x = x whenever f (x) = f (x ), and
hence f is injective.
The statement regarding im f is immediate.
We shall refer to an injective homomorphism as an embedding. The following lemma
is immediate:
Lemma 2.5.10. Let f : G → G be an embedding. Then f induces an isomorphism of
G onto im f .
Definitions 2.5.11. Let G be a group and let n ∈ Z. We say that g ∈ G has exponent
n if g n = e.
If there is a positive integer n which is an exponent for g, we say that g has finite
order, and define the order, o(g), of g to be the smallest positive integer which is an
exponent for g.
If g does not have a positive exponent, we say that o(g) = ∞.
If one knows the order of an element, one can determine all possible exponents for it.
Lemma 2.5.12. Let G be a group and let g ∈ G be an element of finite order. Let n be
any integer. Then g has exponent n if and only if o(g) divides n.
Proof Of course, g mo(g) = (g o(g) )m = em = e, so if o(g) divides n, then g has exponent
n.
Conversely, if g has order k and exponent n, write n = qk + r with 0 ≤ r < k.
Then e = g n = g qk g r = g r , since both n and qk are exponents for g. But this says g
has exponent r. If r were positive, this would contradict the fact that k is the smallest
positive exponent for g. Thus, r = 0, and n = qk, and hence k divides n, as claimed.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
32
Information about exponents sometimes leads to a determination that a particular
element must be the identity.
Corollary 2.5.13. Let m and n be integers that are relatively prime. Suppose that m
and n are both exponents for an element g ∈ G. Then g = e.
Proof The order of g has to divide both m and n, and hence it must also divide their
g.c.d. Since the order of an element is positive, g has order 1, and hence g 1 = e.
Let fg : Z → G be the unique homomorphism with fg (1) = g. Then fg (n) = g n for
all n ∈ Z. Thus, an integer n is in the kernel of fg if and only if g has exponent n. Also,
im fg is precisely g. The next result now follows immediately from Proposition 2.3.2.
It includes, in part, a slicker argument for Lemma 2.5.12.
Proposition 2.5.14. Let G be a group and let g ∈ G. Let fg : Z → G be the unique
homomorphism with fg (1) = g. Then the kernel of fg is o(g), if o(g) is finite, and is 0
otherwise.4 The image of fg is g, the cyclic subgroup of G generated by g.
Lemma 2.5.10 now gives us the following corollary.
Corollary 2.5.15. If o(g) = ∞, then fg : Z → g is an isomorphism.
Corollary 2.5.16. Every nonzero subgroup of Z is isomorphic to Z.
Proof Since mn = 0 if and only if either m or n is 0, no nonzero element of Z has finite
order. But every subgroup of Z is cyclic.
Next, we study the homomorphisms out of Zn . We shall make use of the notion of
commutative diagrams, as defined in Definition 1.2.1.
Proposition 2.5.17. Let G be a group and let g ∈ G. Then there is a homomorphism
hg : Zn → G with hg (1) = g if and only if g has exponent n.
If g does have exponent n, then there is only one such homomorphism, and it has the
property that the following diagram commutes
fg
/ G,
Z@
>
@@
}
@@
}}
}
@
π @@
}}
}} hg
Zn
meaning that hg ◦ π = fg . Here π : Z → Zn is the canonical map, while fg : Z → G
is the unique homomorphism that takes 1 to g. The homomorphism hg must satisfy the
formula
hg (m) = g m
for all m ∈ Zn .
If the homomorphism hg exists, then its image is g. Moreover, the induced map
hg : Zn → g is an isomorphism if and only if o(g) = n.
In particular, if g ∈ G has finite order, then g ∼
= Zo(g) .
4 Recall that, since we are using additive notation for Z, we write 0 for the trivial subgroup consisting
of the identity element alone.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
33
Proof Suppose there exists a homomorphism hg : Zn → G with hg (1) = g. Then
hg ◦ π is a homomorphism from Z to G which takes 1 to hg (π(1)) = hg (1) = g. Thus,
hg ◦ π = fg , and hence hg (m) = hg (π(m)) = fg (m) = g m for all m ∈ Zn . Thus, hg is
unique if it exists. Moreover, the existence of hg gives e = hg (0) = hg (n) = g n , since
n = 0 in Zn , and hence g has exponent n. Clearly, im hg = g.
Conversely, if g has exponent n, we wish to show that hg (m) = g m gives a well defined
function from Zn to G. If m = l, then m−l = sn for some integer s, and hence m = l+sn.
But then g m = g l (g n )s = g l es = g l by the rules of exponents. Thus, hg is well defined.
But hg (q + r) = g q+r = g q g r = hg (q)hg (r), and hence hg is a homomorphism.
It suffices to show that hg is injective if and only if o(g) = n. Suppose first that
o(g) = n. Since g has exponent n, o(g) divides n. Thus, 0 < o(g) < n, and hence
0 = o(g) ∈ Zn . Since o(g) ∈ ker hg , hg is not injective.
On the other hand, if o(g) = n, let k ∈ ker hg . But this says that g k = e, and hence
g has exponent k. But that means that k is divisible by o(g) = n, and hence k = 0 in
Zn . In particular, ker hg = 0, as desired.
Since Zn has order n, we obtain the following corollary.
Corollary 2.5.18. Let G be a group and let g ∈ G. Then the order of g is equal to the
order of the group g.
Corollary 2.5.19. Let f : G → H be a homomorphism, and suppose that g ∈ G has
finite order. Then the order of f (g) divides the order of g.
Proof Let hg : Zo(g) → G be the homomorphism which carries 1 to g. Then f ◦ hg :
Zo(g) → H is a homomorphism that takes 1 to f (g). By Proposition 2.5.17, f (g) has
exponent o(g), so the order of f (g) divides o(g).
Because the elements of monoids do not all have inverses, the definition of homomorphism must be modified for it to be useful for monoids.
Definitions 2.5.20. A homomorphism of monoids is a function f : M → M such that
1. f carries the identity element of M to that of M .
2. For m, m ∈ M , f (mm ) = f (m)f (m ).
An isomorphism of monoids is a homomorphism f : M → M which is bijective.
Exercises 2.5.21.
1. Show that any group with one element is isomorphic to the trivial group.
2. Show that any group with two elements is isomorphic to Z2 .
3. List the homomorphisms from Z9 to Z6 .
4. List the homomorphisms from Z5 to Z6 .
5. Show that the order of m in Zn is n/(m, n). Deduce that the order of each element
divides the order of Zn . Deduce that every non-identity element of Zp has order p,
for any prime p.
† 6. Let G be a group and let x ∈ G. Show that o(x−1 ) = o(x). (Hint: We may assume
G = x.)
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
34
7. Show that a group G is cyclic if and only if there is a surjective homomorphism
f : Z → G.
† 8. Let f : G → G be a homomorphism.
(a) Let H ⊂ G be a subgroup. Define f −1 (H ) ⊂ G by
f −1 (H ) = {x ∈ G | f (x) ∈ H }.
Show that f −1 (H ) is a subgroup of G which contains ker f .
(b) Let H ⊂ G be a subgroup. Define f (H) ⊂ G by
f (H) = {f (x) | x ∈ H}.
Show that f (H) is a subgroup of G . If ker f ⊂ H, show that f −1 (f (H)) = H.
(c) Suppose that H ⊂ im f . Show that f (f −1 (H )) = H . Deduce that there is a
one-to-one correspondence between the subgroups of im f and those subgroups
of G that contain ker f . What does this tell us when f is onto?
9. Let f : G → G be a homomorphism and let x ∈ G. Show that f (x) = f (x).
10. Show that every subgroup of Zn is cyclic.
11. Show that Zn has exactly one subgroup of order d for each d dividing n, and has
no other subgroups. (In particular, among many other important implications, this
shows that the order of every subgroup of Zn divides the order of Zn .) Deduce the
following consequences.
(a) Show that a group containing a subgroup isomorphic to Zk × Zk for k > 1
cannot be cyclic.
(b) Show that if a prime p divides n, then Zn has exactly p − 1 elements of order
p.
(c) Let H and K be subgroups of Zn . Show that H ⊂ K if and only if |H| divides
|K|.
(d) Show that if d divides n, then Zn has exactly d elements of exponent d.
12. Let H and K be subgroups of Zpr , where p is prime. Show that either H ⊂ K or
K ⊂ H.
13. Let p and q be distinct prime numbers and let n be divisible by both p and q.
Find a pair of subgroups H, K ⊂ Zn such that neither H nor K is contained in the
other.
14. Show that a group G is abelian if and only if the function f : G → G given by
f (g) = g 2 is a homomorphism.
15. Show that a group G is abelian if and only if the function f : G → G given by
f (g) = g −1 is a homomorphism.
16. Let G be an abelian group and let n be an integer. Show that the function f : G →
G given by f (g) = g n is a homomorphism.
17. Show that the exponential function exp(x) = ex gives an isomorphism from the
additive group of real numbers to R×
+.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
35
18. Show that a group is finite if and only if it has finitely many subgroups.
19. Let i : Gln (R) → Mn+1 (R) be defined by setting i(A) to be the matrix whose first
n rows consist of the matrix A with a column of 0’s added on the right, and whose
last row consists of n 0’s followed by a 1, for all A ∈ Gln (R). Pictorially,
⎞
⎛
0
⎜
.. ⎟
⎜
. ⎟
A
i(A) = ⎜
⎟.
⎝
0 ⎠
0 ... 0 1
Show that i(A) lies in Gln+1 (R), and that i defines an embedding from Gln (R)
into Gln+1 (R).
20. Show that there is a monoid with two elements which is not a group. Show that
any other monoid with two elements which is not a group is isomorphic to this one.
21. Let f : M → M be a homomorphism of monoids. Show that f restricts to a group
homomorphism f : Inv(M ) → Inv(M ) between their groups of invertible elements.
In particular, if m ∈ M is invertible, then so is f (m).
22. Give an example to show that condition 1 in the definition of monoid homomorphism does not follow from condition 2.
2.6
The Classification Problem
Definition 2.6.1. By a classification of the groups with a certain property, we mean
the identification of a set of groups {Gi | i ∈ I}5 such that the following conditions hold.
1. Each group Gi has the stated property.
2. If Gi is isomorphic to Gj , then i = j.
3. Each group with the stated property is isomorphic to one of the Gi .
A classification of groups with a given property is often referred to as a classification
“up to isomorphism” of the groups in question.
The property by which we collect the groups to be classified is often their order.
This is a natural criterion, as isomorphic groups must have the same order. However,
other criteria are often used. For instance, the Fundamental Theorem of Finite Abelian
Groups, which we give in Section 4.4, gives a classification of all finite abelian groups.
Examples 2.6.2.
1. Problem 2 of Exercises 2.5.21 shows that every group of order 2 is isomorphic to
Z2 . This classifies the groups of order 2.
2. Corollary 2.5.15 and Proposition 2.5.17 show that every cyclic group is isomorphic
to exactly one of the groups in {Z, e, Zn | n ≥ 2}. Thus, we obtain a classification
of the cyclic groups.
5 Here
the set I is called the indexing set for the set {Gi | i ∈ I} of groups.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
36
Classification can be phrased in more general terms using the language of category
theory. (See Chapter 6.) In general, questions of classification are the most fundamental
and important questions that can be asked in mathematics.
It may come as a surprise to some readers that the finite groups have not yet been
classified. It is often astounding that questions which may be so easily phrased can be
so difficult to answer.
Modern mathematics abounds with such questions. A beginning student of mathematics has no conception of the number of important, yet really basic, mathematical
questions that have not yet been solved, despite the number of astute mathematicians
that have been working on them over the years. The surprise that comes later is how
many interesting open questions are amenable to solutions, given the right mixture of
ideas, luck, and persistence.
The main goal of the group theory portion of this book is to be able to obtain as
much information as possible about the classification of groups of relatively small order.
In particular, for a given integer n, say with n ≤ 63, we’d like to classify all groups of
order n.
There is some language we can use to make our discussion clearer.
Definition 2.6.3. We say that two groups are in the same isomorphism class if the
groups are isomorphic to one another.
Isomorphism classes are analogous to equivalence classes, where the equivalence relation used is that of isomorphism. The point is that the collection of all groups of a given
order is not a set, as it has too many elements. (There are too many groups of order 1,
for instance, to form a set, even though there is only one isomorphism class of groups
of order 1.) Thus, we are looking at an analogue for classes (as opposed to sets) of an
equivalence relation.
The interesting thing is that classification, as defined above, can only occur in situations where the isomorphism classes themselves form a set. Indeed, the next lemma is
an immediate consequence of the definition of a classification.
Lemma 2.6.4. The set {Gi | i ∈ I} classifies the collection of groups with a given
property if and only if each isomorphism class of the groups with the stated property
includes exactly one of the groups Gi . In particular, the collection of isomorphism
classes of the groups in question forms a set which is in one-to-one correspondence with
{Gi | i ∈ I}.
Notice that if X is a finite set, then the set of possible group multiplications we could
put on X is a subset of the set of functions from X × X to X. But the set of all functions
from one finite set to another is finite, by Corollary 1.7.6. Thus, there are finitely many
possible group multiplications on a finite set. Different multiplications will in many cases
give rise to isomorphic groups, but the number of isomorphism classes of groups with
|X| elements must be finite.
Proposition 2.6.5. Let n > 0. Then there are finitely many isomorphism classes of
groups of order n.
A countable union of finite sets is countable.
Corollary 2.6.6. The set of isomorphism classes of finite groups is countable.
Exercises 2.6.7.
1. Let X = {x, y} be a set with two elements. Which of the functions from X × X to
X are group multiplications? Which are monoid multiplications?
2. Classify the groups of order 3.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
2.7
37
The Group of Rotations of the Plane
Here, we describe an important subgroup of Gl2 (R), the group, SO(2), of rotations of
the plane.
Definition 2.7.1. For θ ∈ R, we define the rotation matrix Rθ by
cos θ − sin θ
Rθ =
.
sin θ
cos θ
Here, the sine and cosine are taken with respect to radian measure.
Elements of M2 (R) act as transformations of the plane in the usualway. We identify
the plane with the space of column vectors v = ( xy ). A matrix A = ac db acts on the
plane by the usual multiplication of matrices on columns:
ax + by
A · v =
.
cx + dy
To display the geometric effect of the transformation induced by a rotation matrix, we
make use of polar coordinates and the sum formulæ for sines and cosines, as developed
in calculus.
Proposition 2.7.2. The transformation of the plane induced by the rotation matrix Rθ
is the counterclockwise rotation of the plane through the angle θ.
Proof We write the vector v = ( xy ) in polar coordinates, setting x = r cos φ and
y = r sin φ, where r = x2 + y 2 is the length of v and φ is the angle that starts at the
positive x-axis and ends at v . The product formula now gives
r(cos θ cos φ − sin θ sin φ)
r cos(θ + φ)
Rθ · v =
=
,
r(cos θ sin φ + sin θ cos φ)
r sin(θ + φ)
by the sum formulæ for the sine and cosine. Thus, the length of Rθ ·v is the same as that
of v , but the angle that it makes with the positive x-axis has been increased by θ.
As geometric intuition would suggest, the product (composite) of two rotations is
again a rotation:
Lemma 2.7.3. For any real numbers θ and φ, we have
Rθ · Rφ = Rθ+φ .
Proof The definition of matrix multiplication gives
cos θ cos φ − sin θ sin φ − cos θ sin φ − sin θ cos φ
Rθ · R φ =
.
cos θ sin φ + sin θ cos φ
cos θ cos φ − sin θ sin φ
Once again, the result follows from the sum formulæ for the sine and cosine.
Definition 2.7.4. Let SO(2) ⊂ Gl2 (R) be the set of all rotation matrices Rθ , as θ
ranges over all real numbers. We call it the second special orthogonal group.
By Lemma 2.7.3, SO(2) is closed under multiplication in Gl2 (R). Moreover, since
cos 0 = 1 and sin 0 = 0, we have R0 = I2 , the identity element of Gl2 (R). Since
Rθ · R−θ = R0 , R−θ is the inverse of Rθ .
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
38
Lemma 2.7.5. SO(2) is a subgroup of Gl2 (R).
Notice that since addition of real numbers is commutative, Lemma 2.7.3 shows that
Rθ ·Rφ = Rφ ·Rθ . Thus, unlike Gl2 (R), SO(2) is an abelian group. Lemma 2.7.3 actually
shows more:
Proposition 2.7.6. There is a homomorphism, exp, from R onto SO(2), defined by
exp(θ) = Rθ . The kernel of exp is 2π = {2πk | k ∈ Z}. Thus, Rθ = Rφ if and only if
θ − φ = 2πk for some k ∈ Z.
Proof That exp is a homomorphism is the content of Lemma 2.7.3. Suppose that
θ ∈ ker exp. Since the identity element of SO(2) is the identity matrix, the definition of
Rθ forces cos θ to be 1 and sin θ to be 0. From calculus, we know this happens if and
only if θ is a multiple of 2π.6 By Lemma 2.5.9, we have that exp θ = exp φ if and only if
−φ + θ ∈ ker(exp), so the result follows.
We conclude with a calculation of the order of the elements of SO(2).
Proposition 2.7.7. The rotation Rθ has finite order if and only if θ = 2πk/n for some
integers k and n. The order of R2πk/n is n/(k, n).
Proof Suppose that Rθ has finite order. Then (Rθ )n = Rnθ = I2 for some integer n.
Then Proposition 2.7.6 shows that nθ = 2πk for some integer k, so that θ = 2πk/n.
If θ = 2πk/n, the order of Rθ will be the smallest positive integer m for which mθ
is an integer multiple of 2π. Thus, m is the smallest positive integer such that mk is
divisible by n. But a glance at the prime decompositions of n and k shows that this
smallest positive integer is precisely n/(n, k).
In particular, R2π/n has order n for all n ≥ 1.
Corollary 2.7.8. There are embeddings of Zn in SO(2) for all n ≥ 1.
Since SO(2) gives transformations of the plane, this says that Zn may be used to
produce geometric transformations of a familiar space. A certain school of thought
would summarize this by saying that Zn “exists in nature.”
2.8
The Dihedral Groups
Let n > 0, we construct a group, D2n , called the dihedral group of order 2n. These
groups are nonabelian for n > 2, and provide our first examples of nonisomorphic groups
of the same order.
We give a definition here via matrices, which relies on calculus. A direct definition
as an abstract group is also possible, but is a little awkward notationally without some
further discussion. We shall give such a derivation in Section 4.6.
Let b, or bn , if more than one dihedral group is being discussed, be the rotation of
the plane through the angle 2π/n: b = R2π/n ∈ SO(2) in the notation of the last section.
Then by Proposition 2.7.7, b has order n in Gl2 (R). Let a ∈ Gl2 (R) be the matrix
1
0
a=
.
0 −1
6 Actually, one of the nicer ways to prove this fact from calculus is to use derivatives to show that the
kernel of exp is a discrete subgroup of R and apply Problem 5 of Exercises 2.3.18. One may then define
2π as the smallest positive element in ker(exp) and use numerical methods to compute its value. Here,
we may take the sine and cosine to be defined by their Taylor series, and derive their basic properties,
such as their derivatives and their sum formulæ, by manipulating the series.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
39
The matrix a is said to reflect the plane through the x-axis, and we say that a matrix
is a reflection matrix if it is a product a · c with c ∈ SO(2). (It is possible to give a
direct geometric definition of reflection using dot products, and then to prove that these
geometric operations are induced by precisely the matrices just stated. We shall treat
this approach in exercises below, and content ourselves here with the algebraic definition
just given.)
The most important piece of information about rotations and reflections is the way
they interrelate. They don’t commute with each other, but they come close to it:
Lemma 2.8.1. For θ ∈ R, we have
Rθ · a = a · R−θ .
Thus, for b = R2π/n as above, bk · a = a · b−k for any integer k.
Proof Multiplying matrices, we get
Rθ · a =
cos θ
sin θ
sin θ
− cos θ
.
This is equal to the matrix product a · R−θ , since cos(−θ) = cos θ and sin(−θ) = − sin θ.
The statement about powers of b follows since b±k = R±k·2π/n .
The following lemma may be verified by direct computation.
Lemma 2.8.2. The matrix a has order 2.
Definition 2.8.3. Let n ≥ 1. Then the dihedral group of order 2n, D2n ⊂ Gl2 (R), is
given as follows:
D2n = {bk , abk | 0 ≤ k < n},
with a and b = bn as above.
Of course, this definition needs a bit of justification:
Proposition 2.8.4. The dihedral group D2n is a subgroup of Gl2 (R). Moreover, the
elements {bk , abk | 0 ≤ k < n} are all distinct, so that D2n has order 2n.
Proof We first show that the 2n elements bk and abk with 0 ≤ k < n are all distinct. If
abk = bl , then multiplication by b−k on the right gives a = bl−k , a rotation matrix. But
rotation matrices all commute with one another, and Lemma 2.8.1 shows that a fails to
commute with most rotations. Thus, abk = bl .
If abk = abl , then multiplying both sides on the left by a−1 gives bk = bl , which
implies that k ≡ l mod n by Proposition 2.5.17. With 0 ≤ k, l < n, this gives k = l by
Lemma 2.4.4.
We now show that D2n is closed under multiplication in Gl2 (R). By Lemma 2.8.1,
we have abk · abl = a2 bl−k = bl−k , as a has order 2. Since b has order n, any power of b
is equal to one in the list.
Similarly, bk · abl = abl−k is in D2n . The remaining cases are covered by the fact that
b is closed under multiplication.
The fact that D2n is closed under multiplication implies that it is a subgroup of
Gl2 (R), since D2n is finite (Proposition 2.2.13).
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
40
The dihedral groups have geometric meaning as groups of transformations of regular
polygons in the plane. In particular, for n ≥ 3, there is a regular n-gon whose vertices
are the points
cos(2πk/n)
vk =
sin(2πk/n)
for 1 ≤ k ≤ n. This means that the n-gon is the set of convex
of the vertices,
combinations
n
where a convexcombination of v1 , . . . , vn is a vector sum i=1 aivi with 0 ≤ ai ∈ R for
n
1 ≤ i ≤ n and i=1 ai = 1.
If c is a matrix in D2n , then it is easy to see that for each i with 1 ≤ i ≤ n, the matrix
product c · vi is equal to vj for some j with 1 ≤ j ≤ n. In other words, multiplication
by c carries each vertex of our n-gon to another vertex of the n-gon. But the linearity of
matrix multiplication now shows that multiplication by c carries convex combinations of
the vertices to convex combinations of the vertices, and hence c gives a transformation
of the n-gon. In fact, it can be shown that the elements of D2n give all possible linear
transformations from the n-gon onto itself.
Exercises 2.8.5.
1. What is the order of abk ∈ D2n ?
2. Show that D4 and Z4 are non-isomorphic abelian groups of order 4.
3. Show that D2n is nonabelian if and only if n ≥ 3. Deduce that D2n and Z2n are
non-isomorphic groups of order 2n for all n ≥ 3.
4. Show that there is a subgroup of Gl2 (R) consisting of all rotation matrices Rθ ,
together with the matrices a · Rθ . (This subgroup is generally called O(2), the
second orthogonal group.) What is the order of a · Rθ ?
5. Show that if k divides n, then there is a subgroup of D2n which is isomorphic to
D2k .
† 6. Let G be a group. Show that there is a one-to-one correspondence between the
homomorphisms from D2n to G and pairs of elements x, y ∈ G such that o(x)
divides 2, o(y) divides n, and x−1 yx = y −1 .
7. Show that if k divides n, then there is a homomorphism of D2n onto D2k .
8. Let be a line through the origin in the plane. Standard vector geometry shows
there is a vector u of length 1 such that is the set of all vectors perpendicular to
u. (Note this does not define u uniquely: −u would do as well.) Then the reflection
through is the function f defined by f (w)
=w
− 2(w
· u)u, where · is the dot
product. Show that the reflection through is the transformation induced by the
matrix aRθ for some θ. Show also that the transformation induced by aRθ is the
reflection through some line for any θ ∈ R.
2.9
Quaternions
We construct yet another family of groups via matrices. They are the quaternionic groups
Q4n ⊂ Gl4 (R) of order 4n. First, we give a generalization of the rotation matrices. For
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
θ ∈ R, let
⎛
cos θ − sin θ
⎜ sin θ
cos θ
Sθ = ⎜
⎝
0
0
0
0
0
0
cos θ
sin θ
41
⎞
0
0 ⎟
⎟.
− sin θ ⎠
cos θ
The geometric effect of Sθ on an element of the space R4 of column vectors is to simultaneously rotate the first two coordinates and rotate the last two coordinates, treating
these pairs of coordinates as independent planes.
The following lemma may be verified directly by multiplying the matrices.
Lemma 2.9.1. Let θ and φ be real numbers. Then Sθ · Sφ = Sθ+φ . Thus, there is a
homomorphism f : R → Gl4 (R) defined by f (θ) = Sθ . Moreover, as in the case of
rotations of the plane, Sθ = Sφ if and only if θ − φ = 2πk for some k ∈ Z.
Once again, we shall give a matrix a that will interact with these “rotations” in an
interesting way. It is given by
⎛
0
⎜ 0
a=⎜
⎝ −1
0
0 1
0 0
0 0
1 0
⎞
0
−1 ⎟
⎟.
0 ⎠
0
Lemma 2.9.2. The matrix a has order 4. The lower powers of
⎛
⎞
⎛
−1
0
0
0
0
⎜ 0 −1
⎟
⎜ 0
0
0
2
3
⎟ and a = ⎜
a = −I4 = ⎜
⎝ 0
⎝ 1
0 −1
0 ⎠
0
0
0 −1
0
a are given by
⎞
0 −1 0
0
0 1 ⎟
⎟.
0
0 0 ⎠
−1 0 0
Proof The square of a is as stated by direct calculation. We then have that a3 =
a2 a = −a, which is the displayed matrix. Clearly, this is not the identity matrix, but
(−I4 )2 = a4 is the identity matrix. Thus, the 4-th power of a is the smallest that will
produce the identity, so that a has order 4 and a3 = a−1 .
Once again, there is a deviation from commutativity between the Sθ and a. The
following lemma may be verified by direct computation.
Lemma 2.9.3. We have Sθ · a = a · S−θ .
Because Sθ and Sφ commute for any real numbers θ and φ, we obtain the following
corollary.
Corollary 2.9.4. The matrix a is not equal to Sθ for any θ.
Write b (or b2n if more than one quaternionic group is under discussion) for S2π/2n .
Then b has order 2n. The quaternionic group of order 4n is the subgroup of Gl4 (R)
generated by a and b:
Definition 2.9.5. The quaternionic group of order 4n, Q4n ⊂ Gl4 (R), is given as follows:
Q4n = {bk , abk | 0 ≤ k < 2n},
with a and b as above.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
42
Once again we need some justification.
Proposition 2.9.6. For each n ≥ 1, Q4n is a subgroup of Gl4 (R). Moreover, the matrices {bk , abk | 0 ≤ k < 2n} are all distinct, so that Q4n does indeed have order 4n.
Proof The key here is that a2 = −I4 = Sπ = bn , which is why b was chosen to have
even order. Thus, abk · abl = a2 bl−k = bn bl−k = bn+l−k . Since bk · abl = abl−k and since
b has order 2n, Q4n is closed under multiplication. Since it is finite, it is a subgroup by
Proposition 2.2.13.
It remains to show that the stated elements are all distinct. As in the dihedral case,
it suffices to show that abk = bl , and hence that a = bk−l for any k and l. But once again
this follows from the fact that a cannot equal Sθ for any θ.
Exercises 2.9.7.
1. What is the order of abk ∈ Q4n ?
2. Show that Q4 is isomorphic to Z4 .
3. For n ≥ 2, show that Q4n is isomorphic to neither D4n nor Z4n .
4. Show that if k divides n, then there is a subgroup of Q4n which is isomorphic to
Q4k .
† 5. Let G be a group. Show that there is a one-to-one correspondence between the
homomorphisms from Q4n to G and pairs of elements x, y ∈ G such that o(x)
divides 4, o(y) divides 2n, x2 = y n , and x−1 yx = y −1 .
6. Show that there is a homomorphism of Q4n onto D2n .
7. Show that if n is an odd multiple of k, then there is a homomorphism of Q4n onto
Q4k .
8. Show that every subgroup of Q8 other than e contains a2 . (Hint: The cyclic
subgroups are always the smallest ones in a group.)
9. Let n be any power of 2. Show that every subgroup of Q4n other than e contains
a2 .
2.10
Direct Products
We give a way of constructing new groups from old. It satisfies an important universal
property with respect to homomorphisms.
Definition 2.10.1. Let G and H be groups. Then the product (or direct product) of
G and H is the group structure on the cartesian product G × H which is given by the
binary operation
(g1 , h1 ) · (g2 , h2 ) = (g1 g2 , h1 h2 ).
We denote this group by G × H.
Once again, we must justify our definition:
Lemma 2.10.2. The direct product G × H is a group.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
43
Proof Associativity follows directly from the associativity of G and H. The identity
element is the ordered pair (e, e). The inverse of (g, h) is (g −1 , h−1 ).
The reader should note that while this is not the only interesting group structure we
can put on the product of G and H (we shall discuss semidirect products in Section 4.6),
it is the only one we shall refer to as G × H.
The next result is immediate from Proposition 1.6.4.
Corollary 2.10.3. Let G and H be finite groups. Then G × H is finite, and
|G × H| = |G| · |H|.
The reader should supply the verification for the next lemma.
Lemma 2.10.4. There are subgroups G × e = {(g, e) | g ∈ G} and e × H = {(e, h) | h ∈
H} of G × H. Any element of G × e commutes with any element of e × H. There are
homomorphisms ι1 : G → G × H and ι2 : H → G × H defined by ι1 (g) = (g, e) and
ι2 (h) = (e, h). These induce isomorphisms G ∼
= G × e and H ∼
= e × H.
There are also homomorphisms π1 : G × H → G and π2 : G × H → H defined by
π1 (g, h) = g and π2 (g, h) = h.
The homomorphisms ιi are called the canonical inclusions of G and H in G × H. The
homomorphisms π1 and π2 are called the projections of G × H onto the first and second
factors, respectively.
Recall that a function into a product is defined uniquely by its component functions:
f : X → G × H is defined by f (x) = (f1 (x), f2 (x)), where f1 : X → G and f2 : X → H.
In fact, the component functions are nothing other than the composites fi = πi ◦ f ,
where the πi are the projections in the above lemma. The group product in G × H is
the unique product that makes the following statement true:
Lemma 2.10.5. Let G, H, and K be groups. A function f : K → G × H is a homomorphism if and only if its component functions are homomorphisms.
We can also study homomorphisms out of a product. If f : G × H → K is a
homomorphism, so are f ◦ ι1 : G → K and f ◦ ι2 : H → K.
Proposition 2.10.6. Let f : G × H → K be a homomorphism. Then every element of
the image of f ◦ ι1 commutes with every element of the image of f ◦ ι2 . Moreover, f is
uniquely determined by its restrictions to G × e and e × H.
Conversely, given homomorphisms g1 : G → K and g2 : H → K such that each
element in the image of g1 commutes with each element in the image of g2 , then there is
a unique homomorphism f : G × H → K with f ◦ ιi = gi for i = 1, 2.
Proof Since (g, h) = (g, e) · (e, h), f (g, h) = f ◦ ι1 (g) · f ◦ ι2 (h). Thus, f is determined
by the f ◦ ιi , which are determined by the restriction of f to the stated subgroups. The
commutativity assertion follows since if two elements commute, so must their images
under any homomorphism.
Given the gi as above, define f : G × H → K by f (g, h) = g1 (g) · g2 (h). Then
f (g, h) · f (g , h ) = g1 (g)g2 (h)g1 (g )g2 (h ) = g1 (g)g1 (g )g2 (h)g2 (h ) = g1 (gg )g2 (hh ),
with the key step being given by the fact that g1 (g ) commutes with g2 (h). But the last
term is clearly f ((g, h) · (g , h )).
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
44
It is valuable to know when a group breaks up as the product of two of its subgroups.
Definition 2.10.7. Let H and K be subgroups of the group G. We say that G is the
internal direct product of H and K if there is an isomorphism μ : H × K → G such
that μ ◦ ι1 and μ ◦ ι2 are the inclusions of H and K, respectively, as subgroups of G.
Explicitly, this says μ(h, k) = hk.
We now give a characterization of internal direct products.
Proposition 2.10.8. Let H and K be subgroups of the group G. Then G is the internal
direct product of H and K if and only if the following conditions hold.
1. H ∩ K = e.
2. If h ∈ H and k ∈ K, then h and k commute in G.7
3. G is generated by the elements of H and K.
Proof Suppose that G is the internal direct product of H and K. Then the function
μ : H × K → G defined by μ(h, k) = hk is an isomorphism of groups. Since μ is a
homomorphism and since μ ◦ ι1 (h) = h and μ ◦ ι2 (k) = k, each h ∈ H must commute
with each k ∈ K by Proposition 2.10.6. Since μ is onto, each element of G must have
the form hk, with h ∈ H and k ∈ K, so the elements of H and K must generate G.
Finally, let x ∈ H ∩ K. Then (x, x−1 ) ∈ H × K, and μ(x, x−1 ) = xx−1 = e. Since
μ is injective, this says that (x, x−1 ) is the identity element of H × K, which is (e, e).
Thus, x = e, and hence H ∩ K = e.
Conversely, suppose that the three stated conditions hold. Since the elements of H
commute with those of K, Proposition 2.10.6 shows that the function μ : H × K → G
given by μ((h, k)) = hk is a homomorphism. Suppose that (h, k) is in the kernel of μ.
Then hk = e, and hence h = k −1 is in H ∩ K. But H ∩ K = e, and hence h = e, and
hence k = h−1 = e also. Thus, (h, k) = (e, e), and hence μ is injective.
Since H and K generate G, any element in G may be written as a product g1 . . . gn
where gi is an element of either H or K for i = 1, . . . , n. Since elements of H commute
with those of K, we can rewrite this as a product h1 . . . hr k1 . . . ks with hi ∈ H for
i = 1, . . . , r and ki ∈ K for i = 1, . . . , s, by moving the elements in H past those in K.
But this is just μ(h1 . . . hr , k1 . . . ks ), so μ is onto.
We shall see in Proposition 3.7.12 that the second condition of Proposition 2.10.8
may be replaced with the statement that both H and K are what’s known as normal
subgroups (to be defined in Section 3.7) of G.
Now let us generalize the material in this section to the study of products of more
than two groups.
Definition 2.10.9. Suppose given groups G1 , . . . , Gk , for k ≥ 3. The product (or direct
product), G1 × · · · × Gk , of the groups G1 , . . . , Gk is the set of all k-tuples (g1 , . . . , gk )
with gi ∈ Gi for 1 ≤ i ≤ k, with the following multiplication:
(g1 , . . . , gk ) · (g1 , . . . , gk ) = (g1 g1 , . . . , gk gk ).
The reader should be able to supply the proof of the following proposition.
7 Of
course, neither H nor K need be abelian. Thus, if h, h ∈ H, then h and h must commute with
every element of K, but they need not commute with each other.
CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES
45
Proposition 2.10.10. The direct product G1 × · · · × Gk is a group under the stated
multiplication. There are homomorphisms πi : G1 × · · · × Gk → Gi and ιi : Gi → G1 ×
· · · × Gk for 1 ≤ i ≤ k, defined by πi (g1 , . . . , gk ) = gi , and ιi (g) = (e, . . . , e, g, e, . . . , e),
where g appears in the i-th coordinate.
Given homomorphisms fi : H → Gi for 1 ≤ i ≤ k, there is a unique homomorphism
f : H → G1 × · · · × Gk such that πi ◦ f = fi for 1 ≤ i ≤ k. Similarly, for each collection
hi : Gi → H of homomorphisms such that each element of the image of hi commutes
with each element of the image of hj whenever i = j, there is a unique homomorphism
h : G1 × · · · × Gk → H with h ◦ ιi = hi for 1 ≤ i ≤ k.
Similarly, we can discuss internal direct products of several subgroups.
Definition 2.10.11. Let H1 , . . . , Hk be subgroups of G. We say that G is the internal direct product of H1 , . . . , Hk if the function μ : H1 × · · · × Hk → G defined by
μ(h1 , . . . , hk ) = h1 . . . hk is an isomorphism of groups.
We shall not give a characterization here of the internal direct product of multiple
subgroups. (I.e., we shall not give a generalization of Proposition 2.10.8 to the case of
more than two subgroups.) The reader may wish to formulate one after reading the proof
of Proposition 4.4.9.
Exercises 2.10.12.
1. Show that the order of (g, h) ∈ G × H is the least common multiple of the orders
of g and h.
† 2. More generally, show that the order of (g1 , . . . , gk ) ∈ G1 × · · · × Gk is the least
common multiple of the orders of g1 , . . . , gk .
3. Show that Z6 is isomorphic to Z3 × Z2 .
† 4. Let m and n be relatively prime. Show that Zmn is isomorphic to Zm × Zn .
5. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Show that
Zn ∼
= Zpr11 × · · · × Zprkk .
6. Show that if m and n are not relatively prime, then Zm × Zn is not cyclic.
7. Show that D4 is isomorphic to Z2 × Z2 .
8. Show that D12 is isomorphic to D6 × Z2 .
9. Show that if k is odd, then D4k is isomorphic to D2k × Z2 .
10. Show that G × H ∼
= H × G.
11. Show that (G × H) × K ∼
= G × (H × K) ∼
= G × H × K.
12. Show that G × e ∼
= G, where e is the trivial group.
13. Show that the group multiplication μ : G × G → G is a homomorphism if and only
if G is abelian.
14. Let fi : Gi → Hi be a homomorphism for 1 ≤ i ≤ k. Define f1 × · · · × fk :
G1 × · · · × Gk → H1 × · · · × Hk by f1 × · · · × fk (g1 , . . . , gk ) = (f1 (g1 ), . . . , fk (gk )).
Show that f1 × · · · × fk is a homomorphism. In addition, show that f1 × · · · × fk
is an isomorphism if and only if each fi is an isomorphism.
15. Show that not every homomorphism f : G1 × G2 → H1 × H2 has the form f1 × f2 .
Chapter 3
G-sets and Counting
Many of the examples in Chapter 2 were obtained from matrices, and hence give symmetries of n-dimensional Euclidean space, Rn . Here, we study the symmetries of sets,
obtaining a more combinatorial view of symmetry.
In the first section, we define the symmetric group on n letters, Sn , and prove Cayley’s
Theorem, which shows that every finite group is a subgroup of Sn for some n. Later,
in Chapter 10, we will show that Sn embeds in Gln (R). In this manner, we obtain the
assertion of the last chapter that every finite group is a subgroup of Gln (R).
The symmetric groups are interesting in their own right as examples of groups. Via
the technique of cycle decomposition, which we give in Section 3.5, one can say quite a
bit about its structure.
Also, there are natural subgroups An ⊂ Sn , called the alternating groups, which are
the easiest examples of nonabelian simple groups. We define An in Section 3.5 in terms
of the sign homomorphism ε : Sn → {±1} to be defined there. We shall define simple
groups in Section 4.2, and show that An is simple for n ≥ 5. We may then deduce in
Chapter 5 that Sn is not what’s known as a solvable group for n ≥ 5.
The non-solvability of Sn will allow us, in Chapter 11, to prove Galois’s famous
theorem that there is no formula for solving polynomials of degree ≥ 5. This should be
enough to arouse one’s interest in the symmetric groups.
We also introduce a very important concept in group theory: that of the set G/H of
left cosets of a subgroup H in G. One of the first consequences of the study of cosets is
Lagrange’s Theorem: if G is finite, then the order of any subgroup of G must divide the
order of G. We give Lagrange’s Theorem in Section 3.2, and give many more applications
of the study of cosets throughout the material on group theory.
The sets of cosets G/H form an important example of the notion of a group acting on a
set. We formulate the notion of G-sets in Section 3.3 and give a generalization of Cayley’s
Theorem in which a one-to-one correspondence is established between the actions of G
on a set X and the homomorphisms from G to the group of symmetries of X. This
allows us to associate to a finite G-set (such as G/H, if G is finite) a homomorphism
from G into Sn for appropriate n. These homomorphisms are quite useful in determining
the structure of a group G, and will figure prominently in the classification results we
get in Chapter 5. Section 3.3 also contains a couple of important counting arguments,
including what’s known as the G-set Counting Formula.
In the final two sections of the chapter, we study actions of one group on another.
The point is that not all symmetries of a group H preserve the group structure. The ones
that do are the automorphisms of H, which are defined to be the group isomorphisms
46
CHAPTER 3. G-SETS AND COUNTING
47
H → H. These form a subgroup, Aut(H), of the group of symmetries of H. We study
the actions of G on H such that the action of each g ∈ G induces an automorphism of
H.
Actions through automorphisms play an important role in constructing and classifying
the extensions of one group by another (Section 4.7), and therefore, as we shall see, in
the classification of groups.
Closer at hand, there is a very important action of a group G on itself through
automorphisms, called conjugation. Many of the important results in group theory follow
from the study of conjugation, including the Noether Isomorphism Theorems of Chapter 4
and the Sylow Theorems of Chapter 5. We shall begin that study in this chapter.
3.1
Symmetric Groups: Cayley’s Theorem
Here, we show that every finite group embeds in the group of symmetries of a finite set.
Recall that a function is bijective if it is one-to-one and onto.
Definitions 3.1.1. Let X be a set. By a permutation of X we mean a bijective function
σ : X → X.
We write S(X) for the set of all permutations of X. Note that S(X) has a binary
operation given by composition of functions: for σ, τ ∈ S(X), σ · τ is the composite
σ ◦ τ . As we shall see shortly, S(X) is a group under this operation, called the group of
symmetries of X.
Recall that the identity function 1X of a set X is the function defined by 1X (x) = x
for all x ∈ X.
Recall from Proposition 1.1.6 that every bijective function σ : X → Y has an inverse
function σ −1 : Y → X, with the property that σ −1 ◦ σ = 1X and σ ◦ σ −1 = 1Y . Here,
the inverse function σ −1 is defined by setting σ −1 (y) to be the unique x ∈ X such that
σ(x) = y.
Lemma 3.1.2. Let X be a set. Then S(X) forms a group under the operation of composition of functions. The identity element of S(X) is the identity function 1X , and the
inverse of an element σ ∈ S(X) is its inverse function, σ −1 .
We shall see below that if X is infinite, then not only is S(X) an infinite group, it
also contains elements of infinite order. On the other hand, if X is finite, then there are
only finitely many elements in S(X), as, in fact, there are only finitely many functions
from X to X by Corollary 1.7.6.
We shall calculate the order of S(X), for X finite, in Corollary 3.3.11.
Definition 3.1.3. Let f : X → Y be a bijection of sets. Define the induced map
f∗ : S(X) → S(Y ) by
f∗ (σ) = f ◦ σ ◦ f −1
for σ ∈ S(X).
Since two sets are considered equivalent if they may be placed in bijective correspondence, the following lemma should come as no surprise.
Lemma 3.1.4. Let f : X → Y be a bijection of sets. Then the induced map f∗ : S(X) →
S(Y ) is an isomorphism of groups.
CHAPTER 3. G-SETS AND COUNTING
48
Proof It is easy to see that f∗ gives a group homomorphism from S(X) to S(Y ). Moreover, (f −1 )∗ : S(Y ) → S(X) gives an inverse function for f∗ , so f∗ is an isomorphism.
Note that there is no preferred isomorphism between S(X) and S(Y ), but rather an
isomorphism for each choice of bijection from X to Y .
Of special interest here is the case when X is finite. Our preferred model for a set
with n elements is {1, . . . , n}.
Definition 3.1.5. We write Sn for S({1, . . . , n}), and call it the symmetric group on n
letters.
Fixed points are an important feature of a permutation:
Definition 3.1.6. Let σ be a permutation of X. Then σ is said to fix an element x ∈ X
if σ(x) = x. We call such an x a fixed point of σ.
We should now construct some examples of elements of Sn . Recall that if Y is a
subset of X, then the complement of Y in X is the set of all elements of X that are not
in Y .
Definition 3.1.7. Let i1 , . . . , ik be k distinct elements of {1, . . . , n}, k ≥ 2. We write
(i1 i2 · · · ik ) for the permutation of {1, . . . , n} which carries ij to ij+1 for 1 ≤ j < k,
carries ik to i1 , and is the identity map on the complement of {i1 , . . . , ik } in {1, . . . , n}.
We call (i1 i2 · · · ik ) a k-cycle, or a cycle of length k.
Note that a 2-cycle (i j) just interchanges i and j and fixes all other elements of
{1, . . . , n}. Another name for a 2-cycle is a transposition.
Some people consider a 1-cycle (i) to be another way of writing the identity map.
While this is consistent with the notation above, our cycles will all have length ≥ 2.
Note that the permutations (i1 i2 · · · ik ) and (ik i1 · · · ik−1 ) are equal, as they have
the same effect on all elements of {1, . . . , n}. Thus, there is more than one way to denote
a given k-cycle.
We can calculate the order and inverse of a k-cycle.
Lemma 3.1.8. A k-cycle (i1 i2 · · · ik ) has order k. Its inverse is (ik · · · i2 i1 ).
Proof Let σ = (i1 i2 · · · ik ). An easy induction shows that for 1 ≤ j < k, σ j (i1 ) =
ij+1 = i1 , while σ k does fix i1 . Thus, σ j = e for 1 ≤ j < k. But since σ may also be
written as (i2 · · · ik i1 ), σ k fixes i2 by the preceding argument. Another induction shows
that σ k fixes each of the ij . Since it also fixes the complement of {i1 , . . . , ik }, we have
σ k = e.
The permutation (ik · · · i1 ) is easily seen to reverse the effect of σ.
One of the reasons for studying the symmetric groups is that every finite group is
isomorphic to a subgroup of Sn for some n. Recall that an embedding of groups is an
injective homomorphism.
Theorem 3.1.9. (Cayley’s Theorem) Let G be a group. For g ∈ G write g : G → G
for the function g (g ) = gg . Then g is a permutation of G. Moreover, if we define
Λ : G → S(G) by Λ(g) = g , then Λ is an embedding. In consequence, if G is finite, then
it is isomorphic to a subgroup of S|G| .
CHAPTER 3. G-SETS AND COUNTING
49
Proof To show that g is injective, suppose g (g ) = g (g ), and hence gg = gg .
Multiplying both sides on the left by g −1 and cancelling, we get g = g . And g is
surjective because g = g (g −1 g ) for all g ∈ G. Thus, g is a permutation of G.
Now Λ(g) · Λ(g ) is the permutation g ◦ g . We wish to show this coincides with
Λ(gg ), which is the permutation gg . We show they have the same value on each g ∈ G.
(
g ◦ g )(g ) = g(g g ) = (gg )g = gg (g ).
Thus, Λ is a homomorphism. To show it is an injection, we consider its kernel.
Suppose that g = e. Thus, g (g ) = g for all g ∈ G, i.e., gg = g for all g ∈ G.
But multiplying both sides of the equation on the right by (g )−1 , we get g = e. Thus,
ker Λ = e, the trivial subgroup, so Λ is an embedding by Lemma 2.5.9.
Recall that a set is countably infinite if it may be put in one-to-one correspondence
with the positive integers, Z+ . It is easy to show that Z is countably infinite. Since
every nonzero element of Z has infinite order, the next corollary follows from Cayley’s
Theorem.
Corollary 3.1.10. Let X be a countably infinite set. Then S(X) contains elements of
infinite order.
Cayley’s Theorem shows that any finite group G is isomorphic to a subgroup of S|G| .
But most groups are actually isomorphic to a group of permutations of a smaller number
of letters.
As we shall see in a moment, there is a natural choice of embedding of Sn in Sn+1 .
Thus, if G embeds in Sn , then it also embeds in Sn+1 .
In fact, if Y is a subset of X, there is a natural embedding of S(Y ) in S(X):
Definition 3.1.11. Let Y be a subset of X, and let i : Y ⊂ X be the inclusion map.
Define i∗ : S(Y ) → S(X) by
σ(x) if x ∈ Y
(i∗ (σ))(x) =
x
otherwise.
We shall refer to i∗ as the canonical inclusion of S(Y ) in S(X).
Lemma 3.1.12. Let i : Y ⊂ X be the inclusion of a subset, Y , of a set X. Then
the canonical inclusion i∗ : S(Y ) → S(X) is an embedding onto the subgroup of S(X)
consisting of those elements of S(X) which fix all elements of the complement of Y in
X.
Proof Write X − Y for the complement of Y in X. It is easy to see that i∗ is an
embedding and that elements in the image of i∗ fix all elements of Y − X. Thus, it
suffices to show that any element of S(X) that fixes the elements of Y − X lies in the
image of i∗ .
Thus, suppose that τ ∈ S(X) fixes the elements of of X − Y . Note that it suffices
to show that τ restricts to a bijection from Y to itself, as then this restriction defines
an element of S(Y ), which we shall denote by τ . Since τ is the identity on X − Y and
agrees with τ on Y , we obtain that i∗ (τ ) = τ , as desired.
We first show that τ (Y ) ⊂ Y . Suppose not. Then we can find y ∈ Y such that
τ (y) = x ∈ Y − X. But since τ fixes Y − X, we have τ (x) = x, so x and y have the same
image under τ . This is impossible, as τ is injective.
CHAPTER 3. G-SETS AND COUNTING
50
Thus, τ restricts to a function from Y to Y , which is injective, because restrictions
of injective functions are always injective. But note that if x ∈ X − Y , then τ (x) cannot
lie in Y . Thus, since τ is onto, we must have τ (Y ) = Y . Thus, τ restricts to a bijection
from Y to itself, and the result follows.
Taking i to be the inclusion map {1, . . . , n − 1} ⊂ {1, . . . , n}, we get the canonical
inclusion Sn−1 ⊂ Sn . The next corollary is now immediate from Lemma 3.1.12.
Corollary 3.1.13. The canonical inclusion of Sn−1 in Sn identifies Sn−1 with the subgroup of Sn consisting of those permutations that fix n.
Note that the inclusion of {1, . . . , n} in the positive integers, Z+ , induces a canonical
inclusion Sn ⊂ S(Z+ ) which is compatible with the inclusion of Sn in Sn+1 .
Definition 3.1.14. Write i∗ : Sn ⊂ S(Z+ ) for the canonical inclusion, and define S∞ ⊂
S(Z+ ) by
S∞ =
i∗ (Sn ).
n≥1
Lemma 3.1.15. S∞ is a subgroup of S(Z+ ).1
Proof Since each (i∗ (x))−1 = i∗ (x−1 ), S∞ is closed under inverses, and it suffices to
show that S∞ is closed under multiplication.
For x, y ∈ S∞ , there are m, n ≥ 1 such that x ∈ i∗ (Sm ) and y ∈ i∗ (Sn ). Suppose,
then, that m ≤ n. Since the canonical inclusions of the Sk in S(Z+ ) are compatible
with their canonical inclusions in one another, we have i∗ (Sm ) ⊂ i∗ (Sn ), and hence
x, y ∈ i∗ (Sn ). But i∗ (Sn ) is a subgroup of S(Z+ ), and hence xy ∈ i∗ (Sn ) ⊂ S∞ .
In practice, we identify each Sn with its image under i∗ , and treat Sn as a subgroup
of S∞ (and hence also of S(Z+ )).
Exercises 3.1.16.
1. Show that S2 is isomorphic to Z2 .
2. Show that S(X) is nonabelian whenever X has more than two elements.
3. Show that the square of a 4-cycle is not a 4-cycle.
4. What is the order of (1 2) · (3 4) in S4 ?
5. What is the order of (1 2 3) · (4 5) in S5 ?
6. Show that there is an embedding f : D6 → S3 induced by setting f (a) = (1 3) and
f (b) = (1 2 3). We shall show in Section 3.3 that S3 has order 6. We may then
deduce that f is an isomorphism.
7. Define x, y ∈ S4 by x = (1 4) · (2 3) and y = (1 2 3 4). Show that there is an
embedding f : D8 → S4 induced by setting f (a) = x and f (b) = y.
8. Show that D12 embeds in S5 .
9. Show that D2n embeds in Sn for all n ≥ 3.
1 Those
who are acquainted with category theory should note that we can identify S∞ with the direct
limit lim
−→ Sn .
CHAPTER 3. G-SETS AND COUNTING
51
10. Show that Z2 × Z2 embeds in S4 .
11. Show that Zm × Zn embeds in Sm+n .
12. Show that the subgroup S∞ ⊂ S(Z+ ) is an infinite group whose elements all have
finite order.
13. Show that an element of S(Z+ ) lies in S∞ if and only if it fixes all but finitely
many of the elements of Z+ .
14. Let G be a group and let g ∈ G. Define rg : G → G by rg (x) = xg. Show that
rg ∈ S(G). Show that the function R : G → S(G) defined by R(g) = rg−1 is an
embedding.
‡ 15. Let f : Y → Z be any injective function, and define f∗ : S(Y ) → S(Z) by
f (σ(y)) if z = f (y)
(f∗ (σ))(z) =
z
if z is not in the image of f .
(a) Show that f∗ is an embedding onto the subgroup of S(Z) consisting of those
permutations that fix all elements of Z which are not in f (Y ).
(b) If f is a bijection, show that this f∗ coincides with the isomorphism f∗ given
in Definition 3.1.3.
(c) If f is the inclusion of a subset Y ⊂ Z, show that this f∗ coincides with the
embedding f∗ given in Definition 3.1.11.
(d) Let g : Z → W be another injection. Show that g∗ ◦ f∗ = (g ◦ f )∗ .
(e) If X is finite, and if f : X → Z+ is injective, show that the image of f∗ lies in
S∞ . Thus, there is an induced embedding f∗ : S(X) → S∞ .
16. Let X be a set. Show that the set of all functions f : X → X forms a monoid
under the operation of composition of functions. Show that S(X) is its group of
invertible elements.
17. Show that the set of functions from {1, 2} to itself which take 1 to itself gives, via
composition of functions, the unique monoid with two elements that is not a group.
3.2
Cosets and Index: Lagrange’s Theorem
We here begin the study of a very important concept in algebra: the left cosets of a
subgroup H ⊂ G. One of the consequences of this consideration is the famous theorem
of Lagrange: If H is a subgroup of the finite group G, then the order of H divides the
order of G.
Lagrange’s Theorem is the very first step in trying to classify groups up to isomorphism.
Definition 3.2.1. Let H be a subgroup of G and let x ∈ G. Then the left coset xH of
H in G is the following subset of G:
xH = {xh | h ∈ H}.
CHAPTER 3. G-SETS AND COUNTING
52
Thus, as x ranges over the elements of G, we get a collection of different subsets
xH ⊂ G, which are the left cosets of H in G.
Note that eH is just H ⊂ G. The coset H is special in that it is the only left coset
of H in G which is a subgroup. (The next lemma shows that e ∈ xH if and only if
xH = H.) We study the cosets to understand how H relates to the other elements of G.
Example 3.2.2. Let G be a cyclic group of order 4, written multiplicatively, with generator b: G = {e, b, b2 , b3 }. Let H be the cyclic subgroup generated by b2 : H = b2 =
{e, b2 }. Then an easy inspection shows that eH = b2 H = {e, b2 }, while bH = b3 H =
{b, b3 }. Since we have inspected xH for each element x ∈ G, we see that these are all of
the left cosets of H in G.
Recall that cosets are defined to be subsets of G. Thus, since bH and b3 H have the
same elements, they are equal as cosets. In particular, for this choice of H and G, there
are exactly two left cosets of H in G.
We now show that two left cosets that have a single element in common must be
equal.
Lemma 3.2.3. Let H be a subgroup of G and let x, y ∈ G. Then the following statements
are equivalent.
1. xH = yH.
2. xH ∩ yH = ∅.
3. x−1 y ∈ H.
4. y −1 x ∈ H.
Proof Clearly, the first statement implies the second. Suppose, then, that the second
holds. Let z ∈ xH ∩ yH. Then there are h, h ∈ H with z = xh = yh . But then
x−1 y = h(h )−1 ∈ H. Thus, the second statement implies the third.
The third and fourth statements are equivalent, as y −1 x is the inverse of x−1 y. Suppose, then, that the fourth statement holds. Thus, y −1 x = k ∈ H, and hence x = yk.
But then xh = ykh ∈ yH for all h ∈ H, and hence xH ⊂ yH. But also, y = xk −1 , so
that yh = xk −1 h for all h ∈ H, and hence yH ⊂ xH. Thus, xH = yH, and we see that
the fourth statement implies the first.
Since x = x · e we have x ∈ xH for each x ∈ G. But any two left cosets of H that
have an element in common must be equal by Lemma 3.2.3.
Corollary 3.2.4. Let H be a subgroup of G. Then each element x ∈ G is contained in
exactly one left coset of H, and that coset is xH. In particular, the elements x and y lie
in the same left coset of H if and only if xH = yH.
If the operation of G is written additively, we have an additive notation for cosets.
Notation 3.2.5. Suppose that G is abelian and that we use additive notation for the
group operation in G. Then we also use additive notation for the left cosets of a subgroup
H ⊂ G. For x ∈ G, we write x + H for the coset which contains x:
x + H = {x + h | h ∈ H}.
It is very useful in group theory to consider the set of all left cosets of H in G.
CHAPTER 3. G-SETS AND COUNTING
53
Definitions 3.2.6. Let H be a subgroup of G. Write G/H for the set whose elements
are the cosets xH of H in G. Thus, if xH = yH as subsets of G, then xH and yH are
equal as elements of G/H. If the number of elements of G/H (which is equal to the
number of distinct left cosets of H in G) is finite, we say that H has finite index in G,
and write [G : H], the index of H in G, for the number of elements in G/H. If the set
G/H is infinite, we write [G : H] = ∞.
We define the canonical map, π : G → G/H, by π(x) = xH for x ∈ G.
Remarks 3.2.7. Let H be a subgroup of G. We put a relation ∼ on G by setting x ∼ y
if x−1 y ∈ H. Then ∼ is an equivalence relation: it is reflexive because (x−1 y)−1 = y −1 x,
and is transitive because x−1 y · y −1 z = x−1 z.
But x−1 y ∈ H if and only if y ∈ xH, so the equivalence class of x under ∼ is precisely
xH. Thus, G/H is the set of equivalence classes of G under ∼, and the canonical map
π : G → G/H, as defined above, takes each x ∈ H to its equivalence class under ∼.
Thus, π agrees with the canonical map of Definition 1.4.5.
The following corollary is immediate from Corollary 3.2.4.
Corollary 3.2.8. Let H be a subgroup of G and let π : G → G/H be the canonical map.
Then π(x) = π(y) if and only if x and y are in the same left coset of H.
Proposition 3.2.9. (Lagrange’s Theorem) Let H be a subgroup of G and suppose that
H is finite. Then every left coset of H has |H| elements.
Given that H is finite, if either of G or [G : H] is finite, then so is the other, and we
have
|G| = [G : H] · |H|.
In particular, if G is a finite group and H is a subgroup of G, then the order of H
divides the order of G.
Proof For x ∈ G, the permutation x defined in the last section (by x (g) = xg)
restricts to a bijection from H to xH (with inverse induced by x−1 ). Thus, if H is finite,
each coset xH has the same number of elements as H.
Since each element of G lies in exactly one left coset of H, the order of G, if finite, is
equal to the sum of the “orders” of the distinct left cosets of H. Since each left coset has
|H| elements, this says that |G| must be equal to the number of left cosets of H times
the order of H.
Of course, if both H and [G : H] are finite, then G is the union of a finite number of
the finite sets xH, and hence G is finite as well.
Recall that the order of a given element of a group is equal to the order of the cyclic
subgroup generated by that element (Corollary 2.5.18).
Corollary 3.2.10. Let G be a finite group. Then the order of any element of G must
divide the order of G.
We should point out that the converse of Lagrange’s Theorem is false: if k divides
the order of G, it is not necessarily the case that G will have a subgroup of order k. For
instance, in Problem 2 of Exercises 4.2.18, we shall see that the alternating group A4 ,
which has order 12, has no subgroup of order 6.
We shall spend some time investigating partial converses to Lagrange’s Theorem, as
they are quite useful in obtaining classification results. In Cauchy’s Theorem, we will
CHAPTER 3. G-SETS AND COUNTING
54
show that if a prime p divides the order of a group, then G has a subgroup of order p.
Then, in the Sylow Theorems, we will show that G has a subgroup whose order is the
highest power of p that divides the order of G.
We can get even better partial converses to Lagrange’s Theorem if we put extra
assumptions on the group G. For instance, Hall’s Theorem (Theorem 5.6.2) gives a
result that holds for what’s known as solvable groups: if G is solvable and |G| = nk
with (n, k) = 1, then G has a subgroup of order n. Even here, we do not get a full
converse, as the alternating group A4 is solvable. But the full converse does hold in
finite abelian groups, or, more generally, in the finite nilpotent groups, which we shall
study in Section 5.7.
Lagrange’s Theorem allows us to classify the groups of order 4.
Corollary 3.2.11. Let G be a group of order 4 which is not cyclic. Then G is isomorphic
to Z2 × Z2 .
Proof Let a = b be elements of G, neither of which is the identity element. Neither
a nor b can have order 4, as then G would be cyclic. Thus, both have order 2. Now
ab = a, as that would imply that b = e, and ab = b, because then a would be the identity
element. And ab = e, as then a and b would be inverse to one another. But an element
of order 2 is its own inverse, so this would mean a = b. Thus, the four elements of G are
e, a, b, and ab.
A similar argument shows that ba cannot equal e, a, or b. But this forces ab = ba, so
that a and b commute. But now Proposition 2.10.8 shows that G is the internal direct
product of a and b.
There are a couple of very important consequences of Lagrange’s Theorem.
Corollary 3.2.12. Let H and K be subgroups of G such that |H| and |K| are relatively
prime. Then H ∩ K = e.
Proof Let x ∈ H ∩ K. Then Corollary 3.2.10 shows that o(x) must divide both |H|
and |K|. Since |H| and |K| are relatively prime, this forces o(x) = 1.
Corollary 3.2.13. Let G and G be groups such that |G| and |G | are relatively prime.
Then the only homomorphism from G to G is the trivial homomorphism, which sends
each x ∈ G to e ∈ G .
Proof Let x ∈ G. Then o(f (x)) divides o(x) (Corollary 2.5.19), which in turn divides
|G|. But o(f (x)) also divides |G |, as f (x) ∈ G . Since |G| and |G | are relatively prime,
we must have o(f (x)) = 1.
So far, we have studied the left cosets of a subgroup H of G. There is an analogous
notion of right cosets, which we shall study in the exercises below:
Definition 3.2.14. Let H be a subgroup of the group G. The right cosets of H in G
are the subsets
Hx = {hx | h ∈ H}
for x ∈ G.
Clearly, if G is abelian, then xH = Hx, and the right and left cosets of H coincide.
For some nonabelian groups, the left and right cosets of a given subgroup can be totally
different subsets of G. However, we shall see that the left and right cosets of H agree if
H is what’s known as a normal subgroup of G.
CHAPTER 3. G-SETS AND COUNTING
55
Exercises 3.2.15.
1. Let p be a prime number. Show that any group of order p is isomorphic to Zp .
2. For a prime number p, show that the only subgroups of Zp are e and Zp itself.
3. Let H be a subgroup of the finite group G and let K be a subgroup of H. Show
that [G : K] = [G : H] · [H : K].
† 4. Generalize the preceding problem to the case where G is infinite, but [G : K] is
finite.
5. State and prove the analogue of Lemma 3.2.3 for right cosets.
6. Show that every subgroup of Q8 other than Q8 itself is cyclic. List all of the
subgroups.
7. Find all of the left cosets of a ⊂ D6 . Now find all of the right cosets and show
that they are different subsets of D6 .
8. Repeat the preceding exercise with D8 in place of D6 .
9. Repeat the preceding exercise for D2n with n arbitrary.
10. What are the right cosets of b ⊂ D2n ?
11. Find all of the left cosets of a ⊂ Q12 .
12. Let H be a subgroup of G. Show that there is a one-to-one correspondence between
the left cosets of H and the right cosets of H. Deduce that H has finite index in G
if and only if there are finitely many distinct right cosets of H in G, and that the
number of right cosets, if finite, is [G : H].
13. Show that the order of a submonoid of a finite monoid need not divide the order
of the larger monoid.
3.3
G-sets and Orbits
Definitions 3.3.1. Let G be a group. An action of G on a set X is a rule that assigns
to each ordered pair (g, x) with g ∈ G and x ∈ X an element g · x ∈ X,2 such that
1. e · x = x for all x ∈ X
2. The following analogue of the associative law holds:
g1 · (g2 · x) = (g1 g2 ) · x
for all g1 , g2 ∈ G and x ∈ X.
A set X together with an action of G on X is called a G-set. If X and Y are G-sets,
then a G-map, or G-equivariant map, f : X → Y is a function f from X to Y such that
f (g · x) = g · f (x) for all g ∈ G and x ∈ X.
A G-isomorphism is a G-map which is a bijection.
2 In other words, the passage from the ordered pair (g, x) to the product g · x defines a function from
G × X to X.
CHAPTER 3. G-SETS AND COUNTING
56
We will often abbreviate g · x by gx. We now note that we may identify two G-sets
which are G-isomorphic.
Lemma 3.3.2. Let f : X → Y be a G-isomorphism. Then so is the inverse function
f −1 : Y → X.
Examples 3.3.3.
1. G itself is a G-set, under the action obtained by setting g · g equal to the product
gg of g and g in G. We call this the standard action of G on G.
2. Let X be a set with one element: X = {x}. Then there is a unique action of G on
X: g · x = x for all g ∈ G.
3. Let H be a subgroup of G. We let G act on G/H by g · xH = (gx)H for g, x ∈ G.
As the reader may verify, this gives a well defined action. Moreover, the canonical
map π : G → G/H is easily seen to be a G-map, where G is given the action of
the first example. The G-sets G/H are sometimes called “homogeneous spaces” of
G. Here, “spaces” comes from the fact that a similar construction is studied for
topological groups G, where H is a subgroup of G which is also a closed subspace
with respect to the topology of G.
4. Let G be any group and let X be any set. Then there is at least one action of G
on X: just set g · x = x for all g ∈ G and x ∈ X. We call this the trivial action of
G on X. Clearly, it does not tell us very much about either G or X.
5. Let X be a set. Then S(X) acts on X by σ · x = σ(x). Indeed, the facts that
1X is the identity of S(X) and that the group product in S(X) is composition of
functions are precisely what is needed to show that this specifies an action of S(X)
on X. We give a generalization of this example in Proposition 3.3.16.
6. Let X be any set. Let G be a cyclic group of order 2, written multiplicatively:
G = {e, a}, where a has order 2. Then there is an action of G on X × X, obtained
by setting a · (x, y) = (y, x) for each ordered pair (x, y) ∈ X × X.
As Example 3 suggests, the study of G-sets encodes information about the subgroups
of G. We will use G-sets as one of our main tools in understanding the structure of
groups.
Definitions 3.3.4. Let X be a G-set and let x ∈ X. Then the isotropy subgroup of x,
denoted Gx , is the set of elements of G which fix x:
Gx = {g ∈ G | g · x = x}.
The orbit of x, denoted G · x, is the set of all elements obtained from x via multiplication
by elements of G:
G · x = {g · x | g ∈ G} ⊂ X.
We say that G acts transitively on X if X = G · x for some x ∈ X.
The next lemma is needed to justify the preceding definitions.
Lemma 3.3.5. Let X be a G-set and let x ∈ X. Then the isotropy subgroup Gx is a
subgroup of G.
The orbit, G · x, of x is the smallest sub-G-set of X which contains x.
CHAPTER 3. G-SETS AND COUNTING
57
Proof Clearly, Gx is a submonoid of G, and we must show that Gx is closed under
inverses. But if g ∈ Gx , then g −1 · x = g −1 · (g · x) = (g −1 g) · x = x, and hence g −1 ∈ Gx .
We leave the rest to the reader.
As a first example of these concepts, let us consider their application to the “homogeneous spaces” G/H.
Lemma 3.3.6. In the G-set G/H, the isotropy subgroup of eH is H. Also, the orbit of
eH is all of G/H, and hence G/H is a transitive G-set.
Proof Here g · eH = gH, so that every element in G/H is in the orbit of eH. Also, g
is in the isotropy subgroup of eH if and only if gH = eH. By Lemma 3.2.3, this is the
case if and only if e−1 g ∈ H. Thus, the isotropy subgroup of eH is H, as claimed.
G-maps are of course important in the understanding of G-sets. The reader should
verify the following lemma.
Lemma 3.3.7. Let f : X → Y be a G-map and let x ∈ X. Then we have an inclusion
of isotropy subgroups Gx ⊂ Gf (x) . If, in addition, f is injective, then Gx = Gf (x) .
In the study of G-sets, orbits play a role similar to (but simpler than) the role played
by cyclic subgroups in the study of groups. Just as the cyclic groups Zn form models
for the cyclic subgroups, every orbit turns out to be G-isomorphic to one of the G-sets
G/H:
Lemma 3.3.8. Let X be a G-set and let x ∈ X. Let H be a subgroup of G. Then there
is a G-map fx : G/H → X with fx (eH) = x if and only if H ⊂ Gx . If fx exists, it is
unique, and has image equal to the orbit G · x. Moreover, fx induces a G-isomorphism
from G/H to G · x if and only if H = Gx .
In particular, for X any G-set and x ∈ X, G · x is G-isomorphic to G/Gx . Thus, a
G-set is transitive if and only if it is G-isomorphic to G/H for some subgroup H ⊂ G.
Proof Suppose given a G-map fx : G/H → X with fx (eH) = x. Then since H is
the isotropy subgroup of eH in G/H, it must be contained in the isotropy subgroup
of fx (eH) = x. In other words, we must have H ⊂ Gx . Also, since gH = g · eH,
fx (gH) = g · fx (eH) = gx, so fx is uniquely determined by its effect on eH. Clearly, its
image is G · x.
Conversely, if H ⊂ Gx , we use the above formula to define fx : G/H → X: fx (gH) =
gx. We must show this is well defined: i.e. if gH = g H, we must show that gx = g x. By
Lemma 3.2.3, gH = g H if and only if g = g h for some h ∈ H. But then gx = g hx = g x,
since hx = x. Thus, fx is a well defined function. To see that fx is a G-map, note that
fx (g · gH) = fx ((g g)H) = g gx = g fx (gH).
If fx is a G-isomorphism, then the isotropy subgroups of eH and fx (eH) must be
equal. But the isotropy subgroup of eH in G/H is H, while the isotropy subgroup of x
is Gx , so if fx is a G-isomorphism, then H = Gx .
Conversely, suppose that H = Gx . To show that fx induces an isomorphism of G/Gx
onto G · x, it suffices to show that fx is injective. Suppose that fx (gH) = fx (g H),
and hence that gx = g x. Multiplying both sides by g −1 and cancelling, we see that
x = g −1 g x, so that g −1 g is in the isotropy subgroup of x. Thus g −1 g ∈ Gx = H, so
that gH = g H by Lemma 3.2.3.
The next corollary now follows from Lagrange’s Theorem.
CHAPTER 3. G-SETS AND COUNTING
58
Corollary 3.3.9. Let X be a finite G-set and let x ∈ X. Then the number of elements
in the orbit G · x is equal to [G : Gx ]. In particular, since [G : Gx ] is finite, if either G
or Gx is finite, so is the other, and G · x has |G|/|Gx | elements.
As noted in Example 5 of Examples 3.3.3, S(X) acts on X for any set X, via σ · x =
σ(x), for σ ∈ S(X) and x ∈ X. Thus, we may use the notions of orbit and isotropy to
obtain information about S(X). We are primarily concerned with the case where X is
finite.
Recall from Corollary 3.1.13 that the canonical inclusion of Sn−1 in Sn identifies Sn−1
with the subgroup of Sn consisting of those elements that fix n. We restate this in terms
of isotropy groups:
Lemma 3.3.10. Consider the standard action of Sn on {1, . . . , n}. Under this action,
the isotropy subgroup of n is the image of the canonical inclusion Sn−1 ⊂ Sn .
Corollary 3.3.9 now allows us to give an inductive calculation of the order of Sn .
Corollary 3.3.11. The order of Sn is n! (n-factorial).
Proof Since the 2-cycle (i n) takes n to i for each 1 ≤ i < n, each element of {1, . . . , n}
is in the orbit of n under the standard action of Sn . In other words, Sn acts transitively
on {1, . . . , n}. Since this orbit has n elements, and since the isotropy subgroup of n is
Sn−1 , we have n = |Sn |/|Sn−1 | by Corollary 3.3.9. The result follows by induction on n,
starting with the obvious fact that S1 is the trivial group.
It is often desirable to count the elements in a G-set X by adding up the numbers of
elements in its orbits. To do that, we shall need the following lemma.
Lemma 3.3.12. Let X be a G-set and let x, y ∈ X. Then y ∈ G · x if and only if
G · y = G · x. In consequence, if x, z ∈ X, then G · x = G · z if and only if G · x ∩ G · z = ∅.
Proof Since y ∈ G · x, y = gx for some g ∈ G. Thus, g y = (g g)x ∈ G · x for all g ∈ G,
so G · y ⊂ G · x. But x = g −1 y, so G · x ⊂ G · y by the same reasoning.
Notice that the analogy between the orbits in a G-set and the cyclic subgroups of a
group has just been left behind, as we can find lots of examples (e.g., in cyclic groups)
of inclusions K ⊂ H of cyclic subgroups which are not equal.
Definition 3.3.13. Let X be a G-set. Then a set of orbit representatives in X is a
subset Y ⊂ X such that
1. If x, y ∈ Y with x = y, then G · x ∩ G · y = ∅.
2. X = x∈Y G · x.
For a general G-set, existence of a set of orbit representatives turns out to be an
important special case of the Axiom of Choice. But the case for finite G-sets is much
easier.
Lemma 3.3.14. Any finite G-set has a set of orbit representatives.
Proof Let X be our G-set. We argue by induction on the number of elements in X. Let
x ∈ X. If G · x = X, then {x} is the desired set. Otherwise, let X be the complement
of G · x in X. By Lemma 3.3.12, if g ∈ G and x ∈ X , gx must lie in X . Thus, X is a sub-G-set of X. By induction, X has a set, Y , of orbit representatives. But then
Y = Y ∪ {x} is a set of orbit representatives for X.
CHAPTER 3. G-SETS AND COUNTING
59
For infinite sets, one can prove the existence of a set of orbit representatives by using
Zorn’s Lemma (a form of the Axiom of Choice) to perform what amounts to an infinite
induction argument. We shall use similar techniques in the material on ring theory in
Chapters 7 through 12.
The following corollary is now immediate from Corollary 3.3.9.
Corollary 3.3.15. (G-set Counting Formula) Let Y be a set of orbit representatives for
the finite G-set X. Then the number, |X|, of elements in X is given by
[G : Gx ].
|X| =
x∈Y
An understanding of G-sets is vital to an understanding of permutation groups because of the following generalization of Cayley’s Theorem.
Proposition 3.3.16. Let X be a G-set. For g ∈ G, define g : X → X by g (x) = gx.
Then g is a permutation of X. Moreover, the function Λ : G → S(X) defined by
Λ(g) = g is a group homomorphism, whose kernel is the intersection of the isotropy
subgroups of the elements of X:
ker Λ =
Gx .
x∈X
Conversely, if Λ : G → S(X) is any group homomorphism, then X is a G-set under
the action defined by g · x = Λ(g)(x).
In fact, the correspondence which takes a G-set X to the homomorphism Λ : G →
S(X) gives a one-to-one correspondence between the actions of G on X and the homomorphisms from G to S(X).
Proof Let X be a G-set and let g ∈ G. We wish to show that g is a permutation of
X. It is onto, because x = g (g −1 x) for all x ∈ X. It is one-to-one, since if g (x) = g (y),
then gx = gy, and, multiplying both sides by g −1 , we see that x = y.
Now Λ is a homomorphism if and only if g ◦ g = gg for all g, g ∈ G. But clearly,
(
g ◦ g )(x) = gg x = gg (x) for all x ∈ X, so Λ is indeed a homomorphism. The kernel
of Λ is the set of elements in G which act on X by the identity map, and hence fix each
element in X. These then are the group elements which lie in the isotropy group of every
element of X.
Conversely, given the homomorphism Λ, and defining g · x to be equal to Λ(g)(x), we
have that ex = x because Λ(e) is the identity permutation, while g1 (g2 x) = Λ(g1 )(Λ(g2 )(x)) =
(Λ(g1 ) ◦ Λ(g2 ))(x) = Λ(g1 g2 )(x) = (g1 g2 )x.
Finally, the passage from the G-set X to the homomorphism Λ is easily seen to be
inverse to the passage from a homomorphism Λ : G → S(X) to an action on X.
Note that if the G-set X is G (with the usual action), then the map Λ : G → S(G)
given above is precisely the one given in Cayley’s Theorem.
Definitions 3.3.17. Let X be a G-set.
1. We say that the action of G on X is effective, or that X is an effective G-set, if
x∈X Gx = e. In words, G acts effectively on X if the only element of G which
fixes every element of X is e.
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60
2. We say that the action of G on X is free, or that X is a free G-set, if the isotropy
subgroup of every element of X is e. In other words, X is free if for any g ∈ G
and x ∈ X, gx = x implies that g = e.
Clearly, any free action is effective. However, there are many effective actions that
are not free.
Examples 3.3.18.
1. The standard action of G on G is free. Indeed, if gx = x, then right multiplication
by x−1 gives g = e.
2. The standard action of S(X) on X is effective, since the only function X → X
which fixes every element of X is the identity function. Note, however, that if X
has more than two elements, then this action is not free. For instance, a 2-cycle
(i j) fixes every element other than i and j.
Via the map Λ : G → S(X) of Proposition 3.3.16, we may interpret effective actions
in terms of embeddings into symmetric groups.
Corollary 3.3.19. An action of a group G on a set X is effective if and only if the
induced map Λ : G → S(X) is an embedding.
Thus, there is an embedding of G into Sn if and only if there is an effective action of
G on a set with exactly n elements.
Proof For any G-set
X, Proposition 3.3.16 tells us that the kernel of theinduced map
Λ : G → S(X) is x∈X Gx . Since effective actions are those for which x∈X Gx = e,
they are precisely those actions for which Λ is an embedding.
Thus, the key to discovering whether a given group embeds in Sn lies in understanding
the isotropy groups that appear in the various orbits of G-sets. We shall take an ad hoc
approach to these questions at this time, but the reader may want to come back to this
material after reading Section 3.7.
Definition 3.3.20. By a minimal subgroup of G we mean a subgroup H = e with the
property that the only subgroups of H are e and H.
This last is, of course, an abuse of language. Such a subgroup should more properly
be called a minimal nontrivial subgroup. But the trivial subgroup is rather uninteresting,
so we make the above definition for the purposes of economy of language. (Similarly,
in our discussion of commutative rings, we shall use the expression “maximal ideal” for
an ideal which is maximal in the collection of all ideals which are not equal to the ring
itself.)
Note that not all groups have minimal subgroups. For instance, it is easy to see
that every nontrivial subgroup of Z contains a subgroup which is strictly smaller, but
nontrivial. But this cannot happen in a finite group.
Lemma 3.3.21. Every nontrivial finite group G has a minimal subgroup.
Proof We argue by induction on the order of G. If G has no subgroups other than e
and itself (e.g. if |G| = 2), then G is a minimal subgroup of itself. Otherwise, there is a
nontrivial subgroup H = G. But then |H| < |G|, so that H has a minimal subgroup by
induction. However, a minimal subgroup of H is also minimal in G.
CHAPTER 3. G-SETS AND COUNTING
61
Most finite groups have several different minimal subgroups, which are not necessarily
isomorphic to each other. However, in the case that there is a unique minimal subgroup,
we obtain a strong consequence for the embedding problem.
Proposition 3.3.22. Let G be a finite group with exactly one minimal subgroup. Then
any effective G-set has at least |G| elements. In particular, n = |G| is the smallest integer
such that G embeds in Sn .
Proof Let H be the minimal subgroup of G. We claim that every nontrivial subgroup
of G contains H.
To see this, let K be a nontrivial subgroup of G. Then there is a minimal subgroup
K ⊂ K. But any minimal subgroup of K is also a minimal subgroup of G. Since G has
only one minimal subgroup, we must have K = H, and hence H ⊂ K.
Now suppose given an effective G-set X. We claim there must be at least one x ∈ X
with Gx = e.
Suppose not.
Then each Gx is a nontrivial subgroup, and hence H ⊂ Gx for each x.
But then H ⊂ x∈X Gx , and hence X is not effective.
Thus, there is an x ∈ X with Gx = e. But then G · x is G-isomorphic to G/e, and
hence X contains an orbit which has |G| elements in it. In particular, X has at least |G|
elements.
Exercises 3.3.23.
1. Show that the standard action on G is G-isomorphic to G/e and that the one-point
G-set is G-isomorphic to G/G.
‡ 2. Let G = D6 . Show that G acts effectively (but not freely) on G/a, and hence
that G embeds in S3 . Deduce from Corollary 3.3.11 that this embedding is an
isomorphism.
3. Generalize the preceding problem to obtain an embedding from D2n into Sn . Since
the order of the two groups is different for n > 3, this will not be an isomorphism
for these values of n.
4. Let p be a prime and let r ≥ 1. Show that Zpr cannot embed in Sn for n < pr .
5. Show that Q8 cannot embed in Sn for n < 8.
6. Show that Q2r cannot embed in Sn for n < 2r .
7. Find the minimal value of n such that D2r embeds in Sn . Do the same for D2pr ,
where p is an odd prime.
8. Let X and Y be G-sets. By X × Y , we mean the G-set given by the action of G
on the cartesian product of X and Y in which g · (x, y) = (gx, gy), for all g ∈ G,
x ∈ X, and y ∈ Y . What is the isotropy subgroup of (x, y) ∈ X × Y ?
9. Let G = Z6 . Let H, K ⊂ G be the subgroups of order 2 and 3, respectively. Show
that G/H × G/K is a free G-set.
10. Show that the action of G on X×Y need not be transitive, even if G acts transitively
on both X and Y .
11. Show that if G acts freely on X, then it also acts freely on X × Y for any Y .
CHAPTER 3. G-SETS AND COUNTING
62
12. There is a construction in
set theory called the disjoint union. The disjoint union
of X and Y, written X Y , is a set containing both
X and Y in the following
way: in X Y we have X ∩ Y = ∅, and X ∪ Y = X Y . We shall give a formal
construction of the disjoint union in Chapter 6.
Show that if X and Y are G-sets, then G acts on X Y in such a way that X and
Y are G-subsets.
13. Let G = Z6 . Let H, K ⊂ G be the subgroups of order 2 and 3, respectively. Show
that G acts effectively (but not freely) on G/H G/K. Deduce that Z6 embeds
in S5 .
14. Let m and n be relatively prime. Show that Zmn embeds in Sm+n .
15. Show that Q12 embeds in S7 .
16. What is the isotropy subgroup of gH ∈ G/H for an arbitrary g ∈ G?
17. Let X be a G-set and let H be a subgroup of G. The set of H-fixed points of X,
written X H , is defined to be the set of x ∈ X such that hx = x for all h ∈ H.
Show that x ∈ X H if and only if H ⊂ Gx .
‡ 18. Let X be a G-set and let H be a subgroup of G. Show that there is a one-to-one
correspondence between the G-maps from G/H to X and the elements of X H .
‡ 19. Let X be a G-set. Show that there is an equivalence relation on X defined by
setting x ∼ y if y = gx for some g ∈ G. Show that the equivalence classes of this
relation are precisely the orbits G · x for x ∈ X.
The set of equivalence classes thus obtained is called the orbit space of X, and
denoted X/G. (Thus, the elements of X/G are the orbits G · x in X.) We write
π : X → X/G for the canonical map, which takes x to its equivalence class.
20. A G-map f : X → Y is said to be isovariant if Gx = Gf (x) for each x ∈ X. Show
that a G-map f : X → Y is isovariant if and only if its restriction to each orbit is
injective.
21. Recall that a G-set Y has the trivial action if gy = y for all y ∈ Y and g ∈ G.
Show that if X is any G-set, and if we give the orbit space X/G the trivial action,
then the canonical map π : X → X/G is a G-map. Show also that if Y is any G-set
with the trivial action, and if f : X → Y is a G-map, then there is a unique G-map
f : X/G → Y such that f ◦ π = f .
22. Show that if G acts trivially on X and if Y is any G-set, then the G-maps from
G × X to Y are in one-to-one correspondence with the functions from X to Y .
(Here, G in the above product is the standard free G-set G/e.) In this context, we
call G × X the free G-set on the set X.
‡ 23. Let X be any set and write X n for the product X × · · · × X of n copies of X. Show
that X n is an Sn -set via σ · (x1 , . . . , xn ) = (xσ−1 (1) , . . . , xσ−1 (n) ), for σ ∈ Sn and
(x1 , . . . , xn ) ∈ X n .
† 24. Let G be a finite group whose order is divisible by a prime number p. In this
exercise, we shall prove Cauchy’s Theorem (also given by a different proof as Theorem 5.1.1), which says that such a group G must have an element of order p.
CHAPTER 3. G-SETS AND COUNTING
63
(a) Define σ : Gp → Gp by σ(g1 , . . . , gp ) = (gp , g1 , . . . , gp−1 ). Show that σ gives
a permutation of Gp of order p (i.e., σ ∈ S(Gp ) with σ p = e). Deduce that
there is an action of Zp on Gp , given by k · (g1 , . . . , gp ) = σ k (g1 , . . . , gp ) for
all k ∈ Zp . (In fact, if we identify Zp with the subgroup of Sn generated by
the p-cycle (1 2 · · · p), this action is just the restriction to Zp of the Sn action
given in Problem 23.)
(b) Define X ⊂ Gp by
X = {(g1 , . . . , gp ) | g1 · · · gp = e}.
In other words, the product (in order) of the gi is the identity element of G.
Show that X is a sub-Zp -set (but not a subgroup, unless G is abelian) of Gp .
(c) Show that the isotropy group of (g1 , . . . , gp ) ∈ X is the trivial subgroup of Zp
unless g1 = g2 = · · · = gp = g for some g ∈ G with g p = e. Deduce that there
is a one-to-one correspondence between the fixed-point set X Zp and the set
{g ∈ G | g p = e}. Deduce that G has an element of order p if and only if X Zp
has more than one element.
(d) Note that if x ∈ X is not in X Zp , then the orbit Zp · x has p elements. Since
two orbits are either disjoint or equal, X is the disjoint union of X Zp with the
orbits which lie outside of X Zp . Deduce that
|X| = |X Zp | + some multiple of p.
Show that |X| = |G| , so that |X| is divisible by p. Deduce that |X Zp | is
divisible by p, and hence G has an element of order p.
p−1
25. We say that G acts on the right on a set X (or that X is a right G-set) if for all
x ∈ X and g ∈ G there is an element xg ∈ X such that
(xg)g = x(gg )
and
xe = x
for all x ∈ X and all g, g ∈ G. Show that if X is a right G-set, then there is
an induced left action of G on X given by g · x = xg −1 . Deduce that there is a
one-to-one correspondence between the left G-sets and the right G-sets.
26. Let H be a subgroup of G. Then H acts on the right of G by ordinary multiplication.
Show that we may identify the set of left cosets G/H with the orbit space (see
Problem 19) of this action.
27. Let G act on the right on X and act on the left on Y . Then G acts on the left on
X × Y by g · (x, y) = (xg −1 , gy). We call the orbit space of this action the balanced
product of X and Y , which we denote X ×G Y .
Let H be a subgroup of G and let X be a (left) H-set. Then using the right action
of H on G by ordinary multiplication, we can form the balanced product G ×H X.
Show that the standard left action of G on itself induces a left action of G on
G ×H X.
Write [g, x] for the element of G×H X represented by the ordered pair (g, x) ∈ G×X.
(In other words, [g, x] is the orbit of (g, x) under the above action of H on G × X.)
Show that the isotropy subgroup G[e,x] of [e, x] under the above G-action coincides
with the isotropy subgroup Hx of x under the original action of H on X.
CHAPTER 3. G-SETS AND COUNTING
64
28. Let H be a subgroup of G and let X be a (left) H-set. Let Y be a (left) G-set.
Then H acts on Y by restricting the action of G. Show that there is a one-to-one
correspondence between the H-maps from X to Y and the G-maps from G ×H X
to Y .
29. Let M be a monoid. Define an M -set by substituting M for G in the definition
of G-set. Find a monoid M and an M -set X such that the function m : X → X
defined by m (x) = m · x need not be a permutation for all values of m.
30. Show that if you delete the requirement that e · x = x from the definition of G-set,
then the functions g need not be permutations.
3.4
Supports of Permutations
We isolate here some of the more elementary considerations for the study of cycle structure in Section 3.5. We give some consequences in the exercises.
Definition 3.4.1. Let σ ∈ S(X). The support of σ, Supp(σ), is the following subset of
X:
Supp(σ) = {x ∈ X | σ(x) = x}
In particular, σ is the identity outside of its support.
It is easy to calculate the support of a k-cycle.
Lemma 3.4.2. The support of the k-cycle (i1 · · · ik ) is {i1 , . . . , ik }.
Here is another way to think of supports.
Lemma 3.4.3. Let Y be a subset of X and let σ ∈ S(X). Then Supp(σ) ⊂ Y if and
only if σ is the identity on the complement of Y . In particular, if i : Y → X is the
inclusion of Y in X, then Supp(σ) ⊂ Y if and only if σ is in the image of the canonical
inclusion i∗ : S(Y ) ⊂ S(X).
Proof The support is the complement of the subset of points fixed by σ, hence the first
statement. For the second, recall from Lemma 3.1.12 that the image of i∗ is the set of
permutations that fix the complement of Y .
Applying Lemma 3.4.3 to Y = Supp(σ), we obtain the following corollary.
Corollary 3.4.4. Let σ be a permutation of the set X. Then σ restricts to a bijection
σ : Supp(σ) → Supp(σ).
Definition 3.4.5. We say the permutations σ, τ ∈ S(X) are disjoint if Supp(σ) ∩
Supp(τ ) = ∅.
Disjoint permutations interact well with one another:
Proposition 3.4.6. Let σ and τ be disjoint permutations of X. Then σ and τ commute
with each other.
Moreover, the product στ is given as follows:
⎧
⎨ σ(x) if x ∈ Supp(σ)
τ (x) if x ∈ Supp(τ )
(στ )(x) =
⎩
x
otherwise.
CHAPTER 3. G-SETS AND COUNTING
65
In particular, the support of στ is the union Supp(σ) ∪ Supp(τ ).
Finally, στ has finite order if and only if both σ and τ have finite order, in which
case the order of στ is the least common multiple of the orders of σ and τ .
Proof We wish to show that both στ and τ σ have the effect specified by the displayed
formula.
Away from Supp(σ) ∪ Supp(τ ), both σ and τ are the identity, and hence both στ and
τ σ are also.
On Supp(σ), τ is the identity, as Supp(σ) ∩ Supp(τ ) = ∅. Thus, for x ∈ Supp(σ),
(στ )(x) = σ(x). Since x ∈ Supp(σ), σ(x) is also in Supp(σ) by Corollary 3.4.4, and
hence (τ σ)(x) = σ(x) as well.
A similar argument shows that both στ and τ σ act as τ on Supp(τ ). Thus, στ = τ σ,
and the displayed formula holds.
An easy induction based on the displayed formula now shows:
⎧ k
⎨ σ (x) if x ∈ Supp(σ).
τ k (x) if x ∈ Supp(τ ).
For k > 0, (στ )k (x) =
⎩
x
otherwise.
Thus, (στ )k = e if and only if both σ k = e and τ k = e, and the result follows.
Exercises 3.4.7.
1. Show that an element σ ∈ S(Z+ ) lies in S∞ if and only if Supp(σ) is finite.
‡ 2. Suppose given subsets Y, Z ⊂ X, with Y ∩ Z = ∅. Show that there is an embedding
μ : S(Y )×S(Z) → S(X) whose restrictions to S(Y ) = S(Y )×e and S(Z) = e×S(Z)
are the canonical inclusions of S(Y ) and S(Z), respectively, in S(X). Show that
the image of μ is the set of permutations σ ∈ S(X) such that the following two
properties hold:
(a) Supp(σ) ⊂ Y ∪ Z
(b) σ(Y ) = Y .
‡ 3. Combinatorics Consider the action of Sn on {0, 1}n obtained by permuting the
factors (see Problem 23 of Exercises 3.3.23). We can use it to derive some of the
standard properties of the binomial coefficients. First, for 0 ≤ k ≤ n, let us write
xk = (1, . . . , 1, 0, . . . , 0 ) ∈ {0, 1}n .
k times n−k times
(a) Show that the number of elements in the orbit Sn · xk is equal to the number
of ways to choose k things out of n things (which we may identify with the
number of k-element subsets
of an n-element set), which we shall denote by
the binomial coefficient nk .
(b) Show that the isotropy subgroup of xk is isomorphic to the direct product
Sk × Sn−k . Deduce that
n
n!
=
,
k
k! · (n − k)!
and that the right-hand side must therefore be an integer.
CHAPTER 3. G-SETS AND COUNTING
66
(c) Show that the elements x0 , . . . , xn form a set of orbit representatives for the
Sn action on {0, 1}n . Deduce that
n
2 =
n n
k=0
k
.
(d) Deduce that there are exactly 2n subsets of an n-element set.
(e) Show that the number of ordered k-element subsets of an n-element set is k!
times the number of unordered k-element subsets of that set. Deduce that the
number of ordered k-element subsets is n!/(n − k)!.
4. It can be shown that if X is any infinite set, then the disjoint union X X may
be placed in bijective correspondence with X. Deduce from Problem 2 that for any
infinite set X, we may embed S(X) × S(X) in S(X).
5. Reconsider
the preceding problem for X = Z+ . Show that for any choice of injection
Z+ Z+ → Z+ , the induced embedding S(Z+ ) × S(Z+ ) ⊂ S(Z+ ) restricts to an
embedding of S∞ × S∞ in S∞ . (These embeddings turn out to be important
in studying the cohomology of the group S∞ , an endeavor which falls under the
heading of homological algebra, but has applications to topology.)
6. In this problem we assume a familiarity with infinite disjoint unions and with
infinite products, as discussed in Chapter 6.
(a) Suppose given a !
family of sets {X
i | i ∈ I}. Show that we may embed the
infinite product i∈I S(Xi ) in S( i∈I Xi ). Deduce that there is a permutation of Z+ which may be thought of as the disjoint product of infinitely
many cycles, with one k-cycle being given for each k ≥ 2. Deduce that there
are permutations in S(Z+ ) whose orbits are all finite, but which have infinite
order.
(b) Construct a permutation of Z+ which is a product of infinitely many cycles
of the same order. Deduce that there are elements of S(Z+ ) which have finite
order but do not lie in S∞ .
3.5
Cycle Structure
Here, using the ideas of G-sets and orbits, we develop the notion of cycle decompositions
for permutations. This gives a powerful tool for studying permutation groups that has
some analogies to prime decompositions in Z.
Using cycle decomposition, we shall construct the sign homomorphism ε : Sn → {±1}
and define the alternating groups, An . We shall encounter further applications of cycle
decomposition in all of our subsequent study of permutation groups, particularly in
Section 4.2.
Recall that the permutations σ and τ are disjoint if Supp(σ) ∩ Supp(τ ) = ∅, where
Supp(σ) is the complement of the set of points fixed by σ.
Definition 3.5.1. By a product of disjoint permutations, we mean a product σ1 · · · σk ,
such that σi and σj are disjoint for i = j.
We are particularly interested in products of disjoint cycles: i.e., products σ1 · · · σk
of disjoint permutations in Sn , such that each σi is a cycle.
CHAPTER 3. G-SETS AND COUNTING
67
We now generalize Proposition 3.4.6.
Corollary 3.5.2. Let σ1 · · · σk ∈ S(X) be a product of disjoint permutations. Then the
effect of σ1 · · · σk is given as follows: on Supp(σi ), the effect of σ1 · · · σk is the same
k
as that of σi for 1 ≤ i ≤ k. Away from i=1 Supp(σi ), σ1 · · · σk is the identity. In
k
particular, Supp(σ1 · · · σk ) = i=1 Supp(σi ).
In addition, if each σi has finite order, then the order of σ1 · · · σk is the least common
multiple of the orders of the σi .
Proof We argue by induction on k. The case k = 2 is given
k−1by Proposition 3.4.6.
For k > 2, the induction hypothesis gives Supp(σ1 · · · σk−1 ) = i=1 Supp(σi ), and hence
σ1 · · · σk−1 is disjoint from σk . The result now follows by applying Proposition 3.4.6 (and
the induction hypothesis) to the product (σ1 · · · σk−1 ) · σk .
The following proposition gives the cycle decomposition for a permutation, σ.
Proposition 3.5.3. Let σ ∈ Sn for n ≥ 2, with σ = e. Then σ may be written uniquely
as a product of disjoint cycles: σ = σ1 · · · σk . Here, uniqueness means that if σ may also
be written as the product τ1 · · · τl of disjoint cycles, then k = l, and there is a permutation
φ ∈ Sk such that σi = τφ(i) for 1 ≤ i ≤ k.
Proof Let G = σ, the cyclic subgroup of Sn generated by σ. We wish to understand
X = {1, . . . , n} as a G-set. Note that if the orbit of x ∈ X has only one element, then
the isotropy subgroup of x has index 1 in G by Corollary 3.3.9. This means that the
isotropy subgroup of x must be all of G. If this were true for all x ∈ X, then G would
act as the identity on X. Since σ = e, this cannot be the case.
Suppose, then, that y ∈ X has an orbit with more than one element. Let r be the
smallest positive integer with the property that σ r (y) = y. (Such an integer exists,
because σ o(σ) (y) = y.) Note that r > 1, as otherwise y is fixed under σ and all of its
powers, and hence the orbit of y under G would have only one element.
We claim first that the elements y, σ(y), . . . , σ r−1 (y) are all distinct. To see this, note
that otherwise there are integers s and t with 0 ≤ s < t < r such that σ s (y) = σ t (y) =
σ s (σ t−s (y)). Thus, y and σ t−s (y) have the same image under σ s . Since σ s is injective,
this gives σ t−s (y) = y. Since 0 < t − s < r, this contradicts the minimality of r.
Thus, there is an r-cycle σ1 = (y σ(y) · · · σ r−1 (y)). Let Y1 = Supp(σ1 ) = {y, σ(y), . . . , σ r−1 (y)}.
Then, by inspection, we see that σ has the same effect as σ1 on each element of Y1 . Thus,
σ restricts to a bijection Y1 → Y1 . By passage to powers, we see that Y1 is a sub-G-set
of X. Since G acts transitively on Y1 , we see that Y1 is the orbit of y under the action
of G.
Now let Y1 , . . . , Yk be the set of all distinct orbits in X with more than one element.
By Lemma 3.3.12, Yi ∩ Yj = ∅ for i = j. Repeating the argument above, we can find
cycles σi for 2 ≤ i ≤ k such that σi has the same effect on Yi is σ does, and Supp(σi ) = Yi .
In particular, σ1 · · · σk is a product of disjoint cycles.
We claim that σ = σ1 · · · σk . By Corollary 3.5.2, the effect of σ1 · · · σk on Yi =
Supp(σi ) is the same as that of σi . But σi was chosen to have the same effect on Yi as
k
σ. Thus, σ agrees with σ1 · · · σk on i=1 Yi = Supp(σ1 · · · σk ). Since σ is the identity on
k
all one-element orbits, it is the identity off i=1 Yi , and σ = σ1 · · · σk , as claimed.
We show uniqueness by reversing the above argument. In a product τ = τ1 . . . τl of
disjoint cycles, the effect of τ agrees with that of τi on Supp(τi ). As the cycle notation
shows, τi acts transitively on Supp(τi ), so τ acts transitively on Supp(τi ) as well.
CHAPTER 3. G-SETS AND COUNTING
68
Thus, there are l distinct orbits under the action of τ which have more than one
element.
Thus, if τ is equal to the permutation σ, above, then k = l and, equating the orbits
of σ and τ , we may relabel the τi so that Supp(τi ) = Supp(σi ) for i = 1, . . . , k. But
since σ = σi and τ = τi on this common support, τi and σi are equal as permutations.
Remarks 3.5.4. Note the concluding statement of the last proof. A cycle is determined
by its effect as a permutation, not by its notation. However, it’s easy to see that there
are precisely r different ways to write an r-cycle in cycle notation. It depends solely on
which element of its support is written first.
Proposition 3.5.3 shows that every permutation is a product of cycle permutations.
But we can actually write a permutation as a product of even simpler terms as long as
we no longer insist that these terms be disjoint. Recall that a transposition is another
word for a 2-cycle. The following lemma may be verified by checking the effect of both
sides of the proposed equality on the elements {i1 , . . . , ik }. Here, we should emphasize
the fact that if τ, τ ∈ Sn and j ∈ {1, . . . , n}, then (τ τ )(j) = τ (τ (j)).
Lemma 3.5.5. Let σ be the k-cycle σ = (i1 · · · ik ). Then σ is the product of k − 1
transpositions:
σ = (i1 i2 )(i2 i3 ) · · · (ik−1 ik ).
Since the cycle permutations generate Sn , we see that every element of Sn is a product
of transpositions.
Corollary 3.5.6. Sn is generated by the transpositions.
We wish to define the sign of a permutation to be ±1 in such a way that the sign will be
+1 if the permutation may be written as a product of an even number of transpositions,
and is −1 otherwise. Our initial definition will be more rigid than this.
Definition 3.5.7. The sign of a k-cycle is (−1)k−1 . Thus, the sign is −1 if k is even
and is +1 if k is odd.
The sign of a product of disjoint cycles is the product of the signs of the cycles. The
sign of the identity map is +1.
Thus, the sign of every permutation is defined, via Proposition 3.5.3. Notice the
relationship between the sign of a cycle and the number of transpositions in the product
given in Lemma 3.5.5. Multiplying out the signs over a product of disjoint cycles, we see
that if the sign of a permutation, σ, is +1, then there is at least one way to write σ as
the product of an even number of transpositions. If the sign of σ is −1, then there is at
least one way to write σ as the product of an odd number of transpositions.
The next proposition is the key to the utility of signs. Notice that {±1} forms a
group under multiplication. Since it has two elements, it is isomorphic to Z2 .
Proposition 3.5.8. For any σ, τ ∈ Sn , n ≥ 2, the sign of στ is the product of the signs
of σ and τ .
Thus, there is a homomorphism ε : Sn → {±1} obtained by setting ε(σ) equal to the
sign of σ. In particular, ε has the property that ε(τ ) = −1 for any transposition τ .
CHAPTER 3. G-SETS AND COUNTING
69
Proof Since we may write τ as a product of m transpositions in such a way that the
sign of τ is (−1)m , we may argue by induction, assuming that τ is a transposition, say
τ = (i j).
Write σ as a product of disjoint cycles: σ = σ1 · · · σk . There are four cases to consider,
where a non-trivial orbit of σ is one which has more than one element:
1. Neither i nor j lies in a non-trivial orbit of σ.
2. j lies in a non-trivial orbit of σ and i does not.
3. i and j lie in different non-trivial orbits of σ.
4. Both i and j lie in the same non-trivial orbit of σ.
We analyze the decomposition of στ as a product of disjoint cycles in each of these
cases. The first case is simplest, as then σ and τ are disjoint. Thus, τ is one of the cycles
in the decomposition of στ , and the others are those from the decomposition of σ. The
result follows in this case from the definition of sign.
In case 2, we may assume that σ is an r-cycle, σ = (i1 · · · ir ), with ir = j. Then
στ = (i i1 · · · ir ), an (r + 1)-cycle.
In case 3, we may assume that σ is the product of two disjoint cycles, σ1 = (i1 · · · ir ),
with ir = i, and σ2 = (j1 · · · js ), with js = j. Here, we see that στ is the (r + s)-cycle
στ = (i1 · · · ir j1 · · · js ). The signs work out correctly once again.
Case 4 is the most complicated. We may assume that σ is a single cycle. The
analysis falls naturally into sub-cases. In the first, σ = τ = (i j), and here στ = e, as
transpositions have order 2. The next case is where σ is an r-cycle with r ≥ 3 and i
and j appear next to each other in the cycle representation of σ, say σ = (i1 · · · ir ) with
ir−1 = i and ir = j. Here στ is the (r − 1)-cycle στ = (i1 · · · ir−1 ).
In the final case, i and j cannot be written next to each other in the cycle representation of σ. Thus, we may write σ = (i1 · · · ir j1 · · · js ), with ir = i, js = j, and with r
and s both greater than or equal to 2. Here, στ is given by στ = (i1 · · · ir )(j1 · · · js ).
The sign homomorphism gives rise to an important subgroup of Sn .
Definitions 3.5.9. The n-th alternating group, An , is the kernel of the sign homomorphism ε : Sn → {±1}.
We say that a permutation is even if it lies in An (i.e., if its sign is 1). A permutation
is said to be odd if its sign is −1.
The fact that ε is a homomorphism allows us to determine the sign of a permutation
from any representation of it as a product of transpositions.
Corollary 3.5.10. Let σ ∈ Sn . Then σ ∈ An if and only if σ may be written as a
product of an even number of transpositions. If σ ∈ An , then whenever σ is written as
a product of transpositions, the number of transpositions must be even.
Proof The homomorphism ε carries any product of an odd number of transpositions
to −1 and carries any product of an even number of transpositions to 1. Thus, given
any two presentations of a permutation σ as a product of transpositions, the number of
transpositions in the two presentations must be congruent mod 2.
We have made use of the fact that Sn is generated by the transpositions. Using
Corollary 3.5.10, we may obtain a similar result for An .
CHAPTER 3. G-SETS AND COUNTING
70
Corollary 3.5.11. For n ≥ 3, the alternating group An is generated by 3-cycles. Thus,
since each 3-cycle lies in An , a permutation σ ∈ Sn lies in An if and only if σ is a
product of 3-cycles.
Proof Corollary 3.5.10 shows that a permutation is in An if and only if it is the product
of an even number of transpositions. Thus, it suffices to show that the product of any
two transpositions is a product of 3-cycles.
If the two transpositions are not disjoint, then either they are equal, in which case
their product is the identity, or the product has the form
(a b) · (b c) = (a b c),
a 3-cycle.
The remaining case is the product of two disjoint cycles, (a b) and (c d). But then we
have
(a b) · (c d) = (a b c) · (b c d),
the product of two 3-cycles.
Exercises 3.5.12.
1. Write the product (1 2 3 4) · (3 4 5) as a product of disjoint cycles. What is the order
of this product?
2. Consider the function from Z7 to itself induced by multiplication by 2. (Recall that
multiplication mod n gives a well-defined binary operation on Zn .) Identifying the
elements of Z7 with {1, . . . , 7} (by identifying k with k for 1 ≤ k ≤ 7), show
that multiplication by 2 induces a permutation of {1, . . . , 7}. Write down its cycle
structure.
3. Repeat the preceding problem using multiplication by 3 in place of multiplication
by 2.
4. What are the elements of A3 ?
5. What are the elements of A4 ?
6. Recall from Problem 2 of Exercises 3.3.23 that S3 is isomorphic to the dihedral
group D6 . Which subgroup of D6 corresponds to A3 under this isomorphism?
7. Let p be prime and let r ≥ 1. Show that Sk has an element of order pr if and only
if k ≥ pr . (Hint: Write the element in question as a product of disjoint cycles.
What can you say about the length of the cycles?)
8. For a given n, what is the smallest k such that Zn embeds in Sk ? (Hint: The
answer depends on the prime decomposition of n.)
9. What is the largest value of n such that Zn embeds in S8 ?
‡ 10. Expanding on Remarks 3.5.4, show that the number of k-cycles in Sn is 1/k times
the number of ordered k-element subsets of {1, . . . , n}. Deduce from Problem 3 of
n!
Exercises 3.4.7 that there are k·(n−k)!
k-cycles in Sn .
CHAPTER 3. G-SETS AND COUNTING
3.6
71
Conjugation and Other Automorphisms
In addition to the study of actions of the group G on sets, it is useful to study actions of
other groups on G. Of course, S(G) acts on G, but we have seen that the structure of a
symmetric group S(X) depends only on the size of the set X. In particular, the action
of S(G) on G loses all the information about the group structure of G, and captures only
its size.
To capture the group structure on G, a group should act on it through permutations
σ : G → G which are group homomorphisms. These are precisely the group isomorphisms
from G to itself.
Definitions 3.6.1. An automorphism of a group G is a group isomorphism from G to
itself. We write Aut(G) ⊂ S(G) for the set of all automorphisms of G.
The reader may check the details of the following examples, using Problem 6 of Exercises 2.8.5 for the examples involving dihedral groups, and Problem 5 of Exercises 2.9.7
for the examples involving quaternionic groups.
Examples 3.6.2.
1. Let G be an abelian group. Then there is an automorphism f : G → G given by
f (g) = g −1 for all g ∈ G.
2. Let 0 ≤ k < n. Then there is an automorphism f ∈ Aut(D2n ) determined by
f (b) = b, f (a) = abk .
3. There is an automorphism f ∈ Aut(D2n ) determined by f (b) = b−1 , f (a) = a.
4. Let 0 ≤ k < 2n. Then there is an automorphism f ∈ Aut(Q4n ) determined by
f (b) = b, f (a) = abk .
5. There is an automorphism f ∈ Aut(Q4n ) determined by f (b) = b−1 , f (a) = a.
We now state an immediate consequence of Lemma 2.5.7.
Lemma 3.6.3. Let G be a group. Then Aut(G) is a subgroup of S(G).
Definition 3.6.4. We say that the group K acts on G via automorphisms if G is a
K-set in such a way that the action of each k ∈ K, k : G → G (defined as usual by
k (g) = k · g), is an automorphism of G.
The proof of the following proposition is identical to that of Proposition 3.3.16.
Proposition 3.6.5. There is a one-to-one correspondence between the actions of K on
G via automorphisms and the homomorphisms Λ : K → Aut(G). Here, a given action
corresponds to the homomorphism defined by Λ(k) = k .
This formalism is useful, as we are about to define a very important action of G on
itself through automorphisms, called conjugation. But we cannot denote the resulting
G-set by “G,” as we reserve this symbol for the standard action induced by left multiplication. So it is useful to be able to refer to conjugation in terms of a homomorphism
from G to Aut(G).
Definition 3.6.6. Let G be a group and let x ∈ G. By conjugation by x, we mean the
function cx : G → G specified by cx (g) = xgx−1 .
CHAPTER 3. G-SETS AND COUNTING
72
Lemma 3.6.7. Conjugation by x ∈ G is an automorphism of G. There is an action of
G on itself by automorphisms given by the homomorphism Γ : G → Aut(G) defined by
Γ(x) = cx .
Proof To see that cx is a homomorphism, we have
cx (g)cx (g ) = xgx−1 xg x−1 = xgg x−1 = cx (gg ).
It is a bijection for the same reason that left or right multiplication by any element of
G is a bijection: There is an inverse function given by an appropriate multiplication. In
this case the inverse function is conjugation by x−1 .
Thus, there is a well-defined function Γ : G → Aut(G) with Γ(x) = cx . To see that Γ
is a group homomorphism, we have
(cx ◦ cy )(g) = cx (ygy −1 ) = xygy −1 x−1 = (xy)g(xy)−1 = cxy (g),
because (xy)−1 = y −1 x−1 .
The elements in the image of Γ : G → Aut(G) have a special name.
Definitions 3.6.8. An automorphism f ∈ Aut(G) is said to be an inner automorphism
of G if f = cx , the conjugation by x, for some x ∈ G. We write Inn(G) ⊂ Aut(G) for
the set of inner automorphisms. Note that since Inn(G) = im Γ, Inn(G) is a subgroup of
Aut(G).
The isotropy subgroup of y ∈ G under the conjugation action is by definition the set
{x ∈ G | xyx−1 = y}. It is an important enough subgroup that it merits a name:
Definition 3.6.9. Let y be an element of the group G. Then the centralizer of y in G
is the subgroup
CG (y) = {x ∈ G | xyx−1 = y}.
Of course, if x ∈ H ⊂ G, then we write CH (x) for the centralizer of x in H. When
there’s no ambiguity, we shall simply refer to CG (x) as the centralizer of x.
The next lemma is useful in understanding the effect of conjugation. The proof is left
to the reader.
Lemma 3.6.10. Let x, y ∈ G. Then the following are equivalent.
1. x and y commute.
2. x ∈ CG (y), i.e., xyx−1 = y.
3. xyx−1 y −1 = e.
We now give two useful consequences of Lemma 3.6.10.
Corollary 3.6.11. Suppose that the elements x, y ∈ G commute with each other. Then
every element of x commutes with every element of y.
Proof Since x ∈ CG (y) and since CG (y) is a subgroup of G, x ⊂ CG (y). Thus, y
commutes with xk for all k ∈ Z, and hence y ∈ CG (xk ) as well. So y ⊂ CG (xk ), and
hence y l commutes with xk for all k, l ∈ Z.
CHAPTER 3. G-SETS AND COUNTING
73
Corollary 3.6.12. The element x ∈ G is in the kernel of the conjugation-induced homomorphism Γ : G → Aut(G) if and only if x commutes with every element of G.
Proof We have x ∈ ker Γ if and only if x is in the isotropy subgroup of every y ∈ G.
But this says that x ∈ CG (y) for every y ∈ G, and hence x commutes with every element
of G.
We shall refer to the kernel of Γ as the center, Z(G), of G:3
Definition 3.6.13. The center, Z(G), of a group G is given as follows:
Z(G) = {x ∈ G | xy = yx for all y ∈ G}.
Definition 3.6.14. The conjugacy class of g ∈ G is {xgx−1 | x ∈ G}, which is the orbit
of g under the action of G on itself by conjugation. Its elements are called the conjugates
of g.
Since the isotropy subgroup of g is CG (g), the next result follows from Corollary 3.3.9.
Corollary 3.6.15. The number of conjugates of g ∈ G is equal to [G : CG (g)], the index
in G of the centralizer, CG (g), of g in G.
The fact that CG (g) is the set of all elements that commute with g gives a characterization of Z(G) in terms of conjugacy classes.
Corollary 3.6.16. The conjugacy class of g ∈ G consists of a single element if and only
if g ∈ Z(G).
Proof [G : CG (g)] = 1 if and only if CG (g) = G.
The G-set Counting Formula (Corollary 3.3.15) for the conjugation action on G is
extremely useful. We shall give a variant on it, called the Class Formula. We shall need
some setup for it first.
Definitions 3.6.17. We say that a conjugacy class of G is nontrivial if it has more than
one element.
By a set of class representatives for the nontrivial conjugacy classes we mean a subset
X ⊂ G such that
1. The conjugacy class of each x ∈ X is nontrivial.
2. If x, y ∈ X with x = y, then the conjugacy classes of x and y intersect in the null
set.
3. Every element of G that is not in Z(G) lies in the conjugacy class of some x ∈ X.
The desired Class Formula is now immediate from the G-set Counting Formula.
Corollary 3.6.18. (Class Formula) Let G be a finite group and let X be a set of class
representatives for the nontrivial conjugacy classes in G. Then
[G : CG (x)].
|G| = |Z(G)| +
x∈X
3 The
German word for center is Zentrum.
CHAPTER 3. G-SETS AND COUNTING
74
We can compute the conjugates and the centralizers of the elements in the dihedral
and quaternionic groups by hand, using the explicit information we have about these
groups. We shall leave these calculations for the next set of exercises. The next result
computes the conjugates of a k-cycle in Sn .
Lemma 3.6.19. Let σ be any permutation and let (i1 · · · ik ) be any k-cycle in Sn . Then
σ · (i1 · · · ik ) · σ −1 is a k-cycle:
σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )).
Proof If j is not in {σ(i1 ), . . . , σ(ik )}, then σ −1 (j) is not in {i1 , . . . , ik }, and hence
(i1 · · · ik ) acts as the identity on σ −1 (j). But then σ · (i1 · · · ik ) · σ −1 (j) = j, so that
σ · (i1 · · · ik ) · σ −1 acts as the identity outside {σ(i1 ), . . . , σ(ik )}.
For 1 ≤ j < k, σ · (i1 · · · ik ) · σ −1 (σ(ij )) = σ · (i1 · · · ik )(ij ) = σ(ij+1 ). Similarly,
σ · (i1 · · · ik ) · σ −1 (σ(ik )) = σ(i1 ), so that σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )), as
claimed.
Lemma 3.6.19 gives important information about the behavior of the symmetric
groups. Here is a sample.
Corollary 3.6.20. Suppose that σ ∈ Sn lies in the centralizer of the k-cycle (i1 · · · ik ).
Then σ({i1 , . . . , ik }) = {i1 , . . . , ik }.
Proof Let σ ∈ Sn Then
σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )).
Thus,
Supp(σ · (i1 · · · ik ) · σ −1 )
= {σ(i1 ), . . . , σ(ik )}
= σ({i1 , . . . , ik }).
Thus, if σ · (i1 · · · ik ) · σ −1 = (i1 · · · ik ), then
σ({i1 , . . . , ik }) = Supp((i1 · · · ik )) = {i1 , . . . , ik }.
There is an important action related to conjugation, in which G acts on the set of
homomorphisms from another group into G. We shall use a notation consistent with
that used in our discussion of category theory in Chapter 6.
Notation 3.6.21. Let K and G be groups. We write MorGp (K, G) for the set of group
homomorphisms from K to G.
Definition 3.6.22. Let K and G be groups. For g ∈ G and f ∈ MorGp (K, G), the
conjugate of f by g, written g · f , is the homomorphism (g · f ) ∈ MorGp (K, G) defined
by (g · f )(k) = gf (k)g −1 for k ∈ K.
Two homomorphisms f, f ∈ MorGp (K, G) are said to be conjugate if f = g · f for
some g ∈ G.
Note that g · f is equal to the composite cg ◦ f , where cg : G → G is conjugation by
g. Thus, g · f is a homomorphism, as claimed. The following lemma is now immediate.
CHAPTER 3. G-SETS AND COUNTING
75
Lemma 3.6.23. Let K and G be groups. Then there is an action of G on MorGp (K, G)
obtained by setting the action of g ∈ G on f ∈ MorGp (K, G) to be equal to the conjugate
homomorphism g · f defined above.
Conjugate homomorphisms are important in a number of questions, including the
classification of semidirect products and the cohomology of groups.
Exercises 3.6.24.
1. What are all the conjugacy classes of elements of D6 ? Write down all the elements
in each class.
2. What are all the conjugacy classes of elements of D8 ? Write down all the elements
in each class.
3. What are all the conjugacy classes of elements of Q8 ? Write down all the elements
in each class.
4. What are all the conjugacy classes of elements of D2n for n ≥ 3? (Hint: The answer
depends in a crucial way on the value of n mod 2.)
5. What are all the conjugacy classes of elements of Q4n ?
6. Find the centralizers of each of the elements of Q8 .
7. Find the centralizers of the elements a, b ∈ D2n .
8. Find the centralizers of the elements a, b ∈ Q4n .
9. Let H be a subgroup of G and let x ∈ H. Show that
CH (x) = H ∩ CG (x).
10. Suppose that Sn has an element of order k. Show that D2k embeds in Sn .
11. Consider the 2-cycle (n − 1 n) ∈ Sn . Show that the centralizer of (n − 1 n) in Sn is
isomorphic to Sn−2 × Z2 .
‡ 12. Let a, b, c, d be distinct elements of {1, . . . , n}. Show that any element of Sn which
centralizes both (a b c) and (b c d) must fix both a and d. Deduce that if n ≥ 4,
then the only element which centralizes every 3-cycle is the identity.
13. Show that the centralizer of any k-cycle in Sn is isomorphic to Sn−k × Zk . We shall
see in Section 4.2 (using a straightforward application of Lemma 3.6.19 above) that
n!
any two k-cycles in Sn are conjugate. Assuming this, deduce that there are k·(n−k)!
k-cycles in Sn . (See Problem 10 of Exercises 3.5.12 for a different line of argument.)
14. What is the centralizer of (1 2 3) in A4 ? How many conjugacy classes of 3-cycles
are there in A4 ?
15. Let σ, τ ∈ S(X). Show that Supp(τ στ −1 ) = τ (Supp(σ)).
16. What is the center of D8 ?
17. What is the center of D2n for n ≥ 3? (Hint: The answer depends in a crucial way
on the value of n mod 2.)
CHAPTER 3. G-SETS AND COUNTING
76
18. What is the center of Q8 ?
19. What is the center of Q4n ?
20. What is the center of Sn ? What is the center of An ?
‡ 21. Let H be any group and write H n for the product H × · · · × H of n copies of
H. Show that Sn acts on H n through automorphisms via σ · (h1 , . . . , hn ) =
(hσ−1 (1) , . . . , hσ−1 (n) ).
22. Verify that the specifications given in Examples 3.6.2 do in fact give automorphisms
of the groups in question.
23. Show that the only automorphism of Z4 other than the identity is the one which
takes each element to its inverse.
24. Let G be a finite abelian group and let f : G → G be given by f (g) = g n for all
g ∈ G, where n is an integer that is relatively prime to |G|. Show that f is an
automorphism of G.
If G is cyclic of order 7 and n = 2, what is the order of f in Aut(G)?
25. Show that not every automorphism of D8 is an inner automorphism.
26. Show that every automorphism of D6 is an inner automorphism. (Hint: The
composite of two inner automorphisms is an inner automorphism.) Deduce that
Aut(D6 ) ∼
= D6 .
27. Show that any two nontrivial homomorphisms from Z2 to D6 are conjugate.
28. Show that any two nontrivial homomorphisms from Z3 to D6 are conjugate.
29. What are the conjugacy classes of homomorphisms from Z4 to D8 ?
30. Let G be a group and let f : G → Sn be a homomorphism. Write {1, . . . , n}f for the
G-set obtained from the action on {1, . . . , n} induced by f under the correspondence
of Proposition 3.3.16: for g ∈ G and i ∈ {1, . . . , n}f , g · i = f (g)(i).
Let σ ∈ Sn and let f = σ · f , the conjugate homomorphism obtained by conjugating f by σ. Show that σ induces a G-isomorphism of G-sets, σ : {1, . . . , n}f →
{1, . . . , n}f .
31. Let G be a group. Show that there is a one-to-one correspondence between the
conjugacy classes of homomorphisms from G to Sn and the G-isomorphism classes
of the G-sets with n elements. Here, two G-sets are in the same G-isomorphism
class if and only if they are G-isomorphic.
32. Let X and Y be sets, with X finite. Show that if i : X → Y and j : X → Y are any
two injective functions, then the induced homomorphisms i∗ : S(X) → S(Y ) and
j∗ : S(X) → S(Y ) (as defined in Problem 15 of Exercises 3.1.16) are conjugate.
(Hint: In the case that Y is infinite, set theory shows that Y may be put in
one-to-one correspondence with the complement of any of its finite subsets.)
33. Let i and j be injective functions from {1, . . . , n} to Z+ . Show that the induced
homomorphisms i∗ , j∗ : Sn → S∞ are conjugate.
CHAPTER 3. G-SETS AND COUNTING
3.7
77
Conjugating Subgroups: Normality
We haven’t yet made full use of the fact that conjugation is an action by automorphisms.
The key point is that if f is an automorphism of G and if H ⊂ G is a subgroup, then
f (H) = {f (h) | h ∈ H} is also a subgroup. We get an induced action of Aut(G) on the
set Sub(G) of all subgroups of G, via f · H = f (H). The following lemma is immediate.
Lemma 3.7.1. If K acts on G via automorphisms, then there is an induced action of
K on Sub(G), the set of all subgroups of G.
Let’s consider the action of G on Sub(G) induced by conjugation. Here, for x ∈ G and
H ∈ Sub(G), the subgroup obtained by letting x act on H is xHx−1 = {xhx−1 | h ∈ H}:
Definition 3.7.2. Let H be a subgroup of G, and let x ∈ G. The conjugate of H by x is
xHx−1 = {xhx−1 | h ∈ H}, the subgroup obtained by letting x act on H by conjugation.
Of course, the set of all conjugates of H is the orbit of H under the action of G on
Sub(G) by conjugation.
The isotropy subgroup of H ∈ Sub(G) under the conjugation action is called NG (H),
the normalizer of H in G:
Definition 3.7.3. Let H be a subgroup of G. The normalizer in G of H, NG (H), is
given as
NG (H) = {x ∈ G | xHx−1 = H}.
It is the isotropy subgroup of H under the action of G on Sub(G) by conjugation.
Our study of G-sets now allows us to calculate the size of the orbit of H under
conjugation.
Corollary 3.7.4. The number of conjugates of H ⊂ G is [G : NG (H)], the index of the
normalizer of H in G.
Of course, if x ∈ H, then conjugation by x gives an automorphism of H, and hence
xHx−1 = H.
Corollary 3.7.5. Let H be a subgroup of G. Then H ⊂ NG (H). In particular, the
number of conjugates of H in G divides [G : H].
Proof We just saw that xHx−1 = H for x ∈ H, so H ⊂ NG (H), as claimed. The
last statement follows from Problem 4 of Exercises 3.2.15: If [G : H] is finite, then
[G : H] = [G : NG (H)] · [NG (H) : H]. (The reader should be able to verify this easily in
the case that G is finite, which is the most important case for our purposes.)
Thus, we have H ⊂ NG (H) ⊂ G. In practice, we can find examples where NG (H) =
H, and others where NG (H) strictly contains H and is strictly contained in G. Finally,
there is the case where NG (H) = G, which is important enough to merit a name.
Definitions 3.7.6. A subgroup H ⊂ G is called normal if NG (H) = G (i.e., xHx−1 = H
for all x ∈ G). We write H G to indicate that H is a normal subgroup of G.
A subgroup H ⊂ G is called characteristic if f (H) = H for all automorphisms f of
G.
CHAPTER 3. G-SETS AND COUNTING
78
Clearly, a characteristic subgroup of G is normal in G. However, not all normal
subgroups are characteristic.
For instance, conjugation is trivial in an abelian group, and hence every subgroup of
an abelian group is normal. However, as we shall see in the exercises below, the only
characteristic subgroups of Z2 × Z2 are e and the group itself. Thus, abelian groups can
have subgroups which are not characteristic.
We shall consider many more examples of both normal and characteristic subgroups
in the exercises.
If H G, then we get an action of G on H through automorphisms:
Lemma 3.7.7. Let H be a normal subgroup of G. Then for x ∈ G, conjugation by x
gives an automorphism, cx , of H: cx (h) = xhx−1 for all h ∈ H. Moreover, there is a
homomorphism Γ : G → Aut(H) defined by Γ(x) = cx for all x ∈ G.
Proof Since cx gives an automorphism of G, it restricts to a group isomorphism from
H to cx (H) = xHx−1 . Since xHx−1 = H, this restriction is precisely an automorphism
of H. Now clearly, cx ◦ cy = cxy , and the result follows.
Essentially the same proof gives the following lemma.
Lemma 3.7.8. Let H be a characteristic subgroup of G. Then every automorphism f :
G → G restricts to an automorphism f : H → H. We obtain a restriction homomorphism
ρ : Aut(G) → Aut(H), by setting ρ(f ) to be the restriction f : H → H for each f ∈
Aut(G).
As we shall begin to see in the next chapter, the study of normal subgroups is one
of the most important subjects in group theory. It is also a very complicated subject.
For instance, as we shall see in the exercises below, Gertrude Stein’s Theorem fails for
normal subgroups: we can find subgroups K ⊂ H ⊂ G such that K H and H G,
but K is not normal in G. This failure of Gertrude Stein’s Theorem helps motivate the
study of characteristic subgroups:
Corollary 3.7.9. Suppose given subgroups K ⊂ H ⊂ G such that H G and K is a
characteristic subgroup of H. Then K G.
Proof Let x ∈ G. Then conjugation by x restricts to an automorphism of H, and
hence preserves K.
Recognition of normal subgroups is an important endeavor. Of course, if H is finite,
then any injection from H to H is a bijection, so that xHx−1 ⊂ H if and only if
xHx−1 = H. However, there are examples of infinite groups H ⊂ G and x ∈ G where
xHx−1 is contained in but not equal to H. However, if xHx−1 ⊂ H for all x ∈ G, things
work out nicely:
Proposition 3.7.10. Let H be a subgroup of G. Then H G if and only if xHx−1 ⊂ H
for all x ∈ G. Also, H is characteristic if and only if f (H) ⊂ H for all automorphisms
f of G.
Proof Suppose that xHx−1 ⊂ H for each x ∈ G. To show that H G, we must show
that H ⊂ xHx−1 for each x ∈ G. But x−1 Hx ⊂ H, and hence H = x(x−1 Hx)x−1 ⊂
xHx−1 , as desired.
The proof for the case of characteristic subgroups is nearly identical: if f (H) ⊂ H
for all f ∈ Aut(G), then f −1 (H) ⊂ H, so H = f (f −1 (H)) ⊂ f (H).
CHAPTER 3. G-SETS AND COUNTING
79
Notice that the proof of Proposition 3.7.10 actually shows more than was stated. For
instance, in the case of normality, it shows that if xHx−1 ⊂ H and x−1 Hx ⊂ H, then
xHx−1 = H, and hence x ∈ NG (H). Thus, we have the following proposition.
Proposition 3.7.11. Let H and K be subgroups of G. Suppose that xHx−1 ⊂ H for
all x ∈ K. Then K ⊂ NG (H).
One application of the concept of normality is the following variant of Proposition 2.10.8.
Proposition 3.7.12. Let H and K be subgroups of the group G. Then G is the internal
direct product of H and K if and only if the following conditions hold.
1. H ∩ K = e.
2. Both H and K are normal in G.
3. G is generated by the elements of H and K.
Proof Suppose that G is the internal direct product of H and K. By Proposition 2.10.8,
it suffices to show that H and K are normal in G. But by definition of the internal direct
product, this will follow if we know that the subgroups H × e and e × K are normal in
H × K. The verification of this is left to the reader.
Conversely, suppose the three conditions hold. By Proposition 2.10.8, it suffices to
show that for each h ∈ H and k ∈ K, the elements h and k commute.
By Lemma 3.6.10, it suffices to show that for h ∈ H and k ∈ K, hkh−1 k −1 = e. By
condition 1, this says that hkh−1 k −1 ∈ H ∩ K.
Now K G, so hkh−1 ∈ K, and hence hkh−1 k −1 ∈ K. Similarly, the normality of
H in G gives kh−1 k −1 ∈ H, and hence hkh−1 k −1 ∈ H, so the result follows.
Normality is intimately connected with the relationship between left and right cosets.
The reader may verify the following lemma.
Lemma 3.7.13. Let H be a subgroup of G. Then the following conditions are equivalent.
1. H is normal in G.
2. For each x ∈ G, we have xH = Hx.
3. Every left coset of H in G is equal to a right coset of H in G; i.e., for each x ∈ G
there is an element y ∈ G such that xH = Hy.
Exercises 3.7.14.
1. Let G be a group and let H ⊂ G × G be given by
H = {(g, g)|g ∈ G}.
(H is called the diagonal subgroup of G × G.) Show that H G × G if and only
if G is abelian.
2. Let G be a group and let y ∈ G. Show that y G if and only if xyx−1 ∈ y for
all x ∈ G. Show that y is a characteristic subgroup of G if and only if f (y) ∈ y
for each f ∈ Aut(G).
3. Show that the subgroup b ⊂ D2n is normal.
CHAPTER 3. G-SETS AND COUNTING
80
‡ 4. Show that the subgroup b ⊂ D2n is characteristic for n ≥ 3.
5. Find a non-normal subgroup of D8 .
6. Show that every subgroup of Q8 is normal.
7. Find the unique characteristic subgroup of Q8 which is equal to neither e nor Q8 .
‡ 8. Show that if n ≥ 3, then b ⊂ Q4n is characteristic.
‡ 9. Let H be a subgroup of Zn and let f : Zn → Zn be any homomorphism. Show
that f (H) ⊂ H. Deduce that every subgroup of Zn is characteristic. (Hint: Show
that H = d = {dk | k ∈ Zn } for some d that divides n.)
† 10. Using the fact that An is the subgroup of Sn generated by the 3-cycles, show that
An Sn . Thus Lemma 3.7.7 provides a homomorphism Γ : Sn → Aut(An ), where
Γ(σ) is induced by conjugation by σ for all σ ∈ Sn . Deduce from Problem 12 of
Exercises 3.6.24 that Γ is an injection for n ≥ 4.
11. Show that the only characteristic subgroups of G = Z2 × Z2 are the identity and
G itself.
† 12. Let H be an index 2 subgroup of the finite group G. Show that H G.
13. Show that any index 2 subgroup of an infinite group is normal.
14. For x ∈ G, show that CG (x) ⊂ NG (x).
15. If x ∈ G has order 2, show that CG (x) = NG (x). Deduce that x G if and
only if x ∈ Z(G).
16. Give an example of a group G and an x ∈ G with CG (x) = NG (x).
17. List the distinct subgroups of D2n that are conjugate to a.
18. What is the normalizer of a ⊂ Q16 ?
19. List the distinct subgroups of Q16 that are conjugate to a.
20. What is the normalizer of a ⊂ Q4n ? List the distinct subgroups of Q4n that are
conjugate to a. (Hint: The answer depends in a crucial way on the value of n
mod 2.)
21. Show that characteristic subgroups satisfy Gertrude Stein’s Theorem: If H is a
characteristic subgroup of G, and K is a characteristic subgroup of H, show that
K is a characteristic subgroup of G.
22. Show that S∞ S(Z+ ).
23. Show that S({2, 3, . . . }) and S({3, 4, . . . }) are not conjugate as subgroups of S(Z+ ).
24. Let H be a subgroup of G. Show that H is a normal subgroup of NG (H). Show
that NG (H) is the largest subgroup of G that contains H as a normal subgroup;
i.e., if H ⊂ K ⊂ G and H K, then K ⊂ NG (H).
† 25. Let H and K be subgroups of G. Show that
NG (H) ∩ NG (K) ⊂ NG (H ∩ K).
CHAPTER 3. G-SETS AND COUNTING
81
26. Let H be a subgroup of G and let x ∈ G. Show that
NG (xHx−1 ) = xNG (H)x−1 .
27. Show that Gertrude Stein’s Theorem fails for normal subgroups: i.e., find a group
G with a normal subgroup K G and a subgroup H of K such that H K, but
H is not normal in G. (Hint: Find a group G with a subgroup H such that NG (H)
is normal in G but not equal to G.)
28. Let H be a subgroup of the finite group G. Show that the number of elements in
−1
is less than or equal to [G : NG (H)] · (|H| − 1) + 1. Deduce that no
x∈G xHx
proper subgroup of G contains an element in every conjugacy class of G.
29. Let X be a G-set and let x ∈ X. Show that the isotropy subgroup of gx is gGx g −1 .
30. (See Problem 18 of Exercises 3.3.23.) What are the elements of the H-fixed-point
set (G/H)H ? For which elements yH ∈ (G/H)H is the G-map from G/H to itself
which takes eH to yH a G-isomorphism (i.e., which elements of (G/H)H have
isotropy subgroup exactly equal to H)?
‡ 31. Let H be a subgroup of G and consider the standard action of G on G/H. Let
Λ : G → S(G/H) be the homomorphism induced by this action. Show that
ker Λ =
xHx−1 .
x∈G
32. Let H be a subgroup of G and let X be the set of all subgroups conjugate to H:
X = {xHx−1 | x ∈ G}. Thus, X is the orbit of H under the action of G on Sub(G)
(the set of subgroups of G) by conjugation. Since X is a G-set, we get an induced
homomorphism Λ : G → S(X). Show that
ker Λ =
xNG (H)x−1 .
x∈G
Chapter 4
Normality and Factor Groups
The closing section of Chapter 3 gives the definition of a normal subgroup:
A subgroup H ⊂ G is normal if xHx−1 = H for all x ∈ G. We write H G to denote
that H is a normal subgroup of G.
The normal subgroups of a given group are absolutely vital in understanding its
structure.
Some groups G have no normal subgroups other than the trivial subgroup e and
the group G itself. Such groups are called simple. We shall see, via the technique
of composition series below, that all finite groups are built up out of the finite simple
groups.
The finite simple groups are all known. In one of the triumphs of mathematical
history, the finite simple groups have been classified, according to a plan of attack developed by Daniel Gorenstein that played out over a sequence of papers by several authors.
One of the final steps was the construction, by Robert Griess, of a group that was then
known as the “monster group,” as its order is nearly 1054 . In current parlance, the
monster group is the sporadic simple group F1 .1
We shall not need the classification by Gorenstein et. al., as we are concentrating
here on the groups of relatively low order. As will become apparent in the exercises from
the next two chapters, the only simple groups whose order is ≤ 60 are the cyclic groups
Zp , where p is a prime < 60, together with the alternating group A5 .
In any case, the study of the ways in which finite groups can be put together out of
the simple groups is the most important remaining problem for the classification of finite
groups. It is known as the Extension Problem.
Thus, the Extension Problem is, in the opinion of this author at least, one of the
important unsolved problems in mathematics. The study of normal subgroups and factor
groups is at its core.
We do not yet have the language with which to give a precise statement of the
Extension Problem, but we shall do so within this chapter. In fact, we shall develop
techniques to study certain special cases of the Extension Problem. These special cases
will turn out to be enough, together with the Sylow theorems and the material on G-sets
from Chapter 3, to classify the groups in a fair number of the lower orders.
1 See Finite Simple Groups, An Introduction to Their Classification, by Daniel Gorenstein, Plenum
Press, 1982.
82
CHAPTER 4. NORMALITY AND FACTOR GROUPS
83
The most important aspect of normal subgroups is that if H is normal in G, then
there is a group structure on the set of left cosets G/H such that the canonical map
π : G → G/H is a homomorphism. If G is finite and H is nontrivial, then G/H has a
lower order than G, and hence is, in general, easier to analyze. If G has enough normal
subgroups (e.g. if G is abelian or solvable), then induction arguments on the order of
G may be used to deduce facts about G from facts about the lower order factor groups
G/H.
In Section 4.1, we give the Noether Isomorphism Theorems, which study the properties of the factor groups G/H when H is normal in G. These are truly fundamental in
the understanding of groups, and will open the door to some new classification results.
As an application, we shall show that if G is finite and p is the smallest prime dividing
the order of G, then any subgroup of G of index p is normal in G.
In Section 4.2, we give the technique of composition series, which shows that finite
groups are built up out of simple groups. We also show that the alternating groups An
are simple for n ≥ 5. Then, in Section 4.3, we give the Jordan–Hölder Theorem, which
shows that the simple groups out of which a given group is built are unique, though the
way in which they are put together may not be.
Then, by an induction on order, using the fact that every subgroup of an abelian group
is normal, we give a complete classification of the finite abelian groups, the Fundamental
Theorem of Finite Abelian Groups.
In Section 4.5, we calculate the automorphism group of a cyclic group. This calculation will be of essential value in constructing some new groups by the technique of
semidirect products and in classifying the extensions of a given cyclic group.
Section 4.6 develops the technique of semidirect products, by which new groups may
be constructed from old. Unlike the direct product, many of the semidirect products of
two abelian groups are nonabelian. We shall construct some interesting new groups in
this manner.
Section 4.7 defines and studies the extensions of one group by another. Extensions
describe some of the ways that groups are built up out of smaller groups. We shall state
the Extension Problem here and shall develop some classification techniques, particularly
for groups that are semidirect products. We shall make some significant progress in
obtaining actual classification results.
4.1
The Noether Isomorphism Theorems
Here, we study normal subgroups H G and the group structure that’s induced on the
set of left cosets G/H.
One convenient source of normal subgroups (which will also turn out to give all of
them) is from the kernels of homomorphisms:
Lemma 4.1.1. Let f : G → G be a homomorphism. Then ker f is a normal subgroup
of G.
Proof Let K = ker f . By Proposition 3.7.10, it suffices to show that xKx−1 ⊂ K for
each x ∈ G, i.e., that xkx−1 ∈ K for all x ∈ G and all k ∈ K. But if k ∈ K, then
f (xkx−1 ) = f (x) · f (k) · f (x)−1 = f (x) · e · f (x)−1 = e, and hence xkx−1 ∈ K.
We shall see that im f need not be normal in G . We next explore the relationship
between normality of a subgroup H ⊂ G and group structures on the set of cosets G/H:
CHAPTER 4. NORMALITY AND FACTOR GROUPS
84
Theorem 4.1.2. Let H be a subgroup of G. Then H is normal in G if and only if
there is a group structure on the set G/H of left cosets of H with the property that the
canonical map π : G → G/H is a homomorphism.
If H G, then the group structure on G/H which makes π a homomorphism is
unique: we must have gH · g H = gg H for all g, g ∈ G. Moreover, the kernel of
π : G → G/H is H. Thus, a subgroup of G is normal if and only if it is the kernel of a
homomorphism out of G.
Proof Suppose that there is a group structure on G/H such that π is a homomorphism.
Then the identity element of G/H is π(e) = eH. So g ∈ ker π if and only if gH = eH.
By Lemma 3.2.3, this is true if and only if g ∈ H. Thus, H is the kernel of π, and hence
is normal by Lemma 4.1.1.
Clearly, there is at most one group structure on G/H which makes π a homomorphism,
as, since π(g) = gH, we must have gH · g H = gg H for all g, g ∈ G.
Suppose then that H is normal in G. We wish to show that gH · g H = gg H defines
a group structure on G/H. We first show that it gives a well defined binary operation.
Thus, if gH = xH and g H = x H, we wish to show that gg H = xx H. But Lemma 3.2.3
shows that x = gh and x = g h for some h, h ∈ H. Thus, xx H = ghg h H = ghg H.
By Lemma 3.2.3, this is equal to gg H provided that (gg )−1 ghg ∈ H. But (gg )−1 ghg =
(g )−1 g −1 ghg = (g )−1 hg . And (g )−1 hg ∈ H because H G. Thus, the operation is
well-defined.
That this operation gives a group structure now follows immediately from the group
laws in G: associativity in G gives associativity in G/H, and π(e) = eH is an identity
element for G/H. Finally, the inverse of gH is g −1 H, so G/H is indeed a group.
Definition 4.1.3. When H G, we shall simply write G/H for the group structure on
G/H which makes π a homomorphism (i.e., with the multiplication gH · g H = gg H),
and call it the factor group, or quotient group, of G modulo H.
Example 4.1.4. We’ve already seen examples of factor groups, as the groups Zn are
precisely the factor groups Z/n. The point is that for k ∈ Z, the congruence class of
k modulo n is precisely the coset k + n. (To see this, note that l ≡ k mod n if and
only if l − k ∈ n, i.e., l ∈ k + n.) The canonical map π : Z → Zn that we’ve been
using coincides with the canonical map above, as it takes each element in Z to its coset
modulo n. And, indeed, the group operation in Zn is defined precisely as in the more
general case above.
Another way of seeing that Z/n ∼
= Zn will be given using the Noether Isomorphism
Theorems, below.
Since |G/H| = [G : H], Lagrange’s Theorem gives the following formula.
Lemma 4.1.5. Let H G with G finite. Then
|G| = |H| · |G/H|.
Notation 4.1.6. Let H G. For brevity, we shall generally write g in place of gH ∈
G/H. Thus, every element of G/H has the form g = π(g) for some g ∈ G, and have
g · g = gg . The identity element is e = e = h for any h ∈ H.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
85
Note that |G/H| = [G : H] is less than |G| whenever G is finite and H = e. In
this case, the factor group G/H is less complicated than G, and we may obtain information about G from an understanding of the groups G/H as H ranges over the normal
subgroups of G.
We can characterize the homomorphisms out of G/H in terms of the homomorphisms
out of G.
Proposition 4.1.7. Let H G. Then for any group G and any homomorphism f :
G → G whose kernel contains H, there is a unique homomorphism f : G/H → G that
makes the following diagram commute.
/ G
GC
<
CC
z
CC
zz
z
C
zz
π CC
!
zz f
G/H
f
Conversely, if f : G → G factors by a commutative diagram as above, then H must
be contained in ker f . Thus, the passage from a homomorphism f : G/H → G to the
composite (f ◦π) : G → G gives a one-to-one correspondence between the homomorphisms
from G/H to G and those homomorphisms from G to G whose kernels contain H.
Proof Suppose given f : G → G . If a homomorphism f : G/H → G exists with
f ◦ π = f , then for each h ∈ H, f (h) = f (h) = f (e) = e, and hence H ⊂ ker f . Moreover,
f (g) = f ◦ π(g) = f (g), so f is unique if it exists. It suffices to show that such an f exists
whenever H ⊂ ker f .
Thus, suppose H ⊂ ker f . As we’ve just seen, f , if it exists, must be given by
f (g) = f (g). The key is to show this is a well-defined function from G/H to G . Thus,
suppose that g = g in G/H. By Lemma 3.2.3, this means g = gh for some h ∈ H.
Thus, f (g ) = f (g)f (h) = f (g), since h ∈ ker f . So f is well-defined.
But f (g)f (g ) = f (g)f (g ) = f (gg ) = f (gg ) = f (gg ), and hence f is a homomorphism.
We call f the homomorphism induced by f . In the counting argument below, we
again make use of the fact that |G/H| = [G : H].
Theorem 4.1.8. (First Noether Isomorphism Theorem) Let f : G → G be a homomorphism. Then f induces an isomorphism from G/ ker f to im f . Thus, if G is finite,
|im f | = [G : ker f ], so Lagrange’s Theorem gives
|G| = |ker f | · |im f |.
Proof Let f : G/ ker f → im f be the homomorphism induced by f . Since f (g) = f (g),
f is onto. Thus, it suffices (Lemma 2.5.9) to show that ker f = e. Let g ∈ ker f . Then
f (g) = f (g) = e, so g ∈ ker f . But then g = e in G/ ker f .
Example 4.1.9. Recall the rotation matrices Rθ ∈ Gl2 (R). We saw in Section 2.7
that the collection of all rotation matrices, {Rθ | θ ∈ R}, forms a subgroup, SO(2), of
Gl2 (R). Moreover, as shown in Proposition 2.7.6, there is a surjective homomorphism
exp : R → SO(2) given by exp(θ) = Rθ . The kernel of exp is shown there to be the
cyclic subgroup 2π ⊂ R. Thus, by the first Noether theorem, SO(2) is isomorphic to
R/2π.
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86
The Noether Theorem gives us another way to recognize Zn as Z/n.
Corollary 4.1.10. Let n be a positive integer. Then Zn is isomorphic to the factor
group Z/n.
Proof The canonical map π : Z → Zn is a surjection, with kernel n.
Recall that there is a homomorphism ε : Sn → {±1} defined by setting ε(σ) equal to
the sign of σ for σ ∈ Sn . Since ε is onto, its image has order 2. But An is the kernel of
ε, so we may calculate its order:
Corollary 4.1.11. An has index 2 in Sn , and hence |An | = n!/2.
The next result foreshadows the techniques we shall use in Chapter 5. Recall from
Problem 12 of Exercises 3.7.14 that any index 2 subgroup of a finite group is normal.
Using the First Noether Isomorphism Theorem, we shall give a generalization of this
result. First, note from the example of a ⊂ D6 that a subgroup of prime index in a
finite group need not be normal. Of course, 2 is the smallest prime, so the next result
gives the desired generalization.
Proposition 4.1.12. Let G be a finite group and let p be the smallest prime dividing
the order of G. Then any subgroup of G of index p is normal.
Proof Let H be a subgroup of index p in G, and consider the usual action of G on the
set of left cosets G/H. As in Proposition 3.3.16, we get an induced homomorphism
Λ : G → S(G/H) ∼
= Sp
by setting Λ(g)(xH) = gxH for all g ∈ G and xH ∈ G/H.
As shown in Proposition 3.3.16, the kernel of Λ is the intersection of the isotropy
subgroups of the elements of G/H. Since the isotropy subgroup of eH is H, we see that
ker Λ ⊂ H.
Since |Sp | = p! and since p is the smallest prime dividing |G|, the greatest common
divisor of |G| and |Sp | is precisely p. The First Noether Isomorphism Theorem shows
that |im Λ| must divide |G|. Since |im Λ| also divides |Sp |, the order of im Λ must be
either 1 or p.
Since ker Λ ⊂ H, Λ cannot be the trivial homomorphism, so we must have |im Λ| = p.
By the First Noether Isomorphism Theorem, we see that |ker Λ| = |G|/p = |H|. Since
ker Λ ⊂ H, this forces ker Λ = H. Since the kernel of a homomorphism is always normal,
this gives H G, as desired.
The remaining isomorphism theorems are concerned with a deeper analysis of the
canonical maps. The reader may easily verify the following lemma.
Lemma 4.1.13. Let f : G → G be a homomorphism and let H ⊂ G. Let f : H → G
be the restriction of f to H. Then ker f = H ∩ ker f .
Recall from Problem 8 of Exercises 2.5.21 that if f : G → G is a surjective homomorphism, then there is a one-to-one correspondence between the subgroups of G and
those subgroups of G that contain ker f . Under this correspondence, the subgroup K of
G corresponds to f −1 (K ) ⊂ G, while a subgroup K ⊂ G containing ker f corresponds
to f (K) ⊂ G . Let us reconsider this correspondence in the case where f is the canonical
map π : G → G/H.
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87
Lemma 4.1.14. Let H G. Then the subgroups of G/H all have the form K/H, where
K is a subgroup of G that contains H.
Proof The one-to-one correspondence between the subgroups of G/H and those subgroups of G that contain H is based on the following facts (which we assume were verified
by the reader).
1. If K is a subgroup of G/H, then π(π −1 (K )) = K .
2. If K is a subgroup of G which contains H, then π −1 (π(K)) = K.
Given K ⊂ G/H, let K = π −1 (K ). Then K contains H, and K = π(K). Now π
restricts to give a homomorphism, which we shall call π , from K onto π(K).
Now, the kernel of π : K → π(K) is K ∩ ker π = H (Lemma 4.1.13). The first
Noether theorem now allows us to identify π(K) with K/H.
Example 4.1.15. Let n and k be positive numbers such that k divides n. Then the
unique subgroup of Zn of order n/k is k.
Let π : Z → Zn be the canonical map. Clearly, k = π(k). But since k divides n,
we have ker π = n ⊂ k. Thus, k = π −1 k, and the above gives an isomorphism
from k/n to k ∼
= Zn/k .
We wish to generalize Lemma 4.1.14 to an analysis of the subgroups π(K) where
K ⊂ G does not contain H. First consider this:
Definition 4.1.16. Let H and K be subgroups of G. Then there is a subset KH of G
defined by
KH = {kh | k ∈ K, h ∈ H}.
(Of course, if G is abelian, we write K + H = {k + h | k ∈ K, h ∈ H} for this subset.)
Examples 4.1.17. Since any element of D2n can be written as either bk or as abk for
an appropriate value of k, we see that D2n = ab.
Similarly, Q4n = ab. The fact that some elements may be written in more than
one way as a product ar bs with 0 ≤ r ≤ 3 and 0 ≤ s ≤ 2n doesn’t change this: the set
{ar bs | r, s ∈ Z} is ab, and it includes every element in Q4n .
Lemma 4.1.18. Let H be a subgroup of G and let K be a subgroup of NG (H). Then
KH is a subgroup of G, and H ⊂ KH ⊂ NG (H). In addition, in this case, KH is equal
to the subset HK of G, and is the smallest subgroup of G that contains both H and K.
Proof Since K ⊂ NG (H), for any k ∈ K and h ∈ H, we have k −1 hk = h ∈ H, and
hence hk = kh . This is what we need to show that KH is closed under multiplication:
for k1 , k2 ∈ K and h1 , h2 ∈ H, we have k1 h1 k2 h2 = k1 k2 h1 h2 for some h1 ∈ H.
Similarly, (kh)−1 = h−1 k −1 = k −1 h for some h ∈ H, and hence KH a subgroup of
G.
Since h = eh and k = ke, both H and K are contained in KH. And any subgroup
containing both H and K must contain each product kh, so KH is indeed the smallest
subgroup containing H and K. Since K ⊂ NG (H), this implies HK ⊂ NG (H).
Finally, if h ∈ H and k ∈ K, we have khk −1 = h ∈ H, so that kh = h k ∈ HK, and
hence KH ⊂ HK. Similarly, HK ⊂ KH.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
88
Example 4.1.19. Let K be any subgroup of Sn that is not contained in An (e.g., K =
(1 2)). Since An Sn , KAn is a subgroup of Sn . Moreover, since K is not contained
in An , [KAn : An ] > 1. But [KAn : An ] divides [Sn : An ]. Since [Sn : An ] = 2, we see
that KAn = Sn .
Lemma 4.1.18 sets up the second Noether Theorem.
Theorem 4.1.20. (Second Noether Isomorphism Theorem) Let H and K be subgroups
of G with H G. Write π : K → G/H for the restriction to K of the canonical map
π : G → G/H. Then the image of π is KH/H, while the kernel of π is K ∩ H. Thus,
the First Noether Isomorphism Theorem provides an isomorphism
π : K/(K ∩ H)
∼
=
/ KH/H.
Proof By Lemma 4.1.13, the kernel of π is K ∩ ker π = K ∩ H. Thus, it suffices to
show that the image of π is KH/H.
Of course, the image of π is just π(K). Now for k ∈ K and h ∈ H, kh = k ∈ G/H,
and hence π(kh) = π(k). But this says that π(KH) = π(K). Since H ⊂ KH, π(KH) =
KH/H by Lemma 4.1.14, and the result follows.
Corollary 4.1.21. Let H and K be subgroups of G with H G. Then [K : H ∩ K]
divides [G : H].
Proof [K : H ∩ K] is the order of K/(H ∩ K), which is thus equal to the order of the
subgroup KH/H of G/H. By Lagrange’s Theorem, this divides [G : H], the order of
G/H.
Here is a sample application.
Corollary 4.1.22. Let K be a subgroup of Sn and suppose that K is not contained in
An . Then [K : K ∩ An ] = 2.
Proof Corollary 4.1.21 shows that [K : K ∩ An ] divides [Sn : An ], which is 2. But
[K : K ∩ An ] = 1, as K is not contained in An .
The Second Noether Theorem may be used for yet other calculations of indices.
Corollary 4.1.23. Let f : G → G be a surjective homomorphism between finite groups
and let K be a subgroup of G. Then [G : f (K)] divides [G : K].
Proof Let H be the kernel of f . By the First Noether Isomorphism Theorem, there
is an isomorphism f : G/H → G such that f ◦ π = f , where π : G → G/H is the
canonical map. But then f induces an isomorphism from π(K) to f (K). Thus, we see
that [G : f (K)] = [G/H : π(K)].
By the Second Noether Isomorphism Theorem, π(K) ∼
= K/(H ∩ K). Thus, we obtain
[G : f (K)]
=
=
=
|G|/|H|
|K|/|H ∩ K|
|G|/|K|
|H|/|H ∩ K|
[G : K]
.
[H : H ∩ K]
Since these indices are all integers, the result follows.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
89
We now give a slight refinement of the Second Noether Theorem.
Corollary 4.1.24. Let H and K be subgroups of G, with K ⊂ NG (H). Then H ∩K K
and K/(H ∩ K) ∼
= KH/H.
Proof Just replace G by NG (H) and apply the second Noether Theorem.
Now, we give the final Noether Isomorphism Theorem.
Theorem 4.1.25. (Third Noether Isomorphism Theorem) Suppose given two subgroups,
H and K, of G, which are both normal in G, with H ⊂ K. Then K/H G/H, and the
factor group (G/H)/(K/H) is isomorphic to G/K.
Proof Write πH : G → G/H and πK : G → G/K for the two canonical maps. Then
H ⊂ K = ker πK . By Proposition 4.1.7, there is a homomorphism f : G/H → G/K with
f ◦ πH = πK . Clearly, f is onto. It suffices to show that ker f = K/H.
Let g = gH ∈ G/H. Since f (g) = πK (g) ∈ G/K, we see that g ∈ ker f if and only if
g ∈ K = ker πK . Thus, ker f is precisely K/H.
Exercises 4.1.26.
1. Give an example of a homomorphism f : G → G such that im f is not normal in
G .
† 2. Show that every factor group of a cyclic group is cyclic. (Suggestion: Give at least
two different proofs.)
3. Recall that b D8 . For each subgroup H ⊂ b, list all the subgroups K ⊂ D8
with K ∩ b = H. (Hint: What is the order of K/H = K/(K ∩ b)?)
4. List all the normal subgroups of D8 . What is the resulting factor group in each
case?
5. List all the groups (up to isomorphism) that appear as subgroups of D8 .
6. Recall that b D2n . Let H be a subgroup of b. How many subgroups of D2n are
there whose intersection with b is H? (Hint: The answer depends on [b : H].)
What are these subgroups?
7. List all the normal subgroups of D2n . What is the resulting factor group in each
case?
8. List all the groups (up to isomorphism) that appear as subgroups of D2n .
9. Recall that b Q4n . Let H be a subgroup of b. How many subgroups of
Q4n are there whose intersection with b is H? (Hint: The answer depends on
[b : H].) What are these subgroups?
10. List all the normal subgroups of Q4n . What is the resulting factor group in each
case?
11. List all the groups (up to isomorphism) that appear as subgroups of Q4n .
† 12. Show that every element of Q/Z has finite order. Show also that Q/Z has elements
of order n for every n ≥ 1.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
90
† 13. Show that there is an isomorphism between R/Z and R/x for any 0 = x ∈ R.
Show also that the elements of finite order in R/Z are precisely those lying in the
subgroup Q/Z.
† 14. Let Hi Gi for 1 ≤ i ≤ k. Show that H1 × · · · × Hk G1 × · · · × Gk and that
(G1 × · · · × Gk )/(H1 × · · · × Hk ) ∼
= (G1 /H1 ) × · · · × (Gk /Hk ).
15. What are all the subgroups of Z2 × Z4 ? What is the resulting factor group in each
case?
16. Suppose given an index three subgroup H ⊂ G such that H is not normal in G.
Show that there is a surjective homomorphism from G to S3 .
† 17. Let H and K be subgroups of G with K ⊂ NG (H). Show that NG (K) ∩ NG (H) ⊂
NG (KH).
18. Generalize Corollary 4.1.23 as follows: Show that if f : G → G is a surjective
homomorphism and if [G : K] is finite, then [G : f (K)] is finite and divides
[G : K]. (Hint: Show that [G : f (K)] = [G : KH], where H = ker f .)
19. Let H be a subgroup of G. Show that the factor group NG (H)/H acts on the right
of the set of left cosets G/H via gH · x = gxH for all g ∈ G and x ∈ NG (H).
Show also that the action of each x ∈ NG (H)/H gives a G-isomorphism from G/H
to itself. Finally, show that every G-isomorphism from G/H to itself arises in
this manner, and deduce that the group of G-isomorphisms from G/H to itself is
isomorphic to the factor group NG (H)/H.
20. Let X be any G-set and let H be a subgroup of G. Show that there is an induced
action of the factor group NG (H)/H on the H-fixed point set X H .
4.2
Simple Groups
The finite simple groups, which we define in this section, are the building blocks out of
which finite groups are made. We show this here via the technique of composition series.
We then show that the alternating groups An are simple for n ≥ 5.
Definition 4.2.1. A group G is simple if it has no normal subgroups other than e and
G.
We can characterize the abelian simple groups immediately. We shall give more
complicated examples below.
Lemma 4.2.2. An abelian group is simple if and only if it is a cyclic group of prime
order.
Proof Since every subgroup of an abelian group is normal, we see that an abelian group
is simple if and only if it has no subgroups other than e and itself. Thus, an abelian
group G is simple if and only if every non-identity element in it generates G.
In particular, G must be cyclic, and, since Z has nontrivial proper subgroups, it
must be finite. But our calculation of the orders of the elements in Zn (Problem 5 of
Exercises 2.5.21) shows that Zn is simple if and only if n is prime.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
91
The choice of the word “simple” for the simple groups is perhaps unfortunate, as the
nonabelian simple groups can be much more complicated than, say, solvable ones (see
below). Nevertheless, all finite groups are built up out of simple groups, hence the name.
Lemma 4.2.3. Let G = e be a finite group. Then G has a factor group which is simple
and is not equal to the trivial group.
Proof By the first Noether theorem, the statement is equivalent to saying that each
group G = e admits a surjective homomorphism onto a nontrivial simple group. We
argue by induction on |G|. If |G| = 2, then G itself is simple. Thus, suppose |G| > 2.
If G is simple, we’re done, as G = G/e is a factor group of itself. Otherwise, G has a
normal subgroup H, with H not equal to either e or G. But then 1 < |G/H| < |G|, so by
induction, there is a surjective homomorphism f : G/H → G , where G is a nontrivial
simple group. But then the composite f ◦ π : G → G gives the desired surjection.
The way in which finite groups are built up from simple groups is as follows.
Definition 4.2.4. Let G be a finite group. A composition series for G is a sequence
e = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 Hi , and Hi /Hi−1 is a nontrivial simple
group for 1 ≤ i ≤ k.
Recall that this does not imply that the groups Hi are normal in G.
Proposition 4.2.5. Every nontrivial finite group has a composition series.
Proof Again, we argue by induction on the order of G. Once again, if G is simple, then
there is nothing to prove. Otherwise, Lemma 4.2.3 provides a subgroup H G such that
G/H is a nontrivial simple group. But then |H| < |G|, and hence H has a composition
series e = H0 ⊂ · · · ⊂ Hk−1 = H. But then tacking on Hk = G gives a composition
series for G.
The groups Hi /Hi−1 appearing in a composition series are subquotient groups of G:
Definition 4.2.6. A subquotient of a group G is a group of the form H/K, where
K H ⊂ G.
Composition series are far from unique in many cases. We shall discuss their uniqueness properties in Section 4.3, and shall utilize them here simply to motivate the discussion of simple groups.
Our remaining goal for this section is to prove the following theorem, which produces
our first examples of nonabelian simple groups.
Theorem 4.2.7. The alternating groups An are simple for all n ≥ 5.
In the process of proving this, we shall derive some useful information about symmetric groups. We shall start by identifying the conjugacy classes in Sn . First consider
this:
Definition 4.2.8. We say that the permutations σ, τ ∈ Sn have the same cycle structure
if when you write each of them as a product of disjoint cycles, both have the same number
of k-cycles for each k ≥ 2.
Recall that Lemma 3.6.19 computes the conjugates of a k-cycle.
Proposition 4.2.9. Two permutations in Sn are conjugate if and only if they have the
same cycle structure.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
92
Proof Write σ ∈ Sn as a product of disjoint cycles: σ = σ1 . . . σk . Then for τ ∈ Sn ,
τ στ −1 = τ σ1 τ −1 . . . τ σk τ −1 . By Lemma 3.6.19, each τ σi τ −1 is a cycle of the same length
as σi . Thus, σ and its conjugate τ στ −1 have the same cycle structure, provided we show
that for two disjoint cycles σ1 and σ2 , and any permutation τ , the cycles τ σ1 τ −1 and
τ σ2 τ −1 are disjoint. But that follows from the explicit formula of Lemma 3.6.19.
Thus, it suffices to show that any two permutations with the same cycle structure are
conjugate. Let σ and τ be permutations with the same cycle structure. Thus, we can
write σ and τ as products of disjoint cycles, σ = σ1 . . . σk , and τ = τ1 . . . τk , in such a
way that σj and τj have the same length, length rj , for 1 ≤ j ≤ k. Say σj = (ij1 . . . ijrj ),
and τj = (ij1 . . . ijrj ), for 1 ≤ j ≤ k.
Suppose we can find a permutation μ with μ(ijs ) = ijs for 1 ≤ j ≤ k and 1 ≤ s ≤ rj .
Then μσj μ−1 = τj by Lemma 3.6.19 for all j. Thus, μσμ−1 = τ , so it will suffice to find
such a μ.
Since the permutations σj are disjoint for 1 ≤ j ≤ k, an element of {1, . . . , n} may
occur in at most one of the sets {iji , . . . ,
ijrj }. Thus, setting μ(ijs
all of the
) = ijs for
indices js gives a well-defined function μ : 1≤j≤k {ij1 , . . . , ijrj } → 1≤j≤k {ij1 , . . . , ijrj }.
Moreover, the disjointness ofthe τj shows that this function is bijective.
Now the complement of 1≤j≤k {ij1 , . . . ,
ijrj } in {1, . . . , n} may be placed in one-toone correspondence with the complement of 1≤j≤k {ij1 , . . . , ijrj } in {1, . . . , n}. But any
such one-to-one correspondence extends μ to a permutation of {1, . . . , n}, which then
gives the desired conjugation.
It would make our lives easier in the arguments below if we knew that any two elements
of An with the same cycle structure were conjugates in An (i.e., if the conjugating
permutation μ could be chosen to lie in An ). Unfortunately, this is not always true, so
some care will be required.
Lemma 4.2.10. Let σ, σ ∈ An be conjugate in Sn . Suppose that σ commutes with an
element of Sn which is not in An . Then σ and σ are conjugate in An .
Proof Since σ and σ are conjugate in Sn , there is a permutation τ with τ στ −1 = σ .
If τ ∈ An , there’s nothing to prove. Otherwise, the hypothesis provides a τ ∈ Sn , with
τ not in An (hence ε(τ ) = −1), which commutes with σ. Now ε(τ τ ) = ε(τ )ε(τ ) = +1,
−1
so τ τ ∈ An . And (τ τ )σ(τ τ )−1 = τ (τ στ )τ −1 = τ στ −1 = σ .
And here are some situations where the hypothesis is satisfied.
Lemma 4.2.11. Let σ ∈ An . Suppose that either of the following conditions hold.
1. There are at least two elements of {1, . . . , n} which are left fixed by σ.
2. At least one of the cycles in the cycle structure of σ has even length.
Then σ commutes with an element of Sn which is not in An .
Proof Suppose the first conditions holds, and, say, σ fixes i and j. Then σ is disjoint
from the transposition (i j), and hence commutes with it.
Suppose now that the second condition holds. Then σ commutes with the even
length cycle in question. (Said cycle is disjoint from all of the other cycles in the cycle
decomposition of σ, and hence commutes with them, but it also commutes with itself.)
But even length cycles have sign −1.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
93
Our strategy for proving Theorem 4.2.7 may now be laid out. Recall from Corollary 3.5.11 that An is generated by the 3-cycles. Since n ≥ 5, any 3-cycle commutes with
a transposition in Sn . Thus, by Lemma 4.2.10, any two 3-cycles are conjugate in An . In
particular, if H An and if H contains a 3-cycle, it must contain all of its conjugates
in An , and hence must contain every 3-cycle. Since the 3-cycles generate An , we must
have H = An . Thus, we have verified the following lemma.
Lemma 4.2.12. Let H An for n ≥ 5. If H contains a 3-cycle, then H = An .2
Thus, it suffices to show that if n ≥ 5, then any normal subgroup of An which is not
the identity must contain a 3-cycle. Given a subgroup e = H G, the only information
we have for free is that there is a non-identity element in H. We shall consider various
possibilities for the cycle structure of such an element, and verify, in the end, that the
presence of any non-identity element in H implies the existence of a 3-cycle in H.
We may now generalize Lemma 4.2.12.
Lemma 4.2.13. Let H An for n ≥ 5. Suppose that H contains an element of the
form στ such that
1. σ is a cycle of length ≥ 3.
2. σ and τ are disjoint. (Note that τ may be the identity here.)
3. στ commutes with an element of Sn which is not in An .
Then H = An .
Proof Let σ = (i1 . . . ik ). Since the inverse of an r-cycle is an r-cycle, τ −1 has the same
cycle structure as τ . Thus, στ is conjugate to σ τ −1 , where σ = (ik−1 ik ik−2 . . . i1 ). But
then σ τ −1 στ is in H. Since σ and τ are disjoint, they commute, so this is just σ σ, which
is easily seen to be equal to the 3-cycle (ik−2 ik ik−1 ). So H = An by Lemma 4.2.12.
We would like to be able to eliminate the condition in the preceding lemma which
requires that στ commutes with an element of Sn which is not in An . We shall first
consider another special case, which is actually already covered by Lemma 4.2.13 except
in the cases of n = 5 or 6.
Lemma 4.2.14. Let H An , and suppose that H contains a 5-cycle. Then H = An .
Proof Let σ = (a b c d e) be a 5-cycle in H. Then μ = (a b)(c d) is in An . Since H is
normal in An , μσμ−1 ∈ H, and hence σ · μσμ−1 is also in H. By Lemma 3.6.19,
σ · μσμ−1
= (a b c d e) · (b a d c e)
= (a e c),
a 3-cycle. So H = An by Lemma 4.2.12.
We can now improve on Lemma 4.2.13, provided the cycle σ in the statement there
has length > 3.
Lemma 4.2.15. Let H An . Suppose that H contains an element of the form στ ,
where σ is a cycle of length > 3 and σ and τ are disjoint. Then H = An .
2 This result is still true for n < 5, but with different argumentation. We shall treat these cases in
the exercises below.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
94
Proof First note that the hypothesis forces n ≥ 5, as if σ has length 4, then τ cannot
be the identity element, as 4-cycles are not in An .
Note also that if σ is a 4-cycle, then στ commutes with an element (σ) which is in
Sn but not An , and hence the hypothesis of Lemma 4.2.13 is satisfied.
Thus, we may assume that σ has length ≥ 5. Write σ = (i1 . . . ik ), and let μ =
(i1 i2 )(i3 i4 ). Then μ ∈ An . Moreover, since the support of μ is contained in the support
of σ and since σ and τ are disjoint, μ commutes with τ .
Since μ ∈ An and H An , στ · μ(στ )−1 μ−1 is in H. We shall show that στ ·
μ(στ )−1 μ−1 is a 5-cycle. The result will then follow from Lemma 4.2.14.
Expanding (στ )−1 and associating, we get
στ · μ(στ )−1 μ−1 = σ(τ μτ −1 )σ −1 μ−1 .
But τ commutes with μ, so that τ μτ −1 = μ. Simplifying, we see that
στ · μ(στ )−1 μ−1 = σ · μσ −1 μ−1
= (i1 . . . ik )(ik . . . i5 i3 i4 i1 i2 )
= (i5 i4 i2 i1 i3 ).
Here, the second equality follows from Lemma 3.6.19, and the last by direct calculation.
Thus, there is a 5-cycle in H as claimed, and the result follows.
We can now go one step farther and give a version of Lemma 4.2.13 that eliminates
the third condition.
Proposition 4.2.16. Let H An for n ≥ 5. Suppose there is an element σ ∈ H such
that the decomposition of σ as a product of disjoint cycles contains a cycle of length ≥ 3.
Then H = An .
Proof Suppose that the decomposition of σ as a product of disjoint cycles contains a
cycle of length > 3. Then the hypothesis of Lemma 4.2.15 is satisfied, and H = An as
desired. Thus, we may assume that σ is the product of disjoint cycles whose lengths are
all ≤ 3. But if any of them is a 2-cycle, then σ commutes with an element of Sn which
is not in An , and hence the hypothesis of Lemma 4.2.13 is satisfied.
Thus, we may assume that σ is a product of disjoint 3-cycles. If σ consists of a single
3-cycle, then H = An by Lemma 4.2.12. Thus, we may assume there are at least two
3-cycles in the decomposition of σ as a product of disjoint cycles.
Write σ = (a b c)(d e f )τ , where τ is disjoint from the stated 3-cycles. But an immediate application of Lemma 3.6.19 shows that σ commutes with the permutation
(a d)(b e)(c f ), which lies in Sn but not in An . Thus, σ satisfies the hypothesis of
Lemma 4.2.13, and hence H = An .
Thus, the only remaining possibility for an H An with H = An is for H to consist
only of products of disjoint transpositions (2-cycles). Indeed, A4 has such a normal
subgroup, as we shall see in the exercises below. So n ≥ 5 will be essential in what is to
follow. We begin with an important special case.
Lemma 4.2.17. Let H An for n ≥ 5. Suppose that H contains a product of two
disjoint transpositions. Then H = An .
CHAPTER 4. NORMALITY AND FACTOR GROUPS
95
Proof Let σ = (a b)(c d) ∈ H. Since n ≥ 5, there is a fifth letter, e, to play with.
Since σ commutes with a 2-cycle, it is conjugate in An to any element of the same cycle
structure, by Lemma 4.2.10. In particular, it is conjugate to (b c)(d e).
Thus,
(a b)(c d)(b c)(d e) = (a b d e c)
is in H. The result now follows from Lemma 4.2.14.
We may now complete the proof of the theorem.
Proof of Theorem 4.2.7 Let e = H An for n ≥ 5. Let σ be any non-identity
element of H. If the decomposition of σ as a product of disjoint cycles contains a cycle
of length ≥ 3, then H = An by Proposition 4.2.16. Thus, we may assume that σ is a
product of disjoint transpositions.
Since transpositions have sign −1, there must be an even number of disjoint transpositions in the decomposition of σ. But if σ is the product of two disjoint transpositions,
then H = An by Lemma 4.2.17. Thus, we may assume that σ is the product of four or
more disjoint transpositions.
Thus, we can write σ = (a b)(c d)(e f )τ , where τ is disjoint from the displayed transpositions. Once again, since the decomposition of σ contains a cycle of even length, it is
conjugate in An to any permutation with the same cycle structure. In particular, σ is
conjugate to (b c)(d e)(f a)τ −1 . Thus,
(a b)(c d)(e f )(b c)(d e)(f a)τ τ −1 = (a e c)(b d f )
lies in H. The result now follows from Proposition 4.2.16.
Exercises 4.2.18.
† 1. Show that A4 has exactly four elements whose order is a power of 2. Show that
these elements form a normal subgroup of A4 .
† 2. Show that A4 has no subgroup of order 6. Deduce that A4 is the only subgroup
of S4 which has order 12. (Hint: One way to proceed is as follows. Use the
Noether Theorems to show that if H ⊂ A4 has order 6, then H must have a
normal subgroup, K H of order 2. Show that every element of H outside of K
must have order 3 or 6, and that K ⊂ Z(H). Deduce that H must have an element
of order 6 and derive a contradiction.)
‡ 3. Show that An is a characteristic subgroup of Sn for n ≥ 3. (Hint: Use the simplicity
of An for n ≥ 5. For n = 3 or 4, give a separate argument.)
4. Show that any normal subgroup of A4 containing a 3-cycle must be all of A4 .
5. Show that any normal subgroup of A3 containing a 3-cycle must be all of A3 .
6. Let σ ∈ An . Show that the conjugates of σ in Sn are all conjugate to σ in An if
and only if the centralizer of σ in Sn is not contained in An . (Hint: Recall that
the centralizer, CAn (σ), of σ in An satisfies
CAn (σ) = An ∩ CSn (σ).
What are the possible values for [CSn (σ) : CAn (σ)]? What does this say about
[Sn : CSn (σ)]/[An : CAn (σ)]?)
CHAPTER 4. NORMALITY AND FACTOR GROUPS
4.3
96
The Jordan–Hölder Theorem
Recall that a composition series for a group G is a sequence
e = H0 ⊂ · · · ⊂ Hk = G
such that Hi−1 Hi for i = 1, . . . , k and the factor groups Hi /Hi−1 are all nontrivial
simple groups.
Proposition 4.2.5 shows that every nontrivial finite group has a composition series.
But composition series are not unique. However, the Jordan–Hölder Theorem shows that
the simple subquotient groups Hi /Hi−1 that appear in a composition series for G are
independent of the choice of composition series, though they may occur in a different
order in different composition series:
Theorem 4.3.1. (Jordan–Hölder Theorem) Suppose given two different composition series,
e = H0 ⊂ · · · ⊂ H k = G
and
e = G0 ⊂ · · · ⊂ Gl = G
of the finite group G. Then k = l and there is a permutation σ ∈ Sk such that Hi /Hi−1
is isomorphic to Gσ(i) /Gσ(i)−1 for 1 ≤ i ≤ k.
Example 4.3.2. We have two different composition series for Z6 :
e ⊂ 2 ⊂ Z6
and
e ⊂ 3 ⊂ Z6 .
The successive subquotient groups are Z3 followed by Z2 in the first case, and Z2 followed
by Z3 in the second.
Rather than present the proof from the bottom up, which would lose the intuition
in this case, we shall argue conceptually, leaving the proof (and statement) of the key
lemma for the end.
We first generalize the notion of composition series:
Definition 4.3.3. A subnormal series for a group G is a sequence e = H0 ⊂ · · · ⊂ Hk =
G such that Hi−1 Hi for 1 ≤ i ≤ k.
Some treatments would call this a normal series, but we prefer subnormal since it is
not required that Hi G.
Note that a composition series is a subnormal series for which each of the successive
factor groups Hi /Hi−1 is a nontrivial simple group. Note also that it is possible in a
subnormal series (but not a composition series) to have Hi−1 = Hi .
Definition 4.3.4. A refinement of a subnormal series is a subnormal series obtained by
inserting more subgroups into the sequence.
Thus, every subnormal series is a refinement of the trivial series e ⊂ G.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
97
Definition 4.3.5. Two subnormal series
e = H0 ⊂ · · · ⊂ Hk = G
and
e = G0 ⊂ · · · ⊂ G l = G
are equivalent if k = l and there is a permutation σ ∈ Sk such that Hi /Hi−1 is isomorphic
to Gσ(i) /Gσ(i)−1 for 1 ≤ i ≤ k.
Thus, the Jordan–Hölder Theorem simply states that any two composition series are
equivalent. It will follow rather quickly from the next result.
Theorem 4.3.6. (Schreier’s Theorem) Suppose given two subnormal series for the group
G. Then we may refine the two series in such a way that the refined series are equivalent.
This immediately implies the Jordan–Hölder Theorem because refining a composition
series does nothing other than adding trivial groups to the list of factor groups. So two
composition series equivalent after refinement must have been equivalent in the first
place.
Let us now consider the proof of Schreier’s Theorem. We are given subnormal series
e = H0 ⊂ · · · ⊂ H k = G
and
e = G0 ⊂ · · · ⊂ G l = G
of G. We shall insert between each pair Hi−1 ⊂ Hi a collection of inclusions derived
from the G0 ⊂ · · · ⊂ Gl in such a way that the resulting sequence of k · l inclusions is a
subnormal series. Similarly, we shall insert groups between Gi−1 and Gi .
Since Hi−1 Hi , the group Hi ∩ Gj normalizes Hi−1 for each j. Thus, the product
Hi−1 (Hi ∩ Gj ) is a subgroup of G, and we have inclusions
Hi−1 = Hi−1 (Hi ∩ G0 ) ⊂ · · · ⊂ Hi−1 (Hi ∩ Gl ) = Hi .
Similarly, we have inclusions of subgroups
Gj−1 = Gj−1 (H0 ∩ Gj ) ⊂ · · · ⊂ Gj−1 (Hk ∩ Gj ) = Gj .
Schreier’s Theorem now follows by setting Gj−1 = B, Hi−1 = A, Gj = K, and
Hi = H in the next lemma.
Lemma 4.3.7. (Zassenhaus’ Butterfly Lemma) Let H and K be subgroups of G, and
suppose given normal subgroups A H and B K of H and K. Then A(H ∩ B) A(H ∩ K), B(A ∩ K) B(H ∩ K) and there is an isomorphism
A(H ∩ K)/A(H ∩ B) ∼
= B(H ∩ K)/B(A ∩ K).
Proof We shall show that both of the quotient groups are isomorphic to a third quotient:
H ∩K/(A∩K)(H ∩B). (We’ll justify presently that this is a group.) The diagram below,
CHAPTER 4. NORMALITY AND FACTOR GROUPS
98
which gives this result its name, illustrates the relations between the various subgroups
we shall consider. All arrows are the natural inclusions.
A(H ∩ K)
O
iRRRR
RRRR
R
H ∩O K
l5
llll
llll
B(H ∩ K)
O
A(H ∩ B)
B(A ∩ K)
iRRRR
iRRR
l5
l
ll5
l
RRR
l
R
l
l
RRR
RRR
lll
lll
l
l
RR
l
ll
(A ∩ K)(H ∩ B)
A iRRR
ll5 B
RRR
iRRRR
lll
ll5
RRR
l
l
l
R
l
RRRR
RRR
ll
ll
lll
lll
A∩K
H ∩B
We shall give the proof that H ∩ K/(A ∩ K)(H ∩ B) ∼
= A(H ∩ K)/A(H ∩ B). The
argument that H ∩ K/(A ∩ K)(H ∩ B) ∼
= B(H ∩ K)/B(A ∩ K) is analogous.
First note that since H ∩ B ⊂ H ∩ K ⊂ H ⊂ NG (A), the subsets A(H ∩ K) and
A(H ∩ B) are subgroups of G. We argue by a straightforward application of the Second
Noether Isomorphism Theorem, which we restate as follows (using the form given in
Corollary 4.1.24) to avoid conflicting notation:
If X and Y are subgroups of G with X ⊂ NG (Y ), then X ∩ Y X, and X/(X ∩ Y ) ∼
=
XY /Y .
We apply this with X = H ∩ K and Y = A(H ∩ B). We must first verify that for
this choice of X and Y that X ⊂ NG (Y ), and then must show that the quotients stated
in the Noether theorem are the ones that concern us.
First recall from Problem 25 of Exercises 3.7.14 that for subgroups H1 , H2 ⊂ G, we
have NG (H1 ) ∩ NG (H2 ) ⊂ NG (H1 ∩ H2 ). Also, Problem 17 of Exercises 4.1.26 shows
that if H1 ⊂ NG (H2 ), then NG (H1 ) ∩ NG (H2 ) ⊂ NG (H1 H2 ).
Thus, since Y = A(H ∩ B), the inclusion of X in NG (Y ) will follow if we show that
X ⊂ NG (A) ∩ NG (H) ∩ NG (B). But X ⊂ H, which is contained in both NG (A) and
NG (H), and X ⊂ K, which is contained in NG (B), so X normalizes Y as desired.
It suffices to show that X ∩ Y = (A ∩ K)(H ∩ B) and that XY = A(H ∩ K).
For the former equality, we have X ∩ Y = [H ∩ K] ∩ [A(H ∩ B)]. Clearly, this contains
(A ∩ K)(H ∩ B), so it suffices to show the opposite inclusion, i.e., that [H ∩ K] ∩ [A(H ∩
B)] ⊂ (A ∩ K)(H ∩ B). So let a ∈ A and let b ∈ H ∩ B, and suppose that ab ∈ H ∩ K.
It suffices to show that a ∈ K. But since ab ∈ H ∩ K ⊂ K, and since b ∈ B ⊂ K, we
must have a ∈ K.
Finally, we have XY = [H ∩ K][A(H ∩ B)]. This group clearly contains A(H ∩ K),
so it suffices to show the opposite inclusion. Let a ∈ A, b ∈ H ∩ B, and c ∈ H ∩ K. Since
H ∩ K normalizes A, cab = a cb with a ∈ A. But cb ∈ H ∩ K and the result follows.
4.4
Abelian Groups: the Fundamental Theorem
Here, we classify all finite abelian groups.
Recall that an element x of a group G has exponent n if xn = e. Recall also that if
x has exponent n, then the order of x divides n.
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99
Definitions 4.4.1. We say that a group G has exponent n if every element of G has
exponent n.
Given an arbitrary element x ∈ G, we say that x is a torsion element if x has finite
order. If every element of G has finite order, we call G a torsion group. At the other
extreme, if no element of G other than the identity has finite order, we call G torsion-free.
Finally, let p be a prime number. We say that x ∈ G is a p-torsion element if the
order of x is a power of p. If every element of G is a p-torsion element, we call G a
p-group.
Recall that an element has finite order if and only if it has exponent n for some n > 0.
Since the order then divides all exponents for x, we see that x is a p-torsion element if
and only if it has exponent pr for some r ≥ 0.
Examples 4.4.2.
1. Let G be a finite group. By the corollary to Lagrange’s Theorem (Corollary 3.2.10),
the order of any element of G divides |G|. Thus, the group G has exponent |G|.
2. S∞ (the ascending union of the symmetric groups Sn for all n ≥ 1) has been shown
to be an infinite torsion group. But it does not have an exponent, as it contains
elements of every finite order.
3. As shown in Problem 12 of Exercises 4.1.26, Q/Z is an example of an infinite
abelian torsion group which does not have an exponent.
4. !
The reader acquainted with infinite products should have no trouble verifying that
∞
i=1 Z2 , the product of infinitely many copies of Z2 , has exponent 2. Thus, in
addition to having an exponent, it is an infinite abelian 2-group.
Let us now specialize to the study of abelian groups. To emphasize this specialization,
we shall use additive notation for the group operation in the remainder of this section.
Lemma 4.4.3. Let G be abelian and let x, y ∈ G be torsion elements. Then the least
common multiple of the orders of x and y is an exponent for x + y.
Proof Let n be the least common multiple of the orders of x and y. Then n is an
exponent for both x and y: nx = ny = 0. But since G is abelian, n(x + y) = nx + ny = 0.
It is immediate from Lemma 4.4.3 and the fact that the inverse of an element of finite
order also has finite order (Problem 6 of Exercises 2.5.21) that the torsion elements of
an abelian group form a subgroup of it:
Definition 4.4.4. Let G be an abelian group. Write Tors(G) ⊂ G for the set of all
torsion elements of G. We call it the torsion subgroup of G.
Example 4.4.5. As shown in Problem 13 of Exercises 4.1.26, the torsion subgroup of
R/Z is Q/Z.
Surprisingly, the set of torsion elements in a nonabelian group is not necessarily a
subgroup. For instance, Problem 2 of Exercises 10.8.13 shows that Sl2 (Z), the group of
2 × 2 matrices with integer coefficients that have determinant 1, is generated by torsion
elements, but the matrix
1 1
0 1
is easily seen to have infinite order.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
100
Definition 4.4.6. Let G be an abelian group and let p be prime. The p-torsion subgroup, Gp , of G is the set of p-torsion elements of G.
The justification is, in this case, immediate from Lemma 4.4.3:
Corollary 4.4.7. Let G be an abelian group and let p be prime. Then the p-torsion
subgroup Gp is, in fact, a subgroup of G.
The analogue for nonabelian groups of the preceding corollary is false even for finite
nonabelian groups:
Example 4.4.8. Recall from Corollary 3.5.11 that for n ≥ 3, An is generated by 3cycles. But while 3-cycles are 3-torsion elements, if n ≥ 4 there are elements of An (e.g.,
(1 2)(3 4)) that are not 3-torsion elements. Thus, the 3-torsion elements of An do not
form a subgroup if n ≥ 4.
Let us now specialize to the study of finite abelian groups.
Proposition 4.4.9. Let G be a finite abelian group. Let p1 , . . . , pk be the prime numbers
that divide |G|. Then G is the internal direct product of Gp1 , . . . , Gpk , i.e., the function
μ : Gp1 × · · · × Gpk → G
defined by μ(g1 , . . . , gk ) = g1 + · · · + gk is an isomorphism of groups.
Proof Of course, μ restricts on each Gpi to its inclusion into G. Since G is abelian, the
images of these inclusions commute with each other, so that μ is a homomorphism by
Proposition 2.10.10. Suppose that g = (g1 , . . . , gk ) lies in the kernel of μ. If each gi = 0,
then g is the identity element of Gp1 × · · · × Gpk . Otherwise, let i be the smallest index
with gi = 0. Then gi + (gi+1 + · · · + gk ) = 0, and we must have i < k, as otherwise
gi = μ(g) = 0. By inductive application of Lemma 4.4.3, the element gi+1 + · · · + gk
has an exponent which is a product of powers of pi+1 , . . . , pk . But gi+1 + · · · + gk is the
inverse of gi , and hence has order equal to a power of pi . This forces the order of −gi to
be 1 = p0i , and hence gi = 0. So g had to be 0 in the first place, and hence μ is injective.
It suffices to show μ is surjective. Let g ∈ G, and let n = o(g). We argue by induction
on the number of primes that divide n. If there is only one prime dividing n, then n is
a power of pi for some i, and hence g ∈ Gpi ⊂ im μ. Otherwise, n = piri m for some i,
where ri ≥ 1, and (pi , m) = 1. Thus, we can find integers s and t with spri i + tm = 1.
But then
g = (spri i + tm)g = s(pri i g) + t(mg).
By Problem 5 of Exercises 2.5.21, mg has order pri i , and hence lies in Gpi ⊂ im μ, and
pri i g has order m. Since m has one less prime divisor than n, pri i g ∈ im μ by induction.
Thus, g is the sum of two elements of im μ, and hence lies in im μ as claimed.
Note that if p doesn’t divide the order of G, then Gp = 0 by Lagrange’s Theorem.
Since the cartesian product of any group H with the trivial group is canonically isomorphic to H, we obtain the following immediate extension of Proposition 4.4.9, which
is useful when considering homomorphisms between abelian groups whose orders have
differing prime divisors.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
101
Corollary 4.4.10. Let G be a finite abelian group. Let p1 , . . . , pk be a collection of
primes including all the prime divisors of |G|. Let
μG : Gp1 × · · · × Gpk → G
be defined by μG (g1 , . . . , gk ) = g1 + · · · + gk . Then μG is an isomorphism.
The proof of Proposition 4.4.9 is an elaboration of the proof of the next result, which
could have been used inductively to obtain it. We include both proofs as it may be
instructive to the beginner.
Proposition 4.4.11. Let G be a finite group of order n. Suppose that n = mk, where
m and k are relatively prime. Let H ⊂ G be the set of all elements of exponent m and
let K ⊂ G be the set of all elements of exponent k. Then H and K are subgroups of G,
and G is the internal direct product of H and K.
Proof That H and K are subgroups is a consequence of Lemma 4.4.3. For instance,
if x, y ∈ H, then the prime divisors of o(x + y) must all divide m. Since o(x + y) must
divide |G|, this forces o(x + y) to divide m.
By Proposition 2.10.8, since G is abelian, it suffices to show that H ∩ K = 0, and
that H and K generate G.
Since m and k are relatively prime, any element which has both m and k as an
exponent must have order 1 (Corollary 2.5.13), so that H ∩ K is indeed the trivial
subgroup.
But if g ∈ G with o(g) = n , then we may write n = m k , where m divides m and
k divides k. But then m and k are relatively prime, so that sm + tk = 1 for some
integers s and t. But now g = s(m g) + t(k g). Since m g and k g have orders k and m ,
respectively, g is in H + K as desired.
We shall now give a special case of Cauchy’s Theorem, showing that the groups Gp are
nontrivial for all primes p dividing the order of G. Cauchy’s Theorem has already been
given as Problem 24 of Exercises 3.3.23, and will be given again by a different argument,
using the case we’re about to give, as Theorem 5.1.1. Both proofs are instructive.
Proposition 4.4.12. (Cauchy’s Theorem for Abelian Groups) Let G be a finite abelian
group and let p be a prime number dividing the order of G. Then G has an element of
order p.
Proof We argue by induction on |G|, with the case |G| = p a triviality. Let 0 = g ∈ G.
If o(g) is divisible by p, say o(g) = pq, then qg has order p. But if o(g) is not divisible
by p, then |G/g| is divisible by p. Since |G/g| < |G|, the induction hypothesis gives
an element x ∈ G/g of order p. But since x is the image of x ∈ G under the canonical
homomorphism, Corollary 2.5.19 shows that p must divide the order of x.
This now serves to clarify the situation of Proposition 4.4.11:
Corollary 4.4.13. Let G be a finite abelian group of order n = mk, with (m, k) = 1. Let
H ⊂ G be the set of elements in G of exponent m and let K ⊂ G be the set of elements
in G of exponent k. Then H has order m and K has order k.
Proof By Proposition 4.4.12, the order of H cannot be divisible by any prime that
doesn’t divide m, as then H would have an element whose order does not divide m. But
then |H| must divide m by Lagrange’s Theorem. Similarly, |K| divides k.
But since G ∼
= H × K, |G| = |H| · |K|. Since n = mk, the result must follow.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
102
As a special case, Corollary 4.4.13 gives a calculation of the order of the p-torsion
subgroup of a finite abelian group for any prime p dividing the order of the group.
Corollary 4.4.14. Let G be a finite abelian group of order n = pr11 . . . prkk , where p1 , . . . , pk
are distinct primes. Then the pi -torsion subgroup Gpi has order pri i for 1 ≤ i ≤ k.
The analogue of the preceding statement for nonabelian groups is the second Sylow
Theorem. Its proof is a little deeper than that of the abelian case. We now give another
useful consequence of Corollary 4.4.13.
Corollary 4.4.15. Let G be a finite abelian group and let H be a subgroup. Suppose that
the orders of H and G/H are relatively prime. Then H is the subgroup of G consisting
of all elements of exponent |H|, and G is isomorphic to H × G/H.
Proof Write n, m and k for the orders of G, H, and G/H, respectively. Then n = mk.
Let H ⊂ G be elements of exponent m and let K ⊂ G be the elements of exponent k.
We shall show that H = H and that K is isomorphic to G/H.
To see the former, let i : H ⊂ G be the inclusion. Since the elements of H have exponent m, we must have i(H) ⊂ H . But then i : H → H is an injective homomorphism.
Since H and H both have order m, it must be an isomorphism.
Let π : G → G/H be the canonical map. Then ker π = H. Since |K| and |H| are
relatively prime, we have H ∩ K = 0, and hence π restricts to an injection from K to
G/H. But since K and G/H have the same order, π must induce an isomorphism from
K to G/H.
Next, we show that for any prime p that any finite abelian p-group is a direct product
of cyclic groups. Together with Proposition 4.4.9, this will give the existence part of the
Fundamental Theorem:
Theorem 4.4.16. (Fundamental Theorem of Finite Abelian Groups) Every finite abelian
group is a direct product of cyclic groups of prime power order.
We shall restate it with a uniqueness statement below.
Lemma 4.4.17. Let G be an abelian group of exponent m. Suppose that g ∈ G has
order m. Let x ∈ G/g. Then there exists y ∈ G with y = x, such that the order of y in
G is equal to the order of x in G/g.
Proof Let k = o(x) and let n = o(x). Because x is the image of x under the
canonical map π : G → G/g k divides n (Corollary 2.5.19). So n = kl for l ∈ Z.
Note then that kx has order l. But since x has order k, kx = kx = 0, and hence
kx ∈ ker π = g. Say kx = sg.
Since g has order m, the order of sg is m/(m, s). Since kx = sg, this says l = m/(m, s),
and hence l · (m, s) = m. Since n = kl is the order of x, it must divide the exponent of
G, which is m. Thus, k divides (m, s). But then s = kt, and hence y = x − tg has order
k.
For the next result, we only use the special case of Lemma 4.4.17 in which the exponent
m is a prime power. The reader is invited to re-examine the proof in this simpler case.
Proposition 4.4.18. Let p be a prime. Then any finite abelian p-group is a direct
product of cyclic groups.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
103
Proof By Proposition 4.4.12, |G| is a power of p. We argue by induction on |G|. The
induction begins with |G| = p, in which case G itself is cyclic.
By Lagrange’s Theorem, |G| is an exponent for G. Let pj be the smallest power of
p which is an exponent for G. Then there must be an element g ∈ G of order pj . Let
H = G/g, with π : G → H the canonical map. Then |H| < |G|, so our induction
hypothesis gives an isomorphism ν : Zpr1 × · · · × Zprk → H.
Let xi = ν ◦ ιi (1), where ιi : Zpri → Zpr1 × · · · × Zprk is the inclusion of the i-th
factor of Zpr1 × · · · × Zprk . Then xi has order pri . By Lemma 4.4.17, there are elements
yi ∈ G with π(yi ) = xi , such that yi has order pri as well.
Define ν : Zpr1 × · · · × Zprk × Zpj → G by ν(m1 , . . . , mk+1 ) = m1 y1 + · · · + mk yk +
mk+1 g. Since the orders of the yi and of g are as stated, ν ◦ ιi is a homomorphism for
1 ≤ i ≤ k + 1 by Proposition 2.5.17. Thus, ν is a homomorphism by Proposition 2.10.10.
We claim that ν is an isomorphism.
Suppose that α = (m1 , . . . , mk+1 ) is in the kernel of ν. Then
0
= π ◦ ν(m1 , . . . , mk+1 )
= ν(m1 , . . . , mk ).
Since ν is an isomorphism, m1 = · · · = mk = 0. But then 0 = ν(α) = mk+1 g. Since
o(g) = pj , mk+1 = 0, and hence α = 0. Thus, ν is injective.
Let x ∈ G, and suppose that ν −1 ◦ π(x) = (m1 , . . . , mk ). Then π(x − (m1 y1 + · · · +
mk yk )) = 0. Since ker π = g, x − (m1 y1 + · · · + mk yk ) = mk+1 g for some integer mk+1 .
But then x = ν(m1 , . . . , mk+1 ), and hence ν is onto.
This completes the proof of the existence part of the Fundamental Theorem. We shall
now discuss uniqueness. First consider this.
Lemma 4.4.19. Let f : G → H be a homomorphism of abelian groups. Then for each
prime p, f restricts to a homomorphism fp : Gp → Hp .
Proof The order of f (g) must divide the order of g. If g ∈ Gp , then the order of g is a
power of p, so the same must be true of f (g). Thus, f (g) ∈ Hp .
For those who understand the language of category theory, this implies that passage
to the p-torsion subgroup provides a functor from abelian groups to abelian p-groups.
Even better, the next result shows that the isomorphisms μG of Corollary 4.4.10 are
natural transformations. The proof is obtained by chasing elements around the diagram.
Lemma 4.4.20. Let G and H be finite abelian groups and let p1 , . . . , pk be the collection
of all distinct primes that divide either |G| or |H|. Let f : G → H be a homomorphism.
Then the following diagram commutes.
Gp1 × · · · × Gpk
μG
∼
=
fp1 ×···×fpk
Hp1
× · · · × Hpk
/ G
f
μH
∼
=
/ H
Here, μG and μH are the isomorphisms of Corollary 4.4.10.
Note that if p divides |G| but not |H|, then Hp = 0, so that fp is the constant
homomorphism to 0. Similarly, if p divides |H| but not |G|, then Gp = 0, and fp is
CHAPTER 4. NORMALITY AND FACTOR GROUPS
104
the unique homomorphism from 0 to Hp . Note that since μG and μH are isomorphisms,
the above diagram shows that f is an isomorphism if and only if fp1 × · · · × fpk is an
isomorphism. We obtain an immediate corollary.
Corollary 4.4.21. Let f : G → H be a homomorphism of finite abelian groups. Then
f is an isomorphism if and only if fp : Gp → Hp is an isomorphism for each prime p
dividing either |G| or |H|.
Thus, a uniqueness statement for the Fundamental Theorem will follow if we can prove
the uniqueness of the decomposition (from Proposition 4.4.18) of an abelian p-group as a
product of cyclic groups. Fix a prime p. Then, by ordering the exponents, we can write
any finite abelian p-group as a product Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk .
Proposition 4.4.22. Suppose that
ν : Zpr1 × · · · × Zprk
∼
=
/ Zps1 × · · · × Zpsl ,
where 1 ≤ r1 ≤ · · · ≤ rk , and 1 ≤ s1 ≤ · · · ≤ sl . Then k = l and ri = si for 1 ≤ i ≤ k.
Proof We argue by induction on the order of the group. There is only one way in
which we can get a group of order p, so the induction starts there.
Write Zp ⊂ Zps for the cyclic subgroup ps−1 . Note that the elements of Zp are the
only ones in Zps of exponent p.
By Problem 2 of Exercises 2.10.12, the elements of Zpr1 × · · · × Zprk which have
exponent p are precisely those in the product Zp × · · · × Zp of the subgroups of order
p in each factor. Thus, Zpr1 × · · · × Zprk has pk elements of exponent p, all but one of
which has order p.
Similarly, Zps1 ×· · ·×Zpsl has pl elements of exponent p. Since isomorphisms preserve
order, k = l. Moreover, ν restricts to an isomorphism ν : Zp × · · · × Zp → Zp × · · · × Zp .
Passing to factor groups, we obtain an isomorphism
ν : (Zpr1 × · · · × Zprk )/(Zp × · · · × Zp ) → (Zps1 × · · · × Zpsk )/(Zp × · · · × Zp ).
By Problem 2 of Exercises 4.1.26, Zps /Zp ∼
= Zps−1 . Thus, Problem 14 of the same
exercises shows that we may identify ν with an isomorphism
ν : Zpr1 −1 × · · · × Zprk −1
∼
=
/ Zps1 −1 × · · · × Zpsk −1 .
The result now follows from induction on order.
The full statement of the Fundamental Theorem is now immediate:
Theorem 4.4.23. (Fundamental Theorem, with Uniqueness) Any finite abelian group
may be written uniquely as a direct product of cyclic groups of prime power order. Uniqueness means that given any two such decompositions of a given group, the number of factors
of a given order in the two decompositions must be the same.
Exercises 4.4.24.
1. Let G be an abelian group. Show that G/Tors(G) is torsion-free.
‡ 2. Show that a finite abelian group G is cyclic if and only if the p-torsion subgroup
Gp is cyclic for each prime p dividing the order of G. (Hint: This does not require
the Fundamental Theorem.)
CHAPTER 4. NORMALITY AND FACTOR GROUPS
105
3. Show that a finite abelian group is cyclic if and only if it has no subgroup of the
form Zp × Zp for any prime p. (Hint: Apply the Fundamental Theorem.)
† 4. Use the Fundamental Theorem to show that a finite abelian group is cyclic if and
only if it has at most d elements of exponent d for each d dividing the order of G.
5. Give a direct proof (not using the Fundamental Theorem) of the preceding problem.
(Hint: One direction has already been done in Problem 11 of Exercises 2.5.21. For
the other direction, first show via Problem 2 above that we may assume that the
group has order pr for some prime p and some r ≥ 1. Then show that there must
be a generator.)
6. Show that the full converse to Lagrange’s Theorem holds for finite abelian groups
G: if k divides the order of G, then G has a subgroup of order k.
7. Show that any finite abelian group may be written as a product Zn1 × · · · × Znk ,
where ni divides ni−1 for 1 < i ≤ k.
8. Show that the decomposition of the preceding problem is unique.
4.5
The Automorphisms of a Cyclic Group
We calculate the automorphism group of a cyclic group. Recall that for any group G
and for any g ∈ G whose order divides n, there is a unique homomorphism hg : Zn → G
with hg (1) = g.
Lemma 4.5.1. For n > 1 and m ∈ Z, the homomorphism hm : Zn → Zn is an automorphism if and only if m and n are relatively prime.
Proof The image of hm is the cyclic subgroup generated by m. Since the order of m
is n/(m, n), hm is onto if and only if (m, n) = 1. But any surjection from a finite set to
itself is an injection.
Thus, we’ve identified which homomorphisms from Zn to Zn are automorphisms, but
we have yet to identify the group structure on Aut(Zn ).
Recall that there are two operations defined on Zn , addition and multiplication, and
that multiplication gives Zn the structure of a monoid, with unit element 1. Recall that
the collection of invertible elements in a monoid M forms a group (which we have denoted
Inv(M )) under the monoid operation of M .
Definition 4.5.2. For n > 1 we write Z×
n for the group of invertible elements in the
multiplicative monoid structure on Zn . Thus, m ∈ Z×
n if and only if there is a k ∈ Zn
with km = 1 ∈ Zn .
Z×
n is an example of a general construction in rings. The set of elements in a ring A
which have multiplicative inverses forms a group under the operation of multiplication.
It is called the group of units of A, and may be denoted A× .
We wish to identify the elements of Z×
n . The answer turns out to be quite relevant
to our discussion of automorphisms.
Lemma 4.5.3. Let n > 1 and let m ∈ Z. Then m ∈ Z×
n if and only if m and n are
relatively prime.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
106
Proof Suppose that m is invertible in Zn , with inverse k. Since km is congruent to
1 mod n, we obtain that km − 1 = sn for some integer s. But then km − sn = 1, and
hence (m, n) = 1.
Conversely, if rm + sn = 1, then r is the multiplicative inverse for m in Zn .
Thus, there is a one-to-one correspondence between the elements of Aut(Zn ) and Z×
n.
We next show that this correspondence is an isomorphism of groups.
Proposition 4.5.4. Define ν : Z×
n → Aut(Zn ) by ν(m) = hm , where hm : Zn → Zn is
the homomorphism carrying 1 to m. Then ν is an isomorphism of groups.
Proof It suffices to show that hm ◦ hm = hmm . But it’s easy to see (e.g., by Corollary 2.5.5 and Proposition 2.5.17) that hm (k) = mk, and the result follows.
Since multiplication in Zn is commutative, we obtain the following corollary.
Corollary 4.5.5. For any positive integer n, Aut(Zn ) ∼
= Z× is an abelian group.
n
In contrast, most other automorphism groups, even of abelian groups, are nonabelian.
It will require a bit more work to display the group structures on these groups of
units Z×
n . It will be easiest to do so in the case that n only has one prime divisor. We
will then be able to put this information together, via methods we are about to discuss.
First recall from Section 4.4 that if G is abelian and if Gp is its p-torsion subgroup,
then any homomorphism f : G → G restricts to a homomorphism fp : Gp → Gp (i.e.,
fp (g) = f (g) for all g ∈ Gp ). Moreover, Corollary 4.4.21 shows that f : G → G is an
automorphism if and only if fp is an automorphism for each p dividing |G|. The next
result follows easily.
Lemma 4.5.6. Let G be an abelian group and let p1 , . . . pk be the primes dividing the
order of G. Let Gpi be the pi -torsion subgroup of G for 1 ≤ i ≤ k. Then there is an
isomorphism
∼
=
ϕ : Aut(G)
/ Aut(Gp1 ) × · · · × Aut(Gpk )
defined by ϕ(f ) = (fp1 , . . . , fpk ).
Let n = pr11 . . . prkk with p1 , . . . , pk distinct primes. Then the pi -torsion subgroup of
Zn is isomorphic to Zpri . We obtain the following corollary, which may also be proven
i
using the Chinese Remainder Theorem, below.
Corollary 4.5.7. With n as above, there is an isomorphism
∼
=
ρ : Z×
n
/ Z×r1 × · · · × Z×rk
p
p
1
k
given by ρ(m) = (m, . . . , m).
Proof As the reader may easily check, there is a commutative diagram
Z×
n
ν ∼
=
Aut(Zn )
ρ
ϕ
∼
=
/ Z×r1 × · · · × Z×rk
p1
pk
∼
= ν×···×ν
/ Aut(Zpr1 ) × · · · × Aut(Zprk )
1
k
CHAPTER 4. NORMALITY AND FACTOR GROUPS
107
Thus, the calculation of the automorphism groups of cyclic groups is reduced to the
×
calculation of Z×
pr for p a prime and r ≥ 1. We begin by calculating the order of Zpr .
Lemma 4.5.8. For any prime p and any integer r > 0, the group of units Z×
pr has order
pr−1 (p − 1).
Proof The elements of Zpr that do not lie in Z×
pr are the elements m such that (p, m) =
1. But since p is prime, this means that p divides m, and hence m ∈ p ⊂ Zpr . But p
r
r−1
= pr−1 (p − 1).
has order pr−1 , and hence |Z×
pr | = p − p
Corollary 4.5.7 now allows us to calculate |Z×
n |.
Corollary 4.5.9. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes and the exponents ri are all positive. Then
r1 −1
|Aut(Zn )| = |Z×
. . . (pk − 1)prkk −1 .
n | = (p1 − 1)p1
In particular, Z×
p has order p − 1. In Corollary 7.3.15, we shall show, using some
elementary field theory, that Z×
p is always a cyclic group. However, the proof is theoretical, and doesn’t provide any hint as to what elements generate this group. And indeed
there is no known algorithm to obtain a generator other than trial and error. (Some of
the deeper aspects of number theory are lurking in the background.) So for the practical purpose of identifying the orders of individual elements of Z×
p , direct calculation is
necessary. The reader with some computer expertise may find it interesting to write a
program for this purpose.
On the other hand, when r > 1, the orders of the elements of Z×
p will be the only
open question in our understanding of Z×
.
We
need
to
understand
what
happens to Z×
r
p
pr
s
r
as r varies. Thus, let 1 ≤ s < r. Since p divides p , there is a homomorphism, which
we shall denote by π : Zpr → Zps , with π(1) = 1. In fact, π(m) = m for all m ∈ Z. We
call π the canonical homomorphism. We have m ∈ Z×
pr if and only if (m, p) = 1, which
×
is true if and only if m ∈ Zps . The next lemma is now immediate.
Lemma 4.5.10. For 1 ≤ s < r and for p a prime, the canonical homomorphism π :
Zpr → Zps induces a surjective homomorphism of multiplicative groups
×
π : Z×
pr → Zps .
At this point, the cases p > 2 and p = 2 diverge. One difference that is immediately
×
apparent is that Z×
2 = e, while Zp is nontrivial for p > 2. Thus, for p > 2 we shall define
×
K(p, r) to be the kernel of π : Z×
pr → Zp , while K(2, r) is defined to be the kernel of
×
×
π : Z2r → Z4 (and hence K(2, r) is only defined for r ≥ 2). Note that Z×
4 = {±1} has
order 2.
Lemma 4.5.10 and the First Noether Theorem now give us the order of K(p, r).
Corollary 4.5.11. For p > 2 and r ≥ 1, K(p, r) has order pr−1 . Also, for r ≥ 2,
K(2, r) has order 2r−2 .
For p > 2, the orders of K(p, r) and Z×
p are relatively prime, so Corollary 4.4.15 gives
a splitting of Z×
.
r
p
CHAPTER 4. NORMALITY AND FACTOR GROUPS
108
Corollary 4.5.12. For p > 2 and r ≥ 1, K(p, r) is the p-torsion subgroup of Z×
pr , and
we have an isomorphism
×
∼
Z×
pr = K(p, r) × Zp .
The prime 2 is a little different, in that K(2, r) and Z×
4 are both 2-groups. But we
get a product formula regardless.
Lemma 4.5.13. For r ≥ 2, we have an isomorphism
×
∼
Z×
2r = K(2, r) × Z4 .
×
Proof Let H = {±1} ⊂ Z×
2r . Then H is a subgroup of Z2r , and is carried isomorphically
×
×
onto Z4 by π. We shall show that Z2r is the internal direct product of K(2, r) and H,
i.e., the map μ : H × K(2, r) → Z×
2r given by μ(h, k) = hk is an isomorphism.
The key is that since −1 is not congruent to 1 mod 4, −1 is not in K(2, r), and hence
H ∩ K(2, r) = 1. (The unit element here is denoted 1.) This easily implies that μ is
injective.
But since m · k = mk in both Z2r and Z4 , either m or −m is in K(2, r) for each
m ∈ Z×
2r . Thus, μ is onto.
We shall show that K(p, r) is cyclic for all primes p and all r ≥ 2. (Unlike the case
of Z×
p , we shall provide a criterion to determine the order of each element.) The key is
to understand the order of the elements 1 + ps . As the group operation is multiplication
modulo pr , we can get a handle on this if we can calculate the powers (x + y)n of the
sum of two integers x and y.
The formula for (x + y)n is given by what’s called the Binomial Theorem, which in
turn depends on an understanding of the binomial coefficients.
Definition 4.5.14. Let 0 ≤ k ≤ n ∈ Z. We define the binomial coefficient nk by
n
n!
=
.
k!(n − k)!
k
Here, the factorial numbers are defined as usual: 0! = 1, 1! = 1, and n! is defined
inductively by n! = (n − 1)! · n for n > 1.
The next lemma is useful for manipulating the binomial coefficients.
Lemma 4.5.15. The binomial coefficients are all integers. Moreover, if 1 ≤ k < n we
have
n−1
n−1
n
+
=
.
k−1
k
k
Proof An easy inspection of the definition shows
that n0 = nn = 1 for all n. But
the only other binomial coefficients are the nk with 1 ≤ k < n. Thus, if the displayed
equation is true, then the binomial coefficients are all integers by induction on n, starting
with the case n = 1, which we’ve just verified by inspection.
The left-hand side of the displayed equation expands to
(n − 1)!
(n − 1)!
(n − 1)![k + (n − k)]
+
=
,
(k − 1)!(n − k)! k!(n − k − 1)!
k!(n − k)!
and the result follows.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
109
Our next result is the Binomial Theorem, which is true, with the same proof, not
only in the integers, but in any commutative ring.
Theorem 4.5.16. (Binomial Theorem) Let x and y be integers, and n ≥ 1 be another.
Then
n n k n−k
(x + y)n =
x y
k
k=0
n 2 n−2
= y n + nxy n−1 +
x y
+ · · · + nxn−1 y + xn
2
Proof The second linejust
gives
n an expansion of the preceding expression, using the
= n, in addition to the already known calculations
(easy) calculations that n1 = n−1
n
n
of 0 and n .
We prove the theorem by induction on n. For n = 1, it just says that x + y = x + y.
Thus, assume that n > 1 and the theorem is true for n − 1. Then
#
"n−1 n − 1
xk y n−k−1 (x + y).
(x + y)n = (x + y)n−1 (x + y) =
k
k=0
Applying the distributive law to the last expression and collecting terms, we get
$
%
n−1
n−1
n−1
+
.
xn + y n +
xk y n−k
k−1
k
k=1
But this is exactly the desired sum, by Lemma 4.5.15.
We wish to calculate the order of 1 + ps in K(p, r). By the definitions, the order will
be the smallest integer m such that (1 + ps )m is congruent to 1 mod pr . Since the order
of K(p, r) is a power of p, m must be also. Thus, we shall study the binomial expansion
k
for (1 + ps )p . For this purpose, we shall need to know the largest power of p that divides
k
the binomial coefficients pj .
Lemma 4.5.17. Let p be prime and let k ≥ 1. Let 1 ≤ j ≤ pk and write j = pa b, with
(p, b) = 1. Then
k
p
= pk−a c,
j
with (p, c) = 1.
Proof After making appropriate cancellations, we have
k
p
pk (pk − 1) . . . (pk − j + 1)
=
.
j
1 · 2 . . . (pa b)
Consider the right-hand side as a rational number and rearrange the terms slightly. We
get the product of pk /pa b with the product
k
k
k
p −2
p − (j − 1)
p −1
·
...
1
2
j−1
But for 1 ≤ i < pk , pk − i and i have the same p-divisibility, so that the only contribution
to the p-divisibility of our binomial coefficient comes from pk /pa b. But this gives the
desired result.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
110
We shall now calculate the order of the generic element of K(p, r). Note that if p > 2,
then x ∈ K(p, r) if and only if x = 1 + y for some y divisible by p. Similarly, x ∈ K(2, r)
if and only if x = 1 + y for some y divisible by 4.
Proposition 4.5.18. Let p be a prime and let x = 1 + ps a, where (p, a) = 1. If p = 2,
assume that s > 1. Then x has order pr−s in K(p, r).
Proof As discussed above, the order of x in K(p, r) is the smallest power, pk , of p for
k
which xp is congruent to 1 mod pr . So we want to find the smallest value of k such
k
k
that (ps a + 1)p − 1 is divisible by pr . Consider the binomial expansion for (ps a + 1)p .
k
k
We have (ps a + 1)p = 1 + pk ps a plus terms of the form pj psj aj with j > 1. But then
k
k
(ps a + 1)p − 1 is equal to ps+k a plus the terms pj psj aj with j > 1.
k
It suffices to show that if j > 1, then pj psj aj is divisible by ps+k+1 .
k
By Lemma 4.5.17, if j = pb c, with (p, c) = 1, then pj is divisible by pk−b . Thus,
pk sj j
b
sj+k−b
= psp c+k−b , so it suffices to show that spb c+k−b > s+k.
j p a is divisible by p
But this is equivalent to showing that s[pb c − 1] > b. If b = 0, then c = j > 1, and
the result holds. Otherwise, b > 0, and we may and shall assume that c = 1.
For p > 2, we must show that s[pb − 1] > b for all s ≥ 1. But the general case will
follow from that of s = 1, where we show that pb − 1 > b for all b > 0. For p = 2, our
hypothesis is that s > 1, so it suffices to show that 2b − 1 ≥ b for all b > 0.
For any p, pb − 1 = [(p − 1) + 1]b − 1, so we may use the Binomial Theorem again.
We see that pb − 1 is equal to b(p − 1) plus some other non-negative terms. But this is
greater than b if p > 2, and is ≥ b if p = 2, precisely as desired.
Corollary 4.5.19. For p > 2 and r ≥ 1, K(p, r) is the cyclic group (of order pr−1 )
generated by 1 + p.
For p = 2 and r ≥ 2, K(2, r) is the cyclic group (of order 2r−2 ) generated by 5.
Proof The orders of the groups in question were calculated in Corollary 4.5.11. By
Proposition 4.5.18, the 1 + p has the same order as K(p, r) for p > 2, while 5 has the
same order as K(2, r).
Lemma 4.5.13 now gives an explicit calculation of Z×
2r .
Corollary 4.5.20. For r ≥ 2, Z×
2r is isomorphic to Z2r−2 × Z2 .
In particular, Z×
2r is not cyclic for r ≥ 3.
On the other hand, when we have established Corollary 7.3.15, we will know that Z×
p
is a cyclic group of order p − 1 for all primes p. We shall state the consequence here,
though the reader is advised to read ahead for the proof.
This time we make use of Corollary 4.5.12:
Corollary 4.5.21. Let p be an odd prime and let r ≥ 1. Then Z×
pr is a cyclic group of
order (p − 1)pr−1 .
Exercises 4.5.22.
1. Find generators for Z×
p for all primes p ≤ 17.
2. Find a generator for Z×
3r for any r.
3. Find a generator for Z×
5r for any r.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
111
4. Find a generator for Z×
49 .
5. Find all elements of order 2 in Z×
2r .
×
×
6. Find all elements of order 2 in Z×
15 , Z30 , and Z60 .
7. Write a computer program to calculate the order of an element in Z×
pr , where the
input is p, r, and the element in question. Be sure it checks that the element is a
unit.
8. Write a computer program to calculate the order of an element in Z×
n . Be sure it
checks that the element is relatively prime to n.
4.6
Semidirect Products
The simplest way to construct new groups from old is to take direct products. But there
are ways to construct more interesting groups if we can calculate the automorphisms of
one of the groups in question.
Suppose given an action of K on H through automorphisms. We may represent this
by a homomorphism α : K → Aut(H), so that the action of k ∈ K takes h ∈ H to
α(k)(h). We shall define a group multiplication on the set H × K, to obtain a new
group, which we denote H α K, called the semidirect product of H and K given by α:
Definition 4.6.1. Let α : K → Aut(H) be a homomorphism. By the semidirect product
of H and K with respect to α, written H α K, we mean the set H × K with the binary
operation given by setting
(h1 , k1 ) · (h2 , k2 ) = (h1 · α(k1 )(h2 ), k1 k2 ).
Proposition 4.6.2. H α K, with the above product, is a group. There are homomorphisms π : H α K → K and ι : H → H α K, given by π(h, k) = k, and ι(h) = (h, e).
The homomorphism ι is an injection onto the kernel of π. Finally, there is a homomorphism s : K → H α K given by s(k) = (e, k). This homomorphism has the property
that π ◦ s is the identity map of K.
Proof We first show associativity.
((h1 , k1 ) · (h2 , k2 )) · (h3 , k3 )
= (h1 · α(k1 )(h2 ), k1 k2 ) · (h3 , k3 )
= (h1 · α(k1 )(h2 ) · α(k1 k2 )(h3 ), k1 k2 k3 )
= (h1 · α(k1 )(h2 ) · α(k1 )(α(k2 )(h3 )), k1 k2 k3 ).
Here, the last equality holds because α is a homomorphism, so that α(k1 k2 ) = α(k1 ) ◦
α(k2 ).
Since α(k1 ) is an automorphism of H, we have
α(k1 )(h2 ) · α(k1 )(α(k2 )(h3 )) = α(k1 )(h2 · α(k2 )(h3 )),
and hence
((h1 , k1 ) · (h2 , k2 )) · (h3 , k3 ) = (h1 · α(k1 )(h2 · α(k2 )(h3 )), k1 k2 k3 ).
But this is precisely what one gets if one expands out (h1 , k1 ) · ((h2 , k2 ) · (h3 , k3 )).
CHAPTER 4. NORMALITY AND FACTOR GROUPS
112
The pair (e, e) is clearly an identity element for this product, since α(e) is the identity
homomorphism of H. We claim that the inverse of (h, k) is (α(k −1 )(h−1 ), k −1 ). The
reader is welcome to verify this fact, as well as the statements about π, ι, and s.
Note that if α(k) is the identity automorphism for all k ∈ K (i.e., α : K → Aut(H)
is the trivial homomorphism), then H α K is just the direct product H × K.
Notation 4.6.3. In the semidirect product H α K, we shall customarily identify H
with ι(H) and identify K with s(K), so that the element (h, k) is written as hk. Thus,
H α K = {hk | h ∈ H, k ∈ K}.
Lemma 4.6.4. With respect to the above notations, the product in H α K is given by
h1 k1 · h2 k2 = h1 α(k1 )(h2 )k1 k2
for h1 , h2 ∈ H and k1 , k2 ∈ K.
In particular, this gives khk −1 = α(k)(h). Thus, H α K is abelian if and only if H
and K are both abelian and α : K → Aut(H) is the trivial homomorphism.
Proof The multiplication formula is clear. But this gives kh = α(k)(h)k, which in turn
implies the stated conjugation formula.
Thus, when α is nontrivial, the new group H α K may have some interesting features.
Indeed, this method of constructing groups will give us a wide range of new examples.
But it also gives new ways of constructing several of the groups we’ve already seen. For
instance, as we shall see below, the dihedral group D2n is the semidirect product obtained
from an action of Z2 on Zn , and A4 is the semidirect product obtained from an action
of Z3 on Z2 × Z2 .
We wish to analyze the semidirect products obtained from actions of cyclic groups on
other cyclic groups. Since these are, in general, nonabelian, we shall use multiplicative
notation for our cyclic groups. Thus, we shall begin with a discussion of notation.
×
Recall first the isomorphism ν : Z×
n → Aut(Zn ) of Proposition 4.5.4. Here, Zn is the
group of invertible elements with respect to the operation of multiplication in Zn , and,
for m ∈ Z×
n , ν(m) is the automorphism defined by ν(m)(l) = ml.
In multiplicative notation, we write Zn = b = {e, b, . . . , bn−1 }. Thus, the element
of Zn written as l in additive notation becomes bl . In other words, in this notation, we
have an isomorphism ν : Z×
n → Aut(b), where ν(m) is the automorphism defined by
ν(m)(bl ) = (bml ).
Now write Zk = a in multiplicative notation, and recall from Proposition 2.5.17
that there is a homomorphism hg : Zk → G with hg (a) = g ∈ G if and only if g has
exponent k in G. Moreover, if g has exponent k, there is a unique such homomorphism,
and it is given by the formula hg (aj ) = g j .
Thus, suppose given a homomorphism α : Zk = a → Z×
n , with α(a) = m. Then m
×
k
has exponent k in Zn , meaning that m is congruent to 1 mod n. Moreover, α(aj ) = mj .
By abuse of notation, we shall write α : Zk → Aut(Zn ) for the composite of this α
with the isomorphism ν above. Thus, as an automorphism of Zn = b, α(aj ) is given
j
by α(aj )(bi ) = bm i .
×
Corollary 4.6.5. Let m have exponent k in Z×
n , and let α : Zk → Zn be the homomorphism that takes the generator to m. Then writing Zn = b, Zk = a, and identifying
Z×
n with Aut(Zn ), we obtain
Zn α Zk = {bi aj | 0 ≤ i < n, 0 ≤ j < k},
CHAPTER 4. NORMALITY AND FACTOR GROUPS
113
where b has order n, a has order k, and the multiplication is given by
j bi aj bi aj = bi+m i aj+j .
Moreover, the nk elements {bi aj | 0 ≤ i < n, 0 ≤ j < k} are all distinct.
Proof The multiplication formula comes via Lemma 4.6.4 from the formula α(aj )(bi ) =
j
bm i .
Examples 4.6.6.
1. Let α : Z2 → Z×
n be the homomorphism that carries the generator to −1. Then
in Zn α Z2 , with the notations above, we see that b has order n, a has order 2,
and aba−1 = b−1 . Thus, Problem 6 of Exercises 2.8.5 provides a homomorphism
f : D2n → Zn α Z2 with f (a) = a and f (b) = b. Since a and b generate Zn α Z2 , f
is onto, and hence an isomorphism, as the two groups have the same order. Thus,
semidirect products provide a construction of the dihedral groups that does not
require trigonometry.
2. Let α : Z2 → Z×
8 be the homomorphism that carries the generator to 3. We obtain
a group of order 16,
Z8 α Z2 = {bi aj | 0 ≤ i < 8, 0 ≤ j < 2},
where a has order 2, b has order 8, and the multiplication is given by bi aj bi aj =
j bi+3 i aj+j .
3. Let β : Z2 → Z×
8 be the homomorphism that carries the generator to 5. We obtain
a group of order 16,
Z8 β Z2 = {bi aj | 0 ≤ i < 8, 0 ≤ j < 2},
where a has order 2, b has order 8, and the multiplication is given by bi aj bi aj =
j bi+5 i aj+j .
4. Let α : Z4 → Z×
5 be the homomorphism that carries the generator to 2. We obtain
a group of order 20,
Z5 α Z4 = {bi aj | 0 ≤ i < 5, 0 ≤ j < 4},
where a has order 4, b has order 5, and the multiplication is given by bi aj bi aj =
j bi+2 i aj+j .
5. Let α : Z3 → Z×
7 be the homomorphism that carries the generator to 2. We obtain
a group of order 21,
Z7 α Z3 = {bi aj | 0 ≤ i < 7, 0 ≤ j < 3},
where a has order 3, b has order 7, and the multiplication is given by bi aj bi aj =
j bi+2 i aj+j .
6. Let p be an odd prime. Recall from Proposition 4.5.18 that p + 1 has order p in
×
Z×
p2 . Thus, there is a homomorphism α : Zp → Zp2 that carries the generator of
Zp to p + 1. We obtain a group of order p3 ,
Zp2 α Zp = {bi aj | 0 ≤ i < p2 , 0 ≤ j < p},
where a has order p, b has order p2 , and the multiplication is given by bi aj bi aj =
j bi+(p+1) i aj+j .
CHAPTER 4. NORMALITY AND FACTOR GROUPS
114
Other than the first example, these groups are all new to us. Indeed, the groups of
order 16 in Examples 2 and 3 turn out to be the lowest-order groups that we had not
constructed previously.
The group of Example 5 turns out to be the smallest odd order group that is nonabelian, while for each odd prime p, the group of Example 6 represents one of the two
isomorphism classes of nonabelian groups of order p3 .
We may also construct examples of semidirect products where at least one of the two
groups is not cyclic. For instance, since the groups Z×
n are not generally cyclic, we can
obtain a number of groups of the form Zn α K, where K is not cyclic. We are also
interested in the case where group “H” in the construction of semidirect products is not
cyclic. The simplest case is where H = Zn × Zn .
As above, write Zn × Zn = {bi cj | 0 ≤ i, j < n}, where b and c have order n and
commute with each other. Since every element of Zn × Zn has exponent n, Propositions 2.10.6 and 2.5.17 show that there is a homomorphism f : Zn × Zn → Zn × Zn with
the property that f (b) = bc and f (c) = c: explicitly, f (bi cj ) = bi ci+j .
Lemma 4.6.7. The homomorphism f just described is an automorphism of Zn × Zn . It
has order n in the group of automorphisms of Zn × Zn .
Proof For k ≥ 1 the k-fold composite of f with itself is given by f k (c) = c and
f k (b) = f k−1 (bc) = f k−1 (b)f k−1 (c) = (f k−1 (b))c. An easy induction now shows that
f k (b) = bck . Thus, f k is the identity if and only if k is congruent to 0 mod n.
Since f n is the identity, f is indeed an automorphism, with inverse f n−1 .
Thus, if we write Zn = a, there is a homomorphism
α : Zn → Aut(Zn × Zn )
with α(a) = f . We obtain a construction which is of special interest when n is prime:
Corollary 4.6.8. There is a group G of order n3 given by G = {bi cj ak | 0 ≤ i, j, k < n},
where a, b, and c all have order n, and b commutes with c, a commutes with c, and
aba−1 = bc. Thus,
bi cj ak · bi cj ak = bi+i cj+j
+ki k+k
a
.
There are, of course, at least as many examples of semidirect products as there are
groups with nontrivial automorphism groups. Thus, we shall content ourselves for now
with these examples.
In our studies of the dihedral and quaternionic groups, we have made serious use of
our knowledge of the homomorphisms out of these groups. There is a similar result for
the generic semidirect product.
First, we need some discussion. Let ι : H → H α K and s : K → H α K be the
structure maps: in the ordered pairs notation, ι(h) = (h, e) and s(k) = (e, k). Then
every element of H α K may be written uniquely as ι(h) · s(k) with h ∈ H and k ∈ K.
This is, indeed, the basis for the HK notation.
In particular, H α K is generated by the elements of ι(H) and s(K), so that any
homomorphism f : H α K → G is uniquely determined by its restriction to the subgroups ι(H) and s(K). Since ι and s are homomorphisms, this says that f is uniquely
determined by the homomorphisms f ◦ ι and f ◦ s.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
115
Thus, we will understand the homomorphisms out of H α K completely if we can
give necessary and sufficient conditions for a pair of homomorphisms f1 : H → G and f2 :
K → G to be obtained as f1 = f ◦ ι and f2 = f ◦ s for a homomorphism f : H α K → G.
Proposition 4.6.9. Let α : K → Aut(H) be a homomorphism. Suppose given homomorphisms f1 : H → G and f2 : K → G for a group G. Then there is a homomorphism
f : H α K → G with f1 = f ◦ ι and f2 = f ◦ s if and only if f2 (k)f1 (h)(f2 (k))−1 =
f1 (α(k)(h)) for all h ∈ H and k ∈ K.
Such an f , if it exists, is unique, and is given in the HK notation by f (hk) =
f1 (h)f2 (k).
Proof Suppose first that f exists with f ◦ ι = f1 and f ◦ s = f2 . In the HK notation
we have that hk = ι(h)s(k), so that f (hk) = (f ◦ ι)(h) · (f ◦ s)(k) = f1 (h)f2 (k) for all
h ∈ H and k ∈ K.
As shown in Lemma 4.6.4, khk −1 = α(k)(h) in H α K, for all h ∈ H and k ∈ K.
Applying f to both sides, we obtain f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)), as claimed.
Conversely, suppose that f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K.
We define f : H α K → G by f (hk) = f1 (h)f2 (k). It suffices to show that f is a
homomorphism. But
f (h1 k1 )f (h2 k2 )
= f1 (h1 )f2 (k1 )f1 (h2 )f2 (k2 )
= f1 (h1 )f1 (α(k1 )(h2 ))f2 (k1 )f2 (k2 )
= f1 (h1 α(k1 )(h2 ))f2 (k1 k2 )
= f (h1 k1 h2 k2 ),
for all h1 , h2 ∈ H and k1 , k2 ∈ K, where the second equality follows from the formula
f2 (k1 )f1 (h2 )(f2 (k1 ))−1 = f1 (α(k1 )(h2 )).
This permits a slight simplification when H and K are cyclic.
Corollary 4.6.10. Let α : Zk → Z×
n be a homomorphism, with α(a) = m, where Zk =
a in multiplicative notation. Write Zn α Zk in multiplicative notation, with Zn = b.
Then for any group G, there is a homomorphism f : Zn α Zk → G with f (a) = x and
f (b) = y if and only if the elements x, y ∈ G satisfy the following conditions:
1. x has exponent k
2. y has exponent n
3. xyx−1 = y m .
Proof The conditions are clearly necessary, as a has order k, b has order n, and
aba−1 = bm in Zn α Zk .
Conversely, if the three conditions hold, there are homomorphisms f1 : b → G and
f2 : a → G given by f1 (bi ) = y i and f2 (aj ) = xj . Thus, by Proposition 4.6.9, it suffices
j
to show that xj y i x−j = y m i .
j
Since conjugation by x is a homomorphism, xj y i x−j = (xj yx−j )i , so it suffices by the
j
law of exponents to show that xj yx−j = y m . But since xj yx−j = x(xj−1 yx−(j−1) )x−1 ,
the desired result follows from the third condition by an easy induction argument.
The above is an important ingredient in calculating the automorphism groups of the
semidirect products of cyclic groups. But it does not give good criteria for showing
CHAPTER 4. NORMALITY AND FACTOR GROUPS
116
when the homomorphism f is an isomorphism. For this it will be convenient to use the
language and concepts of extensions, as developed in Section 4.7.
We shall now begin to consider the question of the uniqueness of the semidirect
product: When are two different semidirect products isomorphic? We will not answer
this question completely, but will study some aspects of it. First, what happens if we
switch to a different homomorphism K → Aut(H)?
Recall from Section 3.6 that G acts on the set of homomorphisms from K to G as
follows: if f : K → G is a homomorphism and g ∈ G, then g · f is the homomorphism
defined by (g · f )(k) = gf (k)g −1 for all k ∈ K. By abuse of notation, we shall also
write gf g −1 for g · f . We say that the homomorphisms f and gf g −1 are conjugate
homomorphisms.
So it makes sense to ask whether H α K and H α K are isomorphic when α and
α are conjugate homomorphisms of K into Aut(H). Another fairly minor way in which
we could alter a homomorphism α : K → Aut(H) is by replacing it with α ◦ g, where
g ∈ Aut(K). We wish to study the effect of this sort of change as well.
Since we’re considering two different semidirect products of the same groups, we shall
use different notations for their structure maps. Thus, we write ι : H → H α K,
π : H α K → K, and s : K → H α K for the structure maps of H α K, and write
ι : H → H α K, π : H α K → K, and s : K → H α K for the structure maps of
H α K.
Proposition 4.6.11. Let α and α be homomorphisms of K into Aut(H). Then the
following conditions are equivalent:
1. α is conjugate to α ◦ g, with g ∈ Aut(K).
2. There is an isomorphism ϕ : H α K → H α K such that im(ϕ ◦ ι) = im ι and
im(ϕ ◦ s) = im s .
3. There is a commutative diagram
H
f H
ι
ι
/ H α K o
ϕ / H α K o
s
s
K
g K
in which the vertical maps are all isomorphisms.
Proof Suppose that the second condition holds. Then ϕ restricts to an isomorphism
from ι(H) to ι (H), and hence there exists an isomorphism f : H → H making the
left-hand square of the diagram commute.
Similarly, ϕ restricts to an isomorphism from s(K) to s (K) and there exists an
automorphism g of K such that the right-hand square of the diagram commutes. Thus,
the second condition implies the third.
Suppose then that the third condition holds. Then identifying H and K with their
images under ι and s , we see that ϕ ◦ ι = f and ϕ ◦ s = g. By Proposition 4.6.9, this
says that g(k)f (h)(g(k))−1 = f (α(k)(h)) for all h ∈ H and k ∈ K. But since f (h) ∈ H
and g(k) ∈ K, we have g(k)f (h)(g(k))−1 = α (g(k))(f (h)) by the conjugation formula
that holds in H α K.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
117
Putting these two statements together, we see that α (g(k))(f (h)) = (f ◦ α(k))(h),
and hence α (g(k)) ◦ f = f ◦ α(k) as automorphisms of H. But then as maps of K
into Aut(H), this just says that α ◦ g is the conjugate homomorphism to α obtained by
conjugating by f ∈ Aut(H). Thus, the third condition implies the first.
Now suppose that the first condition holds, and we’re given g ∈ Aut(K) such that α ◦g
is conjugate to α. Thus, there is an f ∈ Aut(H) such that (α ◦ g)(k) = f ◦ α(k) ◦ f −1 for
all k ∈ K. Reversing the steps above, this implies that g(k)f (h)(g(k))−1 = f (α(k)(h))
for all h ∈ H and k ∈ K. Thus, by Proposition 4.6.9, There is a homomorphism
ϕ : H α K → H α K given by ϕ(hk) = f (h)g(k).
Clearly, ϕ restricts to an isomorphism, f , from ι(H) to ι (H) and restricts to an
isomorphism, g, from s(K) to s (K), so it suffices to show that ϕ is an isomorphism.
Since the elements of H α K may be written uniquely as an element of H times an
element of K, f (h)g(k) = e if and only if f (h) = e and g(k) = e. Since f and g are both
injective, so is ϕ. Finally, since f and g are both surjective, every element of H α K
may be written as a product f (h)g(k), and hence ϕ is onto. Thus, the first condition
implies the second.
This is far from giving all the ways in which semidirect products can be isomorphic,
but we shall find it useful nevertheless, particularly when H is abelian. As a sample
application, consider this.
Corollary 4.6.12. Let α, α : Z5 → Z×
11 be homomorphisms such that neither α nor α
is the trivial homomorphism. Then the semidirect products Z11 α Z5 and Z11 α Z5
are isomorphic.
Proof By Corollary 7.3.15, Z×
11 is a cyclic group of order 10. Alternatively, we can
verify it by hand: 2k is not congruent to 1 mod 11 for k ≤ 5. Since the order of 2 in Z×
11
divides 10, we see that 2 has order 10, and hence generates Z×
11 .
Thus, Z×
11 has a unique subgroup H of order 5. Since the images of α and α must
have order 5, they induce isomorphisms of Z5 onto H. Thus, the composite
Z5
α
/ H
α−1
/ Z5
gives an isomorphism, which we’ll denote by g : Z5 → Z5 . But then α = α ◦ g, and
hence Z11 α Z5 ∼
= Z11 α Z5 by Proposition 4.6.11.
Using the Sylow Theorems in Chapter 5, we’ll be able to show that every group
of order 55 is a semidirect product of the form Z11 α Z5 . Together with the above
corollary, this shows that there are exactly two isomorphism classes of groups of order
55: the product Z11 ×Z5 (which is cyclic), and the semidirect product Z11 α Z5 obtained
from the homomorphism α : Z5 → Z×
11 that sends the generator of Z5 to 4.
Exercises 4.6.13.
1. Show that the subgroup s(K) ⊂ H α K is normal if and only if α : K → Aut(H)
is the trivial homomorphism.
2. Consider the groups of Examples 4.6.6. For the groups of Examples 2, 3, 4, and 5,
find the orders of all the elements not contained in b. Deduce that the groups of
Examples 2 and 3 cannot be isomorphic.
3. Show that the group of Example 4 of Examples 4.6.6 contains a subgroup isomorphic to D10 .
CHAPTER 4. NORMALITY AND FACTOR GROUPS
118
4. Let α : Z4 → Z×
3 be the homomorphism that carries the generator of Z4 to the
unique nontrivial element of Z×
3 . Construct an isomorphism f : Q12 → Z3 α Z4 .
5. Consider the group G = Z8 β Z2 of Example 3 of Examples 4.6.6. Show that
the subgroup b is not characteristic in G, by constructing an automorphism of G
which carries b outside of itself. (Of course, b is normal in G, as it is the group
being acted on in the semidirect product.)
6. Show that if m divides the order of an element x ∈ G, then o(x) = m · o(xm ).
Deduce that if f : G → G is a homomorphism and if x ∈ G, then xo(f (x)) ∈ ker f
and o(x) = o(f (x)) · o(xo(f (x)) ). Deduce that if hk ∈ H α K, then (hk)o(k) ∈ H,
and o(hk) = o(k) · o((hk)o(k) ). Thus, the determination of the order of hk depends
only on calculating the element (hk)o(k) ∈ H and on understanding the orders of
the elements in H.
† 7. Let x, y ∈ G, such that xyx−1 = y m . Show that (yx)k = y 1+m+···+m
k ≥ 2.
k−1
xk for all
† 8. Let m > 1 and k > 1 be integers. Show that (1 + m + · · · + mk−1 )(m − 1) = mk − 1,
and hence 1 + m + · · · + mk−1 = (mk − 1)/(m − 1).
9. Find the orders of the elements in the group of Example 6 of Examples 4.6.6. Show
also that the subgroup b is not characteristic. (Hint: Take a look at the proof of
Proposition 4.5.18.)
‡ 10. Show that the group constructed in Corollary 4.6.8 for n = 2 is isomorphic to D8 .
11. Consider the groups constructed in Corollary 4.6.8 for n a prime number p. When
p > 2, show that every element in this group has exponent p. (Why is this different
from the case n = 2?) Recall that in Problem 3 of Exercises 2.1.16, it was shown
that any group of exponent 2 must be abelian. As we see here, this property is not
shared by primes p > 2.
12. Now consider the groups of Corollary 4.6.8 for general values of n. For which values
of n does the group in question have exponent n? If the exponent is not equal to
n, what is it?
13. In this problem we assume a familiarity with the group of invertible 3 × 3 matrices
over the ring Zn . Let
⎫
⎧⎛
⎞
⎬
⎨ 1 x y
G = ⎝ 0 1 z ⎠ x, y, z ∈ Zn .
⎭
⎩
0 0 1
Show that G forms a group under matrix multiplication. Show, using Proposition 4.6.9, that G is isomorphic to the group constructed in Corollary 4.6.8.
† 14. There are three non-identity elements in Z2 × Z2 , which must be permuted by any
automorphism. Show that the automorphism group of Z2 × Z2 is isomorphic to
the group of permutations on these elements, so that Aut(Z2 × Z2 ) ∼
= S3 . (Hint:
Take a look at the proof of Corollary 3.2.11.)
15. Show that any two nontrivial homomorphisms from Z2 to S3 are conjugate. Deduce
from Problem 14 that if α, α : Z2 → Aut(Z2 × Z2 ) are nontrivial, then (Z2 ×Z2 )α
Z2 ∼
= (Z2 ×Z2 )α Z2 . Now deduce from Problem 10 that if α : Z2 → Aut(Z2 × Z2 )
is nontrivial, then (Z2 × Z2 ) α Z2 ∼
= D8 .
CHAPTER 4. NORMALITY AND FACTOR GROUPS
119
16. Show that Aut(Z) ∼
= {±1} and write α : Z2 → Aut(Z) for the nontrivial homomorphism. The semidirect product Z α Z2 is called the infinite dihedral group. Show
that the dihedral groups D2n are all factor groups of the infinite dihedral group.
17. Let α : Z → Aut(Z) be the unique nontrivial homomorphism. The semidirect
product Z α Z is called the Klein bottle group, as it turns out to be isomorphic to
the fundamental group of the Klein bottle. Show that the infinite dihedral group
is a factor group of the Klein bottle group.
18. Let Z[ 12 ] ⊂ Q be the set of rational numbers that may be written as 2mk with
m ∈ Z and k ≥ 0. In other words, if placed in lowest terms, these elements have
denominators that are powers of 2. Show that Z[ 12 ] is a subgroup of the additive
group of rational numbers and that multiplication by 2 induces an automorphism
of Z[ 12 ]. Deduce the existence of a homomorphism μ2 : Z → Aut(Z[ 12 ]) such that
μ2 (k) is multiplication by 2k for all k ∈ Z.
‡ 19. Let G = Z[ 12 ] μ2 Z. Let H ⊂ Z[ 12 ] be the standard copy of the integers in Q. Find
an element g ∈ G such that gHg −1 is contained in H but not equal to H.3
4.7
Extensions
Here, loosely, we study all groups G that contain a given group H as a normal subgroup,
such that the factor group G/H is isomorphic to a particular group K. This is the first
step in trying to classify composition series.
Definitions 4.7.1. Let H and K be groups. An extension of H by K consists of a
group G containing H as a normal subgroup, together with a surjective homomorphism
f : G → K with kernel H.4
Examples 4.7.2.
1. Let f : D2n → Z2 be induced by f (a) = 1 and f (b) = 0. Then f is an extension of
Zn = b by Z2 .
2. The projection map π : Zn × Z2 → Z2 is an extension of Zn by Z2 .
3. The unique nontrivial homomorphism Z2n → Z2 is yet another extension of Zn =
2 by Z2 .
4. Let f : Q4n → Z2 be induced by f (a) = 1 and f (b) = 0. Then f is an extension of
Z2n = b by Z2 .
5. Let α : K → Aut(H) be a homomorphism and let π : H α K → K be the
projection. Then π is an extension of H by K.
3 We could have used the rational numbers Q here in place of Z[ 1 ], obtaining a construction that
2
would be slightly simpler for the beginner. However, by using Z[ 12 ], we obtain a group that is finitely
generated (say, by the generator of H and the generator of the Z that acts), whereas we would not have
if we’d used Q.
4 It is sometimes useful to think of an extension of H by K as a group G, together with an embedding
ι : H → G and a surjective homomorphism f : G → K with kernel ι(H). In this language, the entire
f
ι
sequence H −→ G −→ K constitutes the extension in question. Note that given such data, we can form
an extension in the sense of Definition 4.7.1 by replacing G by the group G obtained by appropriately
renaming the elements in im ι.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
120
Clearly, for appropriate choices of H and K, there may be a number of nonisomorphic
extensions of H by K. Moreover, an understanding of the extensions of one group by
another is a key to understanding groups in general.
Recall from Proposition 4.2.5 that every nontrivial finite group G has a composition
series, which is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 Hi , and
Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ k. But this says that Hi is an extension
of Hi−1 by a nontrivial simple group.
In other words, every finite group may be obtained by a sequence of extensions of lower
order groups by simple groups. But the finite simple groups are all known. Thus, the
next step in a systematic study of finite group theory is to classify all possible extensions
of one finite group by another.
The search for such a classification is known as the Extension Problem, and constitutes
one of the major unsolved problems in mathematics.
Notice that there are various ways in which we could desire to classify the extensions
of H by K. For our purposes here, we’d be most interested in simply determining the
isomorphism classes of groups G that occur as extensions of H by K. Such a classification
takes no account of the precise embedding of H in G, nor of the map f : G → K that
determines the extension. Thus, such a classification forgets some of the data that’s part
of the definition of an extension.
A classification that would forget less data would be to classify the extensions of H
by K up to the equivalence relation obtained by identifying extensions f : G → K and
f : G → K if there is a commutative diagram
H
⊂
/ K
f
g2
g1
H
/ G
⊂
/ G
g3
f
/ K
in which the vertical maps are all isomorphisms. An even more rigid equivalence relation
is the one given by such diagrams where the maps g1 and g3 are required to be the
identity maps of H and K, respectively.
All three equivalence relations are relevant for particular questions. We shall discuss
these issues further.
In any case, we shall not be able to solve all the extension questions that might
interest us, but we can make considerable progress in special cases. These cases will be
sufficient to classify all groups of order ≤ 63.
Here is a useful result for obtaining equivalences between extensions.
Lemma 4.7.3. (Five Lemma for Extensions) Let f : G → K be an extension of H by K
and let f : G → K be an extension of H by K . Suppose given a commutative diagram
of group homomorphisms
H
⊂
g1
H
/ G
f
g2
⊂
/ G
/ K
g3
f
/ K
where g1 and g3 are isomorphisms. Then g2 is also an isomorphism.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
121
Proof Let x ∈ ker g2 . Then (f ◦ g2 )(x) = e, and hence g3 (f (x)) = e by commutativity
of the diagram. Since g3 is injective, f (x) = e, and hence x ∈ ker f = H. But on H ⊂ G,
g2 acts as g1 : H → H , and hence x ∈ ker g1 . But g1 is injective, so we must have x = e.
Thus, ker g2 = e, and hence g2 is injective.
Thus, it suffices to show that each y ∈ G lies in the image of g2 . Since g3 is surjective,
f (y) = g3 (z ) for some z ∈ K. But f is also surjective, so z = f (z) for some z ∈ G.
Thus, f (g2 (z)) = g3 (f (z)) = g3 (z ) = f (y), so that g2 (z) and y have the same image
under f .
But then y(g2 (z))−1 lies in the kernel of f , which is H . Thus, since g1 is surjective,
there is an x ∈ H such that g1 (x) = y(g2 (z))−1 . Thus, y = g1 (x)g2 (z) = g2 (xz), and
hence y ∈ im g2 as desired.
The Five Lemma for Extensions is a special case of a more general result which is
given for abelian groups as the Five Lemma in Lemma 7.7.48. The statement of the
general case involves five vertical maps, rather than the three displayed here, hence the
name. In both cases, the technique of proof is called a diagram chase.
As noted above, if α : K → Aut(H) is a homomorphism, then the projection map
π : H α K → K is an extension of H by K. As a general rule, the semidirect products
are the easiest extensions to analyze, so it is useful to have a criterion for an extension
to be one.
Definition 4.7.4. A section, or splitting, for f : G → K is a homomorphism s : K → G,
such that f ◦ s is the identity map of K. A homomorphism f : G → K that admits a
section is said to be a split surjection.
An extension of H by K is called a split extension if f : G → K admits a section.
The section s is said to split f .
The homomorphism s : K → H α K is clearly a section for π. We also have a
converse.
Proposition 4.7.5. Split extensions are semidirect products. Specifically, let f : G → K
be a split extension of H by K, with splitting map s : K → G. Then s induces a
homomorphism α : K → Aut(H) via α(k)(h) = s(k)h(s(k))−1 . Moreover, there is an
isomorphism ϕ : H α K → G such that the following diagram commutes.
H H
⊂
ι
/ H α K o
ϕ ⊂
/ G o
s1
s
K K
Here, s1 is the standard section for the semidirect product. Explicitly, ϕ is defined by
ϕ(hk) = hs(k).
Proof Recall from Lemma 3.7.7 that, since H G, we have a homomorphism Γ : G →
Aut(H), given by setting Γ(g)(h) = ghg −1 for all g ∈ G and h ∈ H. The above specification for α is just the composite Γ ◦ s : K → Aut(H), and hence is a homomorphism as
claimed.
Recall from Proposition 4.6.9 that if f1 : H → G and f2 : K → G are homomorphisms such that f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K, then there
CHAPTER 4. NORMALITY AND FACTOR GROUPS
122
is a unique homomorphism f : H α K → G such that f ◦ ι = f1 and f ◦ s1 = f2 :
specifically, f (hk) = f1 (h)f2 (k).
We apply this with f1 equal to the inclusion of H in G and with f2 equal to the section
s : K → G. But then the condition we need to verify to construct our homomorphism
is that s(k)h(s(k))−1 = α(k)(h), which holds by the definition of α. Thus, there is a
homomorphism ϕ : H α K → G with ϕ◦ι equal to the inclusion of H in G, and ϕ◦s1 = s:
specifically, ϕ(hk) = hs(k). But this just says that the stated diagram commutes, so it
suffices to show that ϕ is an isomorphism.
Since ϕ(hk) = hs(k), we obtain that (f ◦ ϕ)(hk) = (f ◦ s)(k) = k. But this says that
f ◦ ϕ = π, where π : H α K → K is the projection map. Thus, we have an additional
commutative diagram.
H H
⊂
ι
/ H α K
ϕ ⊂
/ G
π
f
/ K
/ K
But this shows ϕ to be an isomorphism by the Five Lemma for Extensions.
Of course, if α : K → Aut(H) is the trivial homomorphism, then the semidirect
product H α K is simply H × K. Thus, the corollary below is immediate from the way
that α depends on s in the statement of Proposition 4.7.5.
Corollary 4.7.6. Let f : G → K be a split extension of H by K, with splitting map
s : K → G. Suppose that s(k) commutes with each element of H for all k ∈ K. Then
G∼
= H × K.
In particular, if f : G → K is a split surjection and if G is abelian, then G ∼
=
ker f × K.
Semidirect products are the easiest extensions to construct, and hence may be viewed
as the most basic of the extensions of one group by another. For this reason, it is useful
to be able to recognize whether an extension splits. The easiest case is when the two
groups have orders that are relatively prime.
Proposition 4.7.7. Let H and K be groups whose orders are relatively prime and let
f : G → K be an extension of H by K. Then the extension splits if and only if G has a
subgroup of order |K|.
Proof If s : K → G is a section of f , then s is injective, so s(K) has the same order as
K.
Conversely, suppose that K ⊂ G has order |K| and write f1 : K → K for the
restriction of f to K . Then ker f1 = K ∩ H. Since the orders of K and H are relatively
prime, K ∩ H = e, by Lagrange’s Theorem. Thus, f1 is injective. Since K and K have
f −1
1
the same order, f1 is an isomorphism, and the composite K −−
→ K ⊂ G gives a section
for f .
In fact, under the hypotheses above, G always has a subgroup of order |K|.
Theorem (Schur–Zassenhaus Lemma) Let f : G → K be an extension of H by K,
where H and K have relatively prime order. Then G has a subgroup of order |K|.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
123
The proof requires the Sylow Theorems and could be given in Chapter 5, but we
shall not do so. In practice, when considering a particular extension, one can generally
construct the section by hand.
Let us now consider the issue of classifying the split extensions of one group by
another.
Note that the definition of the homomorphism α : K → Aut(H) constructed in
Proposition 4.7.5 makes use of the specific section s. Thus, we might wonder whether a
different homomorphism α would have resulted if we’d started with a different section.
Indeed, this can happen when the group H is nonabelian.
Example 4.7.8. Let ε : Sn → {±1} take each permutation to its sign. Then ε is an
extension of An by Z2 . Moreover, setting s(−1) = τ ∈ Sn determines a section of ε if
and only if τ has order 2 and is an odd permutation.
Since the order of a product of disjoint cycles is the least common multiple of the
orders of the individual cycles, we see that τ determines a section of ε if and only if it
is the product of an odd number of disjoint 2-cycles. For such a τ , we write sτ for the
section of ε with sτ (−1) = τ . Given the variety of such τ , we see that there are many
possible choices of section for ε.
Write ατ : Z2 → Aut(An ) for the homomorphism induced by sτ . By definition,
ατ = Γ ◦ sτ , where Γ : Sn → An is the homomorphism induced by conjugation. Thus,
ατ (−1) = Γ(τ ).
But Problem 10 in Exercises 3.7.14 shows that Γ is injective for n ≥ 4. Thus, if n ≥ 4,
then each different choice for a section of ε induces a different homomorphism from Z2
to Aut(An ).
If τ and τ have the same cycle structure, then they are conjugate in Sn , and hence
sτ and sτ are conjugate as homomorphisms into Sn . But then ατ and ατ are conjugate
as homomorphisms into Aut(An ).
However, as we shall see in the exercises below, if τ is a 2-cycle and τ is a product
of three disjoint 2-cycles, then ατ and ατ are not conjugate as homomorphisms into
Aut(An ) if n > 6. Thus, in this case, Proposition 4.6.11 would not have predicted that
An ατ Z2 and An ατ Z2 would be isomorphic; nevertheless they are isomorphic, as each
one of them admits an isomorphism to Sn . However, the criterion in Proposition 4.6.11 is
the most general sufficient condition we shall give for the isomorphism of two semidirect
products.
Thus, the classification of even the split extensions of a nonabelian group offers substantial difficulty. But we shall be mainly concerned here with the extensions of an
abelian group, a problem for which we can develop some reasonable tools.
Lemma 4.7.9. Let f : G → K be an extension of H by K, where H is abelian. For x ∈
G, write cx : H → H for the automorphism obtained from conjugation by x. Suppose that
f (x) = f (y). Then cx = cy as automorphisms of H. Moreover, there is a homomorphism
αf : K → Aut(H) defined by setting αf (k) equal to cx for any choice of x with f (x) = k.
In particular, if f happens to be a split extension and if s : K → G is a section of f ,
then the homomorphism from K to Aut(H) induced by s coincides with the homomorphism αf above. Thus, if H is abelian, the homomorphism from K to Aut(H) induced
by a section of f is independent of the choice of section.
Proof If f (x) = f (y), then y −1 x ∈ ker f = H, so x = yh for some h ∈ H. But then
cx = cy ◦ ch . Since H is abelian, conjugation of H by h is the identity map, so cx = cy
as claimed.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
124
Thus, there is a well defined function αf : K → Aut(H) obtained by setting αf (k)
equal to cx for any x with f (x) = k. But if f (x) = k and f (y) = k , then f (xy) = kk ,
so that αf (kk ) = cxy = cx ◦ cy = αf (k)αf (k ), so αf is a homomorphism.
If s : K → G is a section of f , then the homomorphism α : K → Aut(H) induced by
this section is obtained by setting α(k) = cs(k) : H → H. Since f (s(k)) = k, this just
says that α = αf .
Thus, we may make the following definition.
Definition 4.7.10. Let f : G → K be an extension of H by K, where H is abelian.
Then the action of K on H induced by f is the homomorphism αf : K → Aut(H)
obtained by setting αf (k) equal to the automorphism obtained by conjugating H by any
given element of f −1 (k).
The existence of the homomorphisms αf gives a valuable method of studying the
extensions of an abelian group: fix a homomorphism α : K → Aut(H) and classify the
extensions whose induced action is α. Do this for all possible homomorphism α, and look
to see if a given group can be written as an extension in more than one way.
We shall illustrate this process by classifying the groups of order 8.
Lemma 4.7.11. Let f : G → Z2 be an extension of Z4 by Z2 . Suppose that the induced
action αf : Z2 → Z×
4 carries the generator of Z2 to −1. Then G is isomorphic to either
D8 or Q8 .
Proof Let Z4 = b and Z2 = a. For any x ∈ G with f (x) = a, we have xbk x−1 =
αf (a)(bk ) = b−k for all k. Also, f (x2 ) = a2 = e, so x2 ∈ ker f = b. Thus, x2 = bk for
some k.
Note that x fails to commute with b and b−1 , so neither of these can equal x2 . Thus,
either x2 = e or x2 = b2 .
If x2 = e, then x has order 2, and hence there is a section of f , specified by s(a) = x.
In this case, G ∼
= Z4 αf Z2 , which, for this value of αf , is isomorphic to the dihedral
group D8 by the first example of Examples 4.6.6.
The remaining case gives x2 = b2 . But then there is a homomorphism h : Q8 → G
with h(a) = x and h(b) = b by Problem 5 of Exercises 2.9.7. As the image of h contains more than half the elements of G, h must be onto, and hence an isomorphism.
(Alternatively, we could use the Five Lemma for Extensions to show that h is an isomorphism.)
Q8 , of course, is not a split extension of b by Z2 , for two reasons. First, we know
that every element of Q8 which is not in b has order 4, and hence it is impossible to
define a section from Q8 /b to Q8 . Second, if it were a split extension, then it would be
isomorphic to D8 , which we know is not so.
We have classified the extensions of Z4 by Z2 corresponding to the homomorphism
×
from Z2 to Z×
4 that takes the generator of Z2 to −1. Since Z4 is isomorphic to Z2 , there
×
is only one other homomorphism from Z2 to Z4 : the trivial homomorphism. Let us
begin with some general observations about extensions whose induced action is trivial.
Lemma 4.7.12. Let f : G → K be an extension of H by K, where H is abelian. Then
the induced homomorphism αf : K → Aut(H) is the trivial homomorphism if and only
if H ⊂ Z(G).
CHAPTER 4. NORMALITY AND FACTOR GROUPS
125
Proof Suppose that αf is the trivial homomorphism. Then for any x ∈ G, let y = f (x).
By the definition of αf , we have xhx−1 = αf (y)(h) for all h ∈ H. But αf is the trivial
homomorphism, so that αf (y) is the identity map of H. Thus xhx−1 = h for all h ∈ H,
and hence each h ∈ H commutes with every element of G. Thus, H ⊂ Z(G).
But the converse is immediate: if H ⊂ Z(G), then conjugating H by any element of
G gives the trivial homomorphism, and hence αf (y) is the identity element of Aut(H)
for every y ∈ K.
The extensions of H containing H in their centers are important enough to merit a
name.
Definition 4.7.13. An extension f : G → K of H by K is a central extension if H ⊂
Z(G).
Example 4.7.14. As one may easily see, the center of D8 is b2 . Moreover, D8 /b2 is
isomorphic to Z2 × Z2 , with generators a and b. Thus, D8 is a central extension of Z2
by Z2 × Z2 .
In particular, a central extension of a cyclic group by an abelian group is not necessarily abelian. But a central extension of an abelian group by a cyclic group is abelian:
Lemma 4.7.15. Let f : G → Zk be a central extension of the abelian group H by the
cyclic group Zk . Then G is abelian.
Proof Write Zk = a, and let x ∈ G with f (x) = a. Let g ∈ G. Then f (g) = ai = f (xi )
for some i, and hence gx−i ∈ ker f = H. But then g = hxi for some h ∈ H.
If g is any other element of G, we may write g = h xj for h ∈ H and j ∈ Z. But
then gg = hxi h xj = hh xi+j , as h ∈ Z(G). But multiplying out g g gives the same
result, and hence g and g commute.
We now return to the groups of order 8.
Lemma 4.7.16. Let f : G → Z2 be a central extension of Z4 by Z2 . Then G is isomorphic to either Z8 or Z4 × Z2 .
Proof Since Z2 is cyclic, Lemma 4.7.15 shows that G is abelian. So we could just
appeal to the Fundamental Theorem of Finite Abelian Groups, which tells us that Z8 ,
Z4 × Z2 , and Z2 × Z2 × Z2 are the only abelian groups of order 8. Clearly, the first two
are the only ones that are extensions of Z4 .
Alternatively, we can argue directly. Write Z4 = b and Z2 = a and let f (x) = a.
Then f (x2 ) = e, and hence x2 ∈ b. Now x has even order, so that o(x) = 2o(x2 ). Thus,
if x2 = b or b−1 , then x has order 8, and hence G is cyclic.
If x has order 2, then s(a) = x gives a splitting of f , and hence G is the semidirect
product obtained from the action of Z2 on Z4 . But in this case, the action is trivial, so
that the semidirect product thus obtained is just the direct product.
If x has order 4, then x2 = b2 . But then b−1 x has order 2, and setting s(a) = b−1 x
gives a section of f , and hence G ∼
= Z4 × Z2 as above.
Once again, we have a nonsplit extension: the unique nontrivial homomorphism
f : Z8 → Z2 is a nonsplit extension of Z4 by Z2 .
We now classify the groups of order 8.
Proposition 4.7.17. Every group of order 8 is isomorphic to exactly one of the following
groups: Z8 , D8 , Q8 , Z4 × Z2 , and Z2 × Z2 × Z2 .
CHAPTER 4. NORMALITY AND FACTOR GROUPS
126
Proof If G has an element of order 4, say o(b) = 4, then b has index two in G, and
hence is normal in G by Proposition 4.1.12 (or by the simpler argument of Problem 12
of Exercises 3.7.14). But then G is an extension of Z4 by Z2 . As Z×
4 = {±1}, there are
,
the
homomorphism
that carries the
only two different homomorphisms from Z2 to Z×
4
generator of Z2 to −1 and the trivial homomorphism. Thus, if G is an extension of Z4
by Z2 , then G is isomorphic to one of Z8 , D8 , Q8 , and Z4 × Z2 by Lemmas 4.7.11 and
4.7.16.
If G does not have an element of order 4, then every element of G has exponent 2.
But then G is abelian by Problem 3 of Exercises 2.1.16. But the fundamental theorem
of finite abelian groups shows Z2 × Z2 × Z2 to be the only abelian group of order 8 and
exponent 2.
The reader may easily check that no two of the listed groups are isomorphic.
We shall classify all groups of order p3 for odd primes p as well, but this will require
theoretical results to be given in the next chapter. We now return to the general theory
of extensions.
Lemma 4.7.18. Let f : G → K and f : G → K be extensions of H by K, where H is
abelian. Suppose given a commutative diagram
H
⊂
f
⊂
/ G
/ K
g3
g2
g1
H
/ G
f
/ K,
where g1 , g2 , and g3 are isomorphisms. Then the actions αf , αf : K → Aut(H) induced
by f and f are related by g1 ◦ αf (k) ◦ g1−1 = αf (g3 (k)), so that αf is conjugate to αf ◦ g3
as a homomorphism from K to Aut(H).
Proof For k ∈ K, let x ∈ G with f (x) = k. Then f (g2 (x)) = g3 (k), so αf (g3 (k))(g1 (h)) =
g2 (x)g1 (h)g2 (x)−1 = g2 (xhx−1 ) = g1 (αf (k)(h)) for all h ∈ H. Thus, αf (g3 (k)) ◦ g1 =
g1 ◦ αf (k), and the result follows.
Remarks 4.7.19. There is a general method that may be used to classify the extensions
of an abelian group H by a group K up to the following equivalence relation: The
extensions f : G → K and f : G → K are equivalent if there is a commutative diagram
H
⊂
/ G
f
/ K
f
/ K.
g
H
⊂
/ G
Here, g is forced to be an isomorphism, by the Five Lemma for Extensions.
Notice that Lemma 4.7.18 implies that if f and f are equivalent, then the homomorphisms αf and αf are equal. Thus, the method under discussion will give a separate
classification for each different homomorphism α : K → Aut(H). Thus, fixing such an
α, we shall write Hα to denote H with this particular action of K.
The method of classification is given using the cohomology of groups, a subtopic
of homological algebra. The classification theorem gives a one-to-one correspondence
CHAPTER 4. NORMALITY AND FACTOR GROUPS
127
between the equivalence classes of extensions corresponding to the homomorphism α
and the elements of the cohomology group H 2 (K; Hα ).5
Nevertheless, a couple of deficiencies in this approach should be pointed out. First,
given an element in H 2 (K; Hα ), it is quite tedious to construct the associated extension.
And it comes in a form that is difficult to analyze.
Second, the classification is too rigid for many of the questions of interest: there are
extensions whose associated actions are both α which are not equivalent in the above
sense, but are equivalent under weaker hypotheses. Thus, there may be fewer isomorphism classes of groups G that may be written as extensions of H by K with action α
than there are elements of the cohomology group H 2 (K; Hα ).
Nevertheless, the homological approach can be useful.
It is sometimes useful to be able to go back and forth between the existence of a
commutative diagram such as that in Lemma 4.7.18 and the existence of a suitable
isomorphism from G to G .
Lemma 4.7.20. Let f : G → K and f : G → K be extensions of H by K, and write
ι : H → G and ι : H → G for the inclusions. Then the following conditions are
equivalent.
1. There is a commutative diagram
H
⊂
ι
f
⊂
ι
/ G
/ K
g3
g2
g1
H
/ G
f
/ K,
where g1 , g2 , and g3 are isomorphisms.
2. There is an isomorphism g : G → G with g(im ι) = im ι .
Proof If the first condition holds, then the second one does also, as we may take g = g2 .
Thus, suppose that the second condition holds, and we are given an isomorphism
g : G → G with g(im ι) = im ι . Then g induces an isomorphism from im ι to im ι . Since
ι and ι are embeddings, there is an isomorphism g1 : H → H such that ι ◦ g1 = g ◦ ι.
Thus, if we set g2 = g, then the left square of the displayed diagram commutes.
We must now construct g3 , which comes from a couple of applications of the First
Noether Theorem. First, we have surjective maps f : G → K and f : G → K, with
kernels given by ker f = H and ker f = H. Thus, the First Noether Theorem gives
∼
∼
=
=
isomorphisms f : G/H −→ K and f : G /H −→ K such that the following diagrams
commute.
/K
GD
DD
z<
z
DD
zz
D
zzf
π DD
z
!
z
G/H
f
f
/K
G E
y<
EE
y
EE
yy
E
yyf π EE
y
y
"
G /H
5 See, for instance, A Course in Homological Algebra, by P.J. Hilton and U. Stammbach, SpringerVerlag, 1971, for an exposition of this classification.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
g
128
π
→ G −
→ G /H is precisely H, so
Since g(im ι) = im ι , the kernel of the composite G −
∼
=
the First Noether Theorem provides an isomorphism k : G/H −→ G /H that makes the
following diagram commute.
H
g1 ∼
=
H
Now set g3 = f ◦ k ◦ f
−1
⊂
ι
⊂
ι
/ G
∼
g =
/ G
π
π
/ G/H
k ∼
=
/ G /H
f
∼
=
f
∼
=
/ K
/ K
.
Note that the group H of Lemma 4.7.20 need not be abelian.
We now wish to explore the question of when two different homomorphisms, α, α :
K → Aut(H) produce isomorphic semidirect products. In this regard, the above lemmas
permit an improvement of the result in Proposition 4.6.11 in the case that H is abelian.
Proposition 4.7.21. Let H be an abelian group and let α and α be homomorphisms of
K into Aut(H). Then α is conjugate to a composite α ◦ g with g ∈ Aut(K) if and only
if there is an isomorphism ϕ : H α K → H α K such that ϕ(im ι) = im ι , where ι
and ι are the canonical inclusions of H in H α K and H α K, respectively.
Proof If α is conjugate to a composite α ◦ g with g ∈ Aut(K), then Proposition 4.6.11
provides an isomorphism ϕ : H α K → H α K with ϕ(im ι) = im ι . Thus, it suffices
to show the converse.
So suppose given an isomorphism ϕ : H α K → H α K with ϕ(im ι) = im ι . Then
Lemma 4.7.20 provides a commutative diagram of the sort appearing in the hypothesis
of Lemma 4.7.18.
By Lemma 4.7.9, the action map απ : K → Aut(H) induced by π : H α K → K is
precisely α. A similar result holds for α . Thus, the output of Lemma 4.7.18 is precisely
that α is conjugate to α ◦ g for some g ∈ Aut(K).
We would like to eliminate the hypothesis that ϕ(im ι) = im ι from the above, and
obtain a necessary and sufficient condition for the semidirect products H α K and
H α K to be isomorphic. We can do this if H and K are finite groups whose orders
are relatively prime.
Corollary 4.7.22. Let H and K be finite groups whose orders are relatively prime, and
suppose that H is abelian. Let α and α be two homomorphisms from K to Aut(H).
Then the semidirect products H α K and H α K are isomorphic if and only if α is
conjugate to α ◦ g for some automorphism g of K.
Proof It suffices to show that any homomorphism from H α K to H α K carries ι(H)
into ι (H), and that the analogous statement holds for homomorphisms in the opposite
direction. But this will follow if we show that in either one of the semidirect products,
the only elements of exponent |H| are those in H itself.
If |H| is an exponent for x, then it is also an exponent for f (x) for any homomorphism
f . Since any element whose order divides both |H| and |K| must be the identity, we see
that any element of either semidirect product which has exponent |H| must be carried
to the identity element by the projection map (π or π ) into K. Since H is the kernel of
the projection map, the result follows.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
129
We now give a generalization of Corollary 4.7.22.
Corollary 4.7.23. Let H be an abelian group and let α, α : K → Aut(H). Suppose
that α has the property that any embedding f : H → H α K must have image equal to
ι(H). Then H α K is isomorphic to H α K if and only if α is conjugate to α ◦ g for
some g ∈ Aut(K).
Proof Again, it suffices to show that ϕ(im ι) = im ι , or, equivalently, that im ι =
im(ϕ−1 ◦ ι ), for any isomorphism ϕ : H α K → H α K. But since ϕ−1 ◦ ι is an
embedding, this follows from the hypothesis.
Corollary 4.7.23 may be applied, for instance, if ι(H) is the only subgroup of H α K
which is isomorphic to H. An example where this occurs is for H = b ⊂ D2n for n ≥ 3.
Here, we make use of the fact that every element of D2n outside of b has order 2. Since
b has order n > 2, every embedding of b in D2n must have image in b. Since b is
finite, this forces the image to be equal to b.
Finally, let us consider the extensions of a nonabelian group H. As shown in Example 4.7.8, if f : G → K is even a split extension of H by K, different choices of splittings
may define different homomorphisms from K into Aut(H). In particular, we do not have
a homomorphism from K to Aut(H) which is associated to the extension itself. But we
can define something a little weaker.
As usual, we write ch : H → H for the automorphism induced by conjugating by
h ∈ H: ch (h ) = hh h−1 for all h ∈ H. Recall that an automorphism f ∈ Aut(H) is
an inner automorphism if f = ch for some h ∈ H. The inner automorphisms form a
subgroup, Inn(H) ⊂ Aut(H).
Lemma 4.7.24. The group of inner automorphisms of a group G is a normal subgroup
of Aut(G).
Proof Let x ∈ G and write cx for conjugation by x. Let f ∈ Aut(G). We wish to show
that f ◦ cx ◦ f −1 is an inner automorphism. But
f ◦ cx ◦ f −1 (g)
= f (xf −1 (g)x−1 )
= f (x)f (f −1 (g))f (x)−1 ,
and hence f ◦ cx ◦ f −1 = cf (x) .
Thus, we may make the following definition.
Definition 4.7.25. We write Out(G) for the quotient group Aut(G)/Inn(G) and call it
the outer automorphism group of G.
Proposition 4.7.26. Let f : G → K be an extension of H by K. Then f determines
a homomorphism α : K → Out(H). Here, for k ∈ K, α(k) is represented by the
automorphism of H given by conjugation by g, for any g ∈ G with f (g) = k.
In particular, if f is a split extension and s is a splitting of f , then α = π ◦ αs , where
αs : K → Aut(H) is the homomorphism induced by s and π is the canonical map from
Aut(H) to Out(H) = Aut(H)/Inn(H). Thus, if s and s are two different sections of f ,
then the homomorphisms αs and αs agree in Out(H).
CHAPTER 4. NORMALITY AND FACTOR GROUPS
130
Proof If f (g) = f (g ) = k, then g = gh for some h ∈ H. But then conjugating by g is the same as conjugating first by h and then by g. But conjugation by h is an inner
automorphism of H, and hence represents the identity in Out(H). Thus, conjugation by
g represents the same element of Out(H) as conjugation by g, and hence the prescription
above gives a well-defined function α : K → Out(H).
It suffices to show that α is a homomorphism. But for k, k ∈ K, if f (g) = k and
f (g ) = k , then f (gg ) = kk . But conjugation by gg is the composite of the conjugations
by g and by g .
We should also note that the above is not always very useful. For instance, for n = 6,
Out(Sn ) is the trivial group.
Exercises 4.7.27.
† 1. Problem 1 of Exercises 4.2.18 provides a normal subgroup H ⊂ A4 of order 4.
Show that H ∼
= Z2 × Z2 and that A4 is a split extension of H by Z3 .
2. Show that any nonabelian group G of order 12 which contains a normal subgroup of
order 4 must be isomorphic to A4 . (Hint: Find a G-set with 4 elements and analyze
the induced homomorphism from G to S4 . Alternatively, show that G is a split
extension of a group of order 4 by Z3 , and, using the calculation of Aut(Z2 × Z2 )
given in Problem 14 of Exercises 4.6.13, show that there is only one such split
extension which is nonabelian.)
† 3. Consider the subgroup H ⊂ An of Problem 1. Show that H is normal in S4 and
that S4 is isomorphic to the semidirect product (Z2 × Z2 ) α S3 , where α : S3 →
Aut(Z2 × Z2 ) is an isomorphism. (Recall the calculation of Aut(Z2 × Z2 ) given in
Problem 14 of Exercises 4.6.13. In the situation at hand, you may enjoy giving a
new proof of this calculation, based on showing that the map α above is injective.)
4. Show that if k is odd, then Q4k is isomorphic to Zk α Z4 for some α : Z4 →
Aut(Zk ). Calculate α explicitly.
5. Find all the central extensions of Z2 by Z2 × Z2 .
6. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced
homomorphism αf : Z2 → Z×
8 carries the generator of Z2 to 5. Show that f
is a split extension, and hence that G is isomorphic to the group Z8 β Z2 of
Example 3 in Examples 4.6.6. (Hint: Write Z8 = b and Z2 = a and let x ∈ G
with f (x) = a. Note that x2 = bk for some k, and use this to calculate the order
of each of the elements bi x.)
7. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced
homomorphism αf : Z2 → Z×
8 carries the generator of Z2 to 3. Show that f is a
split extension, and hence that G is isomorphic to the group Z8 α Z2 of Example 2
in Examples 4.6.6. (Hint: Argue as in the preceding problem. Here, if x2 = bk ,
note that the fact that x must commute with x2 places restrictions on which values
of k can occur.)
8. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced
homomorphism αf : Z2 → Z×
8 carries the generator of Z2 to −1. Show that f is
isomorphic to either D16 or Q16 .
9. Classify those groups of order 16 that contain an element of order 8.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
131
10. Let H be any group and let f : G → Z be an extension of H by Z. Show that f
splits.
11. We show that the inner automorphisms of Sn induced by different sections of the
sign homomorphism are not always conjugate in Aut(Sn ).
(a) Let f ∈ Aut(G) and let Gf = {g ∈ G | f (g) = g}, the fixed points of G under
f . Show that Gf is a subgroup of G. Show also that if f and f are conjugate
in Aut(G), then Gf and Gf are isomorphic as groups.
(b) Let τ ∈ Sn be a 2-cycle and write An τ for the fixed points of An under the
automorphism obtained from conjugating by τ . Show that An τ is isomorphic
to Sn−2 .
(c) Let H ⊂ S6 be the subgroup consisting of those elements of S6 that commute
with (1 2)(3 4)(5 6). Show that H is a split extension of Z2 × Z2 × Z2 by S3 .
(d) Let τ ∈ Sn be a product of three disjoint 2-cycles. Show that the subgroup
An τ fixed by τ , is isomorphic to an index 2 subgroup of H × Sn−6 , where
H ⊂ S6 is the group of the preceding part.
Deduce that if n > 6, then the automorphism of An obtained from conjugation
by a 2-cycle cannot be conjugate in Aut(An ) to the automorphism obtained from
conjugation by the product of three disjoint 2-cycles.
‡ 12. Let α : K → Aut(H) and β : K → Aut(H ) be homomorphisms. Let f : H → H be a group homomorphism such that f (α(k)(h)) = β(k)(f (h)) for all k ∈ K and
h ∈ H (i.e., f is a K-map between the K-sets H and H ). Define f∗ : H α K →
H β K by f∗ (hk) = f (h)k. Show that f∗ is a homomorphism.
13. Let α : K → Aut(H) be a homomorphism. Suppose given f ∈ Aut(H) that
commutes with every element of the image of α. Show that the homomorphism
f∗ : H α K → H α K of Problem 12 is defined and gives an automorphism of
H α K.
‡ 14. In this problem, we calculate the automorphisms of a dihedral group. The case of
D4 is exceptional, and indeed has been treated in Problem 14 in Exercises 4.6.13.
Thus, we shall consider D2n for n ≥ 3.
A key fact is given by either Problem 6 of Exercises 2.8.5 or by Proposition 4.6.10:
The homomorphisms from D2n to a group G are in one-to-one correspondence with
the choices of x = f (a) and y = f (b) such that x has exponent 2, y has exponent
n, and xyx−1 = y −1 .
The reader should first recall from Problem 4 of Exercises 3.7.14 that since n ≥ 3,
the subgroup b ⊂ D2n is characteristic. Thus, Lemma 3.7.8 provides a restriction
homomorphism ρ : Aut(D2n ) → Aut(b), given by setting ρ(f ) equal to its restriction f : b → b. As usual, by abuse of notation, we shall identify Aut(b) with
×
Z×
n , and write ρ : Aut(D2n ) → Zn .
(a) Show that any homomorphism f : D2n → D2n whose restriction to b gives
an automorphism of b must have been an automorphism of D2n in the first
place.
(b) Show that the restriction homomorphism ρ : Aut(D2n ) → Z×
n is a split sur×
jection (i.e., ρ : Aut(D2n ) → Z×
n is a split extension of ker ρ by Zn ).
CHAPTER 4. NORMALITY AND FACTOR GROUPS
132
(c) Show that an automorphism f is in the kernel of ρ if and only if f (b) = b and
f (a) = bi a for some i. Deduce that the kernel of ρ is isomorphic to Zn .
(d) Show that the action of Z×
n on Zn = ker ρ induced by the split extension is
the same as the usual action of Z×
n = Aut(Zn ). Deduce that Aut(D2n ) is
isomorphic to the semidirect product Zn id Aut(Zn ), where id is the identity
map of Aut(Zn ).
15. Show that every automorphism of S3 is inner.
16. Recall from Problem 8 of Exercises 3.7.14 that if n ≥ 3, the subgroup b ⊂ Q4n
is characteristic. Using this as a starting point, give a calculation of Aut(Q4n ),
n ≥ 3, by methods analogous to those of Problem 14 above. Note that the lack of
a semidirect product structure on Q4n necessitates greater care in the construction
of a splitting for ρ than was required in the dihedral case. (We shall treat the
exceptional case of Aut(Q8 ) in Problem 9 of Exercises 5.4.9.)
‡ 17. Recall from Problem 3 of Exercises 4.2.18 that An is a characteristic subgroup
of Sn for n ≥ 3. Thus, Lemma 3.7.8 provides a restriction homomorphism ρ :
Aut(Sn ) → Aut(An ).
Let Γ : Sn → Aut(An ) be the homomorphism induced by conjugation. By Problem 10 in Exercises 3.7.14, Γ is injective for n ≥ 4. Use this fact, together with
Proposition 4.6.9, to show that ρ is injective for n ≥ 4.
18. We compute the automorphism groups of A4 and S4 .
(a) Let H ⊂ A4 be the subgroup of order 4. Show that H is characteristic in
A4 . Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(A4 ) →
Aut(H).
(b) Show that the kernel of ρ is isomorphic to H, and hence to Z2 × Z2 .
(c) Show that ρ is a split surjection. Deduce that the homomorphism Γ : S4 →
Aut(A4 ) induced by conjugation is an isomorphism.
(d) Deduce from Problem 17 that the homomorphism Γ : S4 → Aut(S4 ) induced
by conjugation is an isomorphism.
‡ 19. We consider the automorphism groups of the semidirect products G = Zn α Zk
for selected homomorphisms α : Zk → Aut(Zn ). As usual, we write ι : Zn → G
and s : Zk → G for the standard inclusions and write π : G → Zk for the standard
projection.
The one assumption we shall insist on is that ι(Zn ) be a characteristic subgroup of
G. As the reader may easily verify, this is always the case if n and k are relatively
prime. But we have seen other examples, e.g., D2n = Zn α Z2 or Example 2 of
Examples 4.6.6, where Zn is characteristic in G, and our arguments will apply to
these cases as well. In particular, the material here generalizes that in Problem 14.
As usual, we use multiplicative notation, with Zn = b and Zk = a. As discussed
j
in Section 4.6, we have α(aj )(bi ) = bm i for an integer m with the property that
mk ≡ 1 mod n. The elements of G may be written uniquely as bi aj with 0 ≤ i < n
j and 0 ≤ j < k, and bi aj bi aj = bi+m i aj+j .
Since b is assumed characteristic in G, we have a restriction homomorphism ρ :
Aut(G) → Aut(b) ∼
= Z×
n.
CHAPTER 4. NORMALITY AND FACTOR GROUPS
133
(a) Let f ∈ Aut(G). Show that there is a unique element f ∈ Aut(a) such that
the following diagram commutes:
G
f
π
a
/ G
π
f
/ a
Show that the passage from f to f gives a homomorphism
η : Aut(G) → Aut(a) ∼
= Z×
k.
(b) Show, via Corollary 4.6.10, that the image of η consists of those f ∈ Aut(Zk )
such that the following diagram commutes:
Zk
α
/ Z×
n
α
/ Z× .
n
f
Zk
Show that such automorphisms f ∈ Aut(Zk ) correspond to elements j ∈ Z×
k
such that mj ≡ m mod n. Deduce that if α is injective (i.e., if m has order k
in Z×
n ), then η is the trivial homomorphism.
(c) Consider the map (ρ, η) : Aut(G) → Aut(b) × Aut(a) given by (ρ, η)(f ) =
(ρ(f ), η(f )). Show that (ρ, η) induces a split surjection of Aut(G) onto Aut(b)×
(im η).
(d) Show that an automorphism f is in the kernel of (ρ, η) if and only if f (b) = b
and f (a) = bi a, where i is chosen so that bi a has order k in G. Show, via
Problems 7 and 8 of Exercises 4.6.13, that bi a has order k if and only if
i · (mk − 1)/(m − 1) is divisible by n.
Show also that passage from f to bi induces an isomorphism from the kernel
of (ρ, η) onto a subgroup H ⊂ b.
(e) Let β : Aut(Zn ) × (im η) → Aut(H) be the action induced by the extension (ρ, η). Show that β restricts on Aut(Zn ) to the restriction map r :
Aut(Zn ) → Aut(H) induced by the fact that H is characteristic in Zn . (Recall from Problem 9 of Exercises 3.7.14 that every subgroup of a cyclic group
is characteristic.)
Deduce that if α is injective, then Aut(G) ∼
= H r Aut(Zn ).
(f) Let β : Aut(Zn ) × (im η) → Aut(H) be as above. Compute the restriction of
β to im η.
(g) Show that if n is prime, then H = Zn .
(h) Give an example of α, n, and k where H is not all of Zn .
Chapter 5
Sylow Theory, Solvability, and
Classification
The main theorems of this chapter, e.g., Cauchy’s Theorem, the Sylow Theorems, and
the characterization of solvable and nilpotent groups, could have been given much earlier
in the book. In fact, a proof of Cauchy’s Theorem has appeared as Problem 24 in
Exercises 3.3.23. Indeed, these theorems depend on very little other than the theory of
G-sets (which it might be advisable to review at this time) and the Noether Isomorphism
Theorems.
Our main reason for delaying the presentation of these theorems until now was so
that we could develop enough theory to support their applications. Given our analysis
of abelian groups, extensions, and some automorphism groups, we will be able to use the
results in this chapter to classify the groups of most orders less than 64, and, hopefully,
to bring the study of finite groups into a useful perspective.
Lagrange’s Theorem tells us that if the group G is finite, then the order of every
subgroup must divide the order of G. But there may be divisors of |G| that do not
occur in this manner. For instance, Problem 2 of Exercises 4.2.18 shows that A4 has no
subgroup of order 6.
In this chapter, we shall give some existence theorems for subgroups of particular
orders in a given finite group. The first and most basic such result is Cauchy’s Theorem,
which says that if G is a finite group whose order is divisible by a prime p, then G
has an element (hence a subgroup) of order p. We give this in Section 5.1, along with
applications to classification.
In Section 5.2, we study finite p-groups, which, because of Cauchy’s Theorem, are
precisely the finite groups whose order is a power of the prime p. By a proof similar to
the one given for Cauchy’s Theorem, we show that any p-group has a nontrivial center.
We give some applications, including most of the pieces of a classification of the groups
of order p3 for p odd. The last step will be given via the Sylow Theorems in Section 5.3.
Section 5.3 is devoted to the Sylow Theorems, which study the p-subgroups of a finite
group. Extending the result of Cauchy’s Theorem, the Sylow Theorems show that if p
is prime and if |G| = pj m with (p, m) = 1, then G must have a subgroup of order pj .
Such a subgroup is the highest order p-group that could occur as a subgroup of G, and
is known as a p-Sylow subgroup of G (though that is not the initial definition we shall
give for a p-Sylow subgroup).
134
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
135
The Sylow Theorems are actually a good bit stronger than a simple existence proof
for subgroups whose order is the highest power of p dividing the order of G. They also
show that any two p-Sylow subgroups of G are conjugate and that the number of p-Sylow
subgroups of G is congruent to 1 mod p. This leads to some useful theoretical results
regarding p-Sylow subgroups which are normal, as well as to some valuable counting
arguments for the classification of groups of particular orders. In particular, Sylow
theory can be useful for finding normal subgroups in a particular group.
The Sylow Theorems give a useful point of view for considering finite groups. Finite
p-groups are more tractable than general finite groups, and a useful first step in understanding a given group is to understand its Sylow subgroups and their normalizers. For
many low order groups, an understanding of the entire structure of the group will follow
from this analysis.
Given the Sylow Theorems and the material on semidirect products, extensions, and
the automorphisms of cyclic groups, we are able to classify the groups of many of the
orders less than 64. We give numerous classifications in the text and exercises for Section 5.3.
Section 5.4 studies the commutator subgroup and determines the largest abelian factor group that a group can have. Then, in Section 5.5, and using the commutator
subgroups, we study solvable groups. Solvable groups are a generalization of abelian
groups. In essence, in the finite case, they are the groups with enough normal subgroups
to permit good inductions on order. One characterization we shall give of the finite solvable groups is that they are the finite groups whose composition series have no nonabelian
simple groups in the list of successive simple subquotients.
As expected, the induction arguments available for solvable groups provide for stronger
results than we get for arbitrary groups. In Section 5.6, we prove Hall’s Theorem, which
shows that if G is a finite solvable group and if |G| = mn with (m, n) = 1, then G has a
subgroup of order m. This fact turns out to characterize the solvable groups, though we
won’t give the converse here.
The nilpotent groups form a class of groups that sits between the abelian groups
and the solvable groups. They are defined by postulating properties that always hold in
p-groups, as we showed in Section 5.2. By the end of Section 5.7, we show that the finite
nilpotent groups are nothing other than direct products of finite p-groups over varying
primes p. In the process we will learn some more about the p-groups themselves. We
shall also see that the full converse of Lagrange’s Theorem holds for nilpotent groups: A
nilpotent group G has subgroups of every order that divides the order of G.
We close the chapter with a discussion of matrix groups over commutative rings.
Here, we assume an understanding of some of the material in Chapters 7 and 10. The
reader who has not yet studied this material should defer reading Section 5.8 until having
done so.
One of the reasons for studying matrix groups is that the automorphism group of a
direct product, Zkn , of k copies of Zn is isomorphic to Glk (Zn ), so by studying the latter,
we can obtain classification results for groups containing normal subgroups isomorphic
to Zkn . This situation arises for some low order groups.
We also calculate the order of Gln (Fq ), where Fq is the finite field with q = pr
elements with p prime. We describe some important subquotient groups, PSln (Fq ) of
Gln (Fq ). Most of these projective special linear groups turn out to be simple, but we
shall not show it here.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
5.1
136
Cauchy’s Theorem
We now prove Cauchy’s Theorem. An alternative proof was given as Problem 24 of
Exercises 3.3.23.
Theorem 5.1.1. (Cauchy’s Theorem) Let G be a finite group and let p be a prime
dividing the order of G. Then G has an element of order p.
Proof Note that we’ve already verified the theorem in the case that G is abelian:
Cauchy’s Theorem for abelian groups was given as Proposition 4.4.12. We argue by
induction on the order of G. The induction starts with |G| = p, in which case G ∼
= Zp ,
and the result is known to be true.
Recall that the nontrivial conjugacy classes in G are those with more than one element
and that the conjugacy class of x is nontrivial if and only if x ∈ Z(G). Recall also that a
set of class representatives for the nontrivial conjugacy classes in G is a set X ⊂ G such
that
1. X ∩ Z(G) = ∅.
2. If y ∈ G is not in Z(G), then y is conjugate to some x ∈ X.
3. If x, x ∈ X with x = x , then x and x are not conjugate in G.
Then the class formula (Corollary 3.6.18), shows that if X ⊂ G is a set of class
representatives for the nontrivial conjugacy classes in G, then
[G : CG (x)],
|G| = |Z(G)| +
x∈X
where CG (x) is the centralizer of x in G.
As stated, we already know Cauchy’s Theorem to be true for abelian groups. Thus,
if p divides the order of Z(G), then Z(G) has an element of order p, and hence G does
also.
Thus, we may assume that p does not divide the order of Z(G). Since p does divide
the order of G, the class formula tells us that there must be some x ∈ Z(G) such that
p does not divide [G : CG (x)]. Since |G| = [G : CG (x)] · |CG (x)|, this says that p must
divide the order of CG (x).
Since x ∈ Z(G), [G : CG (x)], which is equal to the number of conjugates of x in G,
cannot equal 1. Thus, |CG (x)| < |G|, so the induction hypothesis provides an element of
order p in CG (x), and hence also in G.
As a sample application, we classify the groups of order 6.
Corollary 5.1.2. Let G be a group of order 6. Then G is isomorphic to either Z6 or
D6 .
Proof By Cauchy’s Theorem, G has an element, b, of order 3. But then b has index 2
in G, and hence b G by Proposition 4.1.12 (or by the simpler argument of Problem 12
of Exercises 3.7.14). Let π : G → G/b be the canonical map.
A second application of Cauchy’s Theorem given an element a ∈ G of order 2. But
then a ∈ b, and hence π(a) = a is not the identity element of G/b. Thus, a generates
G/b. Since a and a both have order 2, setting s(a) = a gives a section of π.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
137
∼ Z× = {±1}, there are
Thus, G is a split extension of Z3 by Z2 . Since Aut(Z3 ) =
3
exactly two homomorphisms from Z2 to Aut(Z3 ): the trivial homomorphism and the
homomorphism that sends the generator of Z2 to −1.
The semidirect product given by the trivial homomorphism is Z3 × Z2 , which is Z6 ,
because 2 and 3 are relatively prime (e.g., by Problem 4 of Exercises 2.10.12). The
semidirect product induced by the nontrivial homomorphism from Z2 to Aut(Z3 ) was
shown to be isomorphic to D6 in the first example of Examples 4.6.6.
Exercises 5.1.3.
1. Show that any group of order 10 is isomorphic to either Z10 or D10 .
2. Assuming that Z×
p is cyclic for any odd prime p (as shown in Corollary 7.3.15),
show that any group of order 2p is isomorphic to either Z2p or D2p .
3. Recall that Cauchy’s Theorem for abelian groups was available prior to this chapter.
Give a proof of the preceding problem that doesn’t use Cauchy’s Theorem for
nonabelian groups. (Hint: If G is a group of order 2p, then all of its elements must
have order 1, 2, p, or 2p. If all nonidentity elements have order 2, then G must be
abelian, and hence Cauchy’s Theorem for abelian groups gives a contradiction. So
G must have an element, b, of order p. Now study the canonical map G → G/b,
and show that G must have an element of order 2.)
5.2
p-Groups
Recall that a p-group is a group in which the order of every element is a power of p.
Lagrange’s Theorem shows that any group whose order is a power of p must be a pgroup. But by Cauchy’s Theorem, if |G| has a prime divisor q = p, then it must have an
element of order q, and hence it is not a p-group. We obtain the following corollary.
Corollary 5.2.1. A finite group is a p-group if and only if its order is a power of p.
The proof used above for Cauchy’s Theorem may be used, with suitable modifications,
to prove some other useful facts. For instance:
Proposition 5.2.2. A finite p-group has a nontrivial center (i.e., Z(G) = e).
Proof Let X be a set of class representatives for the nontrivial conjugacy classes in G
and consider the class formula:
|G| = |Z(G)| +
[G : CG (x)].
x∈X
If x ∈ Z(G), then [G : CG (x)], which is equal to the number of conjugates of x in G,
cannot equal 1. Thus, since [G : CG (x)] divides |G|, we must have [G : CG (g)] = pr for
some r > 0, and hence p divides [G : CG (g)].
Thus, p must divide every term in the class formula other than |Z(G)|, so it must
divide |Z(G)| as well. In particular, |Z(G)| = 1, and hence Z(G) = e.
We first give an immediate consequence of Proposition 5.2.2 to classification.
Corollary 5.2.3. Every group of order p2 is abelian.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
138
Proof We argue by contradiction. If G is not abelian, then Z(G) = G. But then if
|G| = p2 , we must have |Z(G)| = p. But conjugation acts trivially on Z(G), so that
Z(G) G. Thus, G/Z(G) has order p, and hence is cyclic. But now G must be abelian
by Lemma 4.7.15, which gives the desired contradiction.
We now consider some more general corollaries to Proposition 5.2.2.
Corollary 5.2.4. The only finite p-group which is simple is Zp .
Proof Let G be a finite simple p-group. The center of any group is a normal subgroup.
Since the center of G is nontrivial and G is simple, G must equal its center. Thus, G
is abelian. But the only abelian simple groups are the cyclic groups of prime order, by
Lemma 4.2.2.
Recall that a composition series for a group G is a sequence
e = H0 ⊂ H1 ⊂ · · · ⊂ H j = G
such that Hi−1 Hi and Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ j. By
Proposition 4.2.5, every nontrivial finite group has a composition series.
Since a quotient of a subgroup of a p-group is again a p-group, the simple groups
Hi /Hi−1 associated to the composition series for a nontrivial finite p-group must all be
Zp . But a direct proof of this fact will give us a stronger statement: We can construct a
composition series for a p-group G such that all the subgroups in the series are normal
in G itself, and not just normal in the next group in the series.
Corollary 5.2.5. Let G be a finite group of order pj , with p prime and j > 0. Then
there is a sequence
e = H0 ⊂ H1 ⊂ · · · ⊂ Hj = G,
with Hi G and Hi /Hi−1 ∼
= Zp for 1 ≤ i ≤ j.
Proof We argue by induction on |G|, with the result being trivial for |G| = p. By
Proposition 5.2.2, Z(G) = e, so Cauchy’s Theorem provides a subgroup H1 ⊂ Z(G) with
|H1 | = p. But any subgroup of Z(G) is normal in G, so we can form the factor group
G/H1 .
Since |G/H1 | < |G|, the induction hypothesis provides a sequence
e = K1 ⊂ · · · ⊂ Kj = G/H1 ,
such that Ki G/H1 and Ki /Ki−1 ∼
= Zp for 2 ≤ i ≤ j.
Let π : G → G/H1 be the canonical map and set Hi = π −1 (Ki ) for 2 ≤ i ≤ j. Since
the pre-image under a homomorphism of a normal subgroup is always normal, we have
Hi G for all i. Also, since Ki = Hi /H1 for i ≥ 1, the Third Noether Theorem shows
that Hi /Hi−1 ∼
= Ki /Ki−1 ∼
= Zp for 2 ≤ i ≤ j, and the result follows.
Since the subgroup Hi above has order pi , we see that the full converse of Lagrange’s
Theorem holds in p-groups.
Corollary 5.2.6. Let G be a finite p group. Then every divisor of |G| is the order of a
subgroup of G.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
139
We shall now begin to try to classify the groups of order p3 for primes p > 2. We
shall not yet succeed entirely, because not all the groups we must consider are p-groups.
We must also understand the automorphism groups of the groups of order p2 , which in
the case of Zp × Zp will require the Sylow theorems.
As in the case of p = 2, the abelian groups of order p3 are Zp3 , Zp2 × Zp , and
Zp × Zp × Zp . For the nonabelian groups of order p3 , the whole story will depend on
whether there exists an element of order p2 .
Proposition 5.2.7. Let G be a nonabelian group of order p3 where p is an odd prime.
Suppose that G contains an element of order p2 . Then G is isomorphic to the semidirect
1
product Zp2 α Zp , where α : Zp → Z×
p2 is induced by α(1) = p + 1.
Proof Let b ∈ G have order p2 . Then b has index p in G, so b G by Proposition
4.1.12. Thus, G is an extension of Zp2 by Zp , via f : G → Zp . Since G is not abelian,
2
Lemma 4.7.15 shows that the induced homomorphism α : Zp → Z×
p2 = Aut(Zp ) is
nontrivial.
We shall use strongly the fact that Z×
p2 is abelian. Because of this, the p-torsion
elements form a subgroup, which, according to Corollary 4.5.12, is K(p, 2), the subgroup
consisting of those units congruent to 1 mod p. Corollary 7.3.15 then shows that K(p, 2)
is a cyclic group of order p, generated by 1 + p.
×
Since K(p, 2) is the p-torsion subgroup of Z×
p2 , the image of α : Zp → Zp2 must lie in
K(p, 2). Since α is nontrivial and Zp is simple, α must be an isomorphism from Zp to
K(p, 2). But then p + 1 = α(x) for some generator x of Zp .
But we may change our identification of Zp to make x the canonical generator of Zp .
Thus, we may assume that α carries the canonical generator of Zp to 1 + p. Thus, it
suffices to show that f is a split extension, as then G ∼
= Zp2 α Zp , with α as claimed.
Let us continue to write x for our preferred generator of Zp , so that α(x) = p + 1.
Since α is the action induced by the extension f : G → Zp , Lemma 4.7.9 shows that
ghg −1 = h1+p for any h ∈ ker f and any g with f (g) = x. To show that f splits, it
suffices to find g ∈ G of order p such that f (g) = x: The splitting is then defined by
setting s(x) = g.
Let a ∈ G with f (a) = x. Then a cannot have order p3 , as G would then be cyclic.
Alternatively, if a has order p, then a itself gives the desired splitting of f . Thus, we
shall assume that a has order p2 .
For simplicity of notation, let m = p + 1, so that aha−1 = hm for any h ∈ ker f .
j
Moreover, since α(aj ) = mj , aj ha−j = hm for h ∈ ker f . Thus, as shown in Problem 7
of Exercises 4.6.13, an easy induction on k shows that
k−1
(ha)k = h1+m+···+m
ak
for k ≥ 1. Thus, taking k = p, Problem 8 of Exercises 4.6.13 gives
(ha)p = h
mp −1
m−1
ap .
Recall that m = p + 1. Then m − 1 = p, while the proof of Proposition 4.5.18 shows
that mp is congruent to 1 + p2 mod p3 . But then (mp − 1)/(m − 1) is congruent to p
mod p2 , so that (ha)p = hp ap .
Now ap has order p and lies in b = ker f . Thus, ap = bjp for some j relatively prime
to p. But then if we take h = b−j , then the element ha has order p, and gives the desired
splitting for f .
1 Thus,
G is the group of Example 6 of Examples 4.6.6.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
140
Thus, there is no analogue at odd primes of Q8 : Every nonabelian extension of Zp2
by Zp is split when p > 2.
Lemma 5.2.8. Let p be an odd prime and let G be a nonabelian group of order p3 that
does not contain an element of order p2 . Then G is a split extension of Zp × Zp by Zp .
Proof By Corollary 5.2.5, G has a normal subgroup H of order p2 . Since groups of
order p2 are abelian, and since G contains no element of order p2 , H must be isomorphic
to Zp × Zp . Since H has index p, G is an extension of H by Zp . But all elements of G
have exponent p, so any lift of the generator of G/H to G provides a splitting for this
extension.
We have constructed an example of a nonabelian group of order p3 in which every
element has exponent p, in Corollary 4.6.8. We would like to show that every nonabelian
split extension of Zp × Zp by Zp is isomorphic to this one. To show this we first need to
understand the automorphism group of Zp × Zp . We will then need to understand the
conjugacy classes of homomorphisms from Zp into this group of automorphisms. But
the automorphism group in question is nonabelian, so that a complete picture of the
conjugacy classes of its subgroups of order p will depend on the Sylow Theorems of Section 5.3. We will content ourselves here with a calculation of the order of Aut(Zp × Zp ).
Together with the Sylow theorems, this will be sufficient to complete our classification.
Given a basic knowledge of linear algebra, the following result becomes a little more
transparent, and may be extended to a calculation of the order of the automorphism
group of a product of n copies of Zp .
Lemma 5.2.9. The automorphism group of Zp × Zp has order (p2 − 1)(p2 − p). In
particular, this group has order jp, where (j, p) = 1.
Proof Write Zp × Zp = {ai bj | 0 ≤ i, j ≤ p}. Of course, a and b commute and each
has order p. Since every element of Zp × Zp has exponent p, the homomorphisms, f ,
from Zp × Zp to itself are in one-to-one correspondence with all possible choices of the
elements f (a) and f (b). Since there are p2 choices for f (a) and p2 choices for f (b), this
gives p2 · p2 = p4 different homomorphisms from Zp × Zp to itself. We’d like to know
which of these are isomorphisms.
Let f : Zp × Zp → Zp × Zp be a homomorphism. Then the image of f has order 1, p
or p2 . It is an isomorphism if and only if the image has order p2 . So one way to argue
at this point is to count up the homomorphisms whose images have order 1 or p, and
subtract this number from p4 .
Since Zp × Zp has exponent p, every e = x ∈ Zp × Zp is contained in exactly one
subgroup of order p. That subgroup is x. Thus, the number of subgroups of order p
is the number of nonidentity elements in Zp × Zp divided by the number of nonidentity
elements of Zp : we have (p2 − 1)/(p − 1) = p + 1, subgroups of order p in Zp × Zp .
But if H ⊂ Zp × Zp has order p, then the homomorphisms from Zp × Zp to H
are in one-to-one correspondence with the choices of pairs of elements of H, so there
are p2 different homomorphisms from Zp × Zp to H. Only one of these is the trivial
homomorphism, and hence there are p2 − 1 homomorphisms from Zp × Zp onto H.
Multiplying this by the number of subgroups of order p, we see that there are p3 +
2
p − p − 1 homomorphisms from Zp × Zp to itself with an image of order p. Since the
trivial homomorphism is the only homomorphism whose image has order 1, this gives
exactly p3 + p2 − p homomorphisms from Zp × Zp to itself that are not automorphisms.
But then there are exactly p4 − p3 − p2 + p automorphisms of Zp × Zp . But this is easily
seen to factor as (p2 − 1)(p2 − p), as desired.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
141
A perhaps simpler argument is as follows. Choose f (a). Any nonidentity choice will
do, so there are p2 − 1 choices for f (a). Now consider the choice of f (b). As long as f (b)
is not in f (a), f will be an automorphism. Thus, there are p2 − p possible choices for
f (b) once f (a) has been chosen. In particular, there are (p2 − 1)(p2 − p) ways to choose
an automorphism f .
Exercises 5.2.10.
1. Show that every nonabelian group of order p3 is a central extension of Zp by Zp ×Zp .
2. Construct a group of order p4 whose center is isomorphic to Zp2 .
3. Find a group of order 16 whose center has order 2.
4. Construct a group of order 81 whose center has order 3.
5. Let p be an odd prime. Construct a group of order p4 whose center has order p.
6. Show that every group of order pn may be generated by n elements.
5.3
Sylow Subgroups
Cauchy’s Theorem shows that if p divides the order of a finite group G, then G has a
subgroup of order p. But what if |G| is divisible by p2 ? Can we then find a subgroup of
order p2 ?
We cannot argue here by Cauchy’s Theorem and induction, as there may not be a
normal subgroup of order p. (E.g., the group G may be simple.) So we need a new
approach to the question.
Note that there are examples of groups G and numbers n such that |G| is divisible
by n, but G has no subgroup of order n. For instance, we’ve seen in Problem 2 of
Exercises 4.2.18 that A4 has no subgroup of order 6.
Definitions 5.3.1. Let G be a finite group and let p be a prime. A p-subgroup of G is
a subgroup which is a p-group. A p-Sylow subgroup P of G is a maximal p-subgroup of
G, i.e., P is a p-subgroup of G, and any p-subgroup of G containing P must be equal to
P.
Examples 5.3.2. In D6 , b has order 3. Since no higher power of 3 divides the order
of D6 , b cannot be properly contained in a 3-subgroup of D6 . Thus, b is a 3-Sylow
subgroup of D6 .
Similarly, a, ab, and ab2 are all 2-subgroups of D6 . Since 2 itself is the highest
2-power that divides |D6 |, each of these is a 2-Sylow subgroup of D6 .
Lemma 5.3.3. Let H be a p-subgroup of a finite group G. Then H is contained in a
p-Sylow subgroup of G.
Proof Let |G| = pi m with p relatively prime to m, and let |H| = pj . If i = j, then H
is a p-Sylow subgroup of G by Lagrange’s Theorem. Otherwise, we argue by induction
on i − j.
From what we know at this point, H itself might be a p-Sylow subgroup of G even
if j < i. If this is the case, we are done. Otherwise, there is a p-subgroup K of G that
strictly contains H. But then the index of K in G is smaller than that of H, so by
induction, K is contained in a p-Sylow subgroup.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
142
Lemma 5.3.4. Let G be a finite group and let p be a prime dividing the order of G. Let
P be a p-Sylow subgroup of G. Then the index of P in its normalizer is relatively prime
to p.
Proof Recall that P NG (P ). Thus, the order of NG (P )/P is the index of P in NG (P ).
If this order is divisible by p, then Cauchy’s Theorem gives a subgroup H ⊂ NG (P )/P
with |H| = p. But then if π : NG (P ) → NG (P )/P is the canonical map, then π −1 H is a
p-subgroup of G that strictly contains P .
Corollary 5.3.5. Let G be a finite group and let p be a prime dividing the order of G.
Let P be a p-Sylow subgroup of G. Then any p-subgroup of NG (P ) is contained in P .
Proof Let K ⊂ NG (P ) be a p-group. Then K/(K ∩ P ) is isomorphic to a subgroup
of NG (P )/P . Since K is a p-group, so is K/(K ∩ P ). Since |NG (P )/P | is prime to p,
K/(K ∩ P ) = e, and hence K ⊂ P .
Lemma 5.3.6. Let P be a p-Sylow subgroup of the finite group G. Then any conjugate
of P is also a p-Sylow subgroup of G.
Proof Since conjugation gives an automorphism of G, any conjugate of a p-subgroup
of G is again a p-subgroup of G. Now for x ∈ G, suppose that xP x−1 not a p-Sylow
subgroup of G. Then xP x−1 must be properly contained in a p-subgroup P of G. But
then P is properly contained in x−1 P x, contradicting the statement that P is a p-Sylow
subgroup of G.
We now give yet another variant of the proof of Cauchy’s Theorem.
Theorem 5.3.7. (First Sylow Theorem) Let G be a finite group and let p be a prime
dividing the order of G. Then any two p-Sylow subgroups of G are conjugate, and the
number of p-Sylow subgroups of G is congruent to 1 mod p.
Proof Let P be a p-Sylow subgroup of G and let X ⊂ Sub(G) be the orbit of P under
the G-action given by conjugation. (Recall that Sub(G) is the set of subgroups of G.)
By Lemma 5.3.6, each element of X is a p-Sylow subgroup of G.
The G-set X may be considered as a P -set by restricting the action. For P ∈ X, the
isotropy subgroup of P under the action of P on X is given by
PP = {x ∈ P | xP x−1 = P }
= P ∩ NG (P ).
In particular, P is fixed by P if and only if P ⊂ NG (P ). By Corollary 5.3.5, this can
happen if and only if P = P , as both P and P are p-Sylow subgroups of G.
Thus, the action of P on X has exactly one trivial orbit: {P }. (Here, an orbit is
trivial if it has only one element.) The number of elements in the orbit of P ∈ X under
the P -action on X is [P : P ∩ NG (P )], which divides |P |. Thus, since P is a p-group, the
number of elements in any nontrivial orbit is divisible by p (hence congruent to 0 mod
p). In particular, if Y ⊂ X is a set of orbit representatives for the nontrivial orbits of the
action of P on X (i.e., the intersection of each nontrivial P -orbit in X with Y consists
of exactly one element), then the G-set Counting Formula (Corollary 3.3.15) gives
|X| = |{P }| +
[P : P ∩ NG (P )].
P ∈Y
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
143
Since each [P : P ∩ NG (P )] is congruent to 0 mod p and since |{P }| = 1, we see that
the number of elements in X is congruent to 1 mod p.
It remains to show that X contains every p-Sylow subgroup of G. Let P be an
arbitrary p-Sylow subgroup of G, and consider the action of P on X by conjugation. If
P is not in X, then every orbit of this P -action must be nontrivial. But that would
make the number of elements in X congruent to 0 mod p, rather than 1 mod p. So we
must have P ∈ X.
Theorem 5.3.8. (Second Sylow Theorem) Let G be a finite group and let p be a prime
dividing the order of G. Let P be a p-Sylow subgroup of G. Then the index of P in G is
prime to p. Thus, if |G| = pi m, with p and m relatively prime, then |P | = pi .
Moreover, the number of p-Sylow subgroups in G divides m in addition to being congruent to 1 mod p.
Proof The index, m, of P in G is the product of [NG (P ) : P ] and [G : NG (P )]. The
former is prime to p by Corollary 5.3.5. The latter is equal to the number of conjugates
of P , and hence is 1 mod p by the First Sylow Theorem.
Recall from Corollary 5.2.6 that if p is prime and if P is a finite group of order pi ,
then P has a subgroup of every order dividing the order of P . Applying this to a p-Sylow
subgroup of a group G, the Second Sylow Theorem produces the following corollary.
Corollary 5.3.9. Let G be a finite group and let q be any prime power that divides the
order of G. Then G has a subgroup of order q.
We now give the final Sylow Theorem.
Theorem 5.3.10. (Third Sylow Theorem) Let G be a finite group and let p be a prime
dividing the order of G. Let P be a p-Sylow subgroup of G. Then the normalizer of P is
its own normalizer: NG (NG (P )) = NG (P ).
Proof We claim that P is a characteristic subgroup of NG (P ). The point here is that
if x ∈ P and if f is any automorphism of NG (P ), then f (x) must be a p-torsion element.
But by Corollary 5.3.5, every p-torsion element of NG (P ) must be contained in P .
Now suppose that NG (P ) H. Then conjugation by an element of H gives an
automorphism of NG (P ). But this must preserve P , so P is normal in H. But then
H ⊂ NG (P ).
We now summarize some important facts implicit in the above discussion.
Corollary 5.3.11. Let G be a finite group and let p be a prime dividing the order of G.
Suppose there is a p-Sylow subgroup P of G which is normal in G. Then P is the only
p-Sylow subgroup of G, and P is the set of all p-torsion elements in G. Indeed, P is a
characteristic subgroup of G.
Conversely, if there is only one p-Sylow subgroup of G, then it is normal in G.
Proof The p-Sylow subgroups of G are all conjugate. But if P G, then it has no
conjugates other than itself. Moreover, since NG (P ) = G, P contains all the p-torsion
elements of G by Corollary 5.3.5. And since P is a p-group, all its elements are p-torsion,
so that P is, in fact, the set of p-torsion elements of G. But then P is characteristic,
since any automorphism preserves the p-torsion elements.
If there is only one p-Sylow subgroup, P , of G, then P has no conjugates other than
itself. Thus, [G : NG (P )] = 1, and hence P G.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
144
In many cases a normal p-Sylow subgroup is sufficient to unfold the entire structure
of a group. The point is that if P is a normal p-Sylow subgroup of G, then |P | and
|G/P | are relatively prime. Thus, the argument of Proposition 4.7.7 gives the following
corollary.
Corollary 5.3.12. Suppose that G has a normal p-Sylow subgroup, P , and that G has a
subgroup K of order [G : P ]. Then G is the semidirect product P α K, where α : K →
Aut(P ) is induced by conjugation in G.
According to the Schur–Zassenhaus Lemma, which we will not prove, such a subgroup
K always exists. However, if [G : P ] is a prime power, the existence of such a K is
automatic from the second Sylow Theorem.
Corollary 5.3.13. Let G be a group of order pr q s where p and q are distinct primes.
Suppose that the p-Sylow subgroup, P , of G is normal. Let Q be a q-Sylow subgroup of
G. Then G is isomorphic to the semidirect product P α Q, where α : Q → Aut(P ) is
induced by conjugation in G.
The Sylow Theorems are extremely powerful for classifying low-order finite groups.
We shall give a few of their applications now. First, let us complete the classification of
groups of order p3 .
Proposition 5.3.14. Let p be an odd prime and let G be a nonabelian group of order
p3 that contains no element of order p2 . Then G is isomorphic to the group constructed
in Corollary 4.6.8 for n = p.
Proof By Lemma 5.2.8, G is a semidirect product (Zp × Zp ) α Zp for some α : Zp →
Aut(Zp × Zp ). Moreover, α cannot be the trivial homomorphism, as then G would be
abelian by Lemma 4.7.15. But by Lemma 5.2.9, the p-Sylow subgroups of Aut(Zp × Zp )
have order p, so α must be an isomorphism onto one of these p-Sylow subgroups.
But the p-Sylow subgroups are all conjugate, and hence an isomorphism onto one
of them is conjugate as a homomorphism into Aut(Zp × Zp ) to an isomorphism onto
any of the others. Moreover, if α and α are isomorphisms onto the same p-Sylow
subgroup of Aut(Zp × Zp ), then because they are isomorphisms, we have α = α ◦ g for
an automorphism g of Zp . Thus, any two nonabelian semidirect products (Zp ×Zp )α Zp
and (Zp × Zp ) α Zp are isomorphic by Proposition 4.6.11.
Let us return to the theme of discovering normal Sylow subgroups and using them to
split the group as a semidirect product.
Corollary 5.3.15. Let G be a group of order pr m, where m < p, with p prime. Then
the p-Sylow subgroup of G is normal.
Proof The number of p-Sylow subgroups of G is congruent to 1 mod p and divides m.
Since m < p, there must be only one p-Sylow subgroup, which is thus normal.
When used in conjunction with Corollary 5.3.13, we can get some concrete results:
Corollary 5.3.16. Let G be a group of order pr q s , where p and q are distinct primes.
Suppose that q s < p. Then G is a semidirect product P α Q for some α : Q → Aut(P ),
where P and Q are the p-Sylow and q-Sylow subgroups of G, respectively.
This becomes more powerful when used in conjunction with what we know about
automorphisms:
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
145
Corollary 5.3.17. Every group of order 15 is isomorphic to Z15 .
Proof Let |G| = 15. By Corollary 5.3.16, G is a semidirect product Z5 α Z3 for some
×
homomorphism α : Z3 → Z×
5 . But Z5 has order 4, which is relatively prime to 3, and
hence α must be the trivial homomorphism. But then G ∼
= Z5 × Z3 , which is isomorphic
to Z15 .
Now let us consider a deeper sort of application of the Sylow Theorems. Sometimes,
under the assumption that a given Sylow subgroup is nonnormal, we can use the count
of the number of its conjugates to find the number of elements of p-power order. This
will sometimes permit us to find other subgroups that are normal.
Corollary 5.3.18. Let G be a group of order 12 whose 3-Sylow subgroups are not normal. Then G is isomorphic to A4 .
Proof We begin by showing that the 2-Sylow subgroup of G is normal. Let P be a
3-Sylow subgroup of G. Then the number of conjugates of P is 1 mod 3 and divides 4.
Thus, P has either 1 or 4 conjugates. But we’ve assumed that P is nonnormal, so it
must have 4 conjugates.
Note that if P and P are distinct 3-Sylow subgroups, then P ∩ P = e, since P and
P both have order 3. But since each 3-Sylow subgroup has exactly two elements of order
3, there must be 8 elements of order 3 in G.
The Second Sylow Theorem, on the other hand, shows that G has a subgroup, H,
of order 4. Since there are 8 elements of order 3 in G, every element outside of H must
have order 3. Since the elements of H are all 2-torsion, none of their conjugates may lie
outside of H, and hence H must be normal.
By Corollary 5.3.16, G ∼
= H α Z3 for some α : Z3 → Aut(H). Since there are no
nontrivial homomorphisms of Z3 into Z×
4 = {±1} and since G is nonabelian, H must be
isomorphic to Z2 × Z2 .
As shown in either Problem 14 of Exercises 4.6.13 or Problem 3 of Exercises 4.7.27,
the automorphism group of Z2 × Z2 is isomorphic to S3 . It’s easy to see that there are
only two nontrivial homomorphisms from Z3 into S3 ∼
= D6 , and that one is obtained
from the other by precomposing with an automorphism of Z3 . Thus, Proposition 4.6.11
shows that there is exactly one isomorphism class of semidirect products (Z2 × Z2 ) α Z3
which is nonabelian.
As shown in Problem 1 of Exercises 4.7.27, A4 is a semidirect product (Z2 ×Z2 )α Z3 ,
and the result follows.
Thus, the classification of groups of order 12 depends only on classifying the split
extensions of Z3 by the groups of order 4. We shall leave this to the exercises.
Here is a slightly more complicated application of the Sylow Theorems.
Lemma 5.3.19. Let G be a group of order 30. Then the 5-Sylow subgroup of G is
normal.
Proof We argue by contradiction. Let P5 be a 5-Sylow subgroup of G. Then the
number of conjugates of P5 is congruent to 1 mod 5 and divides 6. Thus, there must be
six conjugates of P5 . Since the number of conjugates is the index of the normalizer, we
see that NG (P5 ) = P5 .
Since the 5-Sylow subgroups of G have order 5, any two of them intersect in the
identity element only. Thus, there are 6 · 4 = 24 elements in G of order 5. This leaves
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
146
6 elements whose order does not equal 5. We claim now that the 3-Sylow subgroup, P3 ,
must be normal in G.
The number of conjugates of P3 is congruent to 1 mod 3 and divides 10. Thus, if
P3 is not normal, it must have 10 conjugates. But this would give 20 elements of order
3, when there cannot be more than 6 elements of order unequal to 5, so that P3 must
indeed be normal.
But then P5 normalizes P3 , and hence P5 P3 is a subgroup of G. Moreover, the Second
Noether Theorem gives
(P5 P3 )/P3 ∼
= P5 /(P5 ∩ P3 ).
But since |P5 | and |P3 | are relatively prime, P5 ∩ P3 = e, and hence P5 P3 must have
order 15.
Thus, P5 P3 ∼
= Z15 , by Corollary 5.3.17. But then P5 P3 normalizes P5 , which contradicts the earlier statement that NG (P5 ) has order 5.
Corollary 5.3.20. Every group of order 30 is a semidirect product Z15 α Z2 for some
α : Z2 → Z×
15 .
Proof Since the 5-Sylow subgroup P5 is normal, it is normalized by any 3-Sylow subgroup P3 , and hence the product P3 P5 is a subgroup of order 15, and hence index 2.
Since index 2 subgroups are normal, G is an extension of a group of order 15 by Z2 . By
Cauchy’s Theorem, this extension must be split. Since there is only one group of order
15, the result follows.
We shall classify these semidirect products in Exercises 5.3.22.
Often in mathematics the statement of a theorem is insufficient, by itself, to derive
all the applications of its method of proof. This is, in general, a good thing. It means
that the methods have enough substance to be applied in more varied settings. (The
alternative would be to continually learn or write proofs that one never needed to see
again, which, in the end, would become boring.) Here is an example of a proof based on
the method of proof of the Sylow Theorems.
Proposition 5.3.21. Let G be a group of order 24 that is not isomorphic to S4 . Then
one of its Sylow subgroups is normal.
Proof Suppose that the 3-Sylow subgroups are not normal. The number of 3-Sylow
subgroups is 1 mod 3 and divides 8. Thus, if there is more than one 3-Sylow subgroup,
there must be four of them.
Let X be the set of 3-Sylow subgroups of G. Then G acts on X by conjugation, so we
get a homomorphism f : G → S(X) ∼
= S4 . As we’ve seen in the discussion on G-sets, the
kernel of f is the intersection of the isotropy subgroups of the elements of X. Moreover,
since the action is that given by conjugation, the isotropy subgroup of H ∈ X is NG (H).
Thus,
ker f =
NG (H).
H∈X
For H ∈ X, the index of NG (H) is 4, the number of conjugates of H. Thus, the order
of NG (H) is 6. Suppose that K is a different element of X. We claim that the order of
NG (H) ∩ NG (K) divides 2.
To see this, note that the order of NG (H) ∩ NG (K) cannot be divisible by 3. This is
because any p-group contained in the normalizer of a p-Sylow subgroup must be contained
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
147
in the p-Sylow subgroup itself (Corollary 5.3.5). Since the 3-Sylow subgroups have prime
order here, they cannot intersect unless they are equal. But if the order of NG (H) ∩
NG (K) divides 6 and is not divisible by 3, it must divide 2.
In consequence, we see that the order of the kernel of f divides 2. If the kernel has
order 1, then f is an isomorphism, since G and S4 have the same number of elements.
Thus, we shall assume that ker f has order 2. In this case, the image of f has order
12. But by Problem 2 of Exercises 4.2.18, A4 is the only subgroup of S4 of order 12, so
we must have im f = A4 .
By Problem 1 of Exercises 4.2.18, the 2-Sylow subgroup, P2 , of A4 is normal. But
since ker f has order 2, f −1 P2 has order 8, and must be a 2-Sylow subgroup of G. As
the pre-image of a normal subgroup, it must be normal, and we’re done.
The results we have at this point are sufficient to classify the groups of most orders
less than 64. Almost all of them are semidirect products. We shall treat many examples
in the exercises below.
But for some of the orders less than 64, we must understand more about the subnormal
series of p-groups. We shall develop the material to do this in the next three sections.
Then, we must make a deeper analysis of some automorphism groups. For this, we will
need some information from our analysis of matrix groups over finite fields below.
Exercises 5.3.22.
1. Show that the 2-Sylow subgroup of S4 is isomorphic to D8 . Is it normal in S4 ?
2. What is the 2-Sylow subgroup of S6 ?
3. What is the 2-Sylow subgroup of S8 ?
4. Let H be any group and write H n for the product H × · · · × H of n copies of H.
Recall from Problem 21 of Exercises 3.6.24 that Sn acts on H n through automorphisms via σ · (h1 , . . . , hn ) = (hσ−1 (1) , . . . , hσ−1 (n) ). For any subgroup K ⊂ Sn we
get an induced semidirect product H n α K, called the wreath product of H and
K, and denoted H K.
Write Pr for the 2-Sylow subgroup of S2r .
(a) Show that Pr ∼
= Pr−1 S2 for r ≥ 2.
(b) Suppose that 2r−1 < n < 2r . Show that the 2-Sylow subgroup of Sn is
isomorphic to the product of Pr−1 with the 2-Sylow subgroup of Sn−2r−1 .
(c) Express the 2-Sylow subgroup of Sn in terms of the binary expansion of n.
5. Prove the analogue of the preceding problem for primes p > 2.
6. Let p and q be primes, with q < p. Assume, via Corollary 7.3.15, that Z×
p is cyclic.
Show that there are at most two groups of order pq. When are there two groups,
and when only one?
7. Let p and q be primes. Show that every group of order p2 q has a normal Sylow
subgroup.
8. Give a simpler, direct argument for Corollary 5.3.18 by studying the homomorphism
Λ : G → S(G/P3 ) obtained via Proposition 3.3.16 from the standard action of G
on the left cosets of a 3-Sylow subgroup of G.
9. Classify the groups of order 12. How does Q12 fit into this classification?
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
148
10. Classify the groups of order 20.
11. Classify the groups of order 24 whose 2-Sylow subgroup is either cyclic or dihedral.
12. Classify the groups of order 30.
13. We show here that any group G of order 36 without a normal 3-Sylow subgroup
must have a normal 2-Sylow subgroup. Thus, assume that G has a non-normal
3-Sylow subgroup P3 . Then the action of G on the set of left cosets G/P3 induces
a homomorphism Λ : G → S(G/P3 ) ∼
= S4 .
(a) Show, using Problem 31 of Exercises 3.7.14, that the kernel of Λ must have
order 3.
(b) Deduce from Problem 2 of Exercises 4.2.18 that the image of Λ is A4 . Write
V ⊂ A4 for the 2-Sylow subgroup of A4 and let H = Λ−1 V ⊂ G.
(c) Show that H, as an extension of Z3 by Z2 × Z2 , must be isomorphic to either
D12 or to Z3 × Z2 × Z2 .
(d) Show that every 2-Sylow subgroup of G must be contained in H. Deduce that
if H ∼
= Z3 × Z2 × Z2 , then the 2-Sylow subgroup of G is normal.
(e) Show that if H ∼
= D12 , then G has exactly three 2-Sylow subgroups. Deduce
that if P2 is a 2-Sylow subgroup of G, then |NG (P2 )| = 12. Deduce that Λ
induces an isomorphism from NG (P2 ) to A4 .
(f) Show that the only split extension of Z3 by A4 is the trivial extension Z3 ×A4 .
Deduce that H must have been isomorphic to Z3 × Z2 × Z2 .
14. Classify those groups of order 40 whose 2-Sylow subgroup is isomorphic to one of
the following:
(a) Z8
(b) D8
(c) Z4 × Z2 .
15. Show that any group of order 42 is a split extension of Z7 by a group of order 6.
Classify all such groups.
16. Alternatively, show that any group of order 42 is a split extension of a group of
order 21 by Z2 . Classify these, and relate this classification to the preceding one.
(To do this, you should calculate the automorphism group of the nonabelian group
of order 21. See Problem 19 of Exercises 4.7.27.)
17. Classify the groups of order 45.
18. Show that any group G of order 60 whose 5-Sylow subgroups are nonnormal must
be simple. (Hint: First calculate the orders of the normalizers of the 5-Sylow
and the 3-Sylow subgroups of G and then go through all possible orders for the
subgroups of G and show that no subgroup of that order can be normal. In some
cases, it is possible to show that no subgroup of the stated order can exist at all.
The classifications above of the groups whose order properly divides 60 will be
useful.)
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149
19. Show that any simple group G of order 60 is isomorphic to A5 . (Hint: You definitely
want to use the information derived in the solution of the previous exercise. The
key is finding a subgroup of index 5. It may help to calculate the order of the
centralizer of an element of order 2 in G.)
20. Classify the groups of order 60.
21. Show that every group G of order 90 has a normal 5-Sylow subgroup. Deduce that
G has a subgroup of index two.
22. Classify the groups of order 90 whose 3-Sylow subgroup is cyclic.
5.4
Commutator Subgroups and Abelianization
The commutator of two group elements, a and b, may be thought of as a measurement
of how far a and b are from commuting with each other. The commutators of all pairs of
elements in a group generate a useful normal subgroup called the commutator subgroup
of G, and denoted [G, G]. We shall see that G/[G, G] is the largest abelian quotient
group of G.
Definition 5.4.1. Let a and b be elements of the group G. The commutator of a and
b, written [a, b], is defined by
[a, b] = aba−1 b−1 .
We’ve already used the next lemma repeatedly.
Lemma 5.4.2. The elements a and b commute if and only if [a, b] = e.
The next lemma may be seen by inspection.
Lemma 5.4.3. For a, b ∈ G, [a, b]−1 = [b, a].
Definition 5.4.4. The commutator subgroup of G, written [G, G], is the subgroup generated by all the commutators of the elements of G.
Thus, an element of G is in [G, G] if and only if it is a product of commutators. Note,
however, that not every element of [G, G] is actually a commutator itself.
Lemma 5.4.5. Let G be any group. Then the commutator subgroup is a characteristic
subgroup of G, and hence is normal.
Proof Let f be an automorphism of G. Then clearly, f ([a, b]) = [f (a), f (b)]. Thus,
since f is a homomorphism, it must take a product of commutators to a product of
commutators.
Proposition 5.4.6. The quotient group G/[G, G] is abelian. In addition, any homomorphism f : G → H with H abelian factors through the canonical map π : G → G/[G, G].
Proof In G/[G, G], we have ab(a)−1 (b)−1 = aba−1 b−1 = e, so that each pair of elements
a and b of G/[G, G] commutes.
Let f : G → H with H abelian. To see that f factors through the canonical map
to G/[G, G], it suffices to show that the commutator subgroup [G, G] is contained in the
kernel of f . But f ([a, b]) = [f (a), f (b)], and the latter is the identity element of H, since
f (a) and f (b) commute. But then the generators of [G, G] are contained in the kernel of
f , and hence [G, G] must be also.
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Definition 5.4.7. We write Gab = G/[G, G], and call it the abelianization of G.
Corollary 5.4.8. Every subgroup of G that contains [G, G] is normal. Moreover, if
H G, then H contains [G, G] if and only if G/H is abelian.
Proof If [G, G] ⊂ H, then H = π −1 π(H), where π is the canonical map from G to
Gab . But every subgroup of the abelian group Gab is normal, and hence H is normal as
the pre-image of a normal subgroup.
If G/H is abelian, then the canonical map from G to G/H factors through Gab . But
this happens if and only if [G, G] ⊂ H. Conversely, if [G, G] ⊂ H, then G/H is a quotient
group of Gab by the third Noether Theorem, and hence is abelian.
Exercises 5.4.9.
† 1. Show that a group G is abelian, if and only if [G, G] = e.
† 2. Show that if G is nonabelian and simple, then [G, G] = G, and Gab = e.
3. Show that for nonabelian groups of order p3 , p prime, the commutator subgroup
and center coincide.
4. What is the commutator subgroup of D2n ? What is D2n ab ?
5. What is the commutator subgroup of Q4n ? What is Q4n ab ?
6. What is the commutator subgroup of A4 ? What is A4 ab ?
7. What is the commutator subgroup of S4 ? What is S4 ab ?
8. What is the commutator subgroup of Sn for n ≥ 5? What is Sn ab ?
‡ 9. We calculate the automorphism group of Q8 .
(a) Show that any automorphism f of Q8 induces an automorphism f of Q8 ab .
Show that passage from f to f induces a surjective homomorphism η : Aut(Q8 ) →
Aut(Q8 ab ).
(b) Show η is a split extension whose kernel is isomorphic to Z2 × Z2 .
(c) Deduce that Aut(Q8 ) is isomorphic to S4 .
10. Classify the groups of order 24 whose 2-Sylow subgroups are quaternionic. Show
that there is a unique such group whose 3-Sylow subgroups are not normal. We
shall meet this group again later as the matrix group Sl2 (Z3 ).
5.5
Solvable Groups
The most tractable groups to analyze are the abelian groups. Every subgroup is normal,
and hence lots of induction arguments are available. This is amply illustrated by the
classification of finite abelian groups given in Section 4.4.
At the other extreme, we have the nonabelian simple groups. With no nontrivial
proper normal subgroups available, it is difficult to get our hands on the structure of the
group.
The solvable groups give a generalization of abelian groups that share some of the nice
properties found in the abelian setting. In particular, knowing that a group is solvable
will open up a collection of tools for revealing its structure.
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151
Solvability has long been known to be an important idea. For instance, solvability
plays a key role in Galois’ famous proof that there does not exist a formula based on the
operations of arithmetic and n-th roots for finding solutions of polynomials of degree 5
or higher. We shall give an argument for this in Section 11.11.
Thus, it is of interest to determine whether a given group is solvable. Of particular
importance in this regard is the famous Feit–Thompson Theorem:
Theorem (Feit–Thompson Theorem) Every group of odd order is solvable.
The Feit–Thompson Theorem is considered to be the first major breakthrough along
the way to the classification of the finite simple groups. Its proof is quite lengthy and is
well beyond the scope of this book.2 But the fact that odd order groups are solvable is
an important one to keep in mind for intuition’s sake.
We shall need a more compact notation for commutator subgroups, as we shall be
discussing commutator subgroups of commutator subgroups, etc.
Definition 5.5.1. For i ≥ 0, define G(i) inductively by G(0) = G and G(i) = [G(i−1) , G(i−1) ]
for i ≥ 1. We call G(i) the i-th derived subgroup of G.
Thus, G(1) = [G, G]. Given this, let us make the following definition.
Definition 5.5.2. A group G is solvable if G(k) = e for some k ≥ 0.
The virtue of this definition is that it is easy to work with. We shall give an equivalent
definition below that may be more intuitive. In the meantime, we can give a quick
workup of some basic properties. The next lemma is immediate from Problems 1 and 2
of Exercises 5.4.9.
Lemma 5.5.3. Every abelian group is solvable. A simple group, on the other hand, is
solvable if and only if it is abelian.
Lemma 5.5.4. Let f : G → H be a surjective homomorphism. Then f (G(i) ) = H (i) for
all i ≥ 0.
Proof We shall show that f ([G, G]) = [H, H]. This then shows that f : [G, G] → [H, H]
is surjective, and the result follows by induction on k.
But since f ([a, b]) = [f (a), f (b)], f ([G, G]) ⊂ [H, H]. And since f is surjective, the
generators of [H, H] must lie in f ([G, G]).
Corollary 5.5.5. A quotient group of a solvable group is solvable.
Proof If G(k) = e, so is H (k) for any quotient group H of G.
Lemma 5.5.6. Let H be a subgroup of G. Then H (i) ⊂ G(i) for all i ≥ 0.
Proof Once again, by induction, it suffices to treat the case of i = 1. But since H ⊂ G,
if h, k ∈ H, then [h, k] ∈ [G, G].
Of course, H (i) need not equal G(i) ∩ H.
Corollary 5.5.7. A subgroup of a solvable group is solvable.
2 The proof takes up the entirety of a 255-page journal issue. See W. Feit and J. Thompson, Solvability
of groups of odd order, Pacific J. Math. 13 (1963), 775–1029.
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152
Recall that a subnormal series for G consists of a sequence of subgroups
e = Gk Gk−1 · · · G0 = G.
Here, each Gi+1 is normal in Gi , but need not be normal in G if i > 1. (Note that we’ve
reversed the order of the indices from that used previously.)
Definition 5.5.8. A subnormal series is abelian if the quotient groups Gi /Gi+1 are
abelian for 0 ≤ i < k.
Proposition 5.5.9. A group G is solvable if and only if it has an abelian subnormal
series.
Proof Suppose that G is solvable, with G(k) = e. Consider the sequence
e = G(k) G(k−1) · · · G(0) .
ab
Since G(i+1) = [G(i) , G(i) ], the quotient group Gi /Gi+1 is just (G(i) ) , which is certainly
abelian, so that the sequence above is an abelian subnormal series.
Conversely, given an abelian subnormal series
e = Gk Gk−1 · · · G0 = G,
we claim that G ⊂ Gi for all i, which then forces G(k) = e. Clearly, G(0) = G0 , so assume inductively that G(i−1) ⊂ Gi−1 . But then G(i) ⊂ [Gi−1 , Gi−1 ] by Lemma 5.5.6, so it
suffices to see that [Gi−1 , Gi−1 ] ⊂ Gi . But this follows immediately from Corollary 5.4.8,
since Gi−1 /Gi is abelian.
(i)
The above characterization of solvable groups can be quite useful.
Corollary 5.5.10. Let H be a normal subgroup of G. Then G is solvable if and only if
both H and G/H are solvable.
Proof Suppose that G is solvable. Then H is solvable by Corollary 5.5.7, and G/H is
solvable by Corollary 5.5.5.
Conversely, suppose that H and G/H are solvable. Let e = Hk · · · H0 = H
be an abelian subnormal series for H and let e = Kl · · · K0 = G/H be an abelian
subnormal series for G/H. Then if π : G → G/H is the canonical map, we have an
abelian subnormal series
e = Hk · · · H0 = π −1 Kl π −1 Kl−1 · · · π −1 K0 = G
for G.
Also, the following corollary is immediate from Corollary 5.2.5.
Corollary 5.5.11. Let p be a prime. Then any finite p-group is solvable.
Note that other than this last result, we have made no use of finiteness in the discussion of solvability. Recall that a composition series for a group G is a subnormal
series
e = Gk Gk−1 · · · G0 = G
where the subquotient groups Gi /Gi+1 of G are all nontrivial simple groups. By Proposition 4.2.5, every nontrivial finite group has a composition series. The Jordan–Hölder
Theorem shows that any two composition series for a group G have the same length, and
give the same collection of subquotient groups Gk−1 /Gk , . . . , G/G1 , though possibly in
a different order. However, for economy of means, we shall not use it in the following
proof.
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153
Proposition 5.5.12. A finite group is solvable if and only if it has a composition series
in which the simple subquotient groups are all cyclic of prime order. If G is solvable,
then every composition series for G has that form.
Proof A composition series whose simple subquotient groups are cyclic is certainly an
abelian subnormal series, so any group that has such a composition series is solvable by
Proposition 5.5.9.
Conversely, suppose that G has a composition series in which one of the quotients
Gi /Gi+1 is not cyclic of prime order. Since the only abelian simple groups have prime
order (Lemma 4.2.2), Gi /Gi+1 must be a nonabelian simple group. But if G were solvable,
the subgroup Gi would be also. And this would force the quotient Gi /Gi+1 to be solvable
as well, contradicting Lemma 5.5.3.
Exercises 5.5.13.
1. Show that Sn is not solvable if n ≥ 5.
2. Show that A5 is the only group of order ≤ 60 that is not solvable.
3. Show that the Feit–Thompson Theorem is equivalent to the statement that every
nonabelian simple group has even order.
5.6
Hall’s Theorem
There is a generalization of Sylow theory that works in solvable groups. It is due to
Philip Hall.
Definition 5.6.1. Let G be a finite group. A subgroup H ⊂ G is a Hall subgroup if |H|
and [G : H] are relatively prime.
Our purpose in this section is to prove the following theorem, which shows that a
finite solvable group G has Hall subgroups of all possible orders n dividing |G| such that
(n, |G|/n) = 1.
Theorem 5.6.2. (Hall’s Theorem) Let G be a finite solvable group, and let |G| = nk
with (n, k) = 1. Then G has a subgroup of order n. Moreover, any two subgroups of
order n are conjugate.
In fact, Hall also showed a converse to this. We shall state it here, but its proof is
beyond the scope of this book.
Theorem (The Hall Converse) Let G be a finite group with |G| = pr11 . . . prkk with
p1 , . . . , pk distinct primes. Suppose that G has a Hall subgroup of order |G|/pri i for each
i = 1, . . . , k. Then G is solvable.
Thus, the existence of Hall subgroups of every order n dividing G with (n, |G|/n) = 1
is equivalent to solvability.
There is an interesting special case of the Hall Converse. Suppose that |G| = pr q s
with p and q prime. Then a p-Sylow subgroup of G is a Hall subgroup of order |G|/q s
and a q-Sylow subgroup of G is a Hall subgroup of order |G|/pr . The Hall Converse
produces an immediate corollary.
Corollary (Burnside’s Theorem) Let G be a finite group whose order has only two
prime divisors. Then G is solvable.
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154
In fact, Burnside’s Theorem is one of the ingredients in the proof of the Hall Converse,
and, like the converse itself, is beyond the scope of this book. We shall now concentrate
on proving Hall’s Theorem.
Lemma 5.6.3. Let G be a finite group and let H be a Hall subgroup of G of order n.
Let K G with |K| = m. Then H ∩ K is a Hall subgroup of K of order (m, n) and
HK/K is a Hall subgroup of G/K of order n/(m, n).
Proof Write |G| = nk. Since H is a Hall subgroup, this gives (n, k) = 1. Write
m = n k with n dividing n and k dividing k. Thus, n = (m, n).
Of course, |H ∩ K| divides (m, n) = n , so the order of H/(H ∩ K) is divisible by
n/(m, n) = n/n .
The Noether Isomorphism Theorem gives H/(H ∩ K) ∼
= HK/K, so |H/(H ∩ K)|
divides |G/K| = (n/n ) · (k/k ). But also, |H/(H ∩ K)| divides |H| = n. Since n and k
are relatively prime, |H/(H ∩ K)| must divide n/n .
Thus, |H/(H ∩ K)| = n/(m, n), which forces |H ∩ K| = (m, n).
Recall that a “minimal subgroup” of G was defined to be a nontrivial subgroup
H ⊂ G such that H has no nontrivial proper subgroups. Thus, H is minimal among the
nontrivial subgroups of G. We define minimal normal subgroups by analogy.
Definition 5.6.4. We say that H is a minimal normal subgroup of G if e = H G and
if no nontrivial proper subgroup of H is normal in G.
If G is finite, then the existence of minimal normal subgroups in G is easy. Just
choose the nontrivial normal subgroup of the lowest order.
Lemma 5.6.5. Let G be a finite solvable group and let H be a minimal normal subgroup
of G. Then H is an abelian group of exponent p for some prime p.
Proof The commutator subgroup [H, H] is a characteristic subgroup of H. Since
H G, this gives [H, H] G by Corollary 3.7.9.
Thus, since H is a minimal normal subgroup of G, either [H, H] = H or [H, H] = e.
Suppose that [H, H] = H. Then the derived subgroups H (i) = H for all i > 0. But
solvable groups are by definition those for which H (i) = e for some i. Since H is nontrivial,
it thus cannot be solvable, contradicting the fact that a subgroup of a solvable group is
solvable (Corollary 5.5.7).
Thus, [H, H] = e, and hence H is abelian. Let p be a prime dividing |H| and let Hp be
the p-torsion subgroup of G. Then Hp is a characteristic subgroup of H (Lemma 4.4.19),
and hence Hp G. Since Hp is nontrivial by Cauchy’s Theorem, we see that H must
be an abelian p-group for some prime p. But then the elements of exponent p form a
nontrivial characteristic subgroup of H, so H must have exponent p as well.
We are now ready to prove Hall’s Theorem.
Proof of Theorem 5.6.2 We wish to show that for any divisor n of |G| such that
(n, |G|/n) = 1 there are Hall subgroups of G of order n and that any two such subgroups
are conjugate.
We argue by induction on |G|. The proof breaks up into two parts. First, we assume
that G contains a nontrivial normal subgroup, K, whose order is not divisible by k.
Write |K| = n k with n dividing n and k dividing k. The induction hypothesis
gives a Hall subgroup H/K of G/K of order n/n . Thus, H ⊂ G has order nk . Since
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
155
k < k, the induction hypothesis shows that there’s a Hall subgroup H ⊂ H of order n.
This gives the desired subgroup of G.
Suppose now that H and H are two subgroups of G of order n. By Lemma 5.6.3,
HK/K and H K/K are Hall subgroups of order n/n in G/K, and hence are conjugate
by the induction hypothesis. But if x conjugates HK/K to H K/K, then x conjugates
HK to H K. But then x conjugates H to a Hall subgroup of H K of order n. And
induction shows that xHx−1 is conjugate to H in H K.
Thus, we are left with the case where the order of every nontrivial normal subgroup
of G is divisible by k. Let P G be a minimal normal subgroup. Then Lemma 5.6.5,
shows that P is an abelian group of exponent p for a prime number, p. Since |P | = pr is
divisible by k and since (n, k) = 1, we must have k = pr . Thus, P is a p-Sylow subgroup
of G. Since P G, there is only one subgroup of G of order k = pr , and hence P is the
unique minimal normal subgroup of G.
The quotient group G/P is solvable, so it has a minimal normal subgroup Q/P ,
whose order is q s for some prime number q. Let Q be a q-Sylow subgroup of Q. Then
the canonical map π : G → G/P restricts to an isomorphism Q ∼
= Q/P (Corollary 5.3.12),
so Q = QP .
We claim that NG (Q) is the desired subgroup of G of order n. To show this, it suffices
to show that [G : NG (Q)] = pr .
To see this, first note that, since Q is normal in G, every conjugate of Q in G must
lie in Q. Thus, every conjugate of Q in G is a q-Sylow subgroup of Q, and is conjugate
to Q in Q by the First Sylow Theorem. So Q has the same number of conjugates in Q
as in G, and hence
[G : NG (Q)] = [Q : NQ (Q)].
We claim now that NQ (Q) = Q. Since [Q : Q] = pr , this will complete the proof that
|NG (Q)| = n.
Since P is the p-Sylow subgroup of Q and is normal there, it contains every p-torsion
element of Q. Since Q ⊂ NQ (Q), the only way the two could differ is if |NQ (Q)| is
divisible by p. Thus, it suffices to show that NQ (Q) ∩ P = e.
Note first that if x ∈ P normalizes Q and if y ∈ Q, then xyx−1 y −1 ∈ P ∩ Q = e.
Thus, x commutes with every element of Q. Since P is abelian and Q = QP , we see that
NQ (Q) ∩ P ⊂ Z(Q), so it suffices to show that Z(Q) = e.
Recall that Q/P G/P , and hence Q G. Since the center of a group is a
characteristic subgroup, we see that Z(Q) G. Suppose that Z(Q) is nontrivial. Since
P is the unique minimal normal subgroup of G, this implies that P ⊂ Z(Q). But then P
normalizes Q. Since Q = QP , we must have NQ (Q) = Q, so that Q Q. But Q is the
q-Sylow subgroup of Q, so this implies that Q is characteristic in Q and hence normal
in G, contradicting the assumption that G has no normal subgroups whose order is not
divisible by k. Thus, Z(Q) = e, and hence NG (Q) is a Hall subgroup of G of order n, as
desired.
It now suffices to show that if H is another subgroup of G of order n, then H is
conjugate to NG (Q). Since |H| = |G/P | is prime to |P |, the canonical map induces
an isomorphism H → G/P . Thus, H ∩ Q = Q maps isomorphically to Q/P under the
canonical map, and hence is a q-Sylow subgroup of Q. Thus Q and Q are conjugate in Q,
hence in G. But then NG (Q) and NG (Q ) are conjugate as well. Now Q = (H ∩ Q) H,
so H ⊂ NG (Q ). Since the two groups have the same order, the result follows.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
5.7
156
Nilpotent Groups
Nilpotent groups form a class of groups that lie between the solvable and the abelian
groups. They form a generalization of the properties found among the finite p-groups.
The fact that a finite p-group, P , has a nontrivial center is a very useful one, and it
demonstrates that finite p-groups are, in fact, better behaved than the general run of
solvable groups. And since P/Z(P ) is again a p-group, the center of P/Z(P ) must be
nontrivial as well. Thus, induction arguments are possible. This forms the basic idea
that we will generalize to define nilpotency.
In the end, we shall see that the finite nilpotent groups are nothing more than cartesian products of p-groups. But the abstractions we will develop along the way turn out
to be useful for our understanding of the p-groups, if nothing else.
Definition 5.7.1. Let G be a group. Define the subgroups Z i (G) inductively as follows. Z 0 (G) = e and Z 1 (G) = Z(G). Suppose given Z i−1 (G), which is assumed to
be normal in G, and let π : G → G/Z i−1 (G) be the canonical map. Then Z i (G) =
π −1 (Z(G/Z i−1 (G))).
Of course, as the pre-image of a normal subgroup of G/Z i−1 (G), Z i (G) G, and
the induction continues.
Definition 5.7.2. A group G is nilpotent if Z k (G) = G for some k ≥ 0.
Just so the abstraction doesn’t get away from us here, we prove the following lemma.
Lemma 5.7.3. Finite p-groups are nilpotent.
Proof Suppose that Z i−1 (G) = G. Then G/Z i−1 (G) is a nontrivial p-group, and
must have a nontrivial center. Thus, Z i (G) is strictly larger than Z i−1 (G). But then if
|G| = pj , we must have Z k (G) = G for some k ≤ j.
Since the quotient group Z i (G)/Z i−1 (G) is the center of G/Z i−1 (G), and hence
abelian, the sequence
e = Z 0 (G) · · · Z k (G) = G
is an abelian subnormal series when G is nilpotent.
Proposition 5.7.4. Nilpotent groups are solvable.
The next result gives a useful property of nilpotent groups, which we have not yet
verified for p-groups. (We dealt with a special case in Proposition 4.1.12 by a different,
but related, argument.)
Proposition 5.7.5. Let H be a subgroup of the nilpotent group G, with H = G. Then
H cannot be its own normalizer in G.
Proof Let H ⊂ G be its own normalizer. We shall show that H must equal G. Since
G is nilpotent, it suffices to show that Z i (G) ⊂ H for all i.
We show this by induction on i, with the case i = 0 immediate. Suppose that
Z i−1 (G) ⊂ H, and let π : G → G/Z i−1 (G) be the canonical map. Since Z i−1 (G) ⊂ H,
we have H = π −1 (π(H)). This is easily seen to imply that NG (H) = π −1 (NG/Z i−1 (G) (π(H))),
so π(H) must be its own normalizer in G/Z i−1 (G).
Since Z i (G) = π −1 (Z(G/Z i−1 (G))), it suffices to show that Z(G/Z i−1 (G)) ⊂ π(H).
In other words, it suffices to assume that i = 1, and show that Z(G) ⊂ H. But the
elements of Z(G) normalize every subgroup of G, so that Z(G) ⊂ NG (H) = H, and the
result follows.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
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Corollary 5.7.6. Every subgroup of a finite nilpotent group is part of a subnormal series.
Proof Let H be the subgroup and G the group. If H is not already normal in G, then
its normalizer at least strictly contains it, so that the index of the normalizer is strictly
smaller than the index of H. Thus, the result follows by induction on the index.
We can use this in our efforts to classify the low order groups.
Corollary 5.7.7. Let G be a nonabelian group of order 16 that does not contain an
element of order 8. Then G has an index 2 subgroup H such that H has an element of
order 4.
Proof If every element of G has order 2, then G is abelian by Problem 3 of Exercises 2.1.16. Thus, G must have an element x of order 4. If x is not normal in G,
then we may take H to be its normalizer. Otherwise, take H to be the pre-image of a
subgroup of order 2 in G/x.
Thus, we shall be able to classify the groups of order 16 without first understanding
the structure of Aut(Z2 × Z2 × Z2 ). This is useful, because Aut(Z2 × Z2 × Z2 ) turns
out to be a simple group of order 168.
Lemma 5.7.8. A product H × K is nilpotent if and only if both H and K are nilpotent.
Proof This will follow once we show that Z i (H × K) = Z i (H) × Z i (K) for all i ≥ 0.
We show this by induction on i, with the case i = 0 immediate.
Suppose Z i−1 (H × K) = Z i−1 (H) × Z i−1 (K). Then the product of canonical maps
π × π : H × K → H/Z i−1 (H) × K/Z i−1 (K)
has kernel Z i−1 (H × K). Thus, it suffices to show that
Z( H/Z i−1 (H) × K/Z i−1 (K) ) = Z(H/Z i−1 (H)) × Z(K/Z i−1 (K)).
But it is easy to see that for any groups H and K , Z(H × K ) = Z(H ) × Z(K ).
But now we can characterize finite nilpotent groups.
Proposition 5.7.9. A finite group is nilpotent if and only if it is the internal direct
product of its Sylow subgroups, meaning that if p1 , . . . , pk are the primes dividing |G|,
then the pi -Sylow subgroup, Gpi , of G is normal for all i, and G is the internal direct
product of Gp1 , . . . , Gpk .
Proof If G is isomorphic to Gp1 × · · · × Gpk , then G is nilpotent by Lemmas 5.7.3
and 5.7.8. Thus, we prove the converse. Note that for abelian groups, the converse is
given as Proposition 4.4.9.
Let G be nilpotent and let Gpi be a pi -Sylow subgroup of G. Then the third Sylow
Theorem shows that the normalizer of Gpi is its own normalizer in G. But since G is
nilpotent, Proposition 5.7.5 shows that the normalizer of Gpi must be G itself, so that
Gpi G. We shall show that this is sufficient to obtain the stated product formula.
If 1 ≤ i, j ≤ k and if i = j, let x ∈ Gpi and let y ∈ Gpj . Consider the commutator
[x, y] = xyx−1 y −1 . Since Gpj is normal in G, xyx−1 ∈ Gpj , and hence the commutator
[x, y] is also. But the normality of Gpi shows that yx−1 y −1 and hence also [x, y] are in
Gpi as well. But because the identity is the only element that has prime power order for
more than one prime, Gpi ∩ Gpj = e, and hence [x, y] = e.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
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Thus, the elements of Gpi commute with the elements of Gpj whenever i = j, and we
have a homomorphism μ : Gp1 × · · · × Gpk → G given by μ(g1 , . . . , gk ) = g1 · · · gk . We
claim that μ is an isomorphism. By the second Sylow Theorem, Gp1 × · · · × Gpk has the
same order as G, so it suffices to show that μ is injective.
Suppose that μ(g1 , . . . , gk ) = e. We wish to show that gi = e for all i. Assume by
contradiction that i is the smallest index for which gi = e. We cannot have i = k, as
then gi = e as well. Thus, gi−1 = gi+1 · · · gk . But the order of (e, . . . , e, gi+1 , · · · , gk ) in
Gp1 × · · · × Gpk is the least common multiple of the orders of gi+1 , . . . , gk (by Problem 2
of Exercises 2.10.12), and hence the order of its image, under μ is divisible only by the
primes pi+1 , . . . , pk . But the image in question happens to be gi−1 , which is a pi -torsion
element, and we have our desired contradiction.
We obtain a converse to Lagrange’s Theorem for nilpotent groups.
Corollary 5.7.10. Let G be a finite nilpotent group. Then every divisor of the order of
G is the order of a subgroup of G.
Proof Write |G| = pr11 . . . prkk where p1 , . . . , pk are distinct primes. Then any divisor of
|G| has the form m = ps11 . . . pskk with 0 ≤ si ≤ ri for i = 1, . . . , k.
Let Gpi be the pi -Sylow subgroup of G. Then there is a subgroup Hi of Gpi of
order psi i by Corollary 5.2.6. By Proposition 5.7.9, G has a subgroup isomorphic to
H1 × · · · × H k .
Exercises 5.7.11.
1. Give an example of a solvable group that’s not nilpotent.
2. Classify the groups of order 16 containing a copy of D8 .
3. Classify the groups of order 16 containing a copy of Q8 .
4. Classify the groups of order 16 containing a copy of Z4 × Z2 .
5. Which groups appear in more than one of the three preceding problems? Which of
them contain an element of order 8?
6. Classify the groups of order 24 whose 2-Sylow subgroups are isomorphic to Z4 ×Z2 .
5.8
Matrix Groups
Perhaps the most important missing piece at this point in being able to classify low
order groups is an understanding of the automorphism groups of the Sylow subgroups
that arise in them. So far, we know how to calculate the automorphisms only of cyclic
groups, and groups such as dihedral and quaternionic groups, whose automorphisms are
calculated in Exercises 4.7.27 and 5.4.9.
Here, we shall investigate the automorphisms of groups of the form Zn × · · · × Zn ,
with special attention paid to the case where n is prime. There are a number of groups
of order < 64 which have Sylow subgroups of this form.
Unlike the preceding material, we shall need to make use of some theory we have not
yet developed: basic ring theory and linear algebra. The reader who has not yet been
exposed to this material should come back to this section after reading the appropriate
chapters.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
159
Of course, for any ring A, the cartesian product A × · · · × A of n copies of A is the
underlying abelian group of the free A-module of rank n. Consonant with this, we shall
write
A × · · · × A = An .
n times
Linear algebra gives techniques of studying the A-module homomorphisms from An
to itself. Recall that if N is an A-module, then EndA (N ) denotes the set of A-module
homomorphisms from N to itself, while AutA (N ) denotes the group (under the operation
of composition of homomorphisms) of A-module isomorphisms from N to itself. Recall
also that an A-module homomorphism is an isomorphism if and only if it is bijective.
Since every A-module isomorphism is an isomorphism of abelian groups, forgetting the
A-module structure gives an injective group homomorphism
ϕ : AutA (N ) → Aut(N ).
For A = Zn , Corollary 7.7.16 shows that every group homomorphism between Zn modules is a Zn -module homomorphism. Thus, the next lemma is immediate.
Lemma 5.8.1. For all k ≥ 1, forgetting the Zn -module structure gives an isomorphism
∼
=
ϕ : AutZn (Zkn ) −→ Aut(Zkn ).
Recall that the group of invertible n × n matrices over a ring A is denoted by Gln (A).
Here, the group structure comes from matrix multiplication. Corollary 7.10.9 shows that
allowing matrices to act on the left on column vectors gives a group isomorphism from
Gln (A) to AutA (An ).3 Thus, for any A, allowing matrices to act on the left on column
vectors gives an injective ring homomorphism
ιn : Gln (A) → Aut(An ).
For A = Zn , Lemma 5.8.1 allows us to do even better.
Corollary 5.8.2. The standard left action of matrices on column vectors induces a group
isomorphism
ιk : Glk (Zn ) → Aut(Zkn ).
In particular, the study of matrix groups will give us calculations of automorphism
groups of groups that we care about. Recall that the transpose, M t , of an n × n matrix
M = (aij ) is the matrix whose ij-th entry is aji .
Lemma 5.8.3. For any commutative ring A, there is a group automorphism α : Gln (A) →
Gln (A) defined by α(M ) = (M t )−1 .
Proof It is easy to check that for any pair of matrices M1 , M2 ∈ Mn (A), we have
(M1 M2 )t = M2t M1t . This is enough to show that the transpose of an invertible matrix
is invertible, with (M t )−1 = (M −1 )t . Since inversion of matrices also reverses order, the
result follows.
3 Here,
if A is not commutative, then this applies only to the standard right module structure on An .
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
160
For finite rings A, the set of n × n matrices is finite, and hence so is Gln (A). (If A
is infinite, it is not hard to show that Gln (A) is infinite for n ≥ 2.) The next result will
assist in calculating the order of Gln (A) when A is finite.
Proposition 5.8.4. Let A be a commutative ring. Consider the standard action of
Gln (A) on columns, and let en be the n-th canonical basis vector. Then the isotropy
subgroup of en under this action is isomorphic to An−1 ιn−1 Gln−1 (A).
Proof The isotropy subgroup, Gln (A)en , of en consists of those invertible matrices
whose lastcolumn is en . Thus, we are concerned with the invertible matrices of the form
M 0
, where M is (n−1)×(n−1), X is a 1×(n−1) row matrix, 0 is the (n−1)×1
X 1
column matrix whose coordinates are all 0, and the 1 in the lower corner is the 1 × 1
matrix 1. Note that M must lie in Gln−1 (A), as otherwise the first n − 1 rows of the full
matrix would be linearly dependent.
if M is any element of Gln−1 (A) and X
Conversely,
is any 1 × (n − 1) row matrix,
M 0
M −1
0
is invertible, with inverse
, and hence lies in the isotropy
X 1
−XM −1 1
group Gln (A)en .
Thus, if α is the automorphism of Gln (A) replacing a matrix withthe inverseof its
M X
transpose, then H = α(Gln (A)en ) is the set of all matrices of the form
with
0 1
M ∈ Gln−1 (A) and X ∈ An−1 , where we identify An−1 with the set of (n − 1) × 1 column
matrices. Of course, Gln (A)en is isomorphic to H, so we may study the latter.
Note that the elements of H multiply according to the formula
M X
M Y
MM X + MY
=
.
0 1
0
1
0
1
M X
Thus, there is an isomorphism f : An−1 ιn−1 Gln−1 (A) → H, via f (X, M ) =
.
0 1
We can apply this to calculate the order of the general linear group of a finite field.
It is shown in Section 11.3 that the order of any finite field must have the form q = pr
for p prime, and that for any such q, there is a unique field, which we shall denote by
Fq , of order q. Of course, Fp = Zp .
Corollary 5.8.5. Let Fq be the field of order q = pr . Then for n ≥ 1,
|Gln (Fq )| = (q n − 1)(q n − q) . . . (q n − q n−1 ).
Proof We argue by induction on n. For n = 1, Gl1 (Fq ) ∼
= F×
q has order q − 1, and the
assertion holds.
Suppose the result is true for n − 1. Then the formula for the size of an orbit gives
|Gln (Fq )| = |Gln (Fq ) · en | · |Fn−1
ιn−1 Gln−1 (Fq )|
q
= |Gln (Fq ) · en | · q n−1 · (q n−1 − 1) . . . (q n−1 − q n−2 )
= |Gln (Fq ) · en | · (q n − q) . . . (q n − q n−1 ).
Thus, it suffices to show that the orbit of en under the action of Gln (Fq ) has q n − 1
elements, and hence that for any nonzero element v ∈ Fnq , there is an M ∈ Gln (Fq ) with
M en = v.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
161
But this follows immediately from elementary linear algebra: v is linearly independent
over Fq , so there is a basis of Fnq that contains v.
Matrix groups over finite fields have another important use in the theory of finite
groups. They are a source of finite simple groups. Recall that if A is a commutative ring,
then the determinant induces a surjective group homomorphism
det : Gln (A) → A×
for n ≥ 1. The special linear group Sln (A) is defined to be the kernel of det. Since a
matrix is invertible if and only if its determinant is a unit, we see that Sln (A) consists
of all n × n matrices over A of determinant 1.
Recall that aIn is the matrix whose diagonal entries are all equal to a and whose
off-diagonal entries are all 0. It’s an easy exercise to show that det(aIn ) = an .
Definition 5.8.6. Let A be a commutative ring and define H ⊂ Sln (A) by
H = {aIn | a ∈ A
with an = 1}.
Then the projective special linear group is defined as PSln (A) = Sln (A)/H.
It turns out that the subgroup H above is the center of Sln (A).
Theorem Let Fq be the field with q = pr elements. The projective special linear group
PSln (Fq ) is simple for all q if n ≥ 3 and is simple for q > 3 if n = 2.
The proof is accessible to the techniques of this volume, but we shall not give it. It is
somewhat analogous to the proof that An is simple for n ≥ 5. There, the cycle structure
gave us a handle on the conjugacy classes. Here, the conjugacy classes are determined
by the rational canonical form.
Exercises 5.8.7.
1. Give a new proof that Aut(Z22 ) = Gl2 (Z2 ) is isomorphic to S3 .
2. Show that Gl3 (Z2 ) has a subgroup of index 7 that is isomorphic to S4 .
3. What are the normalizers of all the Sylow subgroups of Gl3 (Z2 )?
4. Show that Gl3 (Z2 ) is simple.
5. Complete the classification of the groups of order 24.
6. Let q = pr with p prime. Show that one of the p-Sylow subgroups of Gln (Fq ) is the
set of matrices with 1’s on the diagonal and 0’s below it (i.e., the upper triangular
matrices with 1’s on the diagonal).
7. Show that the p-Sylow subgroups of Gl3 (Zp ) are isomorphic to the group constructed in Corollary 4.6.8 for n = p.
8. Write Tn ⊂ Gln (Fq ) for the subgroup of upper triangular matrices with 1’s on the
diagonal. Then Tn acts on Fnq through automorphisms by an action ιn obtained by
restricting the action of the full general linear group. Show that Tn is isomorphic
to Fn−1
ιn−1 Tn−1 .
q
9. Show that D8 embeds in Gl2 (A) whenever A has characteristic = 2.
CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION
162
10. Show that any element of order 4 in Z24 is part of a Z4 -basis. Calculate the order
of Gl2 (Z4 ).
11. Classify the groups of order 48 whose 2-Sylow subgroup is Z4 × Z4 .
12. What are the orders of Sln (Fq ) and PSln (Fq )?
13. Show that the 2-Sylow subgroup of Sl2 (Z3 ) is isomorphic to Q8 and is normal in
Sl2 (Z3 ). (Hint: Look at the elements of Sl2 (Z3 ) of trace 0.)
14. Deduce that the 2-Sylow subgroup of Gl2 (Z3 ) is a group of order 16 containing
both D8 and Q8 .
15. Classify the groups of order 18.
16. What is the 2-Sylow subgroup of Sl2 (Z5 )?
17. What is the 2-Sylow subgroup of Sl2 (Z4 )?
18. Show that the subgroup H of Definition 5.8.6 is the center of Sln (A).
Chapter 6
Categories in Group Theory
Category theory, a.k.a. abstract nonsense, is a branch of mathematics that attempts to
identify phenomena that crop up in many different types of mathematical structures.
One may think of it as the study of those things that happen for reasons which are
purely formal. For this reason, the facts that hold for categorical reasons are often not
the deepest ones in a given field. But categorical patterns repeat in so many contexts
that they form a collection of truths that may be instantly recognized and used.
What category theory gives us is a language by which to describe patterns common to
different areas of mathematics. Verifying that these patterns hold in a given context can
take some work. For instance, we will give some concepts regarding commonly occurring
universal properties. In each category of interest, one must make a construction, often
unique to that category, to show that objects satisfying the universal property actually
exist. Once that construction has been made, one can revert to formal arguments in
using it.
Categorical language is well suited to describing the overview of the relationships
between objects in a given context. For instance, classification results may be expressed
in categorical terms. For reasons like this, some of the more important mathematical
questions can be phrased in terms of categorical language. Formal study of category
theory will not generally be useful in solving them, but categorical thinking can sometimes
suggest a viable approach.
In the first two sections, we give the most general categorical concepts: categories
and functors. The next three sections introduce some important kinds of universal properties: initial and terminal objects, products, coproducts, pushouts, and pullbacks. We
give constructions realizing these universal properties in various categories of interest,
particularly in the categories of sets, groups, and abelian groups. The constructions realizing these universal properties in these three categories can be quite different. Thus,
the reader can see that a general categorical notion can behave quite differently in different categories. The insight coming from the categorical concept tells us nothing about
whether or how that concept might be realized in a particular category.
In Section 6.6, we define the notion of a free object on a set in a given category
and construct the free groups and the free abelian groups. Then in Section 6.7, we give
the technique of describing a group by generators and relations. This technique can be
valuable in studying infinite groups, and arises frequently in topology, as the fundamental
group of a CW complex with one vertex is expressed naturally in this form.
We then define direct and inverse limits, giving constructions for them in various
categories. Direct limits will be used in the construction of algebraic closures of fields.
163
CHAPTER 6. CATEGORIES IN GROUP THEORY
164
p.
Inverse limits are used to construct the ring of p-adic integers, Z
Then, generalizing free functors, we define adjoint functors, giving examples and
showing their utility for deducing general results. We then close with a description of
colimits and limits indexed on a partially ordered set or a small category. This material
has applications to sheaves and to various of the constructions in algebraic geometry and
homotopy theory, and generalizes many of the universal properties described earlier in
the chapter.
Because of the number of concepts and constructions here that may be new to the
reader, Sections 6.2, 6.3, 6.4, 6.5, and 6.8 have been divided into subsections, each with
its own set of exercises.
6.1
Categories
The notion of a category is very abstract and formal, and is best understood by going
back and forth between the definition and some examples. The point is that the word
“object” has to be taken as a black box at first, as there are categories with many
different sorts of objects. The objects in a category are the things you want to study.
So the category of sets has sets as objects, the category of groups has groups as objects,
etc.
The morphisms in a category determine the ways in which the objects are related.
For instance, when studying sets, one generally takes the functions between two sets as
the morphisms. With this in mind, here is the abstraction.
Definition 6.1.1. A category, C, consists of a collection of objects, ob(C), together
with morphisms between them. Specifically, if X and Y are objects of C (written X, Y ∈
ob(C)) there is a set of morphisms from X to Y , denoted MorC (X, Y ). Morphisms may
be composed, just as functions or homomorphisms may be composed: If Z is another
object of C, then for any f ∈ MorC (Y, Z) and any g ∈ MorC (X, Y ) there is a composite
morphism f ◦ g ∈ MorC (X, Z).
This composition rule is required to satisfy two properties. First, there’s an associativity law: if h ∈ MorC (W, X), then for f and g as above, we have f ◦ (g ◦ h) = (f ◦ g) ◦ h
as elements of MorC (W, Z). Second, each object has an identity morphism. We write
1X ∈ MorC (X, X) for the identity of X. The defining property of the identity morphisms
is that g ◦ 1X = 1Y ◦ g = g for all g ∈ MorC (X, Y ).
Example 6.1.2. One of the most ubiquitous categories is the category of sets and functions, which we denote by Sets. The objects of Sets consist of all possible sets. The
morphisms between X and Y in Sets are all the functions from X to Y . The identity
function is the usual one, and the properties of composition of functions are easily seen
to satisfy the requirements for this to be a category.
A very large number of the categories we will look at can be thought of as subcategories of the category of sets and functions:
Definition 6.1.3. A category D is a subcategory of C if ob(D) is a subcollection of
ob(C), MorD (X, Y ) is a subset of MorC (X, Y ) for all objects X and Y of D, and the
laws of composition for the two categories coincide, as do the identity morphisms.
It may come as a surprise that not all categories are equivalent (in a formal sense
we shall give below, but which includes the intuitive notion of two categories being
CHAPTER 6. CATEGORIES IN GROUP THEORY
165
isomorphic) to subcategories of Sets.1 But the structure of Sets does provide a very
good intuition for what a category is.
Of course, one can define categories from now until doomsday, and very few of the
ones chosen randomly will be mathematically interesting. In particular, if one wants to
study something like groups, one should not take the morphisms between two groups to
be all functions between their underlying sets. The morphisms in a given category should
convey information about the basic structure of the objects being considered. Without
that, a study of the category will not shed light on the mathematics one wants to do.
Examples 6.1.4.
1. The category in which we’ve been working for the most part here is the category Gp,
of groups and homomorphisms. Thus, the objects are all groups and the morphisms
between two groups are the group homomorphisms between them.
2. An interesting subcategory of Gp is the category of abelian groups and homomorphisms, which we denote by Ab. Thus, the objects of Ab are all abelian groups,
while the morphisms between two abelian groups are the group homomorphisms
between them.
The inclusion of Ab in Gp is an example of what’s called a full subcategory:
Definition 6.1.5. A subcategory D of C is a full subcategory if for any objects X, Y ∈
ob(D), the inclusion MorD (X, Y ) ⊂ MorC (X, Y ) is bijective.
It is sometimes useful to restrict the morphisms in a category, rather than the objects. Thus, we can form a subcategory of Gp whose objects are all groups, but whose
morphisms are the injective homomorphisms. Note that in many cases here, the set of
morphisms between two objects is the null set. But that doesn’t cause any problems as
far as the definition of a category is concerned.
It is sometimes useful to make new categories out of old ones.
Definition 6.1.6. Let C and D be categories. The product category C × D is defined
as follows. The objects of C × D are the ordered pairs (X, Y ) with X ∈ ob(C) and
Y ∈ ob(D). The morphisms of C × D are given by
MorC×D ((X, Y ), (X , Y )) = MorC (X, X ) × MorD (Y, Y ).
Here, the latter product is the cartesian product of sets.
We write f ×g : (X, Y ) → (X , Y ) for the morphism corresponding to f ∈ MorC (X, X )
and g ∈ MorD (Y, Y ), and define composition by
(f × g) ◦ (f × g ) = (f ◦ f ) × (g ◦ g )
for f × g : (X , Y ) → (X , Y ) in C × D.
One of the most important ideas about the relationships between objects in a category
is that of isomorphism:
Definition 6.1.7. An isomorphism in a category C is a morphism f : X → Y in C
which admits an inverse. Here, an inverse for f is a morphism g : Y → X such that
f ◦ g = 1Y and g ◦ f = 1X .
1 One example of a category not equivalent to a subcategory of Sets is the homotopy category encountered in topology.
CHAPTER 6. CATEGORIES IN GROUP THEORY
166
Clearly, a morphism in Sets is an isomorphism in Sets if and only if it is a bijection.
Also, a morphism in Gp (or Ab) is an isomorphism in Gp (resp. Ab) if and only if it is a
group isomorphism in the sense we have been studying.
As in the case of finite group theory, one of the most important problems in studying
a given category is being able to classify the objects.
Definitions 6.1.8. The isomorphism class containing a given object consists of the collection of all objects isomorphic to it. Clearly, any two isomorphism classes having an
object in common are equal.
A classification of the objects that have a given property consists of a determination
of all the individual isomorphism classes with that property.
One example of a classification is the determination of a set of representatives of the
isomorphism classes with a given property. This means a set S = {Xi | i ∈ I} of objects
such that every isomorphism class in C having the given property contains exactly one
of the objects in S.
Not every collection of isomorphism classes has a set of representatives. For instance,
the isomorphism classes of torsion-free abelian groups do not form a set.
Even if there is a classification of the objects with a given property, it can be difficult
to determine which isomorphism class contains a given object. In particular, given two
objects in a given category, there may not be any algorithmic procedure to determine
whether the objects are isomorphic.
For instance, think about the classifications we’ve seen of the groups in different orders. At first glance, it can be quite difficult to decide whether two groups are isomorphic.
And the classifications for different orders can involve very different criteria.
The development of criteria to determine whether two objects are isomorphic is one
of the most important problems in studying a given category. It should be noted that
general category theoretic techniques are rarely useful for solving this problem. However,
the notion of functors, to be studied in Section 6.2, can be useful for showing that two
objects are not isomorphic.
There are partial forms of invertibility which can be useful.
Definitions 6.1.9. A left inverse for a morphism f : X → Y in C is a morphism
g : Y → X in C such that g ◦ f = 1X . A left inverse is often called a retraction, and X
is often called a retract of Y in C.
Similarly, a right inverse for f : X → Y in C is a morphism g : Y → X in C such
that f ◦ g = 1Y . A right inverse for f is often called a section of f .
We shall also consider categories of monoids. We write Mon for the category whose
objects are all monoids and whose morphisms are the homomorphisms of monoids. We
write Amon for the full subcategory whose objects are the abelian monoids.
Exercises 6.1.10.
1. Show that if f : X → Y is an isomorphism in C, then its inverse is unique.
2. Show that every injection in Sets admits a left inverse, and that every surjection
in Sets admits a right inverse. Are these inverses unique?
3. Does every injective group homomorphism have a left inverse in Gp? Does every
surjective group homomorphism have a right inverse in Gp?
4. Let g be a left inverse for f : X → Y in C. Suppose that g admits a left inverse
also. Show then that f is an isomorphism with inverse g.
CHAPTER 6. CATEGORIES IN GROUP THEORY
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5. Suppose that f : G → K is a morphism in Gp that admits a right inverse. Show
then that G is a split extension of ker f by K.
6. Suppose that i : H → G is a morphism in Gp that admits a left inverse. Show that
i must be injective. Show that if i(H) G, then G ∼
= H × G/i(H). What can you
say if i(H) is not normal in G?
7. Let G be a group and let G-sets be the category whose objects are all G-sets
and whose morphisms are all G-maps. Show that the isomorphisms in G-sets are
precisely the G-isomorphisms.
8. Let C be any category and let X ∈ ob(C). Show that MorC (X, X) is a monoid
under composition of morphisms. We call it the monoid of endomorphisms of X,
and write MorC (X, X) = EndC (X). Show that this monoid is a group if and only
if every morphism from X to itself is an isomorphism in C.
9. Let M be a monoid. Show that there is a category which has one object, O,
such that the monoid of endomorphisms of O is M . Thus, monoids are essentially
equivalent to categories with one object.
6.2
Functors
Just as we can learn about a group by studying the homomorphisms out of it, we can
learn about a category by studying its relationship to other categories. But to get
information back, we shall need to define ways to pass from one category to another that
preserve structure. We study covariant functors here, and study contravariant functors
in a subsection below.
Definition 6.2.1. Let C and D be categories. A functor, F : C → D assigns to each object X ∈ ob(C) an object F (X) ∈ ob(D) and assigns to each morphism f ∈ MorC (X, Y )
a morphism F (f ) ∈ MorD (F (X), F (Y )), such that
F (1X ) = 1F (X)
for all X ∈ ob(C), and
F (f ) ◦ F (g) = F (f ◦ g)
for each pair of morphisms f and g in C for which the composite f ◦ g is defined.
Functors are often used in mathematics to detect when objects of a given category
are not isomorphic.
Lemma 6.2.2. Let F : C → D be a functor and suppose that f : X → Y is an isomorphism in C. Then F (f ) : F (X) → F (Y ) is an isomorphism in D. Thus, if Z and W
are objects of C, and if F (Z) and F (W ) are not isomorphic in D, then Z and W must
not be isomorphic in C.
Proof Let g be the inverse of f . Then F (g) ◦ F (f ) = F (g ◦ f ) = F (1X ) = 1F (X) .
Similarly, F (f ) ◦ F (g) = 1F (Y ) . Thus, F (g) is an inverse for F (f ) in D.
CHAPTER 6. CATEGORIES IN GROUP THEORY
168
Indeed, in a lot of ways, the preceding lemma is the main reason for looking at
functors, to the point where the definition of functor is tailored to make it true. The
utility of this approach lies in the fact that we can often find functors to categories that
are much more manageable than the one we started in.
Examples 6.2.3.
1. Let p be a prime. Then for an abelian group G, the set Gp of p-torsion elements of G
is a subgroup of G. As shown in Lemma 4.4.19, if f : G → H is a homomorphism
with H abelian, then f restricts to a homomorphism fp : Gp → Hp . Clearly,
gp ◦ fp = (g ◦ f )p for any homomorphism g : H → K with K abelian, and (1G )p =
1(Gp ) , so there is a functor Fp : Ab → Ab given by setting Fp (G) = Gp for all
abelian groups G, and setting Fp (f ) = fp for f : G → H a homomorphism of
abelian groups.
Notice that while Fp (f ) may be an isomorphism even if f is not, Corollary 4.4.21
shows that if f : G → H is a homomorphism between finite abelian groups, then f
is an isomorphism if and only if Fp (f ) is an isomorphism for each prime p dividing
either |G| or |H|.
Thus, on the full subcategory of Ab whose objects are the finite abelian groups,
the collection of functors Fp , for p prime, together detect whether a morphism is
an isomorphism.
2. Let C be any category and let X be an object of C. We define a functor F : C →
Sets as follows. For an object Y of C, we set F (Y ) = MorC (X, Y ). If f : Y → Z
is a morphism from Y to Z in C, we define F (f ) : MorC (X, Y ) → MorC (X, Z) by
setting F (f )(g) = f ◦ g for all g ∈ MorC (X, Y ). (It is often customary to write f∗
instead of F (f ).) The reader may easily check that F is a functor.
3. Define φ : Gp → Sets by setting φ(G) = G, the underlying set of G. For a
homomorphism f : G → H, we set φ(f ) = f . We call φ the forgetful functor that
forgets the group structure on G.
More generally, any functor that simply forgets part of the structure of a given object
is called a forgetful functor. Thus, there is also a forgetful functor from Gp to Mon.
Definition 6.2.4. A functor C × C → D is often called a bifunctor on C, and may be
thought of as a functor of two variables.
Example 6.2.5. There is a bifunctor F : Sets × Sets → Sets obtained by setting
F (X, Y ) = X × Y and setting F (f × g) = f × g, for f ∈ MorSets (X, X ) and g ∈
MorSets (Y, Y ). Here, the right-hand function f ×g is the usual mapping f ×g : X ×Y →
X × Y given by (f × g)(x, y) = (f (x), g(y)).
Exercises 6.2.6.
1. Show that functors preserve left inverses and right inverses.
2. Show that there is a functor F : Gp → Gp such that F (G) is the commutator
subgroup of G for every group G. Here, if f : G → G is a group homomorphism,
then F (f ) is obtained by restricting f to the commutator subgroup.
3. Give an example of groups H and G such that H is a retract of G, but Z(H) is not
a retract of Z(G). Deduce that there is no functor on Gp that takes each group to
its center.
CHAPTER 6. CATEGORIES IN GROUP THEORY
169
4. Show that there is a functor F : Gp → Ab such that F (G) = Gab for all groups
G, and if f : G → G is a group homomorphism, then F (f ) = f ab , the unique
homomorphism such that the following diagram commutes.
G
πG Gab
f
f ab
/ G
πG
/ ab
G
Here, for any group H, we write πH for the canonical map from H to H ab =
H/[H, H].
5. Show that the inclusion of a subcategory is a functor. In particular the inclusion
of a category in itself is called the identity functor.
6. Let φ : Gp → Sets be the forgetful functor. Suppose that φ(f ) is an isomorphism. Is f then an isomorphism? Alternatively, suppose that φ(G) and φ(G ) are
isomorphic in Sets. Are G and G then isomorphic in Gp?
7. Show that passage from a monoid M to its group of invertible elements Inv(M )
gives a functor from Mon to Gp (or from Amon to Ab).
8. Let G be a group and let G be the category with one object, O, such that the monoid
of endomorphisms of O is G. Show that there is a one-to-one correspondence
between G-sets and the functors from G into Sets.
9. Let M and M be monoids and let M and M be categories with one object whose
monoids of endomorphisms are M and M , respectively. Show that there is a oneto-one correspondence between the monoid homomorphisms from M to M and
the functors from M to M .
10. Let C and D be categories.
(a) Show that there are functors πC : C × D → C and πD : C × D → D obtained
by setting πC (X, Y ) = X, πD (X, Y ) = Y , πC (f × g) = f and πD (f × g) = g.
(b) For an object Y of D, show that there is a functor ιY : C → C × D obtained
by setting ιY (X) = X × Y and ιY (f ) = f × 1Y .
Contravariant Functors
Now we come to the fact that there’s more than one kind of functor. The functors we’ve
considered so far are called covariant functors. The other kind of functors are called
contravariant. They behave very much like covariant ones, except that they reverse the
direction of morphisms.
Definition 6.2.7. A contravariant functor F : C → D associates to each X ∈ ob(C)
an object F (X) ∈ ob(D) and to each morphism f ∈ MorC (X, Y ) a morphism F (f ) ∈
MorD (F (Y ), F (X)), such that
F (1X ) = 1F (X)
CHAPTER 6. CATEGORIES IN GROUP THEORY
170
for all X ∈ ob(C), and
F (f ) ◦ F (g) = F (g ◦ f )
for each pair of morphisms f and g in C for which the composite g ◦ f is defined.
Contravariant functors may be viewed as covariant functors out of a different category.
Definition 6.2.8. Let C be a category. The opposite category, C op , is defined by setting
ob(C op ) = ob(C)
and setting
MorC op (X, Y ) = MorC (Y, X)
for all X, Y ∈ ob(C). Here, if f : X → Y is a morphism in C, we write f op : Y → X for
the associated morphism in C op . We define composition in C op by setting f op ◦ g op =
op
op
(g ◦ f ) , and set the identity morphism of X ∈ ob(C op ) equal to (1X ) .
The point here is that any contravariant functor on C may be viewed as a covariant
functor on C op . The reader may supply the details:
Proposition 6.2.9. For any category C, there is a contravariant functor I : C → C op
defined by setting I(X) = X for X ∈ ob(C) and setting I(f ) = f op for f ∈ MorC (X, Y ).
If F : C → D is a contravariant functor, then there is a covariant functor F op :
op
C → D defined by setting F op (X) = F (X) for X ∈ ob(C) = ob(C op ) and setting
F op (f op ) = F (f ) for f ∈ MorC (X, Y ). Thus, F itself factors as the composite
I
F op
C−
→ C op −−→ D.
Exercises 6.2.10.
1. Let C be any category and let Y be an object of C. For X ∈ ob(C), let F (X) =
MorC (X, Y ). For f ∈ MorC (X, X ), define F (f ) : MorC (X , Y ) → MorC (X, Y ) by
F (f )(g) = g ◦ f . (It is often customary to write f ∗ instead of F (f ).) Show that F
defines a contravariant functor from C to Sets.
2. Let C be a category. Show that there is a functor Hom : C op × C → Sets, given
by setting Hom(X, Y ) = MorC (X, Y ) for each pair of objects X, Y of C. Here, if
f : X → X and g : Y → Y are morphisms in C, we define
Hom(f op , g) : MorC (X, Y ) → MorC (X , Y )
by Hom(f op , g)(h) = g ◦ h ◦ f for all h ∈ MorC (X, Y ).
3. Show that a contravariant functor takes isomorphisms to isomorphisms.
4. Show that a contravariant functor takes morphisms that admit left inverses to
morphisms that admit right inverses and vice versa.
5. Let G be a group and let G be the category with one object, O, such that the monoid
of endomorphisms of O is G. Show that there is a one-to-one correspondence
between right G-sets and the contravariant functors from G into Sets.
CHAPTER 6. CATEGORIES IN GROUP THEORY
6.3
171
Universal Mapping Properties: Products and Coproducts
An object, or the relationship between a family of objects connected by morphisms, is
sometimes uniquely determined by what morphisms exist either in or out of the object
or family. This phenomenon is known as a universal mapping property, and is perhaps
best understood through examples. We treat initial and terminal objects here, and then
give subsections for products, coproducts, and free products.
The simplest examples of universal mapping properties are initial and terminal objects
for a category:
Definitions 6.3.1. Suppose that X ∈ ob(C) has the property that MorC (X, Y ) has
exactly one element for each Y ∈ ob(C). Then X is called an initial object for C.
Alternatively, if MorC (Y, X) has exactly one element for each Y ∈ ob(C), then X is
called a terminal object for C.
If X is both an initial object and a terminal object for C, we call X a zero object for
C.
One example that may not be immediately obvious is that the null set is an initial
object for Sets. The point here is that a function is determined by where it sends the
elements of its domain. Since the null set has no elements, there is exactly one way to
specify where its elements go.
Exercises 6.3.2.
1. Show that any two initial objects for C must be isomorphic.
2. Show that any two terminal objects for C must be isomorphic.
3. Show that a set with one element is a terminal object for Sets.
4. Show that the trivial group e is a zero object in both Gp and in Ab.
5. Let G be a group. Show that the G-set with one element is a terminal object in
G-sets.
6. Let G be any nontrivial group. Show that G-sets does not have an initial object.
Products
Other universal mapping properties express relationships between more than one object. These properties often recur in many different categories, and are given names
appropriate to that fact. The one we’ve seen the most of is products:
Definition 6.3.3. Let X, Y ∈ ob(C). A product for X and Y in C consists of an
object, Z ∈ ob(C), together with two morphisms, πX : Z → X and πY : Z → Y , called
projections. The universal property that makes these a product is as follows: For any
object W ∈ ob(C) and any two morphisms f : W → X and g : W → Y , there is a unique
CHAPTER 6. CATEGORIES IN GROUP THEORY
172
morphism (f, g) : W → Z such that the following diagram commutes:
W/ ?OO
// ??OOOO
// ??? h OOgOO
OOO
// ??
OO'
// Z πY / Y
f //
//
// πX
/ X,
Here, h = (f, g).
Thus, the projection maps are part of the definition of a product, and not just the
object Z itself.
Examples 6.3.4.
1. The cartesian product, together with the usual projections, is the product in Sets:
a function into the product is determined by its component functions, which are
precisely the composites of the original function with the projection maps. Indeed,
a complete argument showing that the cartesian product is the product in the
category of sets is given in Proposition 1.6.5.
2. In Gp, the direct product has been shown in Lemma 2.10.5 to be the categorical
product, and this remains true in Ab.
In Gp and Ab, the product has some extra features (like the inclusion maps) that do
not occur in Sets or more general categories.
The universal mapping property that defines the product has at most one solution (up
to isomorphism) for a given pair of objects X and Y . This is because of the uniqueness of
the map from W to Z that makes the diagram commute in the definition of the product:
Proposition 6.3.5. Suppose given objects and morphisms
π
π
X
Y
X ←−
− Z −−→
Y
π
π
X
Y
− Z −−→
Y
and X ←−
in C such that both (Z, πX , πY ) and (Z , πX
, πY ) satisfy the universal mapping property
for the product of X and Y in C. Let f = (πX , πY ) be the unique morphism from Z to
Z which makes the following diagram commute. (Here, we use the universal mapping
π
π
X
Y
property of X ←−
Y .)
− Z −−→
Z / ?OO
// ??OOOO
// ??? fOOOπOY
OOO
// ??
OOO
// '/
/
Y
Z
πX /
πY
//
//
// πX
X
Then f is an isomorphism.
CHAPTER 6. CATEGORIES IN GROUP THEORY
π
173
π
X
Y
Proof By the universal property of X ←−
− Z −−→
Y , there is a unique morphism
g : Z → Z such that πX ◦ g = πX and πY ◦ g = πY .
Now g◦f : Z → Z is the unique morphism such that πX ◦g◦f = πX and πY ◦g◦f = πY .
Thus, g ◦ f = 1Z .
Similarly, f ◦ g = 1Z , and the result follows.
Exercises 6.3.6.
1. What is the product in G-sets? Is the product of two orbits an orbit? What are
its isotropy groups?
2. Suppose that for each pair (X, Y ) of objects of C we are given a product for X and
Y . Show that passage from a pair of objects to their product gives a functor from
C × C to C.
Coproducts
It is frequent in categorical mathematics to ask what happens to a definition if you turn
all the arrows around. This is called dualization. The dual of a product is called a
coproduct.
Definition 6.3.7. Let X, Y ∈ ob(C). A coproduct for X and Y in C is an object Z
together with morphisms ιX : X → Z and ιY : Y → Z, called the natural or canonical
inclusions, with the following universal property: For any pair of morphisms f : X → W
and g : Y → W , there is a unique morphism f, g : Z → W such that the following
diagram commutes:
Y/
//
//
ιY
//
/g
ιX
/
X OO
Z ? ///
OOO
OOO ??h? //
OOO ?? //
OOO?? /
f
O' W,
Here, h = f, g.
Examples 6.3.8.
1. Proposition 2.10.6, when applied strictly to abelian groups and their homomorphisms, shows that the direct product, together with the canonical inclusions,
ι
ι
1
2
G × H ←−
H,
G −→
is the coproduct in the category Ab of abelian groups.
2. The coproduct in Sets
is called the disjoint union. Here, the disjoint union of X
and Y is denoted X Y . Its elements are the union of those in X and those in Y ,
where the elements from X are not allowed to coincide with any of the elements of
Y . Formally, this may be accomplished as follows. Set theory tells us that there’s
some set, S, containing both X and Y . The disjoint union of X and Y is defined
to be the subset of S × {1, 2} consisting of the elements (x, 1) with x ∈ X and the
elements (y, 2) with y ∈ Y . We then have ιX (x) = (x, 1) and ιY (y) = (y, 2).
CHAPTER 6. CATEGORIES IN GROUP THEORY
174
Now if f : X → W and g : Y → W are functions, we define f, g : X Y → W
by setting f, g(x, 1) = f (x) and setting f, g(y, 2) = g(y). Since ιX and ιY
are disjoint and have X Y as their union, this is well defined and satisfies the
universal property for the coproduct.
Thus, unlike the case of the product, the forgetful functor from Ab to Sets does
not preserve coproducts. As we shall see presently, the forgetful functors Ab → Gp
and Gp → Sets also fail to preserve coproducts. Indeed, the coproducts in these three
categories are given by completely distinct constructions.
Exercises 6.3.9.
1. What is the coproduct in G-sets?
2. Show that the direct product is not the coproduct in Gp, even if the two groups
are abelian.
3. Prove the analogue of Proposition 6.3.5 for coproducts, showing that any two coproducts for a pair of objects are isomorphic via a morphism that makes the appropriate diagram commute.
Free Products
The coproduct in Gp is called the free product.
We shall give the construction now. Let
H and K be groups, and write X = H K for the disjoint union of the sets H and K.
By a word in X we mean a formal product x1 x2 · · · xk , with xi ∈ X for all i. (We
write for the formal product operation to distinguish it from the multiplications in H
and K.) The empty word is the word with no elements.
The elements in the free product H ∗ K will not be the words themselves, but rather
equivalence classes of words. The equivalence relation we shall use is derived from a
process called reduction of words. Words may be reduced as follows.
First, if xi and xi+1 are both in the image of ιH : H → X, then x1 · · · xk reduces
to x1 · · · (xi xi+1 ) · · · xk , where xi xi+1 is the product of xi and xi+1 in H. There is a
similar reduction if xi and xi+1 are both in the image of ιK : K → X. Finally, if xi is the
identity element of either H or K, then x1 · · ·xk reduces to x1 · · ·xi−1 xi+1 · · ·xk .
In particular, for either e ∈ H or e ∈ K, then the singleton word e reduces to the empty
word.
e
We shall refer to the reductions above as elementary reductions, and write w w or
e
w w if the word w is obtained from the word w by one of these elementary reductions.
e
e
e
w1 ... wn of elementary reductions, we say
More generally, given a sequence w0 that w0 reduces to wn , and write w0 wn (or wn w0 ). We shall also declare that
w w, so that is reflexive.
The equivalence relation we shall use is the equivalence relation generated by the
process of reduction, in the sense of Section 1.5. This means that the words w and w
are equivalent, written w ∼ w , if there is a sequence w = w0 , . . . , wn = w of words such
that for i = 0, . . . , n − 1, either wi wi+1 or wi wi+1 .
Note that some words (e.g., h h eK k, where h, h ∈ H, k ∈ K, and eK is the
identity of K) can be reduced in more than one way. (Also, as in the case of words
containing an adjacent pair such as eK k above, two different elementary reduction
operations can produce exactly the same word as a result. However, both cases are
e
represented by the same symbols, w w .) An examination of words which admit more
than one reduction shows the following important fact.
CHAPTER 6. CATEGORIES IN GROUP THEORY
175
Lemma 6.3.10. Suppose given two different elementary reduction operations on a word
w, giving w e w1 and w e w2 . Then either w1 = w2 or there is a word w3 with
e
e
w1 w3 w2 .
It is convenient to be able to represent an element of the free product by what’s
known as a reduced word:
Definition 6.3.11. We say that a word x1 · · · xk is reduced if xi = e for 1 ≤ i ≤ k
and if no adjacent pair xi , xi+1 comes from the same group (H or K). We also declare
the empty word to be reduced.
The next lemma is fundamental for the understanding of free products.
Lemma 6.3.12. Let w be a word in X = H K. Then there is a unique reduced word,
w such that w w .
Proof We argue by induction on the number of letters in w. If w is reduced, then it
cannot be reduced further, and we take w = w.
Otherwise, there is an elementary reduction from w to a word v. By induction, there
is a unique reduced word w with v w . But then w w . For the uniqueness
e
e
e
w1 ... w .
statement, suppose that w is reduced and that w w , say, via w If w1 = v, then w = w by the induction hypothesis. Otherwise, by Lemma 6.3.10, there
e
e
is a word w
) with v w
)
w1 . But if w
) w with w reduced, then the uniqueness
statement of the induction hypothesis gives w = w = w .
Recall that ∼ is the equivalence relation generated by reduction.
Proposition 6.3.13. Every equivalence class under ∼ contains exactly one reduced word.
If w is a reduced word, then every word in the equivalence class of w under ∼ reduces to
w.
Proof Suppose that w ∼ w . Then there is a sequence w = w0 , . . . , wn = w such that
for each i = 0, . . . , n − 1, either wi wi+1 or wi wi+1 . In either case, Lemma 6.3.12
tells us that the unique reduced word w to which wi reduces is equal to the unique
reduced word to which wi+1 reduces. In particular, each wi reduces to w , and there can
be at most one reduced word in such a sequence.
We define the elements of the free product H ∗ K to be the equivalence classes of
words under ∼. Thus, the elements of H ∗ K are in one-to-one correspondence with the
reduced words. We define a binary operation on the words themselves by concatenation:
(x1 · · · xk ) · (y1 · · · yl ) = x1 · · · xk y1 · · · yl .
In word notation, we write w · w for the product in this sense of the words w and w .
But clearly, if w reduces to w1 and w reduces to w1 , then w · w reduces to w1 · w1 .
Thus, concatenation induces a well defined binary operation on the equivalence classes
of words.
Lemma 6.3.14. Concatenation of words induces a group structure on the free product
H ∗ K.
Proof Concatenation is clearly an associative operation on the set of words themselves,
with the empty word as an identity element. Since it preserves equivalence classes, the
free product inherits a monoid structure from the set of words themselves. But the
−1
inverse of x1 · · · xk in H ∗ K is given by x−1
k · · · x1 .
CHAPTER 6. CATEGORIES IN GROUP THEORY
176
The inclusions of H and K in X induce group homomorphisms from H and K to
H ∗ K, which we’ll denote by ιH and ιK .
Proposition 6.3.15. The free product H ∗ K, together with the homomorphisms ιH and
ιK is the coproduct of H and K in the category of groups.
Proof Given group homomorphisms f : H → G and g : K → G, we define f, g on X
to be f on H ⊂ X and to be g on K ⊂ X. Given a word x1 · · · xk in X, we may then
define f, g(x1 · · · xk ) to be (f, g(x1 )) . . . (f, g(xk )). Stipulating that f, g carries
the empty word to the identity element of G, we obtain a monoid homomorphism from
the monoid of words in X into G. Since two words differing by an elementary reduction
are carried to the same element of G (since f and g are homomorphisms), passage to
equivalence classes of words gives a group homomorphism from H ∗ K into G.
Clearly, f, g ◦ ιH = f and f, g ◦ ιK = g. Since the images of ιH and ιK generate
H ∗ K, f, g is the unique homomorphism with this property.
Free products can be somewhat intractable, especially in comparison to the groups we
have considered previously. But they are important in understanding groups in general.
For instance, finitely generated free groups, which we shall study in Section 6.6, are
iterated free products of copies of the integers. We shall see that any finitely generated
group is a factor group of such a free group, and this opens up a new way of looking at
groups, called generators and relations, which we shall study in Section 6.7.
A concrete example of free products “in nature” is PSl2 (Z), which turns out to be
isomorphic to Z2 ∗ Z3 .
It should be noted that not all categories have products, coproducts, or other of the
constructions we shall give.
Exercises 6.3.16.
1. Show that if H and K are both nontrivial, then H ∗ K is infinite.
2. Show that if H and K are both nontrivial, then H ∗ K is nonabelian even if H and
K are both abelian.
3. Show that the homomorphisms ιH : H → H ∗ K and ιK : K → H ∗ K are injective.
4. What is H ∗ e?
5. Show that the same constructions that give the product and coproduct in Gp give
a product and coproduct in Mon.
6. Show that the same constructions that give the product and coproduct in Ab give
a product and coproduct in Amon.
6.4
Pushouts and Pullbacks
We now give a generalization of products and coproducts. These are universal mapping
properties not just of pairs of objects, but of particular collections of objects and morphisms. We treat pullbacks now, and treat pushouts in a separate subsection. In further
subsections, we will construct pushouts in Gp, Ab, and Sets.
CHAPTER 6. CATEGORIES IN GROUP THEORY
177
Definition 6.4.1. Suppose given a pair of morphisms into Z, f : X → Z and g : Y → Z.
This gives us the following diagram:
X
.
f
g
Y
/ Z
A pullback for this diagram consists of the following data: We have an object W and
morphisms g : W → X and f : W → Y such that
1.
W
f
g
/ X
f
/ Z
Y
g
W
g /
X
commutes.
2. For every commutative diagram
f
Y
f
g
/ Z
there is a unique morphism h : W → W such that the following diagram commutes.
W / ?OO
// ?? OOOO // ??? OOOgO
OOO
// ??h?
OOO
// '/
/
W
X
f /
g
//
//
f
// f
g
/ Z
Y
In other words, the pullback is a universal completion of the original 3-object diagram
to a commutative square. When the context is totally unambiguous, one often refers to
the object W as the pullback, but the proper usage is that “pullback” refers to the whole
square.
Exercises 6.4.2.
1. Show that pullbacks in Sets, Gp, and Ab may all be constructed as follows. Given
a diagram
X
f
Y
g
/ Z
,
CHAPTER 6. CATEGORIES IN GROUP THEORY
178
let Y ×Z X ⊂ Y × X be given by
Y ×Z X = {(y, x) | g(y) = f (x)}.
Then the pullback of the original diagram is given by
Y ×Z X
f g
Y
g /
X
f ,
/ Z
where f (y, x) = y and g(y, x) = x. The construction Y ×Z X is sometimes called
the fibre product of Y and X over Z.
2. Show that pullbacks, when they exist, are unique.
3. Show that if C has a terminal object, ∗, then the pullback of the unique diagram
X
/ ∗
Y
is the product of X and Y (i.e., show that the pullback of this diagram satisfies the
same universal property as the product). Thus, pullbacks do generalize products.
4. Suppose given a pullback diagram
W
f
g
/ X
f
Y
g
/ Z
in either Gp or Ab. Show that ker f ∼
= ker f and that ker g ∼
= ker g.
5. In either Gp or Ab, show that the pullback of
X
f
e
/ Y
is ker f .
Pushouts
Pushout is the notion dual to pullback. In Sets and Gp, pushouts are more delicate and
harder to construct than pullbacks, but they can be very important.
CHAPTER 6. CATEGORIES IN GROUP THEORY
179
Definition 6.4.3. The pushout of a diagram
X
f
/ Y
f
/ Y
g
Z
is a commutative diagram
X
g
g
Z
f
/ W
which is universal in the sense that if
f
X
/ Y
g
g
Z
f / W
is another commutative diagram, then there is a unique morphism h : W → W such
that the following diagram commutes.
/ Y
//
//
g
g
//
// f
/g
/
Z OO
W ? //
OOO
OOO ??h? ///
OOO ?? /
OOO ??//
f
OO' W
X
f
As in the case of pullbacks, it is customary to refer to the object W as the pushout
as well.
Exercises 6.4.4.
1. Show that pushouts, when they exist, are unique.
2. Show that if C has an initial object, ∗, then the pushout of the unique diagram
∗
Z
is the coproduct of Y and Z.
/ Y
CHAPTER 6. CATEGORIES IN GROUP THEORY
180
3. Suppose given a pushout diagram
f
X
/ Y
g
g
Z
f
/ W
in C, where f is an isomorphism. Deduce that f is also an isomorphism.
4. Show, using the universal mapping property, that in Sets, Gp, or Ab, the pushout
of a product diagram
πY /
X ×Y
Y
πX X
is the terminal object. (In Sets, we must assume that neither X nor Y is the null
set.)
Pushouts in Gp and Ab
Pushouts are closely related to coproducts, and are often constructed using them. We
first give the construction of pushouts in Ab.
Proposition 6.4.5. The pushout in Ab of a diagram
f /
A
B
g
C
is the diagram
f
A
g
C
/ B
ιB
ιC /
(C × B)/D,
where D = im(g, −f ) = {(g(a), −f (a)) | a ∈ A} ⊂ C × B, and where the maps ι are the
composites of the standard inclusions of B and C in the product with the canonical map
from C × B to its factor group.
Proof First, note that factoring out by D makes the square commute. Now given a
commutative square
f /
A
B
g
g
C
f
/ G
CHAPTER 6. CATEGORIES IN GROUP THEORY
181
in Ab, there is a unique homomorphism h from the coproduct C × B into G such that
h ◦ ιB = g and h ◦ ιC = f . But then h((g(a), −f (a)) = f g(a) − g f (a) = 0 by
commutativity of the diagram. Thus, h factors through the factor group of C × B by D.
But factorizations of homomorphisms through factor groups are unique.
In order to understand pushouts in the category of groups, we will need the following
concept.
Definition 6.4.6. Let G be a group and let S be any subset of G. We write S ⊂ G
for the subgroup generated by T = {gxg −1 | g ∈ G, x ∈ S}, the set of all conjugates
of the elements of S. We call S the normal subgroup generated by S. As usual, if
S = {g1 , . . . , gn }, we write S = g1 , . . . , gn .
From a different viewpoint, if S is contained in the subgroup H G and if the
subgroup S = T above equals H, we say that H is normally generated by S in G.
Lemma 6.4.7. Let S be a subset of the group G. Then S is the smallest normal
subgroup of G that contains S.
Proof Clearly, any normal subgroup containing S must contain T , and hence must
contain the subgroup generated by T . Thus, it suffices to show that S is normal in
G.
A generic element of S is a product of powers of elements of T . Since any conjugate
of an element of T is also in T , the result follows.
The pushout in Gp is called the free product with amalgamation. The construction
proceeds precisely as in the abelian case. We give it as an exercise.
We shall see below that every group may be presented in terms of “generators and
relations,” which is an example of a free product with amalgamation.
Exercises 6.4.8.
1. Show that the pushout in Gp of the diagram
f
G
/ H
g
K
is the diagram
G
g
K
f
/ H
ιH
ιK /
(K ∗ H)/S,
where S = {g(x) · f (x)−1 | x ∈ G}, and where the maps ι are the composites of the
standard inclusions of H and K in the free product with the canonical map from
K ∗ H to its factor group.
CHAPTER 6. CATEGORIES IN GROUP THEORY
182
2. What is the pushout in Gp of a diagram of the following form?
f /
H
G
e
What would the pushout be in Ab if the diagram were also in Ab?
3. Suppose given a pushout diagram
A
f
g
C
/ B
g
f
/ D
in Ab.
(a) If f is injective, show that f is also.
(b) Show that f is surjective if and only if f is.
4. Suppose the diagram of the previous problem is a pushout square in Gp rather
than Ab.
(a) If f is surjective, show that f is also.
(b) Show that f may be surjective when f is not.
Pushouts in Sets
In Sets, the pushout is again a quotient of the coproduct. This time we proceed by
placing an equivalence relation on the coproduct and passing to the set of equivalence
classes. Thus, suppose given a diagram
f /
X
Y
g
Z
in Sets. Our equivalence relation on the coproduct Y Z will be the equivalence relation,
∼, generated by the following notion
of basic equivalence: There is a basic equivalence
between ιY (y) with ιZ (z) in Y Z if there is an element x ∈ X with f (x) = y and
g(x) = z. In this case, we write either ιY (y) ιZ (z) or ιZ (z) ιY (y), as the relation
in question is symmetric.
Since ∼ is the equivalence relation generated by , we see that two elements v, v ∈
Y Z are equivalent under ∼ if and only if there is a sequence
v = v0 v1 . . . vn = v .
We define the pushout object Pto be the set of equivalence classes in Y Z under
∼. We then have a function π : Y Z → P that takes each element to its equivalence
class.
CHAPTER 6. CATEGORIES IN GROUP THEORY
183
Exercises 6.4.9.
1. Show that the pushout in sets of the above diagram is
f /
X
Y
g
Z
ιY
ιZ / P,
where the functions ι are the composites of the standard inclusions into the coproduct with the quotient function π.
2. If f is injective, show that ιZ is also. (In this case, P is often denoted Z ∪g Y .)
3. If f is surjective, show that ιZ is also.
6.5
Infinite Products and Coproducts
We’ve discussed products and coproducts of pairs of objects in a category. This is
sufficient, by induction, to provide products or coproducts for any finite collection of
objects. But in some cases, as we shall see in our discussion of rings and modules, it is
necessary to make use of products or coproducts of infinite families of elements.
We define infinite products and coproducts here and construct the products. We give
constructions of infinite coproducts in a separate subsection.
We shall consider sets of objects {Xi | i ∈ I}. In other words, I is a set, called the
indexing set, and Xi is an object of our category for each i ∈ I.
Definitions 6.5.1. The product of a set {Xi | i ∈ I} of objects of C is an object X,
together with morphisms πi : X → Xi , called projections, with the following universal
property. If fi : Y → Xi is a morphism in C for each i ∈ I, then there is a unique
morphism f : Y → X such that πi ◦ f = fi for all i ∈ I.
The coproduct of a set {Xi | i ∈ I} of objects of C is an object X, together with
morphisms ιi : Xi → X for each i ∈ I, called the natural inclusions, with the following
universal property. If fi : Xi → Y is a morphism in C for each i ∈ I, then there is a
unique morphism f : X → Y such that f ◦ ιi = fi for all i ∈ I.
Once again, the products in Sets, Gp, and Ab come from the same construction. We
shall see in our discussion of adjoint functors in Section 6.9 that this is no coincidence.
!
The construction of products in Sets is as follows. The elements of the product i∈I Xi
are collections (xi | i ∈ I) such that xi ∈ Xi for all i ∈ I. In other words, the element
(xi | i ∈ I) consists of a choice, for each i ∈ I, of an element of Xi . We call (xi | i ∈ I) a
tuple or an I-tuple.
Another way of thinking about the elements of a product is as follows. Let S be a
set that contains each Xi . (The !
existence of such a set is one of the standard axioms of
set theory.) Then an I-tuple in i∈I Xi is a function from I to S whose value on i ∈ I
must
lie in Xi . From this viewpoint, we
! can see that the collection of all such I-tuples,
!
i∈I Xi , is a set. The projection πi :
i∈I Xi → Xi takes (xi | i ∈ I) to xi .
The act of choosing elements xi ∈ Xi for each i ∈ I suggests that the Axiom of Choice
is involved here. This is indeed the case.
Axiom of Choice
(second form) Let {Xi | i ∈ I} be a family of nonempty sets. Then
!
the product i∈I Xi is nonempty.
CHAPTER 6. CATEGORIES IN GROUP THEORY
184
We shall discuss the relationship between the two given forms of the Axiom of Choice
in Exercises 6.5.11.
The following argument does not make
! use of the Axiom of Choice. The output, if one
does not accept that axiom, is that if i∈I Xi = ∅, then ∅ is the product of {Xi | i ∈ I}
in Sets. In particular, any infinite family of sets has a product in Sets.
!
Proposition 6.5.2. The product i∈I Xi , together with its projection maps, is the product of {Xi | i ∈ I} in Sets.
Proof Suppose
given a collection of functions fi : Y → Xi for each i ∈ I. Define
!
f : Y → i∈I Xi by f (y) = (fi (y) | i ∈ I). Then πi ◦ f = fi for all i ∈ I. But two
elements of the product are equal if their coordinates are all equal, so f is the unique
function with this property.
Exercises 6.5.3.
1. Show that the category of finite groups and homomorphisms does not have infinite
products.
!
2. Show that if i∈I Xi is the product of {Xi | i ∈ I} in an arbitrary category!C, then
for each
! Y ∈ ob(C) there is an isomorphism of sets between MorC (Y, i∈I Xi )
and i∈I MorC (Y, Xi ), where the latter product is that in Sets. (In fact, this
isomorphism is natural, in the sense of natural transformations, Section 6.9.)
3. Suppose
given objects {Gi | i ∈ I} in either Gp or Ab. Define a!binary operation
!
on i∈I Gi by (xi | i ∈ I) · (yi | i ∈ I) = (xi yi | i ∈ I). Show that i∈I Gi , with this
multiplication, is the product of {Gi | i ∈ I} in the category in question.
Constructions of Infinite Coproducts
The infinite coproduct in Sets is the infinite disjoint union.
Definition 6.5.4. Let {Xi | i ∈ I} be a family
of sets. We define i∈I Xi as follows.
Let S be a set that contains each Xi . Then i∈I Xi is the subset of S × I consisting of
the pairs (x, i) such that x ∈ Xi . The inclusion ιi : Xi → i∈I Xi takes x ∈ Xi to the
pair (x, i).
Proposition 6.5.5. The disjoint union, together with the structure maps ιi , is the coproduct in the category of sets. Thus, given a family {Xi | i ∈ I} ofsets, together with
functions fi : Xi → Y for all i ∈ I, there is a unique function f : i∈I Xi → Y such
that f ◦ ιi = fi for all i.
Proof Given the
functions fi : Xi → Y , define f : i∈I Xi → Y by f (x, i) = fi (x) for
each pair (x, i) ∈ i∈I Xi . Since (x, i) = ιi (x), f satisfies the requirement that f ◦ ιi = fi
for all i, and is the unique function that does so.
The infinite coproduct in Gp is called the infinite free product. It is constructed by
precise analogy to the
free product of two groups. Thus, given a collection {Gi | i ∈ I} of
groups, we set X = i∈I Gi , the disjoint union of the sets Gi . The infinite free product
i∈I Gi is obtained by a passage to equivalence classes from an equivalence relation, ∼,
on the set of all finite words in X.
Here, as in the case of the free product of two groups, the equivalence relation ∼ is
the equivalence relation generated by a notion of elementary reductions. The elementary
reductions are defined as follows.
∗
CHAPTER 6. CATEGORIES IN GROUP THEORY
Definition 6.5.6. Let X =
i∈I
185
Gi and let x1 · · · xn be a word in X.
1. If xi is the identity in the group Gj that contains it, then
e
x1 · · · xn x1 · · · xi−1 xi+1 · · · xn .
2. If xi and xi+1 both lie in the same group Gj ⊂ X, then
e
x1 · · · xn x1 · · · (xi xi+1 ) · · · xn .
∗
As in the case of the free product of two groups, i∈I Gi is a group under the operation
induced by concatenation of words. The canonical map ιi : Gi → i∈I Gi takes g ∈ Gi
to the equivalence class of the singleton word g. The proof of the following proposition
proceeds precisely as did that of Proposition 6.3.15.
∗
Proposition 6.5.7. Let {Gi | i ∈ I} be any family of groups. Then the infinite free
product i∈I Gi , together with the canonical maps ιi : Gi → i∈I Gi , is the coproduct of
{Gi | i ∈ I} in the category of groups.
∗
∗
Intuition might now suggest that the infinite coproduct in Ab would coincide with
the infinite product. However, this is not the case (mainly because we cannot add an
infinite number of group elements). Rather, the coproduct is a subgroup of the infinite
product, called the direct sum.
Definition
6.5.8. Let {Ai | i ∈ I}!be a family of abelian groups. We define the direct
*
sum i∈I Ai to be the subset of i∈I Ai consisting of *
the tuples (ai | i ∈ I) for which
ai = 0 for all but finitely many i. We define ιi : Ai → i∈I Ai as follows: For a ∈ Ai ,
the coordinates of ιi (a) are all 0except for the i-th coordinate, which
*is a.
We shall
customarily
write
a
for
the
I-tuple
(a
|
i
∈
I)
∈
i
i∈I i
i∈I Ai . Note that
the sum i∈I ai is actually finite, as all but finitely many of the ai are 0.
*
!
i∈I Ai is a subgroup of the product
i∈I Ai , and the maps ιi : Ai →
* Clearly,
i∈I Ai are homomorphisms. The next lemma is immediate from the definitions.
Lemma
6.5.9. Let {Ai | i ∈ I} be a family of abelian groups. Then in the direct sum
*
A
,
we have
i
i∈I
ai =
ιi (ai ),
i∈I
i∈I
where the sum on the right is the sum, with respect to the group operation in
of the finitely many terms ιi (ai ) for which ai = 0.
*
i∈I
Ai ,
Proposition
6.5.10. Let {Ai | i ∈ I} be a family of abelian
*
* groups. Then the direct
sum,
A
,
together
with
the
structure
maps
ι
:
A
→
i
i
i∈I i
i∈I Ai , is the coproduct in
the category of abelian groups.
In other words, if B is an abelian group and if fi *
: Ai → B is a homomorphism for
each i ∈ I, then there is a unique homomorphism f : i∈I Ai → B such that f ◦ ιi = fi
for all i ∈ I.
Explicitly, the homomorphism f is given by
"
#
f
ai =
fi (ai ),
i∈I
i∈I
where the right-hand side is the sum in B of the finite number of terms fi (ai ) for which
ai = 0.
CHAPTER 6. CATEGORIES IN GROUP THEORY
186
Proof That f must satisfy the stated formula is immediate from Lemma 6.5.9 and the
requirement
that f ◦ ιi = fi for all i ∈ I. Moreover, the definition of ιi shows that if
*
f : i∈I → B is defined by the stated formula,
= fi for all i, as required.
then f ◦ ιi =
a
Thus, it suffices
to
show
that
setting
f
i∈I i
i∈I fi (ai ) defines a homo*
A
to
B.
But
that
is
immediate
from
the fact that each fi is a
morphism from
i∈I i
homomorphism.
Coproducts are so important in the study of abelian groups that it is often customary
to refer to the finite products as direct sums. Thus, A1 ⊕· · ·⊕Ak is an alternative notation
for A1 × · · · × Ak .
Exercises 6.5.11.
1. Suppose given a family {Xi | i ∈ I} of sets. Define π : i∈I Xi → I by setting π ◦ ιi
equal
to the constant map from Xi to i, for each i ∈ I. Show that the
!
elements of
X
are
in
one-to-one
correspondence
with
the
functions
s
:
I
→
i
i∈I
i∈I Xi such
that π ◦ s = 1I . Deduce that the two forms of the Axiom of Choice are equivalent.
*
!
2. Show
that the direct sum i∈I Ai is the subgroup of the product i∈I Ai generated
by i∈I ιi (Ai ).
3. What are the infinite coproducts in Mon and Amon?
4. For any category C, show that if X is the coproduct of {Xi | i ∈ I},
! then for any object Y , there is an isomorphism of sets between MorC (X, Y ) and i∈I MorC (Xi , Y ).
6.6
Free Functors
Free functors are actually a special case of the adjoint functors discussed in Section 6.9.
For simplicity, and because they constitute a very important special case, we give a
separate treatment of them.
We give an illustration first. Let X be a set. Since abelian groups are sets with
additional structure, we can consider the functions f : X → A, where A is an abelian
group. The free abelian group on X is an abelian group, F A(X), together with a function
ηX : X → F A(X) which is universal in the sense that if f : X → A is any function from
X to an abelian group, then there is a unique group homomorphism f : F A(X) → A
such that the following diagram commutes.
f
/A
XG
;
GG
xx
x
GG
x
GG
x
xx
ηX GG
xx f
#
F A(X)
As usual, we shall refer to F A(X) as the free abelian group on X, even though that
name more properly applies to the function ηX .
We shall give a construction of free abelian groups below. The following proposition
shows how useful they can be.
Proposition 6.6.1. Let f : A → F A(X) be an extension of the abelian group B = ker f
by the free abelian group F A(X). Suppose that A is also abelian. Then the extension
splits, and A is isomorphic to B × F A(X).
CHAPTER 6. CATEGORIES IN GROUP THEORY
187
Proof We shall see below that the function ηX : X → F A(X) is always injective. For
simplicity, we shall make use of this fact, and treat X as a subset of F A(X).
Group extensions are always onto, so that X is contained in the image of f . Thus,
for each x ∈ X there is an element g(x) ∈ A such that f (g(x)) = x. But this defines a
function g : X → A. Since A is abelian, the universal property of free abelian groups
provides a homomorphism g : F A(X) → A such that g(x) = g(x) for all x ∈ X. We
claim that g is a section of f .
Thus, we wish to show that f ◦ g is the identity map of F A(X). But the universal
property shows that homomorphisms out of F A(X) are determined by their restriction
to X. And f ◦ g(x) = f (g(x)) = x. Thus, f ◦ g has the same effect on X as the identity
map, so that f ◦ g is the identity as desired.
Thus, the extension splits. Since A is abelian, F A(X) acts trivially on B = ker f ,
and hence A is the product of B and F A(X).
We define free functors in general as follows. (More general definitions are possible.)
Definition 6.6.2. Suppose given a category C with a forgetful functor to Sets. (In other
words, the objects of C are sets with additional structure, and passage from the objects
to the underlying sets and from morphisms to the underlying functions provides a functor
φ : C → Sets.) Then a free C-object on a set X consists of an object F (X) ∈ ob(C)
together with a function ηX : X → F (X) from X to the underlying set of F (X) with the
universal property that if f : X → A is any function from X to a C-object, then there is
a unique C-morphism f : F (X) → A such that the following diagram commutes.
/A
XE
z<
EE
z
EE
zz
E
zzf
ηX EE
z
"
z
F (X)
f
The free objects we’re most interested in will be very easy to construct. The key is
the following lemma.
Lemma 6.6.3. Let C be a category with a forgetful functor to Sets. Suppose given a
family of sets {Xi | i ∈ I} such that
1. Each set Xi has a free C-object ηi : Xi → F (Xi ).
2. The collection {F (Xi ) | i ∈ I} has a coproduct in C.
*
Write
i∈I F (Xi ) for the coproduct in C of the F (Xi ) and write η :
i∈I Xi →
*
F
(X
)
for
the
unique
function
that
makes
the
following
diagram
commute
for each
i
i∈I
i.
ηi
/ F (Xi )
Xi
ιi ιi ,
+
η /
Xi
F (Xi )
i∈I
i∈I
Here, the maps ιi are the canonical maps for the coproducts, in Sets on the left and in
C on the right.
Then η is the free C-object on i∈I Xi .
CHAPTER 6. CATEGORIES IN GROUP THEORY
Proof Suppose given a function f :
write fi : Xi → A for the composite
i∈I
ι
i
Xi −→
188
Xi → A, with A and object of C. For i ∈ I,
+
f
Xi −
→ A.
i∈I
Then since ηi is the free C-object on Xi , there is a unique C-morphism f i : F (Xi ) → A
such that f i ◦ ηi = fi .
Now apply *
the universal property of the coproduct in C, obtaining the unique morphism f from i∈I F (Xi ) to A such that f ◦ ιi = f i .
An easy diagram chase shows that f ◦ η = f , and uniqueness follows by composition
of the uniqueness property of the coproduct in C and the uniqueness properties of the
free C-objects ηi .
For any set X, we have X = x∈X {x}. Thus, if C has arbitrary coproducts, it will
have arbitrary free objects, provided that a one-point set has a free object. But we have
seen free objects for the one-point set already in Gp and Ab: if G is a group and if g ∈ G,
there is a unique homomorphism f : Z → G with f (1) = g. Thus, if η : {x} → Z takes
x to 1, then η is the free object on {x} in both Gp and Ab. (One-point sets are the only
sets for which free groups are abelian, and therefore are the only sets for which the free
groups and the free abelian groups coincide.)
Proposition 6.6.4. A group is a free abelian group if and only if it is a direct sum of
copies of Z.
A group is a free group if and only if it is a free product of copies of Z.
*
Proof By the preceding lemma, x∈X Z is the free abelian group on X, while x∈X Z
is the free group on X.
But given a direct sum or free product of copies of Z, just let X be the indexing set
for the sum or free product, and this says that our sum or free product is free in the
appropriate category.
∗
We can turn free objects into free functors as follows. The proof follows from the
universal property and is left to the reader.
Proposition 6.6.5. Let C be a category with a forgetful functor to Sets. Suppose that
for every set X we are given a free C-object ηX : X → F (X). Then for any function
f : X → Y there is a unique C-morphism F (f ) : F (X) → F (Y ) such that F (f ) ◦ ιX =
ιY ◦ f .
The correspondence that takes X to F (X) and takes f to F (f ) is a functor F :
Sets → C.
In practice, we identify X with its image under ιX in any of the categories under
discussion. Thus, the free group on X is given by x∈X x, where each x ∈ X has
infinite order, and the generic element may be written xn1 1 . . . xnk k where x1 , . . . , xk ∈ X
are not necessarily distinct, and n1 , . . . , nk ∈ Z.
There is a similar additive notation for the free abelian group on X. Its elements may
be written in the form n1 x1 + · · · + nk xk , where x1 , . . . , xk ∈ X and n1 , . . . , nk ∈ Z. In
this case, because the group is abelian, we may collect terms, and assume that x1 , . . . , xk
are distinct.
∗
Exercises 6.6.6.
CHAPTER 6. CATEGORIES IN GROUP THEORY
189
1. Show that if f : G → F (X) is an extension of a group H by the free group on X,
then f is a split extension.
2. Show that there exist group extensions f : G → F A(X) which cannot be split,
where F A(X) is a free abelian group.
3. Let f : F (X) → G be a homomorphism, with F (X) the free group on X. Show
that f is onto if and only if G is generated by {f (x) | x ∈ X}.
4. Let X be a set and let F (X) and F A(X) be the free group and the free abelian
group on X, respectively. Let f : F (X) → F A(X) be the unique homomorphism
such that the following diagram commutes.
X?
?? η
η 
??

??


?


f
/
F (X)
F A(X)
Here, the maps η are the canonical inclusions of X in the free group and free abelian
group, respectively.
ab
Show that f factors uniquely through an isomorphism f : F (X) ∼
= F A(X) from
the abelianization of F (X) onto F A(X).
5. Let f : F A(X) → A be a homomorphism, with A an abelian group and with F A(X)
the free abelian group on X. Show that f is onto if and only if G is generated by
{f (x) | x ∈ X}.
6. What is the free G-set on a set X?
7. What is the free monoid on a set X?
8. What is the free abelian monoid on a set X?
6.7
Generators and Relations
Let G be a group and suppose that the set X ⊂ G generates G. Then there is a unique
homomorphism f : F (X) → G from the free group on X whose restriction to X is its
inclusion in G. By Problem 3 of Exercises 6.6.6, f is surjective.
In this way, we see that every group is a factor group of a free group. But that,
by itself, isn’t very useful information. We need to be able to say something about the
kernel, K, of f : F (X) → G before we can get any useful information out of it.
Free groups are very complicated objects, and we shall not attempt to say anything
about the case where X is infinite. Thus, we shall assume that G is finitely generated,
by the set X = {x1 , . . . , xk }. We shall write F (X) = F (x1 , . . . , xk ).
It happens that subgroups of free groups are always free groups.2 But even if the
free group is finitely generated, if it is free on more than one generator, then it turns out
to have many subgroups that are not finitely generated. But some infinitely generated
subgroups of a free group turn out to be normally generated by a finite set.
2 Our favorite proof of this uses the topology of covering spaces of graphs. See Section 3.8 of E.H.
Spanier’s Algebraic Topology, McGraw–Hill, 1966.
CHAPTER 6. CATEGORIES IN GROUP THEORY
190
Definitions 6.7.1. A finite presentation for a group G consists of a finite subset X =
{x1 , . . . , xk } ⊂ G which generates G, together with a finite subset R = {R1 , . . . , Rl } of
the free group F (x1 , . . . , xk ) on x1 , . . . , xk , such that the kernel of the homomorphism
f : F (x1 , . . . , xk ) → G
that’s induced by {x1 , . . . , xn } ⊂ G is normally generated by R in F (x1 , . . . , xk ).
Given the above presentation for G, we say that G has generators X and relations
R. In general, given variables x1 , . . . , xn and elements R1 , . . . , Rm of the free group
F (x1 , . . . , xn ), we write x1 , . . . , xn | R1 , . . . , Rm for the group with generators x1 , . . . , xn
and relations R1 , . . . , Rm . Thus,
x1 , . . . , xn | R1 , . . . , Rm = F (x1 , . . . , xn )/R1 , . . . , Rm .
If a group G admits a finite presentation, then we say that G is finitely presented.
Note that a finite presentation of G is the construction of a pushout diagram
F (r1 , . . . , rm )
/ F (x1 , . . . , xn )
e
/ G.
It turns out that not every finitely generated group is finitely presented. Thus, subgroups of finitely generated free groups can be very complicated indeed.
One can show, via the theory of covering spaces, that the commutator subgroup
of the free group, F (x, y), on two letters is infinitely generated. But recall that the
ab
abelianization, F (x, y) , is naturally isomorphic to F A(x, y), the free abelian group of
x and y. Thus, the next example shows that the commutator subgroup of F (x, y) is
normally generated by a single element, xyx−1 y −1 .
Example 6.7.2. We show here that the free abelian group on two letters has the presentation
F A(x, y) = x, y | xyx−1 y −1 .
First note that since F A(x, y) is abelian, xyx−1 y −1 is in the kernel of the map π :
F (x, y) → F A(x, y) which takes x to x and y to y. Thus, there is an induced map
π : G → F A(x, y), where G = x, y | xyx−1 y −1 = F (x, y)/xyx−1 y −1 .
Note that x and y commute in G. Thus, since x and y generate G, G is abelian.
Thus, the universal property of the free abelian group provides a homomorphism ϕ :
F A(x, y) → G with ϕ(x) = x and ϕ(y) = y.
Now notice that the composites ϕ ◦ π and π ◦ ϕ restrict to the identity map on
generators for the two groups. Thus, ϕ and π are inverse isomorphisms.
The hard part in showing that a given group H is given by a particular presentation
is generally in producing a homomorphism from H to the stated quotient of the free
group on the generators. In the above example, the homomorphism is obtained via the
universal property for the free abelian group.
Exercises 6.7.3.
1. Give a finite presentation for the free abelian group on n generators.
CHAPTER 6. CATEGORIES IN GROUP THEORY
191
2. Show that Zn = x | xn .
3. Show that D2n = a, b | a2 , bn , aba−1 b.
4. Show that Q4n = a, b | b2n , a2 bn , aba−1 b.
5. Find a presentation for A4 .
6. Show that x, y | x2 , y 2 , (xy)3 is a presentation for S3 .
7. Show that x, y | x2 , y 2 , (xy)n is a presentation for D2n for each n ≥ 2.
8. Show that x, y, z | x2 , y 2 , z 2 , (xy)3 , (yz)3 , xzx−1 z −1 is a presentation for S4 , via
setting x = (1 2), y = (2 3), and z = (3 4).
9. Let G = x, y | xn , y m , xyx−1 y k for integers n, m, k with m, n > 0. Show that G is
finite and that y G. Do x and y necessarily have order n and m, respectively?
10. Let G = x, y | x4 , y 4 , (xy)4 . Show that Q8 and Z4 × Z4 are both factor groups
of G. What extra relations, when added to those of G, will give Q8 ? What extra
relations, when added to those of G, will give Z4 × Z4 ? Do any of the relations
originally in G become redundant when we do this?
11. Show that x, y | xyx−1 y −2 is a presentation for the group Z[ 12 ]μ2 Z of Problem 19
of Exercises 4.6.13.
12. Given finite presentations for H and K and a homomorphism α : K → Aut(H),
derive a finite presentation for H α K.
6.8
Direct and Inverse Limits
We consider yet another pair of examples of universal properties that replace one diagram
with another. We treat inverse limits here and treat direct limits in a separate subsection.
Inverse limits are related to pullbacks. We suppose given objects Xi for i ≥ 0 and
morphisms pi : Xi → Xi−1 for i ≥ 1. (The indices i here are integers.) We call this data
an inverse system. Pictorially, this gives
/ X 3 p3 / X 2 p2 / X 1 p1 / X 0 .
···
The inverse limit of this system can be thought of as a universal object to tack on at
−∞. What this means is that the inverse limit is a universal diagram of the following
sort.
X JTTTT
JJ TTTT
JJ TTT
TTTT
JJ
$
TT*
/
···
X1 p / X0 .
1
Definition 6.8.1. Suppose given an inverse system, consisting of objects Xi for i ≥ 0
and morphisms pi : Xi → Xi−1 for i ≥ 1. An inverse limit for this system consists
of an object X = ←
lim
−Xi , together with morphisms πi : X → Xi for i ≥ 0 such that
pi ◦ πi = πi−1 for i ≥ 1, and the following universal property is satisfied:
For each object Y and each collection of morphisms fi : Y → Xi such that pi ◦ fi =
fi−1 for all i ≥ 1, there is a unique morphism f : Y → X such that πi ◦ f = fi for all
i ≥ 0.
CHAPTER 6. CATEGORIES IN GROUP THEORY
192
We give the construction of inverse limits in Sets as follows. Suppose given an inverse
system of sets:
···
/ X3
p3 /
X2
p2 /
X1
p1 /
X0 .
!
Let X be the subset of the infinite product i≥0 Xi consisting of those tuples (xi | i ≥
0) with the property that pi (xi ) = xi−1 for all i ≥ 1, and let πi : X → Xi be the
restriction
to X of the i-th projection map of the product (which we shall also write as
!
πi : i≥0 Xi → Xi ).
Proposition 6.8.2. Using the notations just given, the set X, together with the maps
πi : X → Xi , constitutes the inverse limit of the stated inverse system in Sets.
Proof Suppose given functions fi : Y → Xi , i ≥ 0, such that p!i ◦ fi = fi−1 for i ≥ 1.
Then the universal property of products gives a function f : Y → i≥0 Xi with πi ◦f = fi
for all i. As tuples, this says f (y) = (fi (y) | i ≥ 0) for all y ∈ Y . But since pi ◦ fi = fi−1 ,
we have f (y) ∈ lim
←− Xi for all y ∈ Y . In other words, f takes values in the inverse limit.
By construction, πi ◦ f = fi . Moreover, f is the unique map with this property
because X is a subset of the product, and a map into the product is determined by its
coordinate functions.
We now give a useful way to interpret the construction
! of inverse limits. We shall write
(. . . , xi , . . . , x1 , x0 ) for an element of the product i≥0 Xi , reversing the usual order.
!
!
Then there is a “shift” map s : i≥0 Xi → i≥0 Xi given by s(. . . , xi , . . . , x1 , x0 ) =
(. . . , pi+1 (xi+1 ), . . . , p2 (x2 ), p1 (x1 )). It is easy to see that the inverse limit lim
←− Xi is the
!
set of all x ∈ i≥0 Xi such that s(x) = x. In other words, the inverse limit is the set of
!
fixed points of the shift map on i≥0 Xi .
As was the case for products and pullbacks, the construction of inverse limit that
works in Sets also produces the inverse limit in most of the other categories we’ve been
considering.
Proposition 6.8.3. Suppose given an inverse system
···
/ G3
h3 /
G2
h2 /
G1
h1 /
G0
!
in the category of groups, and let G ⊂ i≥0 Gi be the inverse limit of the induced inverse
system of sets (i.e., G = {(gi | i ≥ 0) | hi (gi ) = gi−1 for all i ≥ 0}). Then G is a subgroup
of the direct product of the Gi , and
! with this multiplication, together with the restrictions
to G of the projection maps πi : i≥0 Gi → Gi , forms the inverse limit of this system in
the category of groups.
In addition, if the groups Gi are abelian, so is G, and in this case, G, together with
the projection maps is the inverse limit of the given system in the category of abelian
groups.
Proof That G is a subgroup of the direct product follows from the fact that the hi
are group homomorphisms. Since the direct product is the product in the category of
groups, G, together with the projection maps, satisfies the universal property for the
inverse limit by exactly the same proof as that given for the category of sets. A similar
argument works for abelian groups.
CHAPTER 6. CATEGORIES IN GROUP THEORY
193
We shall see in Chapter 7 that the same construction gives the inverse limit in the
p , in some
category of rings. We shall study one such example, the p-adic integers, Z
depth. Zp is the inverse limit of the unique system of ring homomorphisms
···
/ Zp3
/ Zp2
/ Zp .
Exercises 6.8.4.
1. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets is
lim
surjective, then so is each function πi : ←
− Xi → Xi .
2. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets is
bijective, then so is each function πi : ←
lim
− Xi → Xi .
3. Show that the construction given for Sets, Gp, and Ab also produces the inverse
limit in G-sets, Mon, and Amon.
4. Show that if X = ←
lim
− Xi is the inverse limit for an inverse system consisting of
objects Xi , i ≥ 0, and morphisms pi : Xi → Xi−1 in a category C, then for any
object Y of C, MorC (Y, X) is the inverse limit in Sets of the MorC (Y, Xi ).
Direct Limits
Direct limit is the concept dual to inverse limit:
Definition 6.8.5. A direct system in C consists of objects Xi for i ≥ 0, together with
morphisms ji : Xi → Xi+1 for i ≥ 0. Pictorially, we get
X0
j0 /
X1
j1 /
X2
j2 /
X3
/ ···
A direct limit for this system consists of an object X = −
lim
→ Xi , together with morphisms
ιi : Xi → X for i ≥ 0, such that ιi+1 ◦ ji = ιi for i ≥ 0 and such that the following
universal property is satisfied. Given an object Y and morphisms fi : Xi → Y such that
lim
fi+1 ◦ji = fi for all i ≥ 0, there is a unique morphism f : −
→ Xi → Y such that f ◦ιi = fi
for all i ≥ 0.
Surprisingly (since the direct limit is related to coproducts and pushouts in the same
way that the inverse limit is related to products and pullbacks), the construction of the
direct limit is the same in all the categories we’ve been discussing. The intuition is that
if the morphisms ji are all injective, then the direct limit is the union of the Xi . (See
Problem 6 of Exercises 6.8.8.) If the ji are not injective, we need to work a little harder
to construct the direct limit.
Assume that we’re working in Sets. We shall construct −
lim
by passage to equiva→X
i
lence classes from an equivalence relation on the disjoint union i≥0 Xi .
Here, we use the equivalence relation generated by the following notion of elementary
equivalence: if x ∈ Xi , we set x equivalent
to ji (x) ∈ Xi+1 . We write x ∼e ji (x). We
then declare two elements x, x ∈ i≥0 Xi to be equivalent, written x ∼ x , if there is
a sequence x = x0 , . . . , xn = x such that for i = 0, . . . , n − 1, either xi ∼e xi+1 or
xi+1 ∼e xi .
We define −
lim
→ Xi to be the set of equivalence classes under ∼, and write π : i≥0 Xi →
lim
−→ Xi are defined to be the
−→ Xi for the quotient map. The structure maps ιi : Xi → lim
CHAPTER 6. CATEGORIES IN GROUP THEORY
194
composites
Xi
where ιi : Xi →
i≥0
ιi
/
+
Xi
i≥0
π / lim X ,
−→ i
Xi is the natural inclusion of Xi in the coproduct.
Proposition 6.8.6. The above construction gives the direct limit in Sets.
Proof Given functions fi : Xi
→ Y with fi+1 ◦ ji = fi for all i ≥ 0, there is a unique
function f from the coproduct i≥0 Xi to Y such that f ◦ ιi = fi for all i, where ιi is
the natural inclusion of Xi in the coproduct. Write ≈ for the equivalence relation on
i≥0 Xi given by setting x ≈ x if f (x) = f (x ). Since fi+1 ◦ ji = fi for all i ≥ 0, we see
that x ∼e x implies that x ≈ x .
By Proposition 1.5.3, if x ∼ x , then f (x) must equal f (x ). Thus, f factors (necessarily uniquely, since π is onto) through a function f : −
lim
→ Xi → Y . But now f clearly
has the desired properties.
Example 6.8.7. Let p be a prime. We shall construct a group by taking direct limits
of cyclic p-groups. Specifically, note the equivalence class of p mod pi has order pi−1 in
the additive group Zpi . Thus, there is a group homomorphism ji−1 : Zpi−1 → Zpi which
takes 1 to p: explicitly, ji−1 (n) = pn. Since p has order pi−1 , ji−1 is injective.
Clearly, the maps {ji | i ≥ 1} form a direct system in Ab. We write Tp for the direct
limit of this system, which by Problem 4 below is an abelian group.
Exercises 6.8.8.
1. For any direct system of sets, show that any element of the direct limit lies in the
image of ιi for some i.
2. In Sets, show that if x, y ∈ Xi are equivalent in −
lim
→ Xi , then there exists k ≥ 0
such that ji+k ◦ · · · ◦ ji (x) = ji+k ◦ · · · ◦ ji (y).
3. Suppose given a direct system in Gp. Show that the direct limit of the underlying
sets has a unique group structure that makes it the direct limit of this system in
Gp.
† 4. Repeat the preceding problem in Ab, G-sets, Mon, and Amon.
5. In Sets, suppose there is a set U such that Xi ⊂ U for all i ≥ 0. Suppose also that
each composite
ji
Xi −→ Xi+1 ⊂ U
coincides with the inclusion
Xi ⊂ U . Show that we may identify the direct limit
lim
X
with
the
union
X
i
i ⊂ U.
i≥0
−→
† 6. More generally, suppose given a direct system of sets such that each function ji :
Xi → Xi+1 is injective. Show that the structure maps ιi : Xi → lim
−→ Xi are also
injective. In addition, show that −
lim
X
is
the
union
of
the
images
of
the Xi .
→ i
7. Let Tp , as defined above, be the direct limit of the cyclic groups Zpi and let ιi :
Zpi → Tp be the structure map for the direct limit. By Problem 6, ιi is an injection.
Show that Tp is a p-torsion group and that the image of ιi the subgroup consisting
of those elements of exponent pi .
CHAPTER 6. CATEGORIES IN GROUP THEORY
195
8. Let p be a prime and let Z[ p1 ] ⊂ Q be the subgroup of the rational numbers
consisting of all quotients of the form pmi , where m ∈ Z and i ≥ 0. Show that Z[ p1 ]
is isomorphic to the direct limit of the system
p
p
Z → Z → Z → ...
where each group in the direct system is Z and each map is multiplication by p.
9. Show that the torsion group Tp above is isomorphic to the factor group Z[ p1 ]/Z.
10. Show that any countable abelian torsion group is a direct limit of finite abelian
groups.
11. Show that if G is finite, then a countable G-set is a direct limit of finite G-sets.
12. Show that if X = −
lim
→ Xi is the direct limit for a direct system consisting of objects
Xi , i ≥ 0, and morphisms ji : Xi → Xi+1 in a category C, then for any object Y
of C, MorC (X, Y ) is the inverse limit in Sets of the MorC (Xi , Y ).
6.9
Natural Transformations and Adjoints
In our study of semidirect products, we found that it was interesting to be able to
determine when two homomorphisms are related in certain ways (e.g., conjugation).
Similarly, it is often of interest to relate the effect of two different functors.
Definition 6.9.1. Let C and D be categories, and let F and G be two functors from C
to D. A natural transformation η : F → G consists of a D-morphism ηX : F (X) → G(X)
for each X ∈ ob(C) such that for each f ∈ MorC (X, Y ), the following diagram commutes.
F (f )
/ F (Y )
F (X)
ηX ηY G(f )
/ G(Y )
G(X)
We say that η is a natural isomorphism if ηX is an isomorphism in D for each object
X of C.
With straightforward modifications, we obtain the definitions of natural transformations or isomorphisms between two contravariant functors.
The value of naturality is not immediately apparent. However, it turns out to be a
strong tool in certain kinds of mathematical argument.
On the theoretical level, naturality permits us to define equivalence of categories.
Definition 6.9.2. An equivalence of categories between C and D consists of functors
F : C → D and G : D → C, together with natural isomorphisms between G ◦ F and the
identity functor of C, and between F ◦ G and the identity functor of D.
One place where equivalence of categories turns out to be useful is in the notion of
skeletal subcategories.
CHAPTER 6. CATEGORIES IN GROUP THEORY
196
Definition 6.9.3. Let D be a subcategory of C, and suppose that the inclusion functor
i : D → C is an equivalence of categories. Then we say that D is a skeleton (or skeletal
subcategory) of C.
We shall illustrate a nice feature of skeletal subcategories presently. First, we have
another definition.
Definition 6.9.4. A small category is a category whose objects form a set.
For instance, consider the category of finite groups. It is not small. In fact, the
collection of all objects that are isomorphic to the trivial group is too big to be a set.
But we can recover most of the benefits of smallness if we can construct a skeleton which
is small.
Recall from Corollary 2.6.6 that there are countably many isomorphism classes of
finite groups. In particular, the isomorphism classes of finite groups form a set. (This
does not happen in Gp, as there is no bound on the cardinality of the underlying sets of
the objects in Gp.)
Proposition 6.9.5. Let D be a full subcategory of C, and suppose that we’re given, for
each X ∈ ob(C) an isomorphism ηX : X ∼
= F (X) for some object F (X) of D. Then D
is a skeleton for C.
Proof We claim that there is a unique way to make F into a functor from C to D in
such a way that η is a natural transformation. But if this is the case, we’re done, since
η then provides a natural isomorphism from the identity functor of C to i ◦ F , as well as
from the identity functor of D to F ◦ i, where i is the inclusion of D in C.
Because the morphisms ηX are all isomorphisms, it is easy to give the functoriality
−1
of F : For f : X → Y in C, we take F (f ) = ηY ◦ f ◦ ηX
. This immediately gives the
naturality diagram for η, and the functoriality of F follows by cancelling inverses.
However, the fact that we have to choose an isomorphism ηX : X ∼
= F (X) for each
object X of C can be problematical from a set-theoretic point of view, as it requires
an extension from sets to classes of the Axiom of Choice. Thus, it is not clear that the
category of finite groups has a small skeleton. For any group G, we could use Cayley’s
Theorem to identify G with a subgroup of S(G), the group of symmetries on G, but
there’s no natural ordering of the elements of G, and hence no natural identification of
S(G) with S|G| .
However, this need not be a serious issue. Any real mathematical question that we
cared about would probably either concern isomorphism classes of objects or concern
information relating to particular sorts of diagrams of objects. The latter questions can
generally be phrased in terms of functors from a small category into the category one
wishes to study. Thus, for the questions we care about, the next result will show that
we do not really need a small skeleton.
The proof of the next proposition is based on the method of proof of Proposition 6.9.5.
It is left to the reader.
Proposition 6.9.6. Let D be a full subcategory of C such that D contains an object in
every isomorphism class of objects of C. Let E be any small category. Then any functor
from E into C is naturally isomorphic to a functor that takes value in D.
Also, since D is full in C, any two functors from E into D that are naturally isomorphic as functors into C are also naturally isomorphic as functors into D.
CHAPTER 6. CATEGORIES IN GROUP THEORY
197
Thus, the category of finite subgroups of S∞ , which is small, contains all the information of interest with regard to the category of finite groups.
Just as group isomorphisms are special examples of homomorphisms, equivalences of
categories are special examples of what are called adjoint functors.
Definitions 6.9.7. We say that a pair of functors R : C → D and L : D → C is an
adjoint pair if for each X ∈ ob(D) and Y ∈ ob(C), there is an isomorphism (of sets)
ΦX,Y : MorC (L(X), Y ) ∼
= MorD (X, R(Y )) which is natural in the following sense: given
morphisms f1 : X → X in D and f2 : Y → Y in C, then the following diagram
commutes.
ΦX,Y
/ MorD (X, R(Y ))
MorC (L(X), Y )
L(f1 )∗ ◦ f2 ∗
f1∗ ◦ R(f2 )∗
ΦX ,Y / MorD (X , R(Y ))
MorC (L(X ), Y )
Here, g∗ (h) = g ◦ h = h∗ (g).
If R and L form an adjoint pair, we call R a right adjoint for L and call L a left
adjoint for R. The pair itself can be called an adjunction.
Another way of expressing the naturality property for adjoint functors is as follows.
Fixing X ∈ ob(D), then filling in the blanks provides two functors, MorC (L(X), −)
and MorD (X, R(−)), from C to Sets, and ΦX,− provides a natural isomorphism between them. At the same time, for each Y ∈ ob(C), we have contravariant functors
MorC (L(−), Y ) and MorD (−, R(Y )) from D to Sets, and Φ−,Y provides a natural isomorphism between them.
It turns out that we’ve already seen several examples of adjoint functors. For instance,
if C comes equipped with a forgetful functor to Sets and if ηX : X → F (X) is a free
C-object for each set X, then the universal property of free objects says precisely that
ηX ∗ : MorC (F (X), Y ) ∼
= MorSets (X, Y )
for each object Y of C. The right adjoint here is the forgetful functor to Sets, which
doesn’t really require notation. Naturality in the variable Y is immediate, while naturality in the first variable is shown in Proposition 6.6.5.
We’ve seen some other examples as well.
Exercises 6.9.8.
1. Show that abelianization is a left adjoint to the inclusion of Ab in Gp.
2. Let H be a subgroup of the group G. Then there is a forgetful functor from G-sets to
H-sets. Show that this forgetful functor has a left adjoint, given by L(X) = G×H X
for any H-set X.
3. Let H be a subgroup of the group G and let R : G-sets → Sets be given by R(X) =
X H , the H-fixed point set of X. Define L : Sets → G-sets by L(X) = (G/H) × X.
Show that these form a pair of adjoint functors.
4. Show that passage from a monoid to its group of invertible elements is a right
adjoint to the forgetful functor from groups to monoids.
CHAPTER 6. CATEGORIES IN GROUP THEORY
198
5. Suppose given an adjunction ΦX,Y : MorC (L(X), Y ) ∼
= MorD (X, R(Y )). For X ∈
ob(D), let ηX : X → R(L(X)) be given by ηX = ΦX,L(X) (1L(X) ). Show that η is a
natural transformation from the identity functor to R ◦ L. It is customarily called
the unit of the adjunction. Show also that this unit determines the isomorphism Φ
as follows: for f ∈ MorC (L(X), Y ), ΦX,Y (f ) = R(f ) ◦ ηX .
6. Suppose given an adjunction ΦX,Y : MorC (L(X), Y ) ∼
= MorD (X, R(Y )). For Y ∈
ob(C), let εY : L(R(Y )) → Y be given by εY = Φ−1
R(Y ),Y (1R(Y ) ). Show that ε
gives a natural transformation from L ◦ R to the identity functor. We call it the
counit of the adjunction. Show that it determines the isomorphism Φ as follows:
for g ∈ MorD (X, R(Y )), Φ−1
X,Y (g) = εY ◦ L(g).
7. Show that left adjoints preserve coproducts, pushouts, and direct limits. This
means, for instance, that if we apply a left adjoint to a pushout diagram, the
resulting diagram is also a pushout diagram.
8. Show that right adjoints preserve products, pullbacks, and inverse limits.
The last two problems should explain some of the results in the preceding sections,
as well as predict some others that we haven’t proven yet. That, indeed is one of the
main points in favor of category theory. And in some cases, knowing about adjoints can
help explain why a theorem is true. For instance, it may be instructive to reconsider the
notion of adjoint when one looks at the Frobenius Reciprocity Theorem in representation
theory.
6.10
General Limits and Colimits
Products, coproducts, pushouts, pullbacks, direct and inverse limits are examples of a
couple of much more general types of construction. In all these cases, we’re given some
kind of diagram in C, and we’re asked to extend it to a larger diagram that satisfies a
universal property.
The key to generalizing is in what constitutes a diagram.
Definition 6.10.1. Let I be a small category and let C be an arbitrary category. By
an I-diagram in C, we mean a functor X : I → C. We shall also refer to X as a diagram
in C with indexing category I.
We’ll show in Exercises 6.10.7 how to interpret the universal constructions we’ve
already seen in terms of functors.
By analogy with the constructions we’ve seen, the limit of an I-diagram in C will
be defined to be a diagram defined on a larger indexing category that satisfies a certain
universal property.
Definition 6.10.2. For a category I, let I + be the category defined as follows. The
objects of I + consist of those in I, together with one new object, denoted ∗. The
morphisms in I + between objects that come from I are the same as those in I (i.e., I is
a full subcategory of I + ). In addition, we declare that MorI + (∗, i) has a single element
for all objects i of I + , while MorI + (i, ∗) = ∅ if i = ∗.
There is, of course, a unique composition law which makes this a category.
Clearly, ∗ is an initial object for I + . Indeed, I + is the result of adjoining an initial
object to I. (If I already had an initial object, that object will no longer be initial in
I + .)
CHAPTER 6. CATEGORIES IN GROUP THEORY
199
Definition 6.10.3. Let I be a small category and let X : I → C be a functor. A limit
+
+
for X, denoted ←
lim
− X, is a functor X : I → C whose restriction to I ⊂ I is X, and
which is universal in the following sense: if Y : I + → C is another such extension of
X to I + , then there is a unique natural transformation η : Y → X such that ηi is the
identity map of Xi whenever i is an object of I.
Thus, the limit of an I-diagram is an I + -diagram.
Colimits are defined in terms of the category obtained by adjoining a terminal object
to I.
Definition 6.10.4. For a category I, let I+ be the category defined as follows. The
objects of I+ consist of those in I, together with one new object, denoted ∗. The
morphisms in I+ between objects that come from I are the same as those in I (i.e.,
I is a full subcategory of I+ ). In addition, we declare that MorI+ (i, ∗) has a single
element for all objects i of I+ , while MorI+ (∗, i) = ∅ if i = ∗.
Colimits are obtained as follows.
Definition 6.10.5. Let I be a small category and let X : I → C be a functor. A colimit
for X, denoted −
lim
→ X, is a functor X : I+ → C whose restriction to I ⊂ I+ is X, and
which is universal in the following sense: if Y : I+ → C is another such extension of X to
I+ , then there is a unique natural transformation η : X → Y such that ηi is the identity
map of Xi whenever i is an object of I.
Limit and colimit constructions indexed by arbitrary small categories can be quite
complicated. It is often desirable to restrict attention to diagrams indexed by the category
associated to a partially ordered set. (Recall that a partial ordering on a set I is a relation,
generally denoted ≤ by default, that is reflexive, transitive, and antisymmetric.)
Definition 6.10.6. Let I be a partially ordered set. The category, I, associated to I is
the category whose objects are the elements of I, and whose morphisms are defined as
follows: for i, j ∈ I, MorI (i, j) is empty unless i ≤ j, in which case it has one element,
which we denote by i ≤ j.
I is a category, as transitivity gives the composition of morphisms, while reflexivity
gives the existence of identity morphisms. Since I is a set, I is small.
Exercises 6.10.7.
1. Show that a small category I comes from a partially ordered set if and only if the
following properties hold.
(a) For each i, j ∈ ob(I), MorI (i, j) has at most one element.
(b) If i = j and if MorI (i, j) = ∅, then MorI (j, i) = ∅.
2. Let I be a set. Then we may consider it to be a partially ordered set in which
i ≤ j if and only if i = j. Let I be the induced category and let C be an arbitrary
category. Note that an I-diagram, X, in C is simply a family {X(i)|i ∈ I} of Cobjects indexed by I. Show that specifying a limit for X is equivalent to specifying
a product for {X(i)|i ∈ I} and that specifying a colimit for X is equivalent to
specifying a coproduct for {X(i)|i ∈ I}.
CHAPTER 6. CATEGORIES IN GROUP THEORY
200
3. Let I be a partially ordered set with three elements, i, j and k, where i ≤ j, i ≤ k,
and there are no other nonidentity morphisms in the induced category I. Show
that ordinary diagrams in a category C which have the form
•
/ •
•
are in one-to-one correspondence with I-diagrams in C. Under this correspondence,
show that specifying a colimit for a functor X : I → C is equivalent to specifying
a pushout for the associated ordinary diagram in C.
4. Give the dual of the preceding exercise.
5. Let N be the non-negative integers with the usual ordering, and with associated
category N . Show that direct systems in a category C are in one-to-one correspondence with functors from N into C. Under this correspondence, show that
specifying a colimit for an N -diagram is equivalent to specifying a direct limit for
the associated direct system.
6. Give the dual of the preceding exercise.
7. Let I be a small category and let C be any category. Show that there is a category,
which we denote C I , whose objects are the I-diagrams in C and whose morphisms
are the natural transformations between them.
8. Suppose we’re given a construction that produces a colimit for every I-diagram in
C. Show that we get an induced functor from C I to C I+ .
9. Suppose we’re given a construction that produces a limit for every I-diagram in C.
+
Show that we get an induced functor from C I to C I .
10. Show that left adjoints preserve colimits of I-diagrams.
11. Show that right adjoints preserve limits of I-diagrams.
Chapter 7
Rings and Modules
The study of rings and their modules is vastly more complicated than that of groups. In
this chapter, long as it is, we shall do little more than set the stage for this study. We
shall give some examples that we shall study in greater depth in later chapters, and shall
give a number of basic definitions, constructions, and questions for further study.
Section 7.1 defines concepts such as zero-divisors, unit groups, and algebras over
commutative rings. Examples are given from matrix rings, the rings Zn , the complex
numbers, C, and its subrings, the division ring H of quaternions, and the p-adic integers,
p.
Z
Section 7.2 defines left, right, and two-sided ideals. Quotient rings are defined, and
their universal property is given. Principal ideals and generators for ideals are discussed.
Operations on ideals are considered, and their use is illustrated via the Chinese Remainder Theorem.
Section 7.3 develops polynomials in one and several variables. Evaluation maps are
studied, with the usual applications to the theory of roots. Finitely generated algebras
are discussed. The cyclotomic extensions Z[ζn ] and Q[ζn ] are featured as examples.
Section 7.4 studies the symmetric polynomials, showing that they form a polynomial
algebra on the elementary symmetric functions. Applications are given to the theory of
discriminants.
Section 7.5 develops group rings and monoid rings. It is shown that polynomials form
an example of the latter. Augmentation ideals are studied, and are generalized to the
notion of augmented algebras. The relationship between Z[Zn ] and Z[ζn ] is studied in
the exercises.
Then, in Section 7.6, the theory of prime and maximal ideals is developed for commutative rings. Krull dimension is introduced. So are the ascending and descending chain
conditions for ideals, in the context of the discussion of maximality principles for families
of ideals.
Section 7.7 introduces modules. Cyclic modules, generating sets, annihilators, and
internal and external operations on modules are studied. The theories of free modules
and of exact sequences are developed. The section closes with a discussion of rings
without identity.
Using the techniques developed for studying modules, the ascending and descending
chain conditions are studied for modules and for left and right ideals in Section 7.8. The
relationship between chain conditions and extensions of modules is studied, as is the
relationship between the Noetherian property and finite generation. The section closes
with the Hilbert Basis Theorem and its application to finitely generated algebras over a
201
CHAPTER 7. RINGS AND MODULES
202
Noetherian ring.
Then, Section 7.9 classifies the vector spaces over a division ring, and shows that the
finitely generated free modules over a commutative ring have a well defined rank.
Section 7.10 gives the relationship between matrix rings and the endomorphism rings
of finitely generated free modules.
We close the chapter with a development of rings and modules of fractions. Local rings
and localization are studied. The field of fractions of an integral domain is constructed,
showing that a ring is an integral domain if and only if it is a subring of a field.
7.1
Rings
We give some basic definitions and examples of rings. We then treat the complex numbers, the quaternions, and the p-adic integers in subsections.
We shall follow the tradition that insists that a ring should have a multiplicative
identity. We give a brief discussion of rings without identity in the last subsection of
Section 7.7.
Definitions 7.1.1. A ring A is a set with two binary operations, addition and multiplication.1 Addition gives A the structure of an abelian group, with identity element
0. Multiplication provides A with a (not necessarily abelian) monoid structure, with
identity element 1. The two operations interrelate via the distributive laws:
a(b + c) = ab + ac
and
(b + c)a = ba + ca
for all a, b, c ∈ A.
If the multiplication does satisfy the commutative law, we call A a commutative ring.
Let A and B be rings. Then a ring homomorphism f : A → B from A to B is a
function such that f (a + b) = f (a) + f (b) and f (ab) = f (a)f (b) for all a, b ∈ A, and
f (1) = 1. As usual, a ring isomorphism is a ring homomorphism that is bijective.
A subring of a ring A is a subgroup B of the additive group of A, such that B is
closed under multiplication and contains the multiplicative identity element 1.
Finally, the invertible elements in the multiplicative monoid of a ring A are called the
units of A. They form a group under multiplication, denoted A× .
Let f : A → B be a ring homomorphism. Then f gives a group homomorphism
between the additive groups of A and B, so we can talk about its kernel. Note that the
kernel of f cannot be a subring of A unless 1 = 0 in B, because f (1) = 1. Nevertheless,
ker f is more than just a subgroup of A. It is a two-sided ideal, a concept we shall study
below.
Images are a different story:
Lemma 7.1.2. Let f : A → B be a ring homomorphism. Then the image of f is a
subring of B.
The next example is the exception, rather than the rule.
Example 7.1.3. There is a ring with only one element. We denote the element by 0 and
set 0 + 0 = 0 and 0 · 0 = 0. In other words, both addition and multiplication are given
by the unique binary operation on a one-element set. The element 0 serves as identity
for both operations. Thus, with the multiplicative identity set equal to 0, the properties
of a ring are satisfied.
We call this ring the 0 ring and denote it by 0.
1 The
use of A for a ring derives from anneau, the French word for ring.
CHAPTER 7. RINGS AND MODULES
203
The next lemma shows that the 0 ring is the only ring in which 0 = 1.
Lemma 7.1.4. Let A be a ring. Then 0 · a = a · 0 = 0 for all a ∈ A.
Proof We have
0a = (0 + 0)a = 0a + 0a.
But then subtracting 0a from both sides gives 0 = 0a. The case of a · 0 is similar.
The distributive laws now give a useful fact.
Corollary 7.1.5. Let A be a ring and let −1 ∈ A be the additive inverse of 1. Then for
any a ∈ A, (−1) · a = a · (−1) is the additive inverse of a.
The next result is an important generalization of the preceding ones. The proof is
left to the reader.
Lemma 7.1.6. Let A be a ring. For m ∈ Z, write m for the m-th power of 1 in the
additive group of A. (Thus, if m is positive, m is the sum of m copies of 1 in A.) Then
for any a ∈ A, m · a = a · m is the m-th power of a in the additive group of A.
Examples 7.1.7. The most obvious examples of rings are the integers, rational numbers, and real numbers, Z, Q, and R, with the usual addition and multiplication. The
latter two examples turn out to have some interesting subrings, but Z does not: Since 1
generates Z as an abelian group, the only subring of Z is Z itself.
We’ve also seen (in Proposition 2.4.7) that multiplication in Zn induces a commutative
ring structure on Zn such that the canonical map π : Z → Zn is a ring homomorphism.
(Since π is surjective, this is the only ring structure that makes π a ring homomorphism.)
The integers play a special role in ring theory:
Lemma 7.1.8. Let A be a ring. Then there is a unique ring homomorphism from Z to
A. Moreover, the image of each m ∈ Z under this homomorphism commutes2 with every
element of A.
Proof Ring homomorphisms are required to preserve the multiplicative identity elements. But recall from Proposition 2.5.6 that for any group G, and any element x ∈ G,
there is a unique group homomorphism from Z to G that carries 1 to x. Thus, there is
at most one ring homomorphism from Z to A: as a homomorphism of additive groups,
it must be the homomorphism f : Z → A for which f (1) = 1. It suffices to show that
this group homomorphism is also a ring homomorphism for any ring A, and that f (m)
commutes with every element of A.
Note that since f is a homomorphism of additive groups and since f (1) = 1, f (n)
must be the n-th power of 1 in the additive group structure on A, for all n ∈ Z. Thus,
Lemma 7.1.6 says that for each n ∈ Z, f (n) commutes with every element of a and that
f (n) · a is the n-th power of a in the additive group structure on A.
Thus, f (m)f (n) is the m-th power of the n-th power of 1 with respect to the additive
group operation in A. So f is a ring homomorphism by the power laws for a group.
2 When we say that two elements of a ring commute, we mean this with respect to the operation of
multiplication. The point is that the addition operation is abelian so that two elements always commute
additively.
CHAPTER 7. RINGS AND MODULES
204
In categorical language, this implies that Z is an initial object for the category of
rings. Clearly, 0 is a terminal object.
We now give some examples of unit groups.
Examples 7.1.9.
1. It is easy to see, e.g., via the order inequalities for multiplication, that Z× = {±1}.
2. In Lemma 4.5.3, it is shown that m ∈ Z×
n if and only if (m, n) = 1. Then, in
Corollary 4.5.7, it is shown that if n = pr11 . . . prkk , where p1 , . . . , pk are distinct
primes, then
×
∼ ×
Z×
n = Zpr1 × · · · × Zprk .
1
k
Finally, the unit groups Z×
pr are explicitly calculated in Corollaries 4.5.20 and 4.5.21.
We now give some language for describing properties of rings and their elements.
Definitions 7.1.10. Suppose we have two elements a, b in a ring A, neither of which is
0, such that ab = 0. Then a and b are said to be zero-divisors in A. More specifically, if
A is noncommutative, a is a left zero-divisor and b is a right zero-divisor.
We say that an element a ∈ A is nilpotent if an = 0 for some integer n. Note that
any nonzero nilpotent element is a zero-divisor.
Generalizing the obvious facts about the integers, we say that an integral domain, or
domain, for short, is a nonzero commutative ring with no zero-divisors.
A nonzero ring is called a division ring if every nonzero element in it is a unit. If the
ring is also commutative, we call it a field.
The next lemma connects some of these definitions.
Lemma 7.1.11. A division ring has no zero-divisors. Thus, every field is an integral
domain.
Proof Let A be a division ring and let a, b ∈ A such that a = 0 and ab = 0. It suffices
to show that b = 0.
Because A is a division ring and a = 0, a has a multiplicative inverse, a−1 . But then
0 = a−1 · 0 = a−1 · ab = (a−1 a)b = b.
Of course, Z is an example of an integral domain, and Q and R are examples of fields.
The following lemma could have been left as an exercise, but we shall make sufficient
use of it that a proof has been given. The reader may wish to investigate alternative
proofs.
Lemma 7.1.12. Let n > 1. Then Zn is an integral domain if and only if n is prime.
Moreover, for any prime number p, Zp is a field.
Proof Suppose that n is not prime. Then we can write n = kl, where k and l are both
greater than 1. But then 1 < k, l < n, so k and l are nonzero in Zn . But k · l = kl = 0,
and hence k and l are zero-divisors in Zn .
On the other hand, if p is prime, then each nonzero element of Zp is shown to be a
unit in Lemma 4.5.3.
In a noncommutative ring, it is useful to keep track of the elements that commute
with all other elements of the ring.
CHAPTER 7. RINGS AND MODULES
205
Definition 7.1.13. Let A be a ring. Then the center of A is the set
{a ∈ A | ab = ba for all b ∈ A}
consisting of those elements of a that commute with every element of A.
Lemma 7.1.14. Let A be a ring. Then the center of A is a subring of A.
Our first examples of noncommutative rings come from matrices.
Definitions 7.1.15. Let A be a ring (not necessarily commutative). We write Mn (A)
for the collection of n × n matrices with coefficients in A. Given elements aij ∈ A with
1 ≤ i ≤ n and 1 ≤ j ≤ n, we write (aij ) ∈ Mn (A) for the matrix whose ij-th entry is aij .
We define addition in Mn (A) so that the ij-th entry of (a
ijn) + (bij ) is aij + bij and
define multiplication so that the ij-th entry of (aij ) · (bij ) is k=1 aik bkj . Note that the
order of the factors in the last expression is important if A is noncommutative.
We write 0 for the matrix whose entries are all 0, and write In for the matrix
1 0 0 whose
off-diagonal entries are all 0 and whose diagonal entries are all 1, e.g., I3 = 0 1 0 .
001
More generally, for a ∈ A, we write aIn for the matrix whose diagonal entries are all
equal to a and whose off-diagonal entries are all 0. We write ι : A → Mn (A) for the map
given by ι(a) = aIn for all a ∈ A.
We summarize the most basic properties of matrix rings:
Lemma 7.1.16. Let A be a ring. Then Mn (A) is a ring under the operations of matrix
addition and matrix multiplication, and the map ι : A → Mn (A) is a ring homomorphism.
Moreover, the elements ι(a) = aIn behave as follows: if M = (aij ) ∈ Mn (A), then
aIn · M is the matrix whose ij-th entry is a · aij for all i, j, while M · aIn is the matrix
whose ij-th entry is aij · a for all i, j.
Thus, if B is the center of A, then ι(B) is contained in the center of Mn (A).
The units in Mn (A) are those matrices M for which there exists an M ∈ Mn (A)
such that M M = M M = In . Note that this precisely says that M is invertible in the
usual sense from linear algebra.
×
Definitions 7.1.17. Let A be a ring. We write Gln (A) for Mn (A) , the group of
invertible n × n matrices. We call it the n-th general linear group of A.
The reader should be warned that if A is not commutative, then the usual sort of
determinant theory fails for matrices over A, and hence invertibility cannot be detected
by the familiar methods.
When A is commutative, on the other hand, there is a good theory of determinants,
which we shall develop in Chapter 10. We shall show that a matrix in Mn (A) is invertible
if and only if its determinant is in A× .
Another example relates to a generalization of matrix rings.
Definition 7.1.18. Let G be an abelian group and let EndZ (G) be the set of all group
homomorphisms from G to itself. We call it the endomorphism ring of G.
Here, the ring operations are defined as follows. For f, g ∈ EndZ (G), we let f + g
be the homomorphism given by (f + g)(x) = f (x) + g(x), where additive notation is
used for the group operation in G. The product, f · g, of f, g ∈ EndZ (G) is simply the
composition f ◦ g.
As usual, the justification for this definition is left to the reader:
CHAPTER 7. RINGS AND MODULES
206
Lemma 7.1.19. Under these operations, EndZ (G) is a ring, with additive identity the
trivial homomorphism, and multiplicative identity the identity map.
We shall generalize this when we study A-modules. If M is an A-module, then the
A-module homomorphisms from M to itself will be shown to form a subring, EndA (M ),
of EndZ (M ). If A is commutative, and if An is the free A-module of rank n, we shall
see that EndA (An ) is isomorphic as a ring to Mn (A). Thus, endomorphism rings do
generalize matrix rings when A is commutative.
A useful concept in ring theory is that of an algebra over a commutative ring A.
Definitions 7.1.20. An algebra over a commutative ring A consists of a ring B together
with a ring homomorphism ν : A → B such that the image of ν is contained in the center
of B. We call ν the structure map for B as an A-algebra. If B is a commutative ring,
we call it a commutative A-algebra.
If B and C are A-algebras, via ν : A → B and μ : A → C, then an A-algebra
homomorphism from B to C is a ring homomorphism f : B → C such that the following
diagram commutes.
ν ooo7 B
ooo
f
A OOO
OOO '
μ
C
Examples 7.1.21.
1. Let B be any ring and let A be the center of B. Then the inclusion of A in B
makes B an A-algebra.
2. Let B be any ring and let A be the center of B. Let ι : B → Mn (B) be the ring
homomorphism given by ι(b) = bIn for b ∈ B. Then Lemma 7.1.16 shows that ι(A)
is contained in the center of Mn (B).3 Thus, Mn (B) is an A-algebra via ι ◦ i, where
i : A ⊂ B is the inclusion, and ι : B → Mn (B) is an A-algebra map.
We now show that the notions of rings and of Z-algebras are equivalent.
Lemma 7.1.22. Every ring A has a unique structure as a Z-algebra, and every ring
homomorphism f : A → B is a Z-algebra homomorphism.
Proof The fact that a ring has a unique Z-algebra structure is just a restatement of
Lemma 7.1.8: For any ring A, there is a unique ring homomorphism from Z to A, and
its image lies in the center of A.
Now if ν : Z → A and μ : Z → B are the unique ring homomorphisms and if
f : A → B is any ring homomorphism, we must have f ◦ ν = μ. Thus, f is a Z-algebra
homomorphism.
As in the case of groups, the direct product gives the simplest method of obtaining
new rings from old.
Definition 7.1.23. Let A and B be rings. Then the direct product A × B is the ring
whose additive group is the direct product of the additive groups of A and B and whose
multiplication is given coordinate-wise. In other words, the underlying set of A × B is
the cartesian product of A and B, and the operations are given by
(a, b) + (c, d) = (a + c, b + d)
3 Indeed,
and
(a, b) · (c, d) = (ac, bd).
Problem 10 of Exercises 7.1.27 shows that ι(A) is equal to the center of Mn (B).
CHAPTER 7. RINGS AND MODULES
207
The reader should supply the justification that A × B is a ring:
Lemma 7.1.24. Let A and B be rings. Then the direct product A × B is a ring and the
projection maps from A × B to A and B are both ring homomorphisms.
Moreover, if C is a ring and if f : C → A × B is a map, say f (c) = (f1 (c), f2 (c)),
then f is a ring homomorphism if and only if f1 : C → A and f2 : C → B are ring
homomorphisms.
In category theoretic terms, Lemma 7.1.24 just says that the direct product is the
product in the category of rings.
The finite direct products A1 × · · · × An are defined analogously, and satisfy a similar universal property. Indeed, the product of an infinite family of rings is a ring via
coordinate-wise multiplication:
!
Definition 7.1.25. Suppose given a family {Ai | i ∈ I} of rings. We write i∈I Ai for
the ring structure on the product of the Ai whose operations are coordinate-wise addition
and multiplication. Thus,
(ai | i ∈ I) + (bi | i ∈ I) = (ai + bi | i ∈ I)
and
(ai | i ∈ I) · (bi | i ∈ I) = (ai bi | i ∈ I).
Again, the reader should justify the definition:
!
Lemma 7.1.26. Let {Ai | i ∈ I} be a family of rings. Then i∈I Ai is a ring under
coordinate-wise addition and multiplication. Here 0 is the I-tuple whose coordinates
are
! all 0, and 1 is the I-tuple whose coordinates are all 1. The projection maps πj :
for all j ∈ I.
i∈I Ai → Aj are ring homomorphisms
!
If B is a ring and if f : B → i∈I Ai is a map, say f (b) = (fi (b) | i ∈ I) for each
b ∈ B, then f is a ring homomorphism if and only if fi is a ring homomorphism for all
i ∈ I.
Exercises 7.1.27.
1. Give the proof of Lemma 7.1.6.
2. Show that every subring of a field is an integral domain.4
3. We say that n ∈ Z is square free if it is not divisible by the square of any prime
number. Show that Zn contains no nonzero nilpotent elements if and only if n is
square free.
4. Suppose given a ring structure on the additive group Zn . Show that the multiplicative identity element must generate Zn as an additive group. Deduce that any ring
structure on Zn is isomorphic to the usual one. (Hint: Let e be the multiplicative
identity for a given ring structure on Zn . Show that e fails to generate Zn additively if and only if there is an integer d > 1 such that d divides n and such that
e is the additive d-th power of some element x ∈ Zn . Now show that no such ring
structure could exist.)
5. Let f : A → B be a ring homomorphism. Show that f restricts to a group
homomorphism f : A× → B × . (Note then that the passage from rings to their
unit groups gives a functor from rings to groups.)
6. Give the proof of Lemma 7.1.16.
4 We
shall see in Section 7.11 that a ring is an integral domain if and only if it is a subring of a field.
CHAPTER 7. RINGS AND MODULES
208
7. Let A and B be rings. Show that Mn (A × B) is isomorphic to Mn (A) × Mn (B).
8. Show that Mn (A) contains nonzero nilpotent elements, and hence zero-divisors, for
any nonzero ring A, provided that n ≥ 2.
9. Let A and B be subrings of a ring C. Let
a c R=
a ∈ A, b ∈ B and c ∈ C .
0 b Show that R is a subring of M2 (C).
† 10. Let B be the center of A. Show that ι : A → Mn (A) restricts to an isomorphism
from B onto the center of Mn (A).
11. Let f : A → B be a ring homomorphism. Show that there is a ring homomorphism
f∗ : Mn (A) → Mn (B) defined by setting f∗ ((aij )) equal to the matrix whose ij-th
entry is f (aij ).
12. Give the proof of Lemma 7.1.19.
13. Show that the unit group of EndZ (G) is the automorphism group Aut(G) of G.
14. Let G be the direct sum of a countably infinite number of copies of Z. Find an
element of EndZ (G) which has a left inverse, but is not a unit.
15. Show that the center of a direct product A × B is the direct product of the centers
of A and B.
16. Show that the group of units of A × B is A× × B × .
‡ 17. Suppose given ring homomorphisms f : A → C and g : B → C. Recall that the
pullback of f and g is the diagram
f /
B
A ×C B
g
g
A
f
/ C
where A ×C B = {(a, b) ∈ A × B | f (a) = g(b)}, and f and g are defined by
f (a, b) = b and g(a, b) = a.
(a) Show that A ×C B is a subring of A × B and that f and g are ring homomorphisms.
(b) Suppose given a ring R and ring homomorphisms f : R → B and g : R → A
such that f ◦ g = g ◦ f . Show that there is a unique ring homomorphism
h : R → A ×C B such that f ◦ h = f and g ◦ h = g . Show that h must satisfy
h(r) = (g (r), f (r)) for all r ∈ R.
(c) Show that the natural map (A ×C B)× → A× ×C × B × is an isomorphism.
18. Show that a direct product A × B of commutative rings is an integral domain if
and only if one of A and B is an integral domain and the other is the zero ring.
CHAPTER 7. RINGS AND MODULES
209
19. Show that a direct product A × B of rings is a division ring if and only if one of A
and B is a division ring and the other is the zero ring.
20. Give the proof of Lemma 7.1.24.
21. Give the proof of Lemma 7.1.26.
22. Let A be a ring. Show that Mm (A) × Mn (A)
to the subring of
is isomorphic
M 0
Mm+n (A) consisting of matrices of the form
, where M is an m × m
0 N
matrix, N is an n × n matrix, and the 0’s are zero-matrices of the appropriate size.
The Complex Numbers
The complex numbers, C, give yet another example of a field. We review the definition
and basic properties of C.
Definitions 7.1.28. The elements of the complex numbers are formal sums a + bi, with
a, b ∈ R. This presentation is unique in the sense that if a, b, c, d ∈ R, then a+bi = c+di
if and only if a = c and b = d. Thus, the elements of C are in one-to-one correspondence
with ordered pairs (a, b) of real numbers. So we may identify C with the plane R2 , which
gives us a geometric picture of the complex numbers.
We identify a ∈ R with the complex number a + 0i. We say that a complex number
is in R if it has this form.
Similarly, we write i for the complex number 0 + 1i, and refer to the numbers ai =
0 + ai, with a ∈ R as the pure imaginary numbers.
If z = a + bi and w = c + di are complex numbers, with a, b, c, d ∈ R, we define their
sum by z + w = (a + c) + (b + d)i, and define their product by zw = (ac − bd) + (ad + bc)i.
For z = a + bi with a, b ∈ R, we define the complex conjugate, z, of z by z = a − bi.
Again, there are verifications left to the reader.
Lemma 7.1.29. The above operations of addition and multiplication give C the structure of a commutative ring. Moreover, the standard inclusion R ⊂ C is a ring homomorphism, and hence the additive and multiplicative identities of C are given by 0 = 0 + 0i
and 1 = 1 + 0i, respectively.
The function C → C that carries each z ∈ C to its complex conjugate, z, is a ring
homomorphism.
For z = a + bi with a, b ∈ R, we have the equality
zz = a2 + b2 ∈ R ⊂ C.
In particular, zz ≥ 0 for all z, and zz = 0 if and only if z = 0.
We can now find a multiplicative inverse for every nonzero element of C.
Corollary 7.1.30. The complex numbers, C, is a field. Explicitly, if 0 = z ∈ C, then z
is a unit in C with inverse
z −1 =
1
· z.
zz
Here, 1/(zz) is the inverse of the nonzero real number zz.
CHAPTER 7. RINGS AND MODULES
210
Proof Since zz =
0, the result follows from the facts that C is commutative and that
R is a subring of C.
The notion of planar distance from analytic geometry may be derived from the complex product zz.
Definition 7.1.31. We define the absolute value | | : C → R by
√
|z| = zz = a2 + b2
for z = a + bi with a, b ∈ R.
A very useful tool in studying the complex numbers is the complex exponential function.
Definition 7.1.32. Let x ∈ R. We define the complex exponential eix ∈ C by
eix = cos x + i sin x.
More generally, if z = x + iy ∈ C, with x, y ∈ R, we define ez ∈ C by
ez = ex eiy = ex cos y + iex sin y,
where ex is the ordinary real exponential function applied to x.
We shall be primarily concerned with the exponentials eix with x ∈ R.
Lemma 7.1.33. Let x, y ∈ R, then ei(x+y) = eix · eiy for x, y ∈ R, where the product
on the right is complex multiplication. Moreover, ei·0 = 1. Thus, there is a group
homomorphism exp : R → C× defined by exp(x) = eix .
Proof Complex multiplication gives
eix · eiy
= (cos x + i sin x) · (cos y + i sin y)
= (cos x cos y − sin x sin y) + i(cos x sin y + sin x cos y)
= cos(x + y) + i sin(x + y),
by the formulæ for the sine and cosine of a sum. But the last line is precisely ei(x+y) .
From the known values of the sine and cosine of 0, we see that ei·0 = 1, as claimed,
so that setting exp(x) = eix gives a monoid homomorphism from R to the multiplicative
monoid of C. But any monoid homomorphism from a group to a monoid must take value
in the group of invertible elements of the monoid, which in this case is C× .
The exponential allows us to define some important elements of C, the standard
primitive n-th roots of unity.
Definition 7.1.34. Let n ≥ 1. The standard primitive n-th root of unity, ζn ∈ C, is
defined by
2π
2π
ζn = ei·2π/n = cos
+ i sin
.
n
n
The fact that 2π is the smallest positive real number whose sine is 0 and whose cosine
is 1 gives us the following lemma.
CHAPTER 7. RINGS AND MODULES
211
Lemma 7.1.35. Let n ≥ 1. Then ζn has order n in C× .
There is a noncommutative division ring, called the quaternions, H, that is given by
a construction similar to that of the complex numbers. We shall treat it in the next set
of exercises. Both C and H are examples of what are called Clifford algebras over R.
We shall treat the general case of Clifford algebras in Chapter 9.
We shall not prove it here, but C and H are the only division rings that are algebras
over R in such a way that the induced real vector space structure is finite dimensional.
Exercises 7.1.36.
1. Write Z2 = T = {1, T }. Show that Z2 acts on C via T · z = z for all z ∈ C.
Show that the fixed point set CZ2 is equal to R.
2. Write Zn = T = {1, T, . . . , T n−1 } for n ≥ 2. Show that Zn acts on C via
T k · z = ζnk · z for all k, where ζn = ei·2π/n is the standard primitive n-th root of
unity in C. Show that 1 + ζn + · · · + ζnn−1 is a fixed point under this action. Deduce
that 1 + ζn + · · · + ζnn−1 = 0.
3. Show that |zw| = |z| · |w| for all z, w ∈ C, that |z| ≥ 0 for all z ∈ C, and that
|z| = 0 if and only if z = 0.
4. Show that |z + w| ≤ |z| + |w| for all z, w ∈ C.
5. Show that i is a square root of −1 in C.
6. Let S 1 = {z ∈ C | |z| = 1}. Show that S 1 is a subgroup of C× .
7. Show that eix ∈ S 1 for all x ∈ R, and hence exp gives a homomorphism from R
to S 1 . What is the kernel of exp?
8. Show that exp : R → S 1 is onto. Deduce that S 1 is isomorphic as a group to R/Z.
9. Show that there is a homomorphism expC : C → C× given by expC (z) = ez .
10. Show that eiπ = −1 ∈ C.
11. Show that ζ1 = 1, ζ2 = −1, and ζ4 = i.
12. Let n = mk with m and k positive integers. Show that ζnk = ζm .
13. Show that if n = 2m with m a positive integer, then ζnk+m = −ζnk .
14. Let m > 0 be an odd integer. Show that the subgroup of C× generated by −ζm
coincides with the subgroup generated by ζ2m .
when a, b ∈ R. Show that φ is an
15. Define φ : C → M2 (R) by φ(a + bi) = ab −b
a
injective ring homomorphism. Show that φ(S 1 ) = SO(2) and that φ(eiθ ) = Rθ ,
the matrix that rotates R2 through the angle θ, for all θ ∈ R.
16. We construct the ring of quaternions. The quaternionic groups Q4n , which we
analyzed in the material on group theory, will be seen to embed naturally in the
group of units of the quaternions.
An alternative name for the quaternions is the Hamiltonians. Because of this,
and the fact that Q is already taken, we shall write H for the quaternions. The
elements of H are formal sums a + bi + cj + dk, where a, b, c, and d are real
CHAPTER 7. RINGS AND MODULES
212
numbers. Here, a + bi + cj + dk = a + b i + c j + d k if and only if a = a ,
b = b , c = c , and d = d , and hence we may identify H with the set R4 of 4tuples (a, b, c, d) of real numbers. We define the addition in H accordingly, so that
(a + bi + cj + dk) + (a + b i + c j + d k) = (a + a ) + (b + b )i + (c + c )j + (d + d )k.
We write 1 = 1 + 0i + 0j + 0k, i = 0 + 1i + 0j + 0k, j = 0 + 0i + 1j + 0k and
k = 0 + 0i + 0j + 1k. For a ∈ R, we write a = a · 1 = a + 0i + 0j + 0k. Thus, we
identify the real multiples of 1 with R.
The multiplication in H is defined so that 1 is the identity element, i2 = j 2 = k 2 =
−1, ij = −ji = k, jk = −kj = i and ki = −ik = j. Assembling this information,
we see that if α = a + bi + cj + dk and α = a + b i + c j + d k, then
αα = (aa − bb − cc − dd ) + (ab + ba + cd − dc )i
+ (ac + ca + db − bd )j + (ad + da + bc − cb )k.
Unlike the complex numbers, we see that multiplication in H is noncommutative.
(a) Show that H is a ring and that the real numbers R ⊂ H form a subring.
(b) Show that the center of H is R, so that H is an R-algebra.
(c) Show that if we set x equal to either i, j, or k, then the set of all elements of
H of the form a + bx with a, b ∈ R forms a subring of H that is isomorphic
to C.
(d) Define conjugation in H as follows: for α = a + bi + cj + dk, we set α =
a − bi − cj − dk. Show that the following equalities hold for all α, α ∈ H.
i. α + α = α + α
ii. α · α = α · α
iii. 1 = 1.
iv. α = α.
(According to definitions we shall give below, the first three of the above
properties say that conjugation is an antiautomorphism of H, and hence H is
a self-opposite ring. This is useful in the study of matrices over H.)
(e) Show that α and α commute for all α ∈ H, and that if α = a + bi + cj + dk,
then αα is the real number a2 + b2 + c2 + d2 .
(f) Define the absolute value | | : H → R by
√
|α| = αα = a2 + b2 + c2 + d2
for α = a + bi + cj + dk with a, b, c, d ∈ R. Show that |α · α | = |α| · |α | for all
α, α ∈ H, that |α| ≥ 0 for all α ∈ H, and that |α| = 0 if and only if α = 0.
(g) Show that every nonzero element α ∈ H is a unit in H (and hence H is a
division ring). Here, if α = 0, show that
α−1 =
1
α.
αα
(h) Show that if we set b = cos(π/n) + i sin(π/n) and set a = j, then the subgroup of H× generated by a and b is isomorphic to Q4n , the quaternionic
group of order 4n. (In particular, note that Q8 is isomorphic to the subgroup
{±1, ±i, ±j, ±k} of H× . When studying Q8 alone, rather than the quaternionic groups in general, it is customary to use the notation ±1, ±i, ±j, ±k
for its elements.)
CHAPTER 7. RINGS AND MODULES
213
(i) Let S 3 = {α ∈ H | |α| = 1}. Show that S 3 is a subgroup of H × containing
the above copies of Q4n . Show also that S 1 embeds as a subgroup of S 3 .
17. Let S = {α = ai + bj + ck ∈ H | a, b, c ∈ R and |α| = 1}.5 Show that α2 = −1 for
all α ∈ S. Deduce that H has infinitely many subrings isomorphic to C.
The p-adic Integers
p . For k ≥ 2, let εk : Zpk → Zpk−1 be the unique ring
We construct the p-adic integers, Z
homomorphism (in fact, the unique group homomorphism that carries 1 to 1) from Zpk
to Zpk−1 . The p-adic integers are the inverse limit of the maps εk .
To avoid a dependence on category theory
here, we shall give the construction explic p is a subring of the infinite product !
itly. Z
k≥1 Zpk . It is customary to write the tuples
in the infinite product in the reverse order from the usual one. Thus, a typical element
looks like (. . . , mk , . . . , m2 , m1 ), where each mk lies in Zpk . Under this convention, we
have
p = {(. . . , mk , . . . , m2 , m1 ) | εk (mk ) = mk−1 for all k ≥ 2}.
Z
Recall that the ring operations in the infinite product are defined via coordinate-wise
addition and multiplication, so the fact that the εk are ring homomorphisms verifies the
assertion above:
p is a subring of !
Lemma 7.1.37. Z
k≥1 Zpk .
The additive and multiplicative identities, of course, are 0 = (. . . , 0, . . . , 0, 0) and
1 = (. . . , 1, . . . , 1, 1), respectively. The next lemma is elementary, but important.
p → Zpk be induced by the projection onto the k-th factor.
Lemma 7.1.38. Let πk : Z
Then πk is a ring homomorphism for all k ≥ 1.
The p-adic integers are important in number theory. Despite the fact that they are
made up from quotient rings of Z, the next lemma will show that the p-adic integers
themselves form a ring of characteristic 0. (See Definition 7.2.12.)
p is an embedding.
Lemma 7.1.39. The unique ring homomorphism ι : Z → Z
Proof We have ι(m) = (. . . , m, . . . , m, m), so m ∈ ker ι if and only if m is divisible by
pk for all k ≥ 0.
p . Recall that a
The structure of the unit group is one of the interesting features of Z
splitting map, or section, for a homomorphism f : G → K is a homomorphism s : K → G
such that f ◦ s is the identity map of K. A map f that admits a section is called a split
surjection. Recall from Corollary 4.7.6 that if f : G → K is a split surjection with G
abelian, then G ∼
= ker f × K.
p is a unit if and only
Proposition 7.1.40. An element a = (. . . , mk , . . . , m2 , m1 ) of Z
if m1 is a unit in Zp . The map
×
→ Z×
π1 : Z
p
p
is a split surjection.
5 Geometrically,
S is a 2-dimensional subsphere of S 3 .
CHAPTER 7. RINGS AND MODULES
214
p . Since the multiplication in Z
p is coordinateProof Let a = (. . . , mk , . . . , m2 , m1 ) ∈ Z
wise, if a is a unit in Zp , then each mk must be a unit in Zpk .
Conversely, suppose that m1 is a unit. According to Lemma 4.5.3, an element m is
a unit in Zpk if and only if (m, p) = 1. In particular, m is a unit in Zpk if and only if
εk (m) = m is a unit in Zpk−1 . Thus, if m1 is a unit, then each mk must be a unit in Zpk .
Let nk be the inverse of mk in Zpk . Since each εk is a ring homomorphism, we must
p
have εk (nk ) = nk−1 . Thus, setting b = (. . . , nk , . . . , n2 , n1 ) specifies an element of Z
that is easily seen to be a multiplicative inverse for a.
× → Z× . For p = 2, there’s nothing
It remains to construct a splitting map for π1 : Z
p
p
to show, as Z×
is
the
trivial
group.
For
p
>
2,
we
consider
the
proof of Corollary 4.5.12.
2
Let ηk : Zpk → Zp be the unique ring homomorphism between these rings. Then the
×
kernel of the induced map ηk : Z×
→ Z×
p has order relatively prime to |Zp | = p − 1.
pk
consisting
Thus, the proof of Corollary 4.4.15 shows that if K is the subgroup of Z×
pk
∼
=
of those elements of exponent p − 1, then ηk : K −→ Z×
p . But this says that there is a
×
unique section sk : Z×
→
Z
of
η
.
k
p
pk
But then εk ◦ sk is a section for ηk−1 : Z×
→ Z×
p , so εk ◦ sk = sk−1 . Thus, setting
pk−1
s(m) = (. . . , sk (m), . . . , s2 (m), m)
gives a well defined function s : Z×
p → Zp , that is easily seen to give a section for
×
×
→Z .
π1 : Z
p
p
Exercises 7.1.41.
p is an integral domain. (Hint: Suppose that ab = 0 in Z
p , with
† 1. Show that Z
a = (. . . , mk , . . . ) and b = (. . . , nk , . . . ). Suppose that mk = 0 in Zpk . What does
this say about the p-divisibility of mk+l and nk+l ?)
× → Z× is isomorphic to the
2. Let p be an odd prime. Show that the kernel of π1 : Z
p
p
×
additive group of Zp . Deduce that the torsion subgroup of Zp maps isomorphically
to Z×
p under π1 .
× → Z× is isomorphic to the additive group of Z
2.
3. Show that the kernel of π2 : Z
2
4
×
maps isomorphically to Z× under π2 .
Deduce that the torsion subgroup of Z
2
4
p → Zp . Show that
† 4. Let a be in the kernel of the additive homomorphism π1 : Z
p . (Hint: Consider the diagram
a = pb for some b ∈ Z
Zpk
εk Zpk−1
⊂ /
Zpk+1
εk+1 ⊂ /
Zpk
where the inclusion maps send 1 to p.)
p . Suppose that mk = 0 in Zpk for all k ≤ r and
† 5. Let a = (. . . , mk , . . . , m2 , m1 ) ∈ Z
×.
that mr+1 = 0 in Zpr+1 . Show that a = pr u, where u ∈ Z
p
CHAPTER 7. RINGS AND MODULES
215
6. Let fk : Ak → Ak−1 be a ring homomorphism for k ≥ 2. Give an isomorphism
×
×
∼
lim
A
= lim
←− k
←−(Ak ).
7. Give Zpk the discrete topology for
! all k. Show that Zp is a closed subspace of the
product (Tikhonov) topology on k≥1 Zpk , and hence is compact by the Tikhonov
p . For this reason, Z
p may
Theorem. Show that ι(Z) is dense in this topology on Z
p.
be thought of as a completion of Z in the topology that Z inherits from Z
7.2
Ideals
The ideals in a ring are a very important part of its structure. We give the most basic
results on ideals in this section, and shall continue to study them throughout the material
on rings.
Definitions 7.2.1. Let A be a ring. A left ideal of A is a subgroup a ⊂ A of the additive
group of A with the additional property that if x ∈ a, then for any a ∈ A, we have ax ∈ a.
(In other words, a is closed on the operation of multiplying its elements on the left by
any element of A.)
From the other side, a right ideal of A is a subgroup a ⊂ A of the additive group of
A with the additional property that if x ∈ a, then for any a ∈ A, we have xa ∈ a.
A two-sided ideal of A is a subgroup of the additive group of A which is simultaneously
a left ideal and a right ideal.
If A is commutative, all ideals are two-sided, and we refer to them simply as ideals.
Examples 7.2.2.
1. Let A be a ring. Then A is a two-sided ideal of itself.
2. Let A be a ring and let 0 ⊂ A be the trivial subgroup. Since a · 0 = 0 · a = 0 for all
a ∈ A, 0 is a two-sided ideal of A.
Another set of examples comes from the principal ideals.
Definition 7.2.3. Let A be a ring and let x ∈ A. The principal left ideal, Ax, generated
by x is defined by
Ax = {ax | a ∈ A}.
Similarly, the principal right ideal generated by x is xA = {xa | a ∈ A}.
If A is commutative, then Ax = xA, and we call it simply the principal ideal generated
by x, and denote it by (x).
Principal ideals play a role in ideal theory similar to the role of cyclic subgroups in
group theory:
Lemma 7.2.4. Let A be a ring and let x ∈ A. Then Ax and xA are the smallest left
and right ideals, respectively, that contain x.
Some commutative rings have no ideals other than the principal ones.
Definitions 7.2.5. A principal ideal ring is a commutative ring in which every ideal is
principal.
A principal ideal domain, or P.I.D., is an integral domain in which every ideal is
principal.
CHAPTER 7. RINGS AND MODULES
216
Indeed, the integers are better behaved than the general integral domain:
Lemma 7.2.6. Every subgroup of Z is a principal ideal. In particular, Z is a P.I.D.
Proof We know (Proposition 2.3.2) that the subgroups of Z all have the form n =
{nk | k ∈ Z} for some n ∈ Z. But n is precisely (n), the principal ideal generated by
n.
We shall see below that if K is a field,6 then the polynomial ring K[X] is a principal
ideal domain.
Principal ideals are useful in recognizing division rings and fields.
Lemma 7.2.7. A nonzero ring A is a division ring if and only if the only left ideals of
A are 0 and A.
Proof Let A be a division ring and let a be a nonzero left ideal of A. Thus, there is an
x ∈ a with x = 0. But then x is a unit in A. Now 1 = x−1 x ∈ a, as a is a left ideal. But
then a = a · 1 ∈ a for the same reason, for all a ∈ A. Thus, a = A.
Conversely, suppose that A is a ring with no left ideals other than 0 and A. Let
0 = x ∈ A. Then Ax is a nonzero left ideal, and hence Ax = A. Thus, there is an
element y ∈ A with yx = 1. Thus, every nonzero element in the multiplicative monoid
of A has a left inverse. As in Problem 13 of Exercises 2.1.16, it is easy to see that every
nonzero element of A is a unit.
Two-sided ideals play a role in ring theory somewhat analogous to that played by
normal subgroups in the theory of groups.
Proposition 7.2.8. Let a be any subgroup of the additive group of a ring A, and let
π : A → A/a be the canonical map. Then A/a has a ring structure that makes π a ring
homomorphism if and only if a is a two-sided ideal of A. Such a ring structure, if it
exists, is unique.
Proof Suppose given a ring structure on A/a such that π is a ring homomorphism.
Then writing a ∈ A/a for π(a), we must have a · b = ab, so the ring structure is indeed
unique. Moreover, for x ∈ a, 1 = 1 + x. Thus, for a ∈ A, a = a · 1 = a · 1 + x = a + ax.
In other words, a + ax represents the same element of A/a as a does. But this says
a + ax − a ∈ a, and hence ax ∈ a for a ∈ A and x ∈ a.
Thus, a has to be a left ideal. But a similar argument, obtained by multiplying 1
and 1 + x on the right by a, shows that a must also be a right ideal, so that a is in fact
two-sided.
For the converse, suppose that a is a two-sided ideal. We wish to show that the
multiplication a · b = ab gives a well defined binary operation on A/a. This will be
sufficient to complete the proof, as then π is a homomorphism with respect to this
operation, so that the verifications that A/a is a ring under this multiplication will
follow from the fact that A is a ring.
Thus, for a, b ∈ A and x, y ∈ a, we must show that ab and (a + x)(b + y) represent the
same element of A/a, and hence that (a + x)(b + y) − ab ∈ a. But (a + x)(b + y) − ab =
ay + xb + xy, which is in a because a is a two-sided ideal.
The rings A/a satisfy a universal property similar to that of factor groups in group
theory. We call them quotient rings of A.
6 The use of K for a field derives from the German Körper , which means field in the mathematical
sense. Amusingly, the German word for a farmer’s field is a cognate of ours: Feld.
CHAPTER 7. RINGS AND MODULES
217
Proposition 7.2.9. Let f : A → B be a ring homomorphism. Then ker f is a two-sided
ideal of A. If a is another two-sided ideal of A, then a ⊂ ker f if and only if there is a
ring homomorphism f : A/a → B which makes the following diagram commute, where
π : A → A/a is the canonical map:
/B
AB
=
BB
||
BB
|
|
B
||
π BB
|| f
!
A/a
f
Moreover, if a ⊂ ker f , then the homomorphism f : A/a → B making the diagram
commute is unique.
When a = ker f , f induces a ring isomorphism between A/a and the image of f .
Proof By Proposition 4.1.7, there is a unique homomorphism of additive groups, f :
A/a → B making the diagram commute if and only if a ⊂ ker f . And if a = ker f , this
homomorphism f gives a group isomorphism onto the image of f by the First Noether
Isomorphism Theorem.
Thus, it suffices to show that ker f is a two-sided ideal and that f is a ring homomorphism whenever a is a two-sided ideal contained in ker f . We leave these verifications to
the reader.
Since π : Z → Zn is a surjective ring homomorphism and has kernel (n), we obtain
the following corollary.
Corollary 7.2.10. There is an isomorphism of rings between Zn and the quotient ring
Z/(n).
The next corollary is another consequence of Proposition 7.2.9.
Corollary 7.2.11. Let f : D → B be a ring homomorphism, where D is a division ring
and B is any nonzero ring. Then f is injective.
Proof The kernel of f is a two-sided ideal, and hence must be either 0 or D. As B is
nonzero, ker f = D. So ker f = 0, and f is injective.
Recall that for any ring B, there is a unique ring homomorphism f : Z → B. We
may obtain useful information about B by applying Proposition 7.2.9 with a = ker f .
Definition 7.2.12. Let A be a ring and let f : Z → A be the unique ring homomorphism
from Z to A. Let ker f = (n), with n ≥ 0. Then we say that A has characteristic n. In
particular, A has characteristic 0 if and only if f is injective.
Examples 7.2.13. Since the kernel of the canonical map π : Z → Zn is (n), Zn has
characteristic n. On the other hand, Z is a subring of Q, R, C, and H, so these have
characteristic 0.
Proposition 7.2.14. Let n > 0. Then a ring A has characteristic n if and only if it has
a subring isomorphic as a ring to Zn . If A does have characteristic n, then the subring
in question is the image of the unique ring homomorphism f : Z → A.
CHAPTER 7. RINGS AND MODULES
218
Proof If A has characteristic n with n > 0, then the image of f is isomorphic to Z/(n)
by Proposition 7.2.9. Conversely, suppose that A has a subring B that is isomorphic as
a ring to Zn . Then if f : Z → B is the unique ring homomorphism, we may identify
f with the canonical map π : Z → Z/(n). In particular, the kernel of f is (n). But if
i : B → A is the inclusion, then i ◦ f is the unique ring homomorphism from Z to A.
Since i is injective, the kernel of i ◦ f , whose generator is by definition the characteristic
of A, is precisely the kernel of f .
Let K be a field. Then the image of the unique ring homomorphism f : Z → K, as
a subring of K, must be an integral domain. Thus, Lemma 7.1.12 shows that if f is not
injective (i.e., if the image of f is finite), then f must have image Zp for some prime p.
Corollary 7.2.15. Let K be a field of nonzero characteristic. Then the characteristic
of K is a prime number.
We can generalize the notion of a principal ideal as follows.
Definitions 7.2.16. Let A be a ring and let S be any subset of A. Then the left ideal
generated by S, written AS, is given as follows:
AS = {a1 x1 + · · · + ak xk | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}.
Similarly, the right ideal generated by S is given by
SA = {x1 a1 + · · · + xk ak | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}.
When A is commutative, of course, AS = SA, and we denote it by (S). If S is finite,
say S = {x1 , . . . , xn }, we write (S) = (x1 , . . . , xn ).
For A noncommutative, the two-sided ideal generated by S is
ASA = {a1 x1 b1 + · · · + ak xk bk | k ≥ 1, ai , bi ∈ A and xi ∈ S for 1 ≤ i ≤ k}.
The next lemma is immediate from the definitions.
Lemma 7.2.17. Let A be a ring and let S be a subset of the center of A. Then the
left, right, and two-sided ideals generated by A all coincide, i.e., AS = SA = ASA. In
particular, if A is commutative, then ASA = (S).
Warning: In the case of left ideals, we may apply the distributive law to reduce the
sums a1 x1 + · · · + ak xk until the elements xi are all distinct. Thus, for instance, A{x} =
Ax = {ax | a ∈ A}, the principal left ideal generated by x. The analogous result holds for
the right ideals. But in the case of two-sided ideals in noncommutative rings, we cannot
make this kind of reduction in general. Thus, in A{x}A, the principal two-sided ideal
generated by x, we must consider all elements of the form a1 xb1 + · · · + ak xbk with k ≥ 1
and with ai , bi ∈ A for 1 ≤ i ≤ k.
Lemma 7.2.18. Let A be a ring and let S ⊂ A. Then AS, SA, and ASA are the
smallest left, right, and two-sided ideals, respectively, that contain S.
This now gives the following universal property.
CHAPTER 7. RINGS AND MODULES
219
Corollary 7.2.19. Let A be a ring and let S ⊂ A, and let a ⊂ A be the two-sided ideal
generated by S. Let f : A → B be a ring homomorphism whose kernel contains S. Then
there is a unique homomorphism f : A/a → B such that the following diagram commutes
/B,
AB
=
BB
||
BB
|
|
B
||
π BB
|| f
!
A/a
f
where π : A → A/a is the canonical map.
Proof The kernel of f is a two-sided ideal of A containing S, so we must have a ⊂ ker f
by Lemma 7.2.18. The result now follows from Proposition 7.2.9.
When a ring homomorphism f : B → A is surjective, we can use information about
A and about ker f to deduce information about B.
An important topic in ideal theory is a discussion of the various ways in which we
can obtain new ideals from old. Recall first (from Lemma 4.1.18) that if G is a group
and if H and K are subgroups of G with K ⊂ NG (H), then KH = {kh | k ∈ K, h ∈ H}
is a subgroup of G. Thus, if G is abelian, written in additive notation, and if H and K
are any subgroups of G, we have a subgroup H + K given by
H + K = {h + k | h ∈ H, k ∈ K}.
The next lemma is elementary.
Lemma 7.2.20. Let a and b be left (resp. right) ideals in the ring A. Then the subgroups
a + b and a ∩ b are also left (resp. right) ideals of A.
We have already seen an important use of sums of ideals. Indeed, the greatest common
divisor of two integers, m and n, was defined to be the generator of the ideal (m) + (n).
We can generalize the notion of the ideals generated by a set.
Definitions 7.2.21. Let A be any ring. Let a be a left ideal of A, let b be a right ideal
of A, and let S be any subset of A. We write
aS = {a1 x1 + · · · + ak xk | k ≥ 1, ai ∈ a and xi ∈ S for 1 ≤ i ≤ k}
Sb = {x1 b1 + · · · + xk bk | k ≥ 1, xi ∈ S and bi ∈ b for 1 ≤ i ≤ k}
Thus, under either of the above definitions,
ab = {a1 b1 + · · · + ak bk | k ≥ 1, ai ∈ a and bi ∈ b for 1 ≤ i ≤ k}.
These constructions behave as follows.
Lemma 7.2.22. Let A be any ring. Let a be a left ideal of A, let b be a right ideal of
A, and let S be any subset of A. Then aS is a left ideal of A and Sb is a right ideal of
A. Combining these two facts, we see that ab is a two-sided ideal of A.
Finally, if a and b are both two-sided ideals of A, then we have an inclusion of twosided ideals ab ⊂ a ∩ b.
CHAPTER 7. RINGS AND MODULES
220
Proof We leave the assertions of the first paragraph to the reader. For the final
statement, consider an element of the form a1 b1 + · · · + ak bk with ai ∈ a and bi ∈ b for
all i. Since b is a left ideal, each term ai bi lies in b. Since a is a right ideal, each term
ai bi lies in a. So their sum lies in a ∩ b.
We shall explore the effects of taking products and intersections of ideals in Z in the
next set of exercises.
The following use of sums and intersections can be very useful in the study of rings,
as well as for the module theory we shall introduce in Section 7.7.
Proposition 7.2.23. (Chinese Remainder Theorem) Suppose given two-sided ideals a1 , . . . , ak
of the ring A such that ai + aj = A whenever i = j. Let
f : A → A/a1 × · · · × A/ak
be given by f (a) = (a, . . . , a). Then f induces a ring isomorphism
∼
=
f : A/b −→ A/a1 × · · · × A/ak ,
where b =
k
i=1
ai .
Proof By the obvious extension of Lemma 7.1.24 to k-fold products,
kf is a ring homomorphism. Thus, it suffices to show that f is surjective, with kernel i=1 ai .
The latter statement is immediate, as the kernel of a map into a product is always
the intersection of the kernels of the associated maps into the factors. Thus, we need
only show that f is surjective.
But the product A/a1 × · · · × A/ak is generated as an abelian group by the images
of the A/ai under the canonical inclusions. Thus, it suffices to show that these images
are contained in the image of f . We show this for i = 1.
Now, we claim that it suffices to show that (1, 0, . . . , 0) is in the image of f . To
see this, suppose that f (x) = (1, 0, . . . , 0). Since f is a ring homomorphism f (ax) =
(a, . . . , a) · (1, 0, . . . , 0) = (a, 0, . . . , 0) for all a ∈ A, and hence A/a1 × 0 × · · · × 0 is indeed
contained in the image of f .
We now make use of the fact that a1 + ai = A for all i = 1. Thus, for 2 ≤ i ≤ k,
there are elements ai ∈ a1 and bi ∈ ai , with ai + bi = 1. Then if πi : A → A/ai is the
canonical map for 1 ≤ i ≤ k, we have π1 (bi ) = 1 and πi (bi ) = 0 for 2 ≤ i ≤ k.
Let b = b2 · · · bk . Since the canonical maps πi are ring homomorphisms, π1 (b) = 1,
while πi (b) = 0 for 2 ≤ i ≤ k. But this just says that f (b) = (1, 0, . . . , 0) as desired.
Exercises 7.2.24.
1. Let A and B be rings. Show that the left ideals of A × B all have the form a × b,
where a is a left ideal of A and b is a left ideal of B.
2. Let A be an integral domain and let a, b ∈ A. Show that (a) = (b) if and only if
there is a unit x ∈ A× such that ax = b.
3. Let A be a ring. Show that there is a ring homomorphism from Zn to A if and
only if the characteristic of A divides n. Show that this ring homomorphism, if it
exists, is unique.
4. Let f : A → B be a ring homomorphism. Show that f −1 (a) is a left (resp. right or
two-sided) ideal of A for each left (resp. right or two-sided) ideal a of B.
CHAPTER 7. RINGS AND MODULES
221
† 5. Let f : A → B be a surjective ring homomorphism. Show that f (a) is a left
(resp. right or two-sided) ideal of B whenever a is a left (resp. right or two-sided)
ideal of A. Deduce that there is a one-to-one correspondence between the ideals of
B and the ideals of A containing ker f .
6. Show that every subgroup of Zn is a principal ideal, and hence that Zn is a principal
ideal ring.
7. Find a ring homomorphism f : A → B that is not surjective and an ideal a ⊂ A
such that f (a) is not an ideal of B.
8. Let a ⊂ A be a left ideal. Show that a = A if and only if a contains an element
which has a left inverse.
9. Show that a principal left ideal Ax = A if and only if x has a left inverse.
10. A division ring has been shown to have no two-sided ideals other than 0 and the
ring itself. Show that the paucity of two-sided ideals does not characterize division
rings: Show that if K is a field and n ≥ 2, then Mn (K) has no two-sided ideals
other than 0 and itself. (Hint: The cleanest way to show this uses linear algebra.
The reader may wish to come back to this problem after reading Sections 7.9 and
7.10.)
† 11. Repeat the preceding problem, replacing K by a division ring D.
† 12. Let A be a ring. Define the opposite ring Aop as follows. As an abelian group, Aop
is just A. For a ∈ A, we write )
a for the associated element of Aop . We define the
) Show that Aop is a ring. Show that the left
multiplication in Aop by )
a · )b = ba.
ideals of A are in one-to-one correspondence with the right ideals of Aop , and that
the right ideals of A are in one-to-one correspondence with the left ideals of Aop .
‡ 13. Let A be a ring. An anti-automorphism of A is an automorphism f : A → A of the
additive group of A, such that f (ab) = f (b)f (a) for all a, b ∈ A and f (1) = 1. Show
that A admits an anti-automorphism if and only if A and Aop are isomorphic.
14. We say that a ring is self-opposite if it admits an anti-automorphism. Show that
commutative rings are self-opposite.
15. Let A be a self-opposite ring. Show that Mn (A) is self-opposite for all n ≥ 1.
(Hint: See Problem 5 of Exercises 7.10.23.)
16. Let m, n ∈ Z. Calculate the generator of the product ideal (m)(n).
17. Let m, n ∈ Z. Calculate the generator of (m) ∩ (n).
18. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Show that Zn is isomorphic
as a ring to the direct product Zpr11 × · · · × Zprk .
k
19. Use the preceding problem to give a quick proof that if n = pr11 . . . prkk , then Aut(Zn )
is isomorphic to Aut(Zpr11 ) × · · · × Aut(Zprk ).
k
20. Suppose given an infinite family {ai | i ∈ I} of left ideals of A. Define the sum of
this infinite family by
ai = {ai1 + · · · + aik | k ≥ 1, i1 , . . . , ik ∈ I, and aij ∈ aij for 1 ≤ j ≤ k}.
i∈I
CHAPTER 7. RINGS AND MODULES
222
(a) Show that i∈I ai is the smallest left ideal that contains each ai .
(b) If S is any subset of A, show that AS = x∈S Ax.
7.3
Polynomials
One of the most important constructions in ring theory is that of polynomial rings. We
study the polynomials in one variable now and study the polynomials in several variables
in a subsection (page 228).
Definitions 7.3.1. Let A be a ring (not necessarily commutative) and let X be a
variable.
We define a ring A[X] as follows. The elements of A[X] are formal sums
n
i
a
X
= an X n + · · · + a1 X + a0 , where n is a nonnegative integer and
ai ∈ A for all
i
i=0
n
i. We often use functional notation for these sums, writing, say, f (X) = i=0 ai X i for
the element above. If an = 0, we say that f (X) has degree n and that an is its leading
coefficient. Otherwise,
m if m is the largest integer for which am = 0, we identify f (X) with
the polynomial i=0 ai X i = am X m + · · · + a1 X + a0 , and hence it has degree m. (In the
case of 0 = 0X 0 , there are no possible identifications with lower degree polynomials, and
we say that 0 has degree −∞.) We say that a polynomial f (X) is monic if its leading
coefficient is 1.
As is implicit in the notations above, we write either a or aX 0 for the degree 0
polynomial whose 0-th coefficient is a. We write ι : A → A[X] for the map given by
ι(a) = aX 0 for all a ∈ A.
We also write X n for 1X n and write X for X 1 .
Note that
ntwo polynomials are equal if and only if their coefficients are equal. (Here,
if f (X) = i=0 ai X i , then the coefficients of the X m for m > n are all implicitly
n 0.)
Addition of polynomials is obtained by adding the coefficients: If f (X) = i=0 ai X i
n
and g(X) = i=0 bi X i (where n is the larger of the degrees of f and g), then
f (X) + g(X) =
n
(ai + bi )X i .
i=0
of polynomials is given as follows: if f (X) =
mMultiplication
i
b
X
,
then
their
product is given by
i
i=0
⎛
⎞
n+m
i
⎝
aj bi−j ⎠ X i .
f (X)g(X) =
i=0
n
i=0
ai X i and g(X) =
j=0
Note that if A is noncommutative, then the order of the products ai bi−j is important.
Finally, we give a special name to some of the simplest polynomials: We shall refer to
a polynomial of the form aX i for some a ∈ A and i ≥ 0 as a monomial. Note that every
polynomial is a sum of monomials, so that the monomials generate A[X] as an abelian
group.
We will occasionally use other letters, such as T or Y for the variable of a polynomial
ring. We shall refer to the variable as an indeterminate element over the ring A.
Again, the reader should justify the definitions:
Lemma 7.3.2. Let A be a ring. Then A[X] is a ring under addition and multiplication
of polynomials, and ι : A → A[X] is a ring homomorphism.
The element X ∈ A[X] commutes with every element of ι(A), and if A is commutative, so is A[X].
CHAPTER 7. RINGS AND MODULES
223
Note that since every polynomial is a sum of monomials, the distributive law shows
that the multiplication in A[X] is determined by the rule aX i · bX j = abX i+j . In particular, we see that if A is a subring of the real numbers R, then the multiplication formula
is exactly the one we get by multiplying polynomial functions over R. So polynomial
rings are familiar objects after all.
Polynomial rings have a very important universal property.
Proposition 7.3.3. Let g : A → B be a ring homomorphism and let b ∈ B be an element
that commutes with every element of g(A). Then there is a unique ring homomorphism
g b : A[X] → B such that g b (X) = b and the following diagram commutes.
g
/B
AD
DD
z=
z
DD
zz
D
zzgb
ι DD
z
z
!
A[X]
Moreover, g b satisfies the formula
" n
#
n
i
gb
ai X =
g(ai )bi .
i=0
i=0
0
In this formula, we adopt the convention that b = 1 for any b.
Proof For the uniqueness statement, note that any ring homomorphism g b that carries
X to b and makes the diagram commute must satisfy the stated formula:
" n
#
n
i
=
gb
ai X
g b (ι(ai ))(g b (X))i
i=0
i=0
=
n
g(ai )bi .
i=0
Also, the stated formula does make the diagram commute and carry X to b. Thus,
it suffices to show that g b , as defined by the formula, gives a ring homomorphism.
That it is a homomorphism of additive groups follows from the distributive law in B.
And g b (1) = ι(1) = 1, so it suffices to show that g b (f1 (X)f2 (X)) = g b (f1 (X))g b (f2 (X))
for all f1 (X), f2 (X) ∈ A[X].
Since the monomials generate A[X] additively, distributivity shows that it is sufficient
to check this for the case that f1 (X) and f2 (X) are monomials. Thus, we must check
that for a, a ∈ A and i, j ≥ 0, we have g(aa )bi+j = g(a)bi g(a )bj . Since b commutes
with any element of g(A), so does bi by induction. Thus, g(a)bi g(a )bj = g(a)g(a )bi bj .
That bi bj = bi+j follows from the power laws for monoids (Problem 5 of Exercises 2.2.18).
Thus, the result follows from the fact that g is a ring homomorphism.
If A is commutative, then so is A[X], and A[X] is an A-algebra via ι : A → A[X]. As
such, it satisfies an important universal property.
Corollary 7.3.4. Let A be a commutative ring and let B be an A-algebra, via ν : A → B.
Then for each b ∈ B, there is a unique A-algebra homomorphism εb : A[X] → B with
the property that εb (X) = b. It is given by the formula
n
n
εb (
ai X i ) =
ν(ai )bi ,
i=0
i=0
CHAPTER 7. RINGS AND MODULES
224
where we set b0 = 1 for any b.
Proof Since ν : A → B is the structure map for B as an A-algebra, each element
of ν(A) commutes with each b ∈ B. Thus, Proposition 7.3.3 provides a unique ring
homomorphism ν b : A[X] → B with ν b ◦ ι = ν (i.e., ν b is an A-algebra map) and
ν b (X) = b. So we set εb = ν b . The formula comes from Proposition 7.3.3.
Definitions 7.3.5. Let B be an A-algebra and let b ∈ B. We shall refer to the unique
A-algebra homomorphism, εb : A[X] → B, that carries X to b as the evaluation map
obtained by evaluating X at b. In consonance with this language, for f (X) ∈ A[X], we
sometimes write εb (f (X)) = f (b), as it is obtained by evaluating f (X) at X = b.
We shall write A[b] for the image of εb . We call it the A-subalgebra of B generated
by b, or the algebra obtained by adjoining b to A.
Note that since εb : A[X] → A[b] is onto, A[b] must be commutative even if B is not.
Examples 7.3.6. Recall that every ring is a Z-algebra in a unique way. Consider the
complex numbers, C, as a Z-algebra, and form the Z-subalgebra, Z[i], obtained by
adjoining i to Z. The ring Z[i] is known as the Gaussian integers.
Note that i has order 4 in the unit group C× . Thus, the powers of i consist of i, −1,
−i, and 1. Collecting terms, we see that every element of Z[i] has the form m + ni, with
m, n ∈ Z, and hence
Z[i] = {m + ni ∈ C | m, n ∈ Z}.
As the addition and multiplication operations are inherited from those in C, we have
(m + ni) + (k + li) = (m + k) + (n + l)i and (m + ni) · (k + li) = (mk − nl) + (ml + nk)i
for m, n, k, l ∈ Z.
Note also that since C is a Q-algebra, the same arguments may be used if we replace
Z by Q: Adjoining i to Q, we obtain
Q[i] = {a + bi ∈ C | a, b ∈ Q}.
Recall from Lemma 7.1.35 that for n ≥ 1, the complex number ζn = ei2π/n has order
n in C× . Also, ζ4 = i. Thus, the next example is a generalization of the previous one.
Examples 7.3.7. The non-negative powers of ζn are 1, ζn , . . . , ζnn−1 . Thus, if A = Z or
Q, the elements of A[ζn ] may all be written as sums a0 + a1 ζn + · · · + an−1 ζnn−1 with
a0 , . . . , an−1 ∈ A. We shall refer to Z[ζn ] as the cyclotomic integers obtained by adjoining
a primitive n-th root of unity and to Q[ζn ] as the cyclotomic extension of Q obtained by
adjoining a primitive n-th root of unity.
As we saw in the case of ζ4 = i, elements of A[ζn ], with A = Z or Q, do not have unique
representations as sums of the form a0 + a1 ζn + · · · + an−1 ζnn−1 with a0 , . . . , an−1 ∈ A.
We shall defer the determination of a unique representation for these elements until
Section 8.8.
We shall show in Proposition 8.2.5 that the rings Q[ζn ] are all subfields of C.
Remarks 7.3.8. Let A be a commutative ring. Note that if A is a subring of the center
of B and if we view B as an A-algebra whose structure map is the inclusion of A in B,
then for b ∈ B, we have
n
n
εb (
ai X i ) =
ai bi ,
i=0
n
A[b] = {
i=0
and hence
i=0
ai bi | n ≥ 0 and ai ∈ A for 0 ≤ i ≤ n}.
CHAPTER 7. RINGS AND MODULES
225
We now give an important property of A[b].
Lemma 7.3.9. Let B be an A-algebra and let b ∈ B. Then A[b] is the smallest Asubalgebra of B that contains b.
Proof An A-subalgebra of B is any subring of B containing ν(A), where ν : A → B
is the structure map of B as an A-algebra.
Thus, if C is an A-subalgebra of B that
n
contains b, it must contain every sum i=0 ν(ai )bi , and hence contains A[b].
Since A[b] is an A-subalgebra of B containing b, the result follows.
We shall now compute the kernels of the evaluation maps εa : A[X] → A, for a ∈ A
and give some applications. The next result is a version of the Euclidean algorithm.
n
i
Proposition 7.3.10. Let A be any ring and let g(X) =
i=0 ai X be a polynomial
whose leading coefficient, an , is a unit in A. Then for any polynomial f (X) ∈ A[X], we
may find polynomials q(X), r(X) ∈ A[X] such that f (X) = q(X)g(X) + r(X) and such
that the degree of r(X) is strictly less than that of g(X).
Proof We argue by induction on the degree of f (X). If f (X) has degree less than n,
then we may simply take q(X) = 0 and take r(X) = f (X), and the statement is satisfied.
m
Thus, suppose that f (X) = i=0 bi X i with bm = 0 and that the result is true for
all polynomials of degree less than m. In particular, we may assume that m ≥ n. But
m−n
then (bm a−1
)g(X) is a polynomial of degree m whose leading coefficient is the
n X
m−n
)g(X) has degree less than m. By
same as that of f (X), and hence f (X) − (bm a−1
n X
m−n
induction, we may write f (X) − (bm a−1
X
)g(X)
=
q1 (X)g(X) + r(X), where r(X)
n
m−n
) and we
has degree less than n. But then we may take q(X) = q1 (X) + (bm a−1
n X
have f (X) = q(X)g(X) + r(X), as desired.
As the reader may recall from calculus, the simplest application for division of polynomials is to the theory of roots.
Definition 7.3.11. Let A be a commutative ring and let f (X) ∈ A[X]. Let B be an
A-algebra. We say that b ∈ B is a root of f if f (X) lies in the kernel of the evaluation
map εb : A[X] → B obtained by evaluating X at b.
In particular, a ∈ A is a root of f if and only if f (a) = 0.
Corollary 7.3.12. Let A be a commutative ring and let a ∈ A. Then the kernel of the
evaluation map εa : A[X] → A is (X − a), the principal ideal generated by X − a. In
particular, a ∈ A is a root of f (X) ∈ A[X] if and only if f (X) = (X − a)q(X) for some
q(X) ∈ A[X].
Proof Since X − a has degree 1 and leading coefficient 1, we may write any polynomial
f (X) ∈ A[X] as f (X) = q(X)(X − a) + r(X) where r(X) has degree ≤ 0. In other words,
f (X) = q(X)(X − a) + r, where r ∈ A.
But the evaluation map is a ring homomorphism, so that f (a) = q(a)(a − a) + r = r.
Thus, f (X) ∈ ker εa if and only if r = 0. Thus, f (X) ∈ ker εa if and only if f (X) =
q(X)(X − a) for some q(X) ∈ A[X]. But f (X) = q(X)(X − a) just says that f (X) is in
the principal ideal generated by X − a.
Corollary 7.3.13. Let A be an integral domain and let f (X) ∈ A[X] be a polynomial
of degree n. Then f has at most n distinct roots in A.
CHAPTER 7. RINGS AND MODULES
226
Proof Suppose that a ∈ A is a root of f . Then f (X) = (X − a)g(X), where g(X) has
degree n − 1. Thus, by induction, there are at most n − 1 roots of g. Thus, it suffices to
show that if b = a is a root of f , then b is a root of g.
But 0 = f (b) = (b − a)g(b). Since b − a = 0 and since A is an integral domain, b must
be a root of g.
We obtain a very useful consequence for field theory.
Corollary 7.3.14. Let K be a field. Then any finite subgroup of K × is cyclic.
Proof Recall from Problem 4 of Exercises 4.4.24 that a finite abelian group is cyclic if
and only if it has at most d elements of exponent d for each d which divides its order.
We claim that K × itself has at most d elements of exponent d for any integer d.
To see this, note that 1 is the identity element of K × , so that if a ∈ K × has exponent
d, we have ad = 1. But this says that a is a root of the polynomial X d − 1. Since X d − 1
has degree d, it can have at most d distinct roots, and hence K × can have at most d
distinct elements of exponent d.
As promised during our discussion of group theory, we obtain the following corollary.
Corollary 7.3.15. Let K be a finite field. Then K × is a cyclic group.
Corollary 7.3.15 is a useful consequence of the determination of the kernel of the
evaluation map εa : A[X] → A. Later on, we shall derive useful consequences from
the determinations of the kernels of the evaluation maps εb : A[X] → B for particular
A-algebras B, and elements b ∈ B. Thus, Corollary 7.3.12 should be thought of as a first
step in an important study.
Exercises 7.3.16.
1. Give the proof of Lemma 7.3.2.
2. Suppose that A is noncommutative. What are the elements of the center of A[X]?
3. Let f (X), g(X) ∈ A[X]. Show that the degree of f (X) + g(X) is less than or equal
to the larger of the degrees of f and g.
4. Let f (X), g(X) ∈ A[X]. Show that the degree of f (X) · g(X) is less than or equal
to the sum of the degrees of f and g.
5. Show that A[X] is an integral domain if and only if A is.
6. Show that the units of A[X] are the image under ι of the units of A.
7. Let
M=
1
0
1
1
∈ M2 (R).
Consider the evaluation map εM : R[X] → M2 (R). Show that a polynomial
f (X) ∈ R[X] lies in the kernel of εM if and only if f (1) = 0 and f (1) = 0, where
f (X) is the derivative of f in the sense of calculus.
CHAPTER 7. RINGS AND MODULES
8. Let
M=
227
a 1
0 a
∈ M2 (R)
for some real number a. Consider the evaluation map εM : R[X] → M2 (R). Show
that a polynomial f (X) ∈ R[X] lies in the kernel of εM if and only if f (a) = 0 and
f (a) = 0, where f (X) is the derivative of f in the sense of calculus.
9. Let p be a prime. What are the elements of Z[1/p] ⊂ Q?
10. What are the elements of A[X 2 ] ⊂ A[X]?
11. What are the elements of A[X n ] ⊂ A[X]?
12. What are the elements of Z[2X] ⊂ Z[X]?
13. Show that C = R[i].
‡ 14. Show that Q[i] is a subfield of C.
√
15. What are the elements of Z[ 2] ⊂ R?
√
16. What are the elements of Q[ 2] ⊂ R?
17. Let m be a positive, odd integer. Show that Z[ζm ] = Z[ζ2m ] as a subring of C.
18. Show that every element of Z[ζ8 ] may be written in the form a + bζ8 + cζ82 + dζ83
with a, b, c, d ∈ Z. Write out the formula for the product (a + bζ8 + cζ82 + dζ83 ) ·
(a + b ζ8 + c ζ82 + d ζ83 ).
19. Let A be a ring and let 1 = a ∈ A. Show that
ak − 1 = (a − 1)(1 + a + · · · + ak−1 )
for k ≥ 1.
‡ 20. Cyclotomic units Let m and k be relatively prime to n. Show that the element
(ζnm − 1)/(ζnk − 1) ∈ Q[ζn ] actually lies in Z[ζn ]. Deduce that (ζnm − 1)/(ζnk − 1) is
a unit in Z[ζn ].
More generally, if (m, n) = (k, n) = n, show that (ζnm − 1)/(ζnk − 1) is a unit in
Z[ζn ]. The elements (ζnm − 1)/(ζnk − 1) are often called the cyclotomic units.
21. Let n > 3 be odd. Show that the cyclotomic unit (ζn2 − 1)/(ζn − 1) has infinite
order in Z[ζn ]× .
22. If (m, n) = 1 and m ≡ ±1 mod n, show that the cyclotomic unit (ζnm − 1)/(ζn − 1)
has infinite order in Z[ζn ]× . (Hint: If 0 < φ < π/2, show that
ei(θ−φ) + ei(θ+φ) = reiθ
for some positive real number r. Show also that r > 1 if φ < π/3.)
23. Let n be an even number greater than 6. Show that Z[ζn ]× is infinite.
24. Show that in the category of rings, Z[X] is the free object on a one-element set.
CHAPTER 7. RINGS AND MODULES
228
25. Show that every element of Z6 is a root of X 3 − X ∈ Z6 [X]. Deduce that the
hypothesis that A be an integral domain in Corollary 7.3.13 is necessary.
26. Show that X 2 + 1 has infinitely many roots in the quaternion ring, H. (The issue
here is not zero-divisors, as H is a division ring. The point here is that evaluating
polynomials at an element α ∈ H does not give a ring homomorphism from H[X]
to H unless α lies in the center of H.)
!∞
27. Show that X p − X has infinitely many roots in i=1 Zp .
28. Let p1 , . . . , pk be distinct primes and let m be the least common multiple of p1 −
1, . . . , pk − 1. Show that every element of Zn is a root of X m+1 − X, where
n = p1 . . . pk .
29. Let n > 1 be an integer and let (n, X) be the ideal of Z[X] generated by n and X.
Show that Z[X]/(n, X) is isomorphic to Zn .
30. Let B be any ring and let A be the center of B. Let a ∈ A. Show that there is a
unique homomorphism εa : B[X] → B such that εa (X) = a and εa ◦ ι = 1B , where
ι : B → B[X] is the standard inclusion. Show that the kernel of εa is the two-sided
ideal of B[X] generated by X − a.
31. Let A be a ring and X a variable. We define the ring of formal power series, A[[X]],
with coefficients in A.
∞
The elements of A[[X]] are formal sums i=0 ai X i , with ai ∈ A for all i ≥ 0. We
shall refer to them as formal power series with coefficients in A. Unlike the case of
polynomial rings, we do not assume that only finitely many of the coefficients are
nonzero; indeed, all of the coefficients may be nonzero.
Here, the sum and product operations are given by
∞
"∞
i=0
i=0
ai X i +
∞
bi X i
i=0
=
∞
(ai + bi )X i ,
and
i=0
⎛
⎞
# "∞
#
∞
i
⎝
=
ai X i ·
bi X i
aj bi−j ⎠ X i .
i=0
i=0
j=0
We have an inclusion map η : A[X] → A[[X]], that identifies a polynomial, f (X),
of degree n with the formal power series whose coefficients of index ≤ n agree with
those of f , and whose higher coefficients are all 0.
(a) Show that A[[X]] is a ring and that η is a ring homomorphism.
(b) Show that
∞the polynomial 1 − X is a unit in A[[X]] (but not in A[X]) with
inverse i=0 X i = 1 + X + X 2 + · · · + X n + · · · .
∞
(c) Let A be commutative. Show that a formal power series i=0 ai X i is a unit
in A[[X]] if and only if a0 ∈ A× .
Polynomials in Several Variables
We can also take polynomials in more than one variable. When considering polynomials
in several variables, the most common case, which is the default if nothing more is said,
is to assume that the variables commute with each other. In fact, we shall assume that
CHAPTER 7. RINGS AND MODULES
229
this is the case whenever we use the word polynomial, but some people would explicitly
call these “polynomials in several commuting variables.”
Definitions 7.3.17. Suppose given n variables X1 , . . . , Xn . A primitive7 monomial in
these variables is a product X1i1 · · · Xnin , with ij ≥ 0 for 1 ≤ j ≤ n. We abbreviate
X1i1 · · · Xnin = X I for I = (i1 , . . . , in ) and write
I = {(i1 , . . . , in ) | i1 , . . . , in ∈ Z, i1 , . . . , in ≥ 0}
for the set of all n-tuples of non-negative integers. (We call the elements of I multiindices.)
For any ring A, we define a monomial in X1 , . . . , Xn with coefficient in A to be a product aX1i1 · · · Xnin where a ∈ A and X1i1 · · · Xnin is a primitive monomial in X1 , . . . , Xn .
A polynomial in X1 , . . . , Xn with coefficients in A is a finite sum of monomials in
X1 , . . . , Xn with coefficients in A.
The elements of the polynomial ring A[X1 , . . . , Xn ] are the polynomials in X1 , . . . , Xn
with coefficients in A. Since there’s no preferred order for the multi-indices of the monomials, we shall write the generic polynomial in the form
f (X1 , . . . , Xn ) =
aI X I ,
I∈I
where all but finitely many of the coefficients aI ∈ A are 0. So f is the sum of the finitely
many monomials whose coefficients are nonzero.
Addition of polynomials is obtained as in the one-variable
case, by adding the co
I
efficients
of
the
monomials:
if
f
(X
,
.
.
.
,
X
)
=
a
X
and g(X1 , . . . , Xn ) =
1
n
I
I∈I
I
b
X
,
then
I
I∈I
f (X1 , . . . , Xn ) + g(X1 , . . . , Xn ) =
(aI + bI )X I .
I∈I
The multiplication in A[X1 , . . . , Xn ] is set up to have the following effect on monomials: If I = (i1 , . . . , in ) and J = (j1 , . . . , jn ), then aX I · bX J = abX I+J , where
I + J = (i1 + j1 , . .
. , in + jn ). For generic polynomials,
this is expressed as follows:
If f (X1 , . . . , Xn ) = I∈I aI X I and g(X1 , . . . , Xn ) = I∈I bI X I , then
⎛
⎞
⎝
aJ bI−J ⎠ X I .
f (X1 , . . . , Xn ) · g(X1 , . . . , Xn ) =
I∈I
J≤I
Here, for I = (i1 , . . . , in ) and J = (j1 , . . . , jn ) in I, J ≤ I means that jk ≤ ik for all
indices k (so that ≤ is only a partial order on I), and we define I − J to be the expected
multi-index (i1 − j1 , . . . , in − jn ).
We write 0 for the multi-index whose coordinates are all 0, and identify a ∈ A
with a · X 0 . We write ι : A → A[X1 , . . . , Xn ] for the map that takes each a ∈ A to
a = aX 0 ∈ A[X1 , . . . , Xn ].
We identify each primitive monomial X I with 1X I and identify each variable Xi with
the primitive monomial X (0,...,0,1,0,...,0) , where the 1 occurs in the i-th place.
When n = 2, we shall customarily write A[X, Y ] in place of A[X1 , X2 ].
We require the usual justifications.
7 The use of the word primitive here has no connection whatever to the notion of a primitive element
in a Hopf algebra.
CHAPTER 7. RINGS AND MODULES
230
Lemma 7.3.18. Suppose given a ring A and variables X1 , . . . , Xn . Then the polynomials A[X1 , . . . , Xn ] form a ring under the above addition and multiplication formulæ, and
ι : A → A[X1 , . . . , Xn ] is a ring homomorphism.
Each variable Xi commutes with every element of ι(A), and the variables Xi and Xj
commute with each other for all i, j. If A is commutative, so is A[X1 , . . . , Xn ]. More
generally, a polynomial f (X1 , . . . , Xn ) lies in the center of A[X1 , . . . , Xn ] if and only if
its coefficients lie in the center of A.
Polynomials in several variables satisfy universal properties similar to those of polynomials in one variable. We leave the proofs to Exercises 7.3.27.
Proposition 7.3.19. Let g : A → B be a ring homomorphism and suppose given elements b1 , . . . , bn ∈ B such that the following two conditions hold.
1. Each bi commutes with every element of g(A).
2. For i, j ∈ {1, . . . , n}, bi commutes with bj .
Then there is a unique ring homomorphism g b1 ,...,bn : A[X1 , . . . , Xn ] → B such that
g b1 ,...,bn (Xi ) = bi for 1 ≤ i ≤ n and the following diagram commutes.
g
/B
A LL
LLL
rr8
r
r
LLL
r
rrr
ι LLL
rrr gb1 ,...,bn
%
A[X1 , . . . , Xn ]
Moreover, g b1 ,...,bn satisfies the formula
"
#
I
=
g b1 ,...,bn
aI X
g(aI )bI .
I∈I
I
Here, if I = (i1 , . . . , in ), we set b =
bi11
I∈I
. . . binn ,
where b0i is set equal to 1 for all i.
As in the case of a single variable, Proposition 7.3.19 admits a useful restatement in
the context of algebras over a commutative ring.
Corollary 7.3.20. Let A be a commutative ring and let B be an A-algebra, via ν : A →
B. Suppose given b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}.
Then there is a unique A-algebra homomorphism εb1 ,...,bn : A[X1 , . . . , Xn ] → B with the
property that εb (Xi ) = bi for 1 ≤ i ≤ n. It is given by the formula
"
#
I
=
εb1 ,...,bn
aI X
ν(aI )bI .
I∈I
I∈I
Here, if I = (i1 , . . . , in ), we set bI = bi11 . . . binn , where b0i is set equal to 1 for all i.
We now codify some terms.
Definitions 7.3.21. Let A be a commutative ring. Let B be an A-algebra and let
b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. We shall refer
to the A-algebra homomorphism εb1 ,...,bn as the evaluation map obtained by evaluating
Xi at bi for 1 ≤ i ≤ n. We write A[b1 , . . . , bn ] for the image of εb1 ,...,bn . We call
A[b1 , . . . , bn ] the A-subalgebra of B generated by b1 , . . . , bn .
Suppose now that B is a commutative A-algebra. We say that B is finitely generated
as an A-algebra (or has finite type as an A-algebra) if B = A[b1 , . . . , bn ] for some collection
of elements b1 , . . . , bn ∈ B.
CHAPTER 7. RINGS AND MODULES
231
A special case of the evaluation maps is of fundamental importance in algebraic
geometry.
Proposition 7.3.22. Let A be a commutative ring and let a1 , . . . , an ∈ A. Let
a = (X1 − a1 , . . . , Xn − an ),
the ideal of A[X1 , . . . , Xn ] generated by X1 − a1 , . . . , Xn − an . Then a is the kernel of
the evaluation map εa1 ,...,an : A[X1 , . . . , Xn ] → A.
Proof Since each Xi −ai lies in the kernel of εa1 ,...,an , Corollary 7.2.19 shows that there
is a unique ring homomorphism εa1 ,...,an : A[X1 , . . . , Xn ]/a → A such that the following
diagram commutes.
εa1 ,...,an
/
A[X1 , . . . , Xn ]
q8 A
RRR
q
q
RRR
qq
RR
qqq
q
π RRRR
q
ε
a
,...,a
n
R(
qq 1
A[X1 , . . . , Xn ]/a
Here, π : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn ]/a is the canonical map. It suffices to show
that εa1 ,...,an is an isomorphism.
Let ι = π ◦ ι : A → A[X1 , . . . , Xn ]/a, where ι is the A-algebra structure map of
A[X1 , . . . , Xn ]. Then
εa1 ,...,an ◦ ι = εa1 ,...,an ◦ π ◦ ι = εa1 ,...,an ◦ ι = 1A ,
where the last equality comes from the fact that εa1 ,...,an is an A-algebra homomorphism.
Thus, it suffices to show that ι is onto.
Since every element of A[X1 , . . . , Xn ] is a sum of monomials and since π is onto, it
suffices to show that π(aX1i1 . . . Xnin ) lies in the image of ι for all a ∈ A and all n-tuples
(i1 , . . . , in ) of nonnegative integers. Since π(Xi − ai ) = 0, we have π(Xi ) = ι(ai ) for all
i. Thus, π(aX1i1 . . . Xnin ) = ι(aai11 . . . ainn ), and the result follows.
The next result could have been left as an exercise in universal properties, but it is
sufficiently important that we give it here.
Proposition 7.3.23. Let X1 , . . . , Xn be indeterminates over the ring A. Then there is
a unique ring homomorphism
α : (A[X1 , . . . , Xn−1 ])[Xn ] → A[X1 , . . . , Xn ]
that carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions of
A in the two sides. Moreover, α is an isomorphism whose inverse is the unique ring
homomorphism
β : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ][Xn ]
which carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions of A in
the two sides.
Thus, A[X1 , . . . , Xn ] may be considered to be the polynomial ring in the variable Xn
with coefficients in A[X1 , . . . , Xn−1 ]. In particular, every polynomial f (X1 , . . . , Xn ) in
n variables over A may be written uniquely in the form
f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 )
for some k ≥ 0, where fi (X1 , . . . , Xn−1 ) ∈ A[X1 , . . . , Xn−1 ] for all i.
CHAPTER 7. RINGS AND MODULES
232
Proof Here, the natural inclusion of A in A[X1 , . . . , Xn−1 ][Xn ] is the composite
ι
ι
A−
→ A[X1 , . . . , Xn−1 ] −
→ A[X1 , . . . , Xn−1 ][Xn ],
where ι is the standard inclusion of A in the polynomial ring in n − 1 variables and ι is
the standard inclusion of A[X1 , . . . , Xn−1 ] in its polynomial ring in the variable Xn .
Notice that the variables Xi lie in the centers of each of the rings under consideration.
Thus, the existence and uniqueness of β is immediate from Proposition 7.3.19. For α,
note that Proposition 7.3.19 provides a unique homomorphism α0 : A[X1 , . . . , Xn−1 ] →
A[X1 , . . . , Xn ] that carries each Xi to Xi and commutes with the natural inclusions of
A in both sides. But now the universal property of a polynomial ring in a single variable
allows us to extend α0 to a map α carrying Xn to Xn .
For the uniqueness of α, we claim that any two ring homomorphisms f, g : A[X1 , . . . , Xn−1 ][Xn ] →
B that agree on A and agree on each of the variables Xi must be equal. To see this,
note first that f and g must agree on A[X1 , . . . , Xn−1 ] by the uniqueness property of
Proposition 7.3.19. But then, since f and g also agree on Xn , they must be equal by the
uniqueness property of Proposition 7.3.3.
But now notice that each of the composites α ◦ β and β ◦ α carries each Xi to Xi
and commutes with the standard inclusions of A. Thus, α ◦ β is the identity by the
uniqueness property of Proposition 7.3.19, while β ◦ α is the identity by the uniqueness
property derived in the preceding paragraph.
If A is commutative and f (X1 , . . . , Xn ) ∈ A[X1 , . . . , Xn ], then there is a function
f : An → A which takes (a1 , . . . , an ) to f (a1 , . . . , an ), the effect of evaluating f at
(a1 , . . . , an ). Indeed, much of our intuition about polynomials will have come from calculus, where polynomials are treated as functions.
Of course, if A is finite, then there are only finitely many functions from An to A,
but infinitely many polynomials in A[X1 , . . . , Xn ]. Thus, in this case, a polynomial is
not determined by its effect as a function on An .
Finiteness is not
!∞the only difficulty in trying to treat polynomials as functions. For
instance, if A = i=1 Z2 , then the polynomial X 2 is easily seen to induce the identity
map of A, which is also induced by X.
This suggests that if we wish to interpret polynomials as functions, then we might
think about restricting attention to integral domains with infinitely many elements. First,
we shall formalize the situation.
Definition 7.3.24. Let X be any set and let A be a ring. The ring of functions from
X to A, written Map(X, A), is the ring whose elements are the functions from X to A
and whose operations are given by
(f + g)(x) = f (x) + g(x)
and
(f · g)(x) = f (x)g(x)
for all f, g ∈ Map(X, A) and x ∈ X. The additive and multiplicative identities are the
constant functions to 0 and 1, respectively.8
The next lemma follows from the fact that the evaluation maps are ring homomorphisms.
8 If X is a compact topological space, then the ring C(X) of continuous functions from X to C plays
an important role in the theory of C ∗ algebras. Of course, C(X) is a subring of Map(X, C).
CHAPTER 7. RINGS AND MODULES
233
Lemma 7.3.25. Let A be a commutative ring and let φ : A[X1 , . . . , Xn ] → Map(An , A)
be the map which takes each polynomial f to the function obtained by evaluating f on
the elements of An . Then φ is a ring homomorphism.
Thus, two polynomials, f and g, induce the same function from An to A if and only
if f − g ∈ ker φ.
Thus, the elements of A[X1 , . . . , Xn ] are determined by their effect as functions if
and only if ker φ = 0. The next proposition gives a slightly stronger result when A is an
infinite domain.
Proposition 7.3.26. Let S be an infinite subset of the integral domain A. Suppose given
a polynomial f (X1 , . . . , Xn ) in n variables over A with the property that f (a1 , . . . , an ) = 0
for all a1 , . . . , an ∈ S. Then f = 0.
Thus, an element of A[X1 , . . . , Xn ] is determined by its effect on S × · · · × S ⊂ An
for any infinite subset S of A.
Proof We argue by induction on n, with the result being true for n = 1, as f would
have infinitely many roots.
Suppose, inductively, that the result is true for polynomials of n − 1 variables, and
let a ∈ S. Then evaluating Xn at a gives a polynomial f (X1 , . . . , Xn−1 , a) in n − 1
variables over A, which vanishes on all (n − 1)-tuples, a1 , . . . , an−1 of elements of S. By
the inductive assumption, f (X1 , . . . , Xn−1 , a) = 0.
Write B = A[X1 , . . . , Xn−1 ]. By Proposition 7.3.23, we may identify A[X1 , . . . , Xn ]
with B[Xn ]. Under this identification, f (X1 , . . . , Xn−1 , a) is the image of f under the
evaluation map
εa : B[Xn ] → B
obtained by evaluating Xn at a. Thus, when f is thought of as an element of B[Xn ],
each a ∈ S is a root of f . Since B is a domain and since f has infinitely many roots in
B, f = 0.
Exercises 7.3.27.
1. Give the proof of Lemma 7.3.18.
2. Give the proof of Proposition 7.3.19.
3. Give the proof of Corollary 7.3.20.
‡ 4. Let B be an A-algebra and let b1 , . . . , bn ∈ B such that bi commutes with bj
whenever i, j ∈ {1, . . . , n}. Show that A[b1 , . . . , bn ] is the smallest A-subalgebra of
B containing b1 , . . . , bn .
5. Let A be a commutative ring. What are the elements of A[X 2 , X 3 ] ⊂ A[X]?
6. Let A be a commutative ring. What are the elements of A[X 3 , X 4 ] ⊂ A[X]?
7. Let A be a commutative ring and let X and Y be indeterminates. Consider the
A-algebra structure on A[X] × A[Y ] given by the structure map
Δ
ι×ι
A −→ A × A −−→ A[X] × A[Y ]
where Δ is the diagonal map Δ(a) = (a, a) and the maps ι are the standard
structure maps for the polynomial rings A[X] and A[Y ]. What are the elements of
the A-subalgebra of A[X] × A[Y ] generated by (X, 0) and (0, Y )?
CHAPTER 7. RINGS AND MODULES
234
8. Show that Z[1/2, 1/3] = Z[1/6] as subrings of Q. (Hint: By Problem 4, it suffices
to show that 1/2, 1/3 ∈ Z[1/6] and that 1/6 ∈ Z[1/2, 1/3].)
9. Show that Q is not finitely generated as a Z-algebra.
10. Let m and n be relatively prime positive integers. Show that Z[ζm , ζn ] = Z[ζmn ]
2π
as subrings of C. Here, ζk = ei· k is the standard primitive k-th root of unity.
11. Let A be a commutative ring. Show that A[X1 , . . . , Xn ] is the free commutative
A-algebra on the set {X1 , . . . , Xn }.
7.4
Symmetry of Polynomials
We consider some issues relating to permutations of the variables in a polynomial ring.
Throughout this section, A is a commutative ring.
Suppose given n variables, X1 , . . . , Xn . Then the symmetric group on n letters, Sn ,
acts on the set of variables in the obvious way: σ · Xi = Xσ(i) . By Corollary 7.3.20, the
induced map
σ : {X1 , . . . , Xn } → {X1 , . . . , Xn } ⊂ A[X1 , . . . , Xn ]
extends uniquely to an A-algebra homomorphism
σ∗ : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn ]
via σ∗ (f (X1 , . . . , Xn )) = f (Xσ(1) , . . . , Xσ(n) ).
Since an A-algebra homomorphism out of A[X1 , . . . , Xn ] is determined by its effect on
the variables, σ ·f = σ∗ (f ) is easily seen to give an action of Sn on A[X1 , . . . , Xn ] through
A-algebra homomorphisms. We first consider the fixed points of this action.
Definition 7.4.1. We write A[X1 , . . . , Xn ]Sn for the fixed points of the action of Sn on
A[X1 , . . . , Xn ]. Thus, a polynomial f (X1 , . . . , Xn ) is in A[X1 , . . . , Xn ]Sn if and only if
σ∗ (f ) = f for all σ ∈ Sn . We shall refer to A[X1 , . . . , Xn ]Sn as the symmetric polynomials
in n variables over A.
The reader may easily check that the symmetric polynomials form an A-subalgebra
of A[X1 , . . . , Xn ]. We shall show that it is a polynomial ring on certain symmetric
polynomials. We shall argue by an induction in which the next lemma will play a part.
Lemma 7.4.2. Let π : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ] be the A-algebra map that
takes Xi to Xi if i < n and takes Xn to 0. Then π carries the symmetric polynomials
in n variables into the symmetric polynomials in n − 1 variables, inducing an A-algebra
homomorphism
π : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 .
Proof Recall from Proposition 7.3.23 that A[X1 , . . . , Xn ] may be identified with the
polynomial ring on A[X1 , . . . , Xn−1 ] in the variable Xn . Thus, any nonzero element of
A[X1 , . . . , Xn ] may be written uniquely in the form
f (X1 , . . . , Xn ) =
m
i=0
fi (X1 , . . . , Xn−1 )Xni
CHAPTER 7. RINGS AND MODULES
235
for some m ≥ 0, with f0 , . . . , fm ∈ A[X1 , . . . , Xn−1 ], and fm = 0.
Now Sn−1 acts as permutations on the first n − 1 variables of the polynomials in n
variables, and if σ ∈ Sn−1 , then
σ∗ (f ) =
m
σ∗ (fi )Xni .
i=0
But by the uniqueness of the representation of polynomials in n variables as polynomials
in Xn with coefficients in A[X1 , . . . , Xn−1 ] we see that if f ∈ A[X1 , . . . , Xn ]Sn , then each
fi must be fixed by each σ ∈ Sn−1 . Thus, if f is a symmetric polynomial in n variables,
then each fi is a symmetric polynomial in n − 1 variables.
But since π(Xn ) = 0, we see that π(f ) = f0 , and the result follows.
We now define the elements we will show to be the polynomial generators of the
symmetric polynomials.
Definition 7.4.3. The elementary symmetric polynomials, or elementary symmetric
functions, on n variables are defined by
⎧
1
if i = 0
⎪
⎨
X
.
.
.
X
if 1 ≤ i ≤ n
j
j
1
i
si (X1 , . . . , Xn ) =
⎪
⎩ j1 <···<ji
0
if i > n.
More generally, if B is a commutative ring and if b1 , . . . , bn ∈ B, then we write
si (b1 , . . . , bn ) ∈ B for the element obtained by evaluating si (X1 , . . . , Xn ) at b1 , . . . , bn .
Note that we may do this because the elementary symmetric functions have integer
coefficients.
Thus, for instance, s1 (X1 , . . . , Xn ) = X1 + · · · + Xn , or at the other extreme,
sn (X1 , . . . , Xn ) = X1 · · · Xn .
Note that the summands of si (X1 , . . . , Xn ) are the products of i distinct variables,
and that each such product occurs exactly once. Thus, if σ ∈ Sn , then σ∗ permutes the
summands of each si (X1 , . . . , Xn ), and hence the elementary symmetric functions are
indeed symmetric polynomials.
The elementary symmetric functions are ubiquitous in mathematics. One of the
reasons for this is that they give the formula for computing the coefficients of a polynomial
with a predetermined collection of roots.
Lemma 7.4.4. Suppose given n elements, a1 , . . . , an ∈ A, not necessarily distinct. Then
the following equality holds in A[X].
(X − a1 ) . . . (X − an ) = X n − s1 X n−1 + · · · + (−1)n sn ,
where si = si (a1 , . . . , an ) for i = 1, . . . , n.
Proof In the expansion of (X − a1 ) . . . (X − an ) from the distributive law, the terms
of degree k in X are those obtained by choosing k of the factors from which to choose
an X, leaving n − k factors of the form (−ai ). Adding these up, we get
⎞
⎛
⎝
(−aj1 ) . . . (−ajn−k )⎠ · X k = (−1)n−k sn−k (a1 , . . . , an ) · X k
j1 <···<jn−k
for the monomial of degree k in the resulting polynomial in X.
CHAPTER 7. RINGS AND MODULES
236
The reader may easily verify the following lemma.
Lemma 7.4.5. Let π : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 be obtained by setting
Xn = 0, as above. Then
si (X1 , . . . , Xn−1 ) if 0 ≤ i < n
π(si (X1 , . . . , Xn )) =
0
if i ≥ n
There are various definitions of degree for a polynomial in several variables. The
notion of total degree is one of them.
Definition 7.4.6. The total degree of a monomial aX1r1 . . . Xnrn is r1 + · · · + rn . The
total degree of a polynomial f ∈ A[X1 , . . . , Xn ] is the maximum of the total degrees of
the nonzero monomials in f .
We are now ready to show that A[X1 , . . . , Xn ]Sn is a polynomial algebra on the
generators s1 (X1 , . . . , Xn ), . . . , sn (X1 , . . . , Xn ).
Proposition 7.4.7. Let αn : A[Y1 , . . . , Yn ] → A[X1 , . . . , Xn ]Sn be the A-algebra homomorphism that carries Yi to si (X1 , . . . , Xn ), where Y1 , . . . , Yn are indeterminates over A.
Then α is an isomorphism.
Proof We argue by induction on n. Since s1 (X) = X, the result is trivial for n = 1.
Thus, we shall assume, inductively, that αn−1 is an isomorphism.
We first show that αn is surjective. Suppose given a symmetric polynomial f (X1 , . . . , Xn ).
We shall argue by induction on the total degree of f that f is in the image of α. The
result is trivial in total degree 0.
Since αn−1 is surjective, there is a polynomial, g, of n − 1 variables such that
π(f ) = g(s1 (X1 , . . . , Xn−1 ), . . . , sn−1 (X1 , . . . , Xn−1 ))
Thus, our calculation of the effect of π on the elementary symmetric functions shows
that if we set
h(X1 , . . . , Xn ) = f (X1 , . . . , Xn ) − g(s1 (X1 , . . . , Xn ), . . . , sn−1 (X1 , . . . , Xn )),
then h lies in the kernel of π.
m
i
Thus, if we write h(X1 , . . . , Xn ) =
i=0 hi (X1 , . . . , Xn−1 )Xn , we see that h0 =
π(h) = 0, and hence h(X1 , . . . , Xn ) = Xn · k(X1 , . . . , Xn ) for some polynomial k.
Let πi : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ] be the A-algebra map obtained by setting
⎧
if j < i
⎨ Xj
0
if j = i
πi (Xj ) =
⎩
Xj−1 if j > i,
and write πi : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 for the induced map. Thus,
π = πn . Since h is in the kernel of π, symmetry shows that it must also be in the kernel
of πi for 1 ≤ i ≤ n − 1. Since h(X1 , . . . , Xn ) = Xn · k(X1 , . . . , Xn ), we have
0 = πn−1 (h(X1 , . . . , Xn )) = πn−1 (Xn )πn−1 (k(X1 , . . . , Xn )).
Since πn−1 (Xn ) = Xn−1 is not a zero divisor in A[X1 , . . . , Xn−1 ], k must lie in the kernel
of πn−1 , and hence k(X1 , . . . , Xn ) = Xn−1 · p(X1 , . . . , Xn ) for a suitable polynomial p,
by the argument above.
CHAPTER 7. RINGS AND MODULES
237
An easy induction now shows that h(X1 , . . . , Xn ) = X1 · · · Xn · q(X1 , . . . , Xn ) for a
suitable polynomial q. Since both h and X1 · · · Xn = sn (X1 , . . . , Xn ) are symmetric, and
since sn (X1 , . . . , Xn ) is not a zero divisor, we see that q must be a symmetric polynomial.
Note that by the known properties of π, h must have total degree less than or equal
to that of f , while q has total degree less than that of h. Thus, q(X1 , . . . , Xn ) is in the
image of αn by induction. But then f must also be in the image of αn .
To see that αn is injective, we argue by contradiction. Let f (Y1 , . . . , Yn ) be an
element
of the kernel of αn with the lowest possible total degree. Write f (Y1 , . . . , Yn ) =
m
i
f
(Y
i
1 , . . . , Yn−1 )Yn . Note that
i=0
0 = (π ◦ αn )(f ) = f0 (s1 (X1 , . . . , Xn−1 ), . . . , sn−1 (X1 , . . . , Xn−1 )).
Since αn−1 is injective, this forces f0 (Y1 , . . . , Yn−1 ) to be 0. But then
f (Y1 , . . . , Yn ) = Yn · g(Y1 , . . . , Yn ),
and hence
αn (f ) = sn (X1 , . . . , Xn ) · αn (g).
Thus, g lies in the kernel of αn . Since its total degree is less than that of f , this contradicts
the minimality of the degree of f .
Another important topic regarding symmetries of polynomials concerns discriminants.
Proposition 7.4.8. Consider the following element of A[X1 , . . . , Xn ]:
.
(Xj − Xi ).
δ(X1 , . . . , Xn ) =
i<j
Then for σ ∈ Sn , we have σ∗ (δ) = ε(σ) · δ, where ε(σ) is the sign of σ.
Proof We may may identify the set of indices i < j in the above product with the
set, S, of all two-element subsets of {1, . . . , n}. Of course, Sn acts on S, via σ({i, j}) =
{σ(i), σ(j)}. Using this action to re-index the factors of δ, we see that
.
(Xmax(σ(i),σ(j)) − Xmin(σ(i),σ(j)) ).
δ(X1 , . . . , Xn ) =
i<j
For i < j, σ∗ (Xj − Xi ) = Xσ(j) − Xσ(i) = ±(Xmax(σ(i),σ(j)) − Xmin(σ(i),σ(j)) ), where the
sign is positive if σ preserves the order of i and j and is negative otherwise. Thus,
σ∗ (δ) = (−1)k δ,
where k is the number of pairs i < j for which σ(j) < σ(i).
Define ψ : Sn → ±1 by ψ(σ) = (−1)k where (−1)k is defined by the equation
σ∗ (δ) = (−1)k δ. Then it follows immediately from the fact that the homomorphisms σ∗
define an action of Sn on A[X1 , . . . , Xn ] that ψ is a homomorphism.
It suffices to show that ψ = ε. Since Sn is generated by transpositions, it suffices to
show that if τ is a transposition, then ψ(τ ) = −1.
Thus, let τ = (i j), with i < j. Then the pairs on which τ reverses order consist of
the pairs i < k and k < j, where i < k < j, together with the pair i < j itself. The
signs associated with i < k < j cancel each other out, leaving a single −1 from the pair
i < j.
Definition
7.4.9. The discriminant, Δ(X1 , . . . , Xn ), is defined by Δ = δ 2 , where δ =
!
i<j (Xj − Xi ), as above.
CHAPTER 7. RINGS AND MODULES
238
Since each σ∗ (δ) = ±δ, we obtain the following corollary.
Corollary 7.4.10. The discriminant, Δ, is a symmetric polynomial in the variables
X1 , . . . , Xn . Thus, there is a unique polynomial Fn (X1 , . . . , Xn ), such that
Δ(X1 , . . . , Xn ) = Fn (s1 (X1 , . . . , Xn ), . . . , sn (X1 , . . . , Xn )).
Example 7.4.11. The discriminant on two variables is given by
Δ(X, Y ) = (Y − X)2 = X 2 + Y 2 − 2s2 (X, Y ).
Now note that
(s1 (X, Y ))2 = (X + Y )2 = X 2 + Y 2 + 2s2 (X, Y ).
Thus, Δ = s21 − 4s2 , and hence F2 (X, Y ) = X 2 − 4Y .
We shall calculate F3 (X, Y, Z) in the next set of exercises. It takes considerably more
work than F2 . Since the polynomials Fn (X1 , . . . , Xn ) have integer coefficients, one may
save considerable time by using a computer algebra package to calculate them.
Recall from Lemma 7.4.4 that if a polynomial f (X) factors as a product of monic
polynomials of degree 1, then the coefficients of f are, up to sign, the elementary symmetric functions on the roots.
Definition 7.4.12. Let f (X) = a0 X n + a1 X n−1 + · · · + an be a polynomial of degree n
in A[X] and let Fn (X1 , . . . , Xn ) be the polynomial determined in Corollary 7.4.10. Then
the discriminant, Δ(f ), of f is the following element of A:
Δ(f ) = Fn (−a1 /a0 , a2 /a0 , . . . , (−1)n an /a0 ).
Of course, we don’t know much at this point about the polynomials Fn , so the following corollary expresses the discriminant in a more familiar setting. It follows immediately
from Lemma 7.4.4 and Corollary 7.4.10.
Corollary 7.4.13. Let f (X) be a monic polynomial in A[X], and suppose that f factors
as
f (X) = (X − b1 ) . . . (X − bn )
in B[X], where B is a commutative ring that contains A as a subring. Then the discriminant, Δ(f ), may be calculated by
Δ(f )
=
=
Δ(b1 , . . . , bn )
.
(bj − bi )2 .
i<j
Exercises 7.4.14.
1. Write X12 + · · · + Xn2 as a polynomial in the elementary symmetric functions on
X1 , . . . , Xn .
CHAPTER 7. RINGS AND MODULES
239
2. Show that πi : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 is injective when restricted
to the A-submodule consisting of the symmetric polynomials of total degree less
than n.
3. Let f (X) = aX 2 + bX + c be a quadratic in A[X]. Show that Δ(f ) = (b2 − 4ac)/a2 .
Deduce that a quadratic f (X) ∈ R[X] has a root in R if and only if Δ(f ) ≥ 0.
4. Calculate the polynomial F3 (X, Y, Z) that’s used to calculate the discriminant of
a cubic.
† 5. Show that if f (X) = X 3 + pX + q is a monic cubic with no X 2 term, then Δ(f ) =
−4p3 − 27q 2 .
7.5
Group Rings and Monoid Rings
Polynomial rings are a special case of a more general construction called monoid rings.
Definitions 7.5.1. Let M be a monoid, and write the product operation of M in multiplicative notation, even if M is abelian. We define A[M
], the monoid ring of M over
A, as follows. The elements of A[M ] are formal sums m∈M am m, such that the coefficient am is zero for all but finitely many m in M . (We abbreviate this by saying that
am = 0 almost
polynomials, addition is given by adding
everywhere.) As
in the case of coefficients: ( m∈M am m) + ( m∈M bm m) = m∈M (am + bm )m.
The multiplication is defined as follows.
⎛
⎞
#"
#
"
⎝
am m
bm m =
ax by ⎠ m.
m∈M
m∈M
m∈M
{(x,y) | xy=m}
The last sum ranges over all ordered pairs (x, y) of elements in M whose product is m.
In the generic monoid, that collection of ordered pairs can be difficult to determine. But
in the examples we are concerned with, there will be no difficulty.
We identify each m ∈ M with the element of A[M ] whose m-th coefficient is 1 and
whose other coefficients are all 0. Thus, we have a natural inclusion M ⊂ A[M ].
Similarly, we identify a ∈ A with the element of A[M ] whose m-th coefficient is 0
for all m = 1, and whose coefficient of 1 is equal to a. Here, we write 1 for the identity
element of M . We write ι : A → A[M ] for the map that gives this identification.
In the case where the monoid is a group G, we call A[G] the group ring (or group
algebra, if A is commutative) of G over A.
Example 7.5.2. Write Z2 = T = {1, T }. Then for any ring A,
A[Z2 ] = {a + bT | a, b ∈ A}.
Here, if a, b, c, d ∈ A, we have (a+bT )+(c+dT ) = (a+c)+(b+d)T and (a+bT )·(c+dT ) =
(ac + bd) + (ad + bc)T .
Lemma 7.5.3. Let A be a ring and let M be a monoid. Then A[M ] is a ring under the
above operations, and ι : A → A[M ] is a ring homomorphism.
The inclusion M ⊂ A[M ] is a monoid homomorphism from M to the multiplicative
monoid of A[M ], and each m ∈ M commutes with every element of ι(A).
If A is commutative, then A[M ] is an A-algebra, via ι : A → A[M ].
CHAPTER 7. RINGS AND MODULES
240
Proof Everything here is straightforward, with the possible exception of the associativity of multiplication in A[M ]. We shall treat that, and leave the rest to the reader.
Thus, suppose given α, β, γ ∈ A[M ], with
α=
am m, β =
bm m and γ =
cm m.
m∈M
m∈M
m∈M
Then
αβ =
dm m, where dm =
ax by .
{(x,y) | xy=m}
m∈M
Thus,
(αβ)γ
=
⎛
⎝
m∈M
=
⎛
⎛
⎝
{(w,z) | wz=m}
⎝
m∈M
dw cz ⎠ m
{(w,z) | wz=m}
⎝
m∈M
=
⎛
⎞
⎞
⎞
ax by ⎠ cz ⎠ m
{(x,y) | xy=w}
⎞
ax by cz ⎠ m.
{(x,y,z) | xyz=m}
Here, the last equality comes from distributing the terms cz through the sums {(x,y) | xy=w} ax by .
For each ordered triple (x, y, z) with xyz = m, there is exactly one summand ax by cz thus
obtained.
As the reader may now check, the expansion of α(βγ) produces exactly the same
result.
Monoid rings have an important universal property.
Proposition 7.5.4. Let f : A → B be a ring homomorphism. Let M be a monoid and
let g : M → B be a homomorphism from M to the multiplicative monoid of B. Suppose
that for each m ∈ M , g(m) commutes with every element of f (A). Then there is a unique
ring homomorphism h : A[M ] → B such that the following diagram commutes:
⊃
ι /
A[M ] o
A?
M
??

??


??

?? h 
g

f
?? 

B
Explicitly, we have
"
h
m∈M
#
am m
=
f (am )g(m).
m∈M
Proof The element m∈M am m is, in fact, a finite sum of terms am m. Moreover, if
j : M ⊂ A[M ] is the inclusion, then am m = ι(am )j(m). Thus, if h : A[M ] → B is a ring
CHAPTER 7. RINGS AND MODULES
241
homomorphism
making the diagram commute, then we must have h
m∈M am m =
m∈M f (am )g(m), and hence h is unique as claimed.
Thus, it suffices to show that if f and g satisfy the stated hypotheses, then h, as
defined by the formula above, is a ring homomorphism. By the distributive law in
B and the fact that f is a homomorphism of abelian groups, we see that h is also a
homomorphism of abelian groups. Also, h(1) = f (1)g(1) = 1, so it suffices to show that
h(αβ) = h(α)h(β) for all α, β ∈ A[M ].
By the distributive laws in both A[M ] and B, it suffices to show that h(am · a m ) =
h(am)h(a m ) for all a, a ∈ A and m, m ∈ M . But h(am · a m ) = h(aa mm ) =
f (aa )g(mm ) = f (a)f (a )g(m)g(m ), while h(am)h(a m ) = f (a)g(m)f (a )g(m ), so
the result follows from the fact that f (a ) commutes with g(m).
In the case of algebras over a commutative ring, the statement becomes simpler.
Corollary 7.5.5. Let A be a commutative ring and let B be an A-algebra. Let M be a
monoid and let g : M → B be a monoid homomorphism from M to the multiplicative
monoid of B. Then there is a unique A-algebra homomorphism h : A[M ] → B such that
h
the composite M ⊂ A[M ] −
→ B is equal to g. Explicitly,
"
#
h
am m =
ν(am )g(m),
m∈M
m∈M
where ν : A → B is the structure map of B as an A-algebra.
Proof If B is an A-algebra with structure map ν, then ν(A) is contained in the center
of B, and hence every element of ν(A) commutes with every element of g(M ). Thus,
Proposition 7.5.4 provides a unique ring homomorphism h, given by the displayed formula, such that the diagram in the statement of Proposition 7.5.4 commutes, with f
replaced by ν. But the left-hand triangle of the diagram commutes if and only if h is
an A-algebra homomorphism, while the right-hand triangle commutes if and only if the
h
composite M ⊂ A[M ] −
→ B is equal to g.
Note that if G is a group, then a monoid homomorphism g : G → B from G to the
multiplicative monoid of B always takes value in B × , the group of invertible elements
in the multiplicative monoid of B. Thus, g restricts to give a group homomorphism
g : G → B × . We obtain the following corollary.
Corollary 7.5.6. Let A be a commutative ring and let B be an A-algebra. Let G be
a group and let f : G → B × be a homomorphism. Then there is a unique A-algebra
h
homomorphism h : A[G] → B such that the composite G ⊂ A[G] −
→ B is equal to f .
Explicitly,
⎛
⎞
h⎝
ag g ⎠ =
ν(ag )f (g),
g∈G
g∈G
where ν : A → B is the structure map of B as an A-algebra.
There is an important homomorphism from a group ring A[G] to A.
Definitions 7.5.7. Let A be a ring and let G be a group. Then the standard augmentation of the group ring A[G] is the unique ring homomorphism (which exists by
CHAPTER 7. RINGS AND MODULES
242
Proposition 7.5.4) ε : A[G] → A with the property that ε(g) = 1 for all g ∈ G and such
that ε◦ι = 1A , where ι is the standard inclusion of A in A[G]. Explicitly, Proposition 7.5.4
gives
⎛
⎞
ε⎝
ag g ⎠ =
ag .
g∈G
g∈G
We define the augmentation ideal, I(G) (or IA (G), if the ring A is allowed to vary),
to be the kernel of ε.
Thus, we can obtain information about A[G] from information about A and information about I(G). We shall study augmentation ideals in the next set of exercises.
In the case that A is commutative, augmentation ideals admit a nice generalization
to the study of certain A-algebras.
Definition 7.5.8. Let A be a commutative ring. An augmented A-algebra consists of
an A-algebra, B, together with an A-algebra homomorphism ε : B → A. Here, since A is
given the A-algebra structure map 1A , this simply means that ε is a ring homomorphism
such that ε ◦ ν = 1A , where ν is that A-algebra structure map of B.
The ideal of the augmentation ε is its kernel.
Examples 7.5.9.
1. Let A be a commutative ring and let a ∈ A. Then the evaluation map εa : A[X] →
A is an augmentation. By Corollary 7.3.12, the ideal of this augmentation is the
principal ideal (X − a).
2. Similarly, if A is commutative and a1 , . . . , an ∈ A, then the evaluation map εa1 ,...,an :
A[X1 , . . . , Xn ] → A is an augmentation. Here, Proposition 7.3.22 tells us the augmentation ideal is (X1 − a1 , . . . , Xn − an ).
3. Let M be any monoid and let A be a commutative ring. Then the trivial monoid
homomorphism from M tothe multiplicative
monoid of A induces an augmentation
map ε : A[M ] → A via ε( m∈M am m) = m∈M am .
Exercises 7.5.10.
1. Complete the proof of Lemma 7.5.3.
2. Show that in the group ring of a group G over A, we may write
ax by =
ax bx−1 g .
{(x,y) | xy=g}
x∈G
3. Let Z2 = T = {1, T }. For any ring A, show that (1 − T )(1 + T ) = 0 in A[Z2 ],
and hence A[Z2 ] has zero-divisors.
4. Show that A[Zn ] has zero-divisors for any ring A and any integer n > 1. Deduce
that if A is any ring and if G is a group with torsion elements of order > 1, then
A[G] has zero-divisors.
5. Let A be a ring and
let G be a finite group. Let Σ ∈ A[G] be the sum of all the
elements of G: Σ = g∈G g. Show that for any x ∈ A[G], xΣ = (ι ◦ ε(x))Σ, where
ε : A[G] → A is the standard augmentation and ι : A → A[G] is the standard
inclusion.
CHAPTER 7. RINGS AND MODULES
243
‡ 6. Let ζn = ei·2π/n ∈ C be the standard primitive n-th root of unity. Let A be any
subring of C and let Zn = T = {1, T, . . . , T n−1 }. Show that there is a unique
A-algebra homomorphism f : A[Zn ] → A[ζn ] such that f (T ) = ζn . Show that f is
onto.
‡ 7. Let Zn = T = {1, T, . . . , T n−1 }. Let f : Z[Zn ] → Z[ζn ] be the homomorphism of
Problem 6.
(a) Show that Σ = 1 + T + · · · + T n−1 lies in the kernel of f , and hence there’s
an induced, surjective ring homomorphism
f : Z[Zn ]/(Σ) → Z[ζn ],
where (Σ) is the principal ideal generated by Σ.
(b) Suppose given integers m, k which are relatively prime to n. Let l > 0 be an
integer such that m ≡ kl mod n, and let
umk = 1 + T k + · · · + (T k )l−1 ,
where the right-hand side is interpreted as lying in Z[Zn ]/(Σ).
Show that if we chose a different integer l with m ≡ kl mod n, then the above
formula gives rise to precisely the same element of Z[Zn ]/(Σ).
Show that umk is a unit in Z[Zn ]/(Σ), and that f (umk ) is the cyclotomic unit
(ζnm − 1)/(ζnk − 1) from Problem 20 of Exercises 7.3.16.
8. Show that there is a commutative diagram of ring homomorphisms
Z[Zn ]
π /
Z[Zn ]/(Σ)
ε
ε
Z
/ Zn
π
where ε is the standard augmentation, the maps π are the canonical maps to the
respective quotient rings, and Σ = 1 + T + · · · + T n−1 as in Problem 7.
(a) Show that the above diagram induces an isomorphism
∼
=
h : Z[Zn ] −→ Z ×Zn (Z[Zn ]/(Σ)) ,
via h(α) = (ε(α), π(α)) for α ∈ Z[Zn ]. (See Problem 17 of Exercises 7.1.27.)
For this reason, we call the diagram a pullback square.
Deduce that there is a pullback square:
Z[Zn ]×
π /
×
(Z[Zn ]/(Σ))
ε
Z×
ε
π
/ Z×
n
CHAPTER 7. RINGS AND MODULES
244
(b) Let m1 , . . . , ms , k1 , . . . , ks be relatively prime to n and let r1 , . . . , rs > 0. Let
u = urm11 k1 . . . urmss ks , the unit of Z[Zn ]/(Σ) constructed in Problem 7. Show
that u lifts to a unit of Z[Zn ] if and only if mr11 . . . mrss ≡ ±k1r1 . . . ksrs mod n.
9. Let X = {X k | k ≥ 0}. Show that the operation X k · X l = X k+l turns X into
a monoid isomorphic to the monoid, N, of non-negative integers under addition.
Show that the monoid ring A[X] is isomorphic to the polynomial ring A[X] for any
ring A.
10. Let X be as in the previous exercise and let Xn = X × · · · × X be the direct
n times
product of X with itself n times. For any ring A, show that the polynomial ring
A[X1 , . . . , Xn ] is isomorphic to the monoid ring A[Xn ].
‡ 11. Let A be a ring and let f : M → M be a monoid homomorphism.
Show that
there
is
a
ring
homomorphism
f
:
A[M
]
→
A[M
]
defined
by
f
(
∗
∗
m∈M am m) =
(
a
)m
.
−1
m
m ∈M
m∈f (m )
‡ 12. Show that if f : A → B is a ring homomorphism and M is a monoid, then there is
a ring homomorphism f∗ : A[M ] → B[M ] defined by
f∗ (
am m) =
f (am )m.
m∈M
m∈M
13. Let G be a group and let A be a commutative ring. Let α = g∈G ag g be an
element of A[G]. Show that α is in the center of A[G] if and only if axgx−1 = ag
for all x, g ∈ G. The reader who has studied module theory should now show that
the center of A[G] is free as an A-module, and give a basis for it.
14. Show that if A is a ring and if M1 and M2 are monoids, then there is an isomorphism between (A[M1 ])[M2 ] and A[M1 × M2 ]. (Hint: One could either argue
directly, or via universal mapping properties. The latter method is quicker, and
less cumbersome notationally.)
15. Let A and B be rings and let M be a monoid. Show that (A × B)[M ] is isomorphic
to A[M ] × B[M ].
† 16. Let A be a ring and let a be the principal two-sided ideal in the polynomial ring
A[X] generated by X n − 1. Show that the quotient ring A[X]/a is isomorphic to
the group ring A[Zn ] of Zn over A. (Hint: There is a direct proof, that uses module
theory, and there is an easier one using universal mapping properties.)
17. Let a be a two-sided ideal of A and let M be a monoid. Let π : A → A/a be the
canonical
map and let
π∗ : A[M ] → (A/a)[M ] be the ring homomorphism given by
π∗ ( m∈M am m) = m∈M π(am )m (see Problem 12). Show that π∗ is a surjection
whose kernel is the two-sided ideal of A[M ] generated by a. (Here, we identify a
with its image under ι : A → A[M ].)
18. Let A be any ring and let Z2 = T = {1, T }. Determine precisely which elements
a+bT of A[Z2 ], a, b ∈ A, lie in the augmentation ideal IA (Z2 ). Deduce that IA (Z2 )
is the principal two-sided ideal generated by T − 1.
CHAPTER 7. RINGS AND MODULES
245
19. Let A be a commutative ring and let G be a group. Let S ⊂ G be a generating set
for G (i.e., G = S as in Definitions 2.2.11). Show that the augmentation ideal
IA (G) is the two-sided ideal of A[G] generated by {x − 1 | x ∈ S}. (Hint: Consider
the proof of Proposition 7.3.22.)
20. Let f : G → K be a split surjection of groups. Find a set of generators (as
an ideal) for the kernel of the ring homomorphism f∗ : A[G] → A[K] given by
f∗ ( g∈G ag g) = k∈K ( g∈f −1 k ag )k, where A is any commutative ring. (This
map was shown to be a ring homomorphism in Problem 11.)
21. Show that a group ring A[G] over a commutative ring A is self-opposite.
22. Show that a group ring A[G] over a self-opposite ring A is self-opposite.
23. Show that if A is commutative, then the passage from monoids M to the monoid
rings A[M ] gives a left adjoint to the forgetful functor from A-algebras to their
multiplicative monoids.
24. Let A be commutative. Construct the free A-algebra on an arbitrary set X.
25. Let A be commutative. Construct the free commutative A-algebra on an arbitrary
set X. (So far, we’ve only handled the case where X is finite.)
26. Show that if A is commutative, then the passage from groups G to the group rings
A[G] provides a left adjoint to the functor that takes an A-algebra to its group of
units.
7.6
Ideals in Commutative Rings
We learned a lot about groups by studying their factor groups. Similarly, quotient rings
will be useful for studying rings. The drawback here, of course, is that while the kernel
of a surjective group homomorphism is again a group, the kernel of a surjective ring
homomorphism is only an ideal and not a ring.
Thus, there is no good analogue for a composition series for a ring. However, we can
mimic the first step by which the composition series for a finite group was constructed.
In that step, we constructed a surjective homomorphism from our finite group to a simple
group.
The analogue for rings of a simple group is a ring with no two-sided ideals other than
0 and the ring itself, and hence has no useful quotient rings.
Definition 7.6.1. A nonzero ring is Jacobson simple if it has no two-sided ideals other
than 0 and the ring itself.
Problem 11 of Exercises 7.2.24 shows that if D is a division ring, then Mn (D) is
a Jacobson simple ring for all n ≥ 1. (We shall give a slicker argument for this in
Chapter 12.) There are other examples of noncommutative Jacobson simple rings more
complicated than these. But in the commutative case, Jacobson simple rings are as nice
as we could ask. The next corollary is immediate from Lemma 7.2.7.
Corollary 7.6.2. A commutative ring is Jacobson simple if and only if it is a field.
We wish to study those quotient rings A/a that are not the 0 ring, and hence, we are
concerned with those ideals a that are not equal to A. We shall give them a name.
CHAPTER 7. RINGS AND MODULES
246
Definition 7.6.3. An ideal a of A is proper if a ⊂ A is a proper inclusion (i.e., if a = A).
We now introduce two important kinds of proper ideals.
Definitions 7.6.4. Let A be a commutative ring. A proper ideal, p, of A is prime if
ab ∈ p implies that either a or b must be in p.
A proper ideal, m, of A is maximal if the only proper ideal containing m is m itself.
Examples 7.6.5. Let A be a commutative ring. Then it is immediate from the definitions that the ideal 0 of A is prime if and only if A is an integral domain.
Note that the ideal 0 is maximal if and only if A has no proper ideals other than 0.
But we’ve just seen in Corollary 7.6.2 that this occurs if and only if A is a field.
These two examples give a characterization of prime and maximal ideals in terms of
the quotient rings they determine.
Lemma 7.6.6. Let A be a commutative ring. Then a proper ideal p of A is prime if
and only if the quotient ring A/p is an integral domain. Also, a proper ideal m of A is
maximal if and only if the quotient ring A/m is a field. Thus, every maximal ideal is
prime.
Proof Let p be a prime ideal of A. We must show that A/p has no zero-divisors. Thus,
let a and b be elements of A/p such that a · b = 0. But this says that ab = 0, and hence
ab ∈ p. But then one of a or b is in p, and hence that element goes to 0 in A/p.
Conversely, if A/p is an integral domain and if ab ∈ p, then ab = 0, so that one of a
and b must be 0 in A/p. But then one of a and b must be in p.
For the case of maximal ideals, the correspondence between the ideals in a quotient
ring A/a and the ideals in A containing a (Problem 5 of Exercises 7.2.24) shows that a
proper ideal m of A is maximal if and only if the 0 ideal is maximal in A/m. As shown
in the preceding example, this occurs if and only if A/m is a field.
Examples 7.6.7.
1. For any prime p, the ideal (p) ⊂ Z is maximal, as Zp is a field. In fact, Lemma 7.1.12
shows that if n > 1, then (n) is a prime ideal of Z if and only if n is a prime number.
Thus, every nonzero prime ideal of Z is maximal.
2. Let K be a field and let a ∈ K. Then the evaluation map εa : K[X] → K
is surjective, so its kernel is a maximal ideal. By Corollary 7.3.12, ker εa is the
principal ideal generated by X − a.
p → Zp be the ring homomorphism from the p-adic integers onto Zp
3. Let π1 : Z
that’s induced by the projection map. (See Lemma 7.1.38.) Then ker π1 , which by
Problem 4 of Exercises 7.1.41 is the principal ideal generated by p, is maximal.
The next lemma is an easy, but important observation.
Lemma 7.6.8. Let f : A → B be a ring homomorphism between commutative rings and
let p be a prime ideal of B. Then f −1 (p) is a prime ideal of A.
Note that the analogue of the preceding lemma for maximal ideals is false: Let
i : Z → Q be the inclusion. Then 0 is a maximal ideal of Q, while f −1 (0) = 0, which is
prime in Z, but not maximal.
Prime ideals satisfy an important primality principle with respect to the ideal product
of Definitions 7.2.21. This would have made a good exercise, but its uses are ubiquitous
enough that we give it here.
CHAPTER 7. RINGS AND MODULES
247
Lemma 7.6.9. Let p be a prime ideal in the commutative ring A and let a and b be
ideals with ab ⊂ p. Then at least one of a and b must be contained in p.
Proof If ab ⊂ p and a ⊂ p, let a ∈ a with a ∈ p. Then ab ∈ p for all b ∈ b. Since a ∈ p,
we must have b ∈ p for all b ∈ b.
By Lemma 7.2.22, ab ⊂ a ∩ b for any ideals a, b of a commutative ring.
Corollary 7.6.10. Let p be a prime ideal in the commutative ring A and let a and b be
ideals such that a ∩ b ⊂ p. Then at least one of a and b must be contained in p.
Prime ideals may be used to define a notion of dimension for a commutative ring.
Definition 7.6.11. Let A be a commutative ring. We say that the Krull dimension of
A is ≤ n if whenever we have a sequence
p0 ⊂ p1 ⊂ · · · ⊂ pk
of proper inclusions (i.e., inclusions that are not onto) of prime ideals of A, we have
k ≤ n. Thus, the Krull dimension of A is finite if there is a maximum length for such
sequences of proper inclusions of prime ideals in A, and if
p 0 ⊂ p1 ⊂ · · · ⊂ pn
is the longest such sequence that can occur in A, then the Krull dimension is n.
Examples 7.6.12.
1. Since a field has only one prime ideal, 0, every field has Krull dimension 0.
2. The first example in Examples 7.6.7 shows that the longest chains of proper inclusions of prime ideals in Z are those of the form 0 ⊂ (p), for p prime. Thus, Z has
Krull dimension 1.
Let us return to the analogy with group theory. As noted above, we’ve made use of
the fact that any nontrivial finite group has a nontrivial simple group as a factor group.
By analogy, we would like to show that any commutative ring has a Jacobson simple ring
as a quotient ring. Since the only commutative rings that are Jacobson simple are fields,
this is equivalent to showing that every commutative ring has a maximal ideal.
To show this, we shall make use of Zorn’s Lemma, which is equivalent to the Axiom
of Choice. The reader can find the proof of this equivalence in any good book on set
theory.9
First we shall need some setup. Recall that a partially ordered set is a set X together
with a relation, ≤, which is reflexive (x ≤ x for all x ∈ X), transitive (if x ≤ y and
y ≤ z, then x ≤ z), and antisymmetric (if x ≤ y and y ≤ x, then x = y). In a partially
ordered set, it is not required that an arbitrary pair of elements in X be comparable by
the relation. In other words, there may exist pairs x, y such that neither x ≤ y nor y ≤ x
is true.
On the other hand, a partially ordered set in which any pair of elements is comparable
is called a totally ordered set.
Let S be a subset of a partially ordered set. We say that x ∈ X is an upper bound
for S if s ≤ x for all s ∈ S. Note that the upper bound x need not lie in S.
Finally, a maximal element of X is an element y such that if x ∈ X with y ≤ x, then
x and y must be equal.
9 There’s a good exposition of Zorn’s Lemma and the Well Ordering Principle in Topology, by James
Dugundji, Allyn and Bacon, 1966.
CHAPTER 7. RINGS AND MODULES
248
Lemma 7.6.13. (Zorn’s Lemma) Let X be a nonempty partially ordered set such that
every totally ordered subset of X has an upper bound in X. Then X has at least one
maximal element.
Corollary 7.6.14. Let A be a commutative ring. Then every proper ideal of A is contained in a maximal ideal.
Proof Let a be a proper ideal of A, and let X be the set of all proper ideals of A that
contain a. Put a partial ordering on X by b ≤ b if b ⊂ b . Then a maximal element of
X with respect to this ordering is precisely a maximal ideal containing a.
Since a ∈ X, X is nonempty, so it suffices to show that any totally ordered subset of
X has an upper bound in X. Let S be a totally ordered subset of X. We claim that the
union b∈S b of the ideals in S is such an upper bound.
Of course, unions of ideals are not generally ideals, but if we’re dealing with a totally
ordered set of ideals, this problem disappears: For x, y ∈ b∈S b, we can find b, b ∈ S
such that x ∈ b and y ∈ b . But since any two elements of S are comparable, either
b ≤ b or b ≤ b. But in either case, x+ y is in the larger of the two, as is ax for any
a ∈ A. But then x + y and ax are in b∈S b, and hence the union is an ideal. Clearly,
it is an upper bound for the ideals in S, so it suffices to show that it is proper. But if
1 ∈ b∈S b, then 1 must be an element of some b ∈ S. But this is impossible, since the
ideals in X are all proper.
If we make extra assumptions on our ring, we can obtain an even stronger sort of
maximality principle for ideals.
Definitions 7.6.15. Let A be a commutative ring. We say that A is Noetherian, or has
the ascending chain condition (a.c.c.) if any ascending sequence of inclusions of ideals of
A is eventually constant. In other words, given a sequence
a 1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · ·
there is some n ≥ 1 such that an = ak for all k ≥ n.
Alternatively, we say that A is Artinian, or has the descending chain condition (d.c.c.)
if any decreasing sequence of ideals is eventually constant. This means that given a
sequence
a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · ·
there is some n ≥ 1 such that an = ak for all k ≥ n.
It turns out that the Artinian condition (in a commutative ring) implies that the ring
is Noetherian. The Artinian condition is a very strong one, and fails to hold in many
rings of interest. For instance, as we shall see in the exercises below, the only Artinian
integral domains are fields.
The descending chain condition for left or right ideals turns out to be extremely
useful for certain topics in noncommutative ring theory. For instance, descending chain
conditions are important in the study of group rings K[G], where K is a field and G is
a finite group.
Here is the strengthened maximality principle that holds in Noetherian rings.
Proposition 7.6.16. Let A be a Noetherian ring and let X be the partially ordered set
of ideals in A (ordered by inclusion). Let S be any nonempty subset of X. Then S, with
the partial ordering inherited from X, has maximal elements.
CHAPTER 7. RINGS AND MODULES
249
Proof We argue by contradiction. Suppose that S has no maximal elements, and let
a1 be an element of S. Since a1 is not a maximal element of S, we can find a proper
inclusion a1 ⊂ a2 , with a2 in S.
Since no element of S is maximal, induction now shows that we may choose a sequence
a 1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · ·
of elements of S, such that each inclusion ak−1 ⊂ ak is proper.10 But this contradicts the
hypothesis that A is Noetherian, and hence S must have maximal elements after all.
We shall study Noetherian rings in much greater detail in the context of module
theory.
Exercises 7.6.17.
1. Let f : A → B be a surjective ring homomorphism. Show that there is a one-to-one
correspondence between the prime ideals of B and those primes of A that contain
ker f . Show also that there is a one-to-one correspondence between the maximal
ideals of B and those maximal ideals of A containing ker f .
2. Let n > 1. Show that every prime ideal of Zn is maximal. Given the prime
decomposition of n, list all the maximal ideals, m, of Zn and calculate the quotient
fields Zn /m.
3. Show that an integral domain has Krull dimension 0 if and only if it is a field.
4. Let p be a prime ideal in the commutative ring A and let A[X] · p be the ideal
generated by p in A[X]. Show that A[X] · p is a prime ideal of A[X].
5. Let A be an integral domain which is not a field. Show that the principal ideal
(X − a) of A[X] (here, a ∈ A) is a prime ideal that is not maximal. Deduce that
A[X] has Krull dimension ≥ 2.
6. Let p be a prime and let n be any integer. Show that the ideal (p, X − n) of Z[X],
generated by p and X − n, is maximal.
7. Let f : Q[X] → Q[i] ⊂ C be the unique Q-algebra map that carries X to i. Recall
from Problem 14 of Exercises 7.3.16 that Q[i] is a field. Show that ker f = (X 2 +1),
and hence (X 2 + 1) is a maximal ideal in Q[X].
8. Let Z4 = T = {1, T, T 2 , T 3 }. Recall from Problem 6 of Exercises 7.5.10 that there
is a unique, surjective, Q-algebra homomorphism f : Q[Z4 ] → Q[i] with f (T ) = i.
Show that ker f = (T 2 + 1), and hence (T 2 + 1) is a maximal ideal in Q[Z4 ].
9. Let X1 , . . . , Xn be indeterminates over a field K. Let a1 ∈ K and for 2 ≤ i ≤
n, let fi (X) ∈ K[X1 , . . . , Xi−1 ] ⊂ K[X1 , . . . , Xn ]. Show that (X1 − a1 , X2 −
f2 (X), . . . , Xn − fn (X)) is a maximal ideal of K[X1 , . . . , Xn ].
10. Let X1 , . . . , Xn be indeterminates over a field K. Find a chain p0 ⊂ · · · ⊂ pn of
proper inclusions of prime ideals of K[X1 , . . . , Xn ]. Deduce that K[X1 , . . . , Xn ]
has Krull dimension ≥ n.
10 Note that the infinite induction here makes use of the Axiom of Choice. Thus, while our conclusion
is stronger than that of the Zorn’s Lemma argument for the existence of maximal ideals, the Noetherian
property has not permitted us to dispense with the use of the choice axiom.
CHAPTER 7. RINGS AND MODULES
250
11. Let A and B be commutative rings. What are the prime ideals of A × B?
Suppose that A and B have finite Krull dimension. Calculate the Krull dimension
of A × B in terms of the dimensions of A and B.
12. Show that Z is Noetherian but not Artinian.
13. Show that Zn is both Noetherian and Artinian.
14. Let A be an Artinian integral domain and let a be a nonzero element of A. Show
that the descending chain condition on
(a) ⊃ (a2 ) ⊃ · · · ⊃ (an ) ⊃ · · ·
implies that a is a unit. Deduce that A must be a field. Deduce that a finite domain
is a field.
15. Let A be an Artinian commutative ring. Show that every prime ideal of A is
maximal.
7.7
Modules
Modules play a role in ring theory analogous to the role of G-sets in understanding
groups. Among other things, module theory will enable us to obtain a better grasp of
the theory of ideals, and hence of the structure of the rings themselves.
We give some general concepts and constructions here, and study free modules and
exact sequences in subsections. We then close the section with a subsection on the notion
of rings without identity elements.
Definitions 7.7.1. Let A be a ring. A left A-module is an abelian group M together
with an operation that assigns to each a ∈ A and m ∈ M an element am ∈ M subject
to the following rules:
a(m + m )
= am + am
(a + a )m = am + a m
1 · m = m,
a(a m) = (aa )m,
for all a, a ∈ A and all m, m ∈ M .
If M is an abelian group, then an operation A × M → M making M an A-module is
called an A-module structure on M .
A submodule of M is a subgroup M ⊂ M of the additive group of M , such that
am ∈ M for all a ∈ A and m ∈ M (i.e., M is closed under multiplication by elements
of A).
An A-module homomorphism between the left A-modules M and N is a homomorphism f : M → N of abelian groups such that f (am) = af (m) for all a ∈ A and all
m ∈ M.
Right modules and their homomorphisms are defined by reversing the order of the
a’s and m’s in the above definitions.
CHAPTER 7. RINGS AND MODULES
251
Note that if K is a field, then a K-module is nothing other than a vector space over K,
and a K-module homomorphism between vector spaces is nothing other than a K-linear
map.
Another case where we understand what it means to be a module is that of Z-modules.
The proof of the next lemma is similar to that of Lemma 7.1.6, which we left to the reader,
so we shall leave this to the reader as well.
Lemma 7.7.2. Every abelian group G is a Z-module. Moreover, the Z-module structure
on G is unique: for n ∈ Z and g ∈ G, ng is the n-th power of g in the group structure
of G. (Thus, if n > 0, ng = g + · · · + g, the sum of n copies of g.) Finally, every group
homomorphism between abelian groups is a Z-module homomorphism.
When A is commutative, any left A-module may be made into a right A-module by
setting m · a = am. This process is clearly reversible, so if A is commutative, then there
is a one-to-one correspondence between the left and right A-modules.
Not only does the above correspondence fail for noncommutative rings, but the theories of left modules and of right modules can be quite different for some rings.
Example 7.7.3. Let A be a ring. Then A is a left A-module via the structure map
a · a = aa . Indeed, the equalities that demonstrate that this is an A-module structure
are all part of the definition of a ring.
The submodules of A under this structure are precisely the left ideals of A.
Note that A is also a right A-module, via a · a = a a. The submodules under this
right module structure are the right ideals of A.
In particular, module theory will help us study ideals, and hence will provide important information about the ring A.
Here is another source of A-modules.
Example 7.7.4. Let f : A → B be a ring homomorphism. Then B is a left A-module
via a · b = f (a)b. Also, B is a right A-module via b · a = bf (a).
For simplicity, we shall restrict attention to left modules for the rest of this section.
Analogues of all the results will be true for right modules. Thus, for the duration, “Amodule” will mean left A-module. We shall, however, use the word “left” to emphasize
certain one-sided properties.
Given the modules from the examples above, we can construct yet more via factor
groups.
Lemma 7.7.5. Let N be a submodule of the A-module M . Then the factor group M/N
is an A-module via a·m = am. Moreover, if f : M → M is an A-module homomorphism,
then there is a factorization
/ M
ME
EE
x<
x
EE
xx
E
xx
π EE
"
xx f
M/N
f
(i.e., an A-module homomorphism f making the diagram commute) if and only if N ⊂
ker f . The factorization f , if it exists, is unique.
Finally, the kernel of an A-module homomorphism is always a submodule, and hence
∼
=
we obtain an isomorphism of A-modules f : M/(ker f ) −→ im f from any A-module
homomorphism f : M → M .
CHAPTER 7. RINGS AND MODULES
252
We now analyze the A-module homomorphisms out of A.
Lemma 7.7.6. Let M be a left A-module. Then for each m ∈ M there is a unique
A-module homomorphism fm : A → M with fm (1) = m.
Proof We must have fm (a) = fm (a · 1) = afm (1) = am, so uniqueness follows, and it
suffices that this formula defines an A-module homomorphism. But that follows immediately from the definition of module.
In category theoretic terms, this says that A is the free A-module on one generator.
There’s another way to express the result which introduces a useful piece of notation.
Definition 7.7.7. Let M and N be A-modules. We write HomA (M, N ) for the set of
A-module homomorphisms from M to N .11
Lemma 7.7.8. Let M and N be A-modules. Then HomA (M, N ) is an abelian group
under the operation that sets (f + g)(m) = f (m) + g(m) for f, g ∈ HomA (M, N ).
Under certain circumstances (e.g., if A is commutative), HomA (M, N ) is naturally an
A-module. We shall discuss the details in Section 9.7. We now give an easy consequence
of Lemma 7.7.6.
Corollary 7.7.9. Let M be an A-module. Then there is an isomorphism of abelian
∼
=
groups, ε : HomA (A, M ) −→ M , given by ε(f ) = f (1).
ε is an example of what’s called an evaluation map, as it is obtained by evaluating
the homomorphisms at 1. As shown in Proposition 9.7.4, ε is an A-module isomorphism
under the appropriate conventions on the group of homomorphisms.
We return to the study of the homomorphisms from A to M .
Definitions 7.7.10. Let m ∈ M and let fm : A → M be the unique A-module homomorphism that carries 1 to m. Then we write Am for the image of fm and write Ann(m)
(or AnnA (m), if there’s more than one ring over which M is a module) for the kernel.
Thus,
Am
Ann(m)
= {am | a ∈ A}
and
= {a ∈ A | am = 0}.
We shall refer to Ann(m) as the annihilator of m, and say that a kills (or annihilates) m
if am = 0.
In general, we say that N is a cyclic module if N = An for some n ∈ N .
Corollary 7.7.11. Let M be a left A-module and let m ∈ M . Then Ann(m) is a left
ideal of A, and the cyclic module Am is isomorphic to A/Ann(m).
There is also the notion of the annihilator of a module.
Definition 7.7.12. Let M be an A-module. Then the annihilator of M , written as
Ann(M ) or AnnA (M ), is given by
Ann(M )
= {a ∈ A | am = 0 for all m ∈ M }
=
Ann(m).
m∈M
11 We
use the same notation for the set of A-module homomorphisms between a pair of right A-modules.
This can, at times, require some specificity of language.
CHAPTER 7. RINGS AND MODULES
253
Unlike the case of the annihilator of an element, we have the following.
Lemma 7.7.13. The annihilator of an A-module is a two-sided ideal of A.
Warning: If A is not commutative, then Ann(Am) may not be equal to Ann(m). In
addition, Ann(m) may not be a two-sided ideal, while Ann(Am) always is.
Definitions 7.7.14. Let f : A → B be a ring homomorphism. Let M be a B-module.
The A-module structure on M induced by f is given by the structure map a · m = f (a)m
for a ∈ A and m ∈ M .
Let N be an A-module. We say that a B-module structure on N is compatible with
the original A-module structure if the A-module structure induced by f from the Bmodule structure is the same as the one we started with (i.e., f (a) · m = am for all a ∈ A
and m ∈ M ).
We wish to study these relationships in the case where f is the canonical map π :
A → A/a, where a is a two-sided ideal of A. Here, if M is an A/a module, then we shall
implicitly consider M to be an A-module under the structure induced by π.
Proposition 7.7.15. Let a be a two-sided ideal in the ring A. Then an A-module M
admits an A/a-module structure compatible with the given A-module structure if and only
if a ⊂ Ann(M ). Such an A/a-module structure is unique if it exists. Finally, if M and N
are A/a-modules, then every A-module homomorphism from M to N is an A/a-module
homomorphism as well.
Proof If M is an A/a-module, then, in the induced A-module structure, if x ∈ a and
m ∈ M , then xm = π(x)m = 0, because π(x) = 0. Thus, a ⊂ Ann(M ).
Conversely, if a ⊂ Ann(M ), given a ∈ A, x ∈ a, and m ∈ M , (a + x)m = am.
Thus, setting a · m = am gives a well defined correspondence from A/a × M to M . And
this correspondence is easily seen to satisfy the conditions required for an A/a-module
structure on M .
Two A/a-module structures that induce the same A-module structure must be equal,
because π : A → A/a is surjective.
Finally, given an A-module homomorphism f : M → N between A/a-modules, we
have f (am) = f (am) = af (m) = af (m), since the A-module structures on both M and
N are induced by the given A/a-module structures.
For A = Z, this gives a characterization of Zn -modules.
Corollary 7.7.16. An abelian group M admits a Zn -module structure if and only if
M has exponent n as a group. The Zn -module structure on M , if it exists, is unique.
Finally, if M and N are Zn -modules, then every group homomorphism f : M → N is a
Zn -module homomorphism.
Proof We use the fact that Zn = Z/(n). We know that every abelian group M has
a unique Z-module structure, in which k · m is the k-th power of m in the abelian
group structure on M for all k ∈ Z and m ∈ M . Thus, k annihilates M if and only if
M has exponent k. But since (n) is the smallest ideal of Z containing n, we see that
n annihilates M if and only if (n) ⊂ Ann(M ). The result now follows directly from
Proposition 7.7.15.
Now we consider operations on submodules. The constructions that work are similar
to those that work for ideals.
CHAPTER 7. RINGS AND MODULES
254
Definitions 7.7.17. Let N1 and N2 be submodules of M and let a be a left ideal of A.
We define N1 + N2 and aN1 by
N1 + N2
aN1
= {n + m | n ∈ N1 , m ∈ N2 }
= {a1 n1 + · · · + ak nk | k ≥ 1, ai ∈ a, and ni ∈ N1 for i = 1, . . . , k}
Lemma 7.7.18. Let N1 and N2 be submodules of the A-module M and let a be a left
ideal of A. Then N1 + N2 , aN1 , and N1 ∩ N2 are all submodules of M .
The modules aM have the following virtue.
Lemma 7.7.19. Let M be an A-module and let a be a two-sided ideal of M . Then
M/aM is an A/a-module. Moreover, if N is any A/a-module and if f : M → N is an
A-module homomorphism, then there is a unique factorization
/N
M G
x;
GG
x
x
GG
xx
G
xx
π GG
xx f
#
M/aM
f
where f : M/aM → N is an A/a-module homomorphism.
Proof Clearly, a ⊂ Ann(M/aM ), and hence M/aM is an A/a-module by Proposition 7.7.15. If f : M → N is an A-module homomorphism, with N an A/a-module,
then a ⊂ Ann(N ), and hence aM must lie in ker f . Thus, there is a unique A-module
homomorphism f : M/aM → N making the diagram commute, and f is an A/a-module
homomorphism by Proposition 7.7.15.
In categorical terms, this says that the passage from M to M/aM is a left adjoint to
the forgetful functor from A/a-modules to A-modules.
We can also consider external operations on modules. Recall that the direct sum
M1 ⊕ · · · ⊕ Mk of a finite collection of abelian groups is an alternative notation for the
direct product.
Definition 7.7.20. Let M1 , . . . , Mk be A-modules. By the direct sum, M1 ⊕ · · · ⊕ Mk ,
of M1 , . . . , Mk , we mean their direct sum as abelian groups, endowed with the A-module
structure obtained by setting a(m1 , . . . , mk ) = (am1 , . . . , amk ), for each a ∈ A and each
k-tuple (m1 , . . . , mk ) ∈ M1 ⊕ · · · ⊕ Mk .
Note that the canonical inclusions ιi : Mi → M1 ⊕ · · · ⊕ Mk are A-module maps, as
are the projection maps πi : M1 ⊕ · · · ⊕ Mk → Mi . The proof of the following lemma is
identical to that of its analogue for abelian groups (see Proposition 2.10.6).
Lemma 7.7.21. Let fi : Mi → N be A-module homomorphisms for 1 ≤ i ≤ k. Then
there is a unique A-module homomorphism f : M1 ⊕ · · · ⊕ Mk → N such that f ◦ ιi = fi
for 1 ≤ i ≤ n. Explicitly,
f (m1 , . . . , mk ) = f1 (m1 ) + · · · + fk (mk ).
CHAPTER 7. RINGS AND MODULES
255
The next lemma will be useful, for instance, in our understanding of the properties
of free modules.
Lemma 7.7.22. Let a be a two-sided ideal in A and let M1 , . . . , Mk be A-modules. Let
M = M1 ⊕ · · · ⊕ Mk . Then there is a natural isomorphism of A/a-modules between
M/aM and (M1 /aM1 ) ⊕ · · · ⊕ (Mk /aMk ).
Proof Recall from Proposition 7.7.15 that any A-module map between A/a-modules
is an A/a-module map. Thus, it suffices to show that aM = (aM1 ) ⊕ · · · ⊕ (aMk ).
Note that the inclusion aM ⊂ (aM1 ) ⊕ · · · ⊕ (aMk ) follows from the fact that if a ∈ a,
then a(m1 , . . . , mk ) = (am1 , . . . , amm ) lies in (aM1 ) ⊕ · · · ⊕ (aMk ) for each k-tuple
(m1 , . . . , mk ). Since (aM1 ) ⊕ · · · ⊕ (aMk ) is closed under addition, the stated inclusion
follows.
For the opposite inclusion, note that (m1 , . . . , mk ) = ι1 (m1 ) + · · · + ιk (mk ). Since
each ιi is an A-module homomorphism, we have ιi (mi ) ∈ aM whenever mi ∈ aMi . Since
aM is closed under addition, we see that (aM1 ) ⊕ · · · ⊕ (aMk ) ⊂ aM , as claimed.
We shall also make use of*
infinite direct sums. Recall that if Mi is
!an abelian group
for i ∈ I, then the direct sum i∈I Mi is the subgroup of the product i∈I Mi consisting
of those I-tuples (mi | i ∈ I) such that
*all but finitely many of the mi are 0.
The homomorphisms ιi : Mi →
i∈I Mi are defined by setting ιi (m) to be the Ituple
whose
i-th
coordinate
is
m
and
whose
other coordinates are all 0. Making use of
m
as
an
alternate
notation
for
(m
|
i
∈
I), Lemma 6.5.9 gives
i
i
i∈I
mi =
ιi (mi ),
i∈I
where the right-hand side is the sum in
for which mi = 0.
*
i∈I
i∈I
Mi of the finite collection of terms ιi (mi )
Definition
7.7.23. Let Mi be an A-module for each i ∈ I. We define the direct sum
*
abelian group
i∈I Mi to be the A-module whose underlying is the direct sum of the
Mi and whose A-module structure is given by a i∈I mi = i∈I ami .
*
The maps ιi : Mi →
i∈I Mi are easily seen to be A-module homomorphisms.
The proof of the next result is identical to that of its analogue for Z-modules (Proposition 6.5.10).
Proposition 7.7.24. Suppose given A-modules Mi for i ∈ I. Then for any collection
of A-module homomorphisms
{fi : Mi → N | i ∈ I}, there is a unique A-module homo*
morphism f : i∈I Mi → N such that f ◦ ιi = fi for each i. Explicitly,
#
"
mi =
fi (mi ).
f
i∈I
i∈I
We also have an infinite sum operation for submodules of a given module.
Definition 7.7.25. Suppose given a set {Ni | i ∈ I} of submodules of an A-module M .
Then the sum of the Ni is the submodule
Ni = {ni1 + · · · + nik | k ≥ 1, ij ∈ I and nij ∈ Nij for j = 1, . . . , k}.
i∈I
Thus,
i∈I
Ni is the set of all finite sums of elements from the various submodules Ni .
CHAPTER 7. RINGS AND MODULES
256
Sums of submodules satisfy the following properties. The proof is left as an exercise.
Proposition 7.7.26. Let M be an A-module. If N1 and N2 are submodules of M , then
N1 + N2 is the smallest submodule of M containing both
N1 and N2 .
Similarly, if Ni is a submodule of M for i ∈ I, then i∈I Ni is the smallest submodule
of M that contains each of the Ni .
If fi : Mi → M *
is an A-module homomorphism for each i ∈ I (with I either finite or
infinite) and if f : i∈I Mi → M is the unique homomorphism such that f ◦ ιi = fi for
all i, then
im fi .
im f =
i∈I
Exercises 7.7.27.
1. Give the proof of Lemma 7.7.13.
2. Give the proof of Lemma 7.7.18.
3. Give the proof of Corollary 7.7.26.
4. Let M be an abelian group, considered as a Z-module. Show that the annihilator
Ann(M ) is nonzero if and only if M has some positive exponent. If Ann(M ) = (n),
with n > 0, show that n is the smallest positive number which is an exponent for
M.
5. Let N1 and N2 be submodules of M . Show that
Ann(N1 + N2 ) = Ann(N1 ) ∩ Ann(N2 ).
6. Let N1 and N2 be submodules of M . Show that
Ann(N1 ) + Ann(N2 ) ⊂ Ann(N1 ∩ N2 ).
Can you give an example where this inclusion is proper?
7. Consider Z8 as a Z-module. What is Ann((2)Z8 )?
8. Let f : A → B be a ring homomorphism. In the left A-module structure on B
that’s induced by f , what is Ann(B)?
9. Let A be a commutative ring. Let m be an A-module and let m ∈ M . Show that
Ann(m) = Ann(Am).
10. Let A be any ring and let x ∈ Mn (A) be the matrix whose 11-entry is 1 and whose
other entries are all 0.
⎛
⎞
1 0 ... 0
⎜ 0 0 ... 0 ⎟
⎜
⎟
a=⎜
⎟
..
⎝
⎠
.
0
0
...
(a) What are the elements of AnnMn (A) (x)?
0
CHAPTER 7. RINGS AND MODULES
257
(b) What are the elements of the principal left ideal Mn (A)x?
(c) What are the elements of AnnMn (A) (Mn (A)x)?
11. Let A be a ring and let
G be a finite group. Let Σ ∈ A[G] be the sum of all the
elements of G: Σ = g∈G g. Show that AnnA[G] (Σ) = I(G), the augmentation
ideal of G.
12. Let A be a ring. Let G be a finite group and let x ∈ G. Show that an element
g∈G ag g ∈ A[G] lies in Ann(x − 1) if and only if for each left coset gx of x, the
coefficients of the elements in gx are all equal (i.e., agxi = agx
j for all i, j ∈ Z).
Deduce that if G = x, then Ann(x − 1) = A[G]Σ, where Σ = g∈G g.
13. Let (m) be an ideal in Z. Show that for n ≥ 1, Zn /(m)Zn is isomorphic to Z(m,n) ,
where (m, n) is the greatest common divisor of m and n. (Recall that Z1 is the
trivial group.)
14. Let M be an abelian group and let r ≥ 0. Show that (pr )M/(pr+1 )M is a vector
space over Zp .
15. Let M be a finite abelian group of exponent pr . Show that (pr−1 )M is a vector
space over Zp .
16. Let M be a finite abelian group whose order is prime to n. Show that M/(n)M = 0.
17. Let M = Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk .
(a) Show that M/(p)M is the product of k copies of Zp .
(b) Calculate, in similar terms, the Zp -vector spaces (pr )M/(pr+1 )M for r ≥ 1.
(c) Calculate Ann(M ).
18. Suppose given a family of submodules {Ni | i ∈ I} of M . Show that the intersection
i∈I Ni is a submodule of M .
19. Let N be a submodule of M . Show that N = x∈N Ax.
20. What is Ann(M1 ⊕ · · · ⊕ Mk )?
† 21. Let A be a ring. Show that a left A-module structure on an abelian group M
determines and is determined by a ring homomorphism from A to EndZ (M ).
22. Use the preceding problem to give a quick proof of Proposition 7.7.15.
23. Let A be a ring. Show that a right A-module structure on an abelian group M
determines and is determined by a ring homomorphism from the opposite ring Aop
to EndZ (M ).
† 24. Let A be a ring and let M be either a right or a left A-module. Let EndA (M ) ⊂
EndZ (M ) be the collection of A-module homomorphisms from M to itself. Show
that EndA (M ) is a subring of EndZ (M ).
25. Let A be a commutative ring and let M be an A-module. Show that EndA (M ) is
an A-algebra via the map ι : A → EndA (M ) defined by ι(a)(m) = am for all a ∈ A
and m ∈ M .
CHAPTER 7. RINGS AND MODULES
258
26. Let A be a commutative ring. Suppose given an A-algebra B and an A-module
M . Show that there is a one-to-one correspondence between the A-algebra homomorphisms from B to EndA (M ) and the B-module structures on M that are
compatible with the original A-module structure.
27. For any ring A and any A-module M , we write AutA (M ) for the group of units
in EndA (M ). Suppose that A is commutative and that G is a group. Show that
there is a one-to-one correspondence between the group homomorphisms from G to
AutA (M ) and the A[G]-module structures on M that are compatible (with respect
to the standard inclusion ι : A → A[G]) with the original A-module structure on
M.
Free Modules and Generators
Finite generation is as useful a concept in module theory as it is in group theory.
Definitions 7.7.28. Let m1 , . . . , mk be elements of the A-module M . By the submodule generated by m1 , . . . , mk , we mean the sum Am1 + · · · + Amk . Clearly, this is the
smallest submodule of M containing the elements m1 , . . . , mk .
The elements of Am1 + · · · + Amk all have the form a1 m1 + · · · + ak mk , for some
a1 , . . . , ak ∈ A. Such a sum is called a linear combination of m1 , . . . , mk .
Similarly, the submodule of M generated by an infinite family {mi | i ∈ I} is the
infinite sum, i∈I Ami , of the submodules Ami . Here, a linear combination of the mi
is a finite sum ai1 mi1 + · · · + aik mik , with i1 , . . . , ik ∈ I.
If M is generated by a finite set m1 , . . . , mk , we say that M is a finitely generated
A-module. Similarly, we say that a left ideal is finitely generated if it is finitely generated
as an A-module.
Clearly, A is generated by 1, while the cyclic module Ax is generated by x. A very
important collection of finitely generated modules is given by the free modules An :
Definitions 7.7.29. We write An for the direct sum of n copies of A, and write ei =
(0, . . . , 0, 1, 0, . . . , 0). for the element of An whose i-th coordinate is 1 and whose other
coordinates are all 0, with 1 ≤ i ≤ n. Note that (a1 , . . . , an ) = a1 e1 + · · · + an en , so that
An is generated by e1 , . . . , en . We call e1 , . . . , en the canonical basis of An . (We adopt
the convention that A0 = 0, the 0-module, whose canonical basis is taken to be the null
set.)
We call An the standard free A-module of rank n.
The next lemma shows that An is the free A-module (in the sense of category theory)
on the set {e1 , . . . , en }.
Lemma 7.7.30. Let M be an A-module and let m1 , . . . , mn ∈ M . Then there is a
unique A-module homomorphism f : An → M with f (ei ) = mi for i = 1, . . . , n. Explicitly,
f (a1 , . . . , an ) = a1 m1 + · · · + an mn .
Proof Let fi : A → M be given by fi (a) = ami . Thus, fi is the unique A-module
homomorphism with fi (1) = mi . By Lemma 7.7.21 there is a unique A-module homomorphism from An to N with the property that f ◦ ιi = fi for 1 ≤ i ≤ n, where ιi : A → An
CHAPTER 7. RINGS AND MODULES
259
is the inclusion of the i-th summand. Since ei = ιi (1), this gives f (ei ) = fi (1) = mi .
The explicit formula of Lemma 7.7.21 gives
f (a1 , . . . , an ) = f1 (a1 ) + · · · + fn (an ) = a1 m1 + · · · + an mn ,
as claimed.
Note that the uniqueness statement, while embedded in the above, also follows from
the fact that the e1 , . . . , en generate An : Since (a1 , . . . , an ) = a1 e1 + · · · + an en , we must
have f (a1 , . . . , an ) = a1 f (e1 ) + · · · + an f (en ), and hence any A-module homomorphism
on An is determined by its values on e1 , . . . , en .
We can generalize this to the study of arbitrary direct sums of copies of A.
*
Definition 7.7.31. Let *
I be any set, and let ei = ιi (1) ∈ i∈I A for i ∈ I. Thus, in
the I-tuples notation for i∈I A, ei is the element whose i-th coordinate is 1 and whose
other
* coordinates are all 0. The elements ei , i ∈ I are called the canonical basis elements
of i∈I A.
*
Note that an element of i∈I A may be written uniquely as a sum i∈I ai ei , where
all but finitely many of the ai are equal to 0.
The proof of the next lemma is identical to that of Lemma 7.7.30, using Proposition 7.7.24 in place of Lemma 7.7.21.
Lemma 7.7.32. Let I be any set. Let M be an A-module
and let mi ∈ M for all i ∈ I.
*
Then there is a unique A-module homomorphism f : i∈I A → M such
* that f (ei ) = mi
for all i ∈ I, where the elements ei are the canonical basis elements of i∈I A. Explicitly,
#
"
f
ai ei =
ai mi .
i∈I
i∈I
For some purposes it is useful to get away from the use of canonical bases.
Definitions 7.7.33. An A-module M is a free module if it is isomorphic to a direct sum
of copies of A.
We say that M is free of finite rank if it is isomorphic to An for some integer n ≥ 0.
We say that a finite sequence x1 , . . . , xn of elements of M is a basis for M if the
unique A-module homomorphism f : An → M with f (ei ) = xi for i = 1, . . . , n is an
isomorphism.12
If I is an infinite set, we say
* that a set {mi | i ∈ I} is a basis for M if the unique
A-module homomorphism f : i∈I A → M with f (ei ) = mi for i ∈ I is an isomorphism.
Note that in the case of infinitely generated free modules, a basis comes equipped
with an indexing set I. In the finitely generated case, we insist that a basis, if non-null,
be indexed by one of the standard ordered sets {1, . . . , n}.
Examples 7.7.34.
12 By abuse of language, we shall speak of a sequence x , . . . , x of elements of M as an ordered subset
n
1
of M . In the case of a basis, the elements x1 , . . . , xn must all be distinct, so that the sequence in question
does specify an ordering on an n-element subset of M .
CHAPTER 7. RINGS AND MODULES
260
1. Let G be a finite group, and consider A[G] as a left A-module via the canonical
inclusion ι : A → A[G]. Then the elements of G, in any order, form a basis
for A[G].
The point is that every element of A[G] may be written uniquely as g∈G ag g, so
the A-module homomorphism from A|G| to A[G] induced by an ordering of the
elements of G is a bijection.
2. For the same reason, if G is an infinite group, then {g | g ∈ G} forms a basis for
the infinitely generated free module A[G].
3. Consider Mn (A) as a left A-module via the canonical inclusion ι : A → Mn (A).
Let eij be the matrix whose ij-th coordinate is 1 and whose other coordinates are
all 0, where 1 ≤ i, j ≤ n. Then the matrices eij , in any order, form a basis for
Mn (A). Here, we make use of the obvious fact that
(aij ) =
aij eij
1≤i,j≤n
for any matrix (aij ) ∈ Mn (A).
The very same elements form bases for A[G] and Mn (A) when we consider them as
right modules via the inclusions ι. Another very important example of free modules
comes from quotients of polynomial algebras.
Proposition 7.7.35. Let A be a commutative ring and let f (X) ∈ A[X] be a polynomial
of degree n > 0 whose leading coefficient is a unit. Then A[X]/(f (X)) is a free Amodule with basis given by the images of 1, X, . . . , X n−1 under the canonical map from
A[X]. Here, (f (X)) is the principal ideal of A[X] generated by f (X), and the A-module
structure on the quotient ring A[X]/(f (X)) comes from the A-algebra structure induced
by the standard algebra structure on A[X].
Proof Write h : An → A[X]/(f (X)) for the A-module homomorphism that carries ei
to X i−1 for i = 1, . . . , n. We show that h is an isomorphism.
For g(X) ∈ A[X], the Euclidean algorithm for polynomials (Proposition 7.3.10) allows
us to write g(X) = q(X)f (X) + r(X), where deg r(X) < deg f (X). But then the images
of g(X) and r(X) under the canonical map π : A[X] → A[X]/(f (X)) are equal, and, if
r(X) = an−1 X n−1 + · · · + a0 , then π(r(X)) = h(a0 , . . . , an−1 ). Thus, h is onto.
Now suppose that (a0 , . . . , an−1 ) ∈ ker h and let g(X) = an−1 X n−1 + · · · + a0 . Then
π(g(X)) = 0, and hence g(X) ∈ (f (X)). Thus, g(X) = q(X)f (X) for some q(X) ∈ A[X].
Since the leading coefficient of f (X) is a unit, the degree of q(X)f (X) must be ≥ deg f
unless q(X) = 0. Thus, q(X) = g(X) = 0, and hence ai = 0 for 0 ≤ i ≤ n − 1. Thus, h
is injective.
The next lemma is almost immediate from the definition, but is useful nonetheless,
especially in understanding matrix groups.
Lemma 7.7.36. Let f : M → N be an A-module homomorphism between the free modules M and N , and let x1 , . . . , xn be any basis of M . Then f is an isomorphism if and
only if f (x1 ), . . . , f (xn ) is a basis for N .
Proof Let g : An → M be the A-module homomorphism with g(ei ) = xi for i =
1, . . . , n. By the definition of basis, g is an isomorphism. The result now follows from
the fact that f ◦ g is the unique A-module homomorphism from An to N which takes ei
to f (xi ) for i = 1, . . . , n.
CHAPTER 7. RINGS AND MODULES
261
Surprisingly, rank is not well defined for free modules over arbitrary rings: There are
rings for which An ∼
= Am for n = m. But such rings turn out to be exotic.
We shall see shortly that rank is well defined over commutative rings. The following
corollary, which is immediate from Lemma 7.7.22 and the definition of An , will help us
do so.
Corollary 7.7.37. Let a be a two-sided ideal in the ring A. Then An /aAn is isomorphic
to (A/a)n as an A/a-module.
While free modules of finite rank may not always have a uniquely defined rank, we
can always tell the difference between the cases of finite and infinite rank.
*
Lemma 7.7.38. Let I be a set. Suppose that the free module M = i∈I A is finitely
generated. Then I must be a finite set. Thus, a finitely generated free module has finite
rank.
Proof Let x1 , . . . , xn be a generating set for M , and write xj = i∈I aij ei for j =
1, . . . , n. Then for each j, all but finitely many of the aij are 0. Thus, if S = {i ∈
I | aij = 0 for some j}, then S must be finite. But then, the submodule of M generated
by x1 , . . . , xn must lie in the submodule generated by {ei | i ∈ S}. But this says that M
itself must be generated by {ei | i ∈ S}.
But clearly, if i is not in S, then ei is not in the submodule of M generated by
{ei | i ∈ S}. Thus, S = I, and hence I is finite.
It is useful to understand what it means to be a basis.
Definitions 7.7.39. Let m1 , . . . , mn be elements of the A-module M . We say that
m1 , . . . , mn are linearly dependent if there are elements a1 , . . . , an of A, not all 0, such
that a1 m1 + · · · + an mn = 0.
Similarly, we say that an infinite set {mi | i ∈ I} is linearly dependent if there is a
relation of the form
ai1 mi1 + · · · + aik mik = 0,
where k > 0, i1 , . . . , ik are distinct elements of I and not all of the aij are 0.
We say that a collection of elements is linearly independent if it is not linearly dependent.
Proposition 7.7.40. Let m1 , . . . , mn be elements of the A-module M and let f : An →
M be the unique A-module homomorphism such that f (ei ) = mi for 1 ≤ i ≤ n. Then
the following conditions hold.
1. The elements m1 , . . . , mn are linearly independent if and only if f is an injection.
2. The image of f is the submodule generated by m1 , . . . , mn . Thus, m1 , . . . , mn
generate M if and only if f is a surjection.
Thus, m1 , . . . , mk is a basis for M if and only if it is linearly independent and generates
M.
*
Similarly, if {mi | i ∈ I} is a family of elements of M , and if f : i∈I A → M is the
unique homomorphism such that f (ei ) = mi for all i, then f is injective if and only if
{mi | i ∈ I} are linearly independent, and is surjective if and only if {mi | i ∈ I} generate
M.
CHAPTER 7. RINGS AND MODULES
262
Proof First, consider the case of a finite subset m1 , . . . , mk . Here, an element of the
kernel of f is precisely an n-tuple (a1 , . . . , an ) such that a1 m1 + · · · + an mn = 0. So
linear independence is equivalent to f being injective.
Also, f (a1 , . . . , an ) = a1 m1 + · · · + an mn . Since this gives the generic element of the
submodule generated by m1 , . . . , mn , the second condition
also holds.
In the infinitely generated case, a nonzero element of i∈I A may be written uniquely
in the form ai1 ei1 + · · · + aik eik , where k > 0, i1 , . . . , ik are distinct elements of I, and
ai = 0 for all i. And f (ai1 ei1 + · · · + aik eik ) = ai1 mi1 + · · · + aik mik . Thus, there is a
nonzero element of the kernel if and only if {mi | i ∈I} is linearly dependent.
Proposition 7.7.26 shows that the image of f is i∈I Ami , so the result follows.
The following corollary is immediate.
Corollary 7.7.41. An A-module M is finitely generated if and only if it is a quotient
module of An for some integer n.
Exercises 7.7.42.
1. Both Z2 and Z3 are Z6 -modules, as they both have exponent 6. Show that Z2 ⊕ Z3
is a free Z6 -module, but that neither Z2 nor Z3 is free over Z6 .
† 2. Show that every A-module is a quotient module of a free module.
3. Show that every quotient module of a finitely generated module is finitely generated.
Exact Sequences
Definition 7.7.43. Suppose given a “sequence,” either finite or infinite, of A-modules
and module homomorphisms:
f
g
· · · → M −
→M −
→ M → · · ·
We say that the sequence is exact at M if the kernel of g is equal to the image of f .
We call the sequence itself exact if it is exact at every module of the sequence that sits
between two other modules.
One sort of exact sequence has special importance.
Definition 7.7.44. A short exact sequence of A-modules is an exact sequence of the
form
f
g
0 → M −
→M −
→ M → 0.
Note that exactness at M shows that g has to be onto, while exactness at M shows
that f is an injection. Thus, M/g(M ) is isomorphic to M , and hence the short exact
sequence presents M as an extension of M by M .
Example 7.7.45. For any pair M, N of A-modules, the sequence
ι
π
0→M −
→M ⊕N −
→N →0
is exact, where ι is the standard inclusion map and π is the standard projection map.
We shall make extensive use of exact sequences, both short and otherwise. For instance, we can use exactness to measure the deviation of a map from being an isomorphism.
CHAPTER 7. RINGS AND MODULES
263
Definition 7.7.46. Let f : M → N be an A-module homomorphism. Then the cokernel
of f is the module C = N/(im f ).
In particular, f is onto if and only if the cokernel of f is 0.
Lemma 7.7.47. Let f : M → N be an A-module homomorphism. Then there is an
exact sequence
f
i
π
0→K−
→M −
→N −
→ C → 0,
where i is the inclusion of the kernel of f and π is the canonical map onto the cokernel
of f . Conversely, given a six term exact sequence
f
0 → K → M −
→ N → C → 0,
f is injective if and only if K = 0 and is surjective if and only if C = 0.
The next result can be very useful for recognizing isomorphisms. The technique used
in the proof is called a diagram chase.
Lemma 7.7.48. (Five Lemma) Suppose given a commutative diagram with exact rows
M1
α1 /
M2
f1
N1
β1
α2 /
M3
α3 /
M4
f2
f3
f4
/ N2
/ N3
/ N4
β2
β3
α4 /
M5
f5
β4 / N5
Then the following properties hold.
1. If f2 and f4 are injective and if f1 is surjective, then f3 is injective.
2. If f2 and f4 are surjective and if f5 is injective, then f3 is surjective.
In particular, if f2 and f4 are isomorphisms, f1 is surjective, and f5 is injective, then f3
is an isomorphism.
Proof Suppose first that f2 and f4 are injective and f1 is surjective. Let m ∈ ker f3 .
Then (f4 ◦ α3 )(m) = 0 by the commutativity of the diagram. In other words, α3 (m) is
in the kernel of f4 . Since f4 is injective, m must be in the kernel of α3 .
By exactness, m = α2 (m ) for some m ∈ M2 . Since the diagram commutes and
m ∈ ker f3 , this shows that f2 (m ) ∈ ker β2 . By the exactness of the lower sequence,
f2 (m ) = β1 (n) for some n ∈ N1 .
Since f1 is surjective, n = f1 (m ) for some m ∈ M1 . But then f2 (α1 (m )) = β1 (n).
But n was chosen so that β1 (n) = f2 (m ), so α1 (m ) has the same image as m under
f2 . Since f2 is injective, m = α1 (m ). But m was chosen so that α2 (m ) = m. Since
α2 ◦ α1 = 0, m must be 0, and hence f3 is injective as claimed.
Now suppose that f2 and f4 are surjective and f5 is injective, and let n ∈ N3 . Since
f4 is surjective, β3 (n) = f4 (m) for some m ∈ M4 . Since β4 ◦ β3 = 0, commutativity of
the diagram shows that (f5 ◦ α4 )(m) = 0. But f5 is injective, and hence α4 (m) = 0.
By the exactness of the upper sequence, m = α3 (m ) for some m ∈ M3 . Recall that
m was chosen so that f4 (m) = β3 (n). By commutativity of the diagram, β3 carries n and
f3 (m ) to the same element, so that n − f3 (m ) = β2 (n ) for some n ∈ N2 by exactness.
But f2 is surjective, so that n = f2 (m ) for some m ∈ M2 . So f3 (m + α2 (m )) =
n − f3 (m ) + f3 (m ) = n, and hence f3 is surjective.
CHAPTER 7. RINGS AND MODULES
264
A useful illustration of the utility of the Five Lemma comes from an analysis of when
a short exact sequence may be used to express the middle term as the direct sum of
its surrounding terms. Recall that a section for a homomorphism g : M → N is a
homomorphism s : N → M such that g ◦ s = 1N . Recall also that a retraction for a
homomorphism f : N → M is a homomorphism r : M → N such that r ◦ f = 1N . We
assume that all homomorphisms under discussion are A-module homomorphisms.
Proposition 7.7.49. Suppose given a short exact sequence
f
g
→M −
→ M → 0.
0 → M −
Then the following additional conditions are equivalent.
1. f admits a retraction.
2. g admits a section.
3. There is an isomorphism h : M ⊕ M → M such that the following diagram
commutes.
0
/ M
ι
/ M ⊕ M π /
M / 0
/ M / 0
h
0
/ M
f
/ M
g
Proof We first show that the existence of a retraction for f implies the existence of a
section for g. Thus, let r be a retraction for f . We claim then that the restriction, g , of
g to the kernel of r is an isomorphism of ker r onto M . Note that this is sufficient to
produce a section, s, for g by setting s to be the composite
M (g )−1
/ ker r ⊂ M.
To prove the claim, let m ∈ ker r. If g(m) = 0, then m ∈ ker g = im f , and hence
m = f (m ) for some m ∈ M . But since r ◦ f = 1M , r(m) = m . But m = 0, since
m ∈ ker r, and hence m = 0 as well. Thus, the restriction of g to ker r is injective, and
it suffices to show that g : ker r → M is onto.
Let m ∈ M . Then m = g(m) for some m ∈ M . Consider the element x =
m − f (r(m)). Then r(x) = r(m) − r(m), since r ◦ f = 1M . Thus, x ∈ ker r. But clearly,
∼
=
g(x) = m , so g : ker r −→ M as claimed, and hence g admits a section.
Now suppose given a section s for g. We define h : M ⊕ M → M by h(m , m ) =
f (m ) + s(m ). This is easily seen to make the diagram that’s displayed in the third
condition commute. But then h is an isomorphism by the Five Lemma, and hence the
third condition holds.
It suffices to show that the third condition implies the existence of a retraction for f .
But this is immediate: just take r = π1 ◦ h−1 , where π1 is the projection of M ⊕ M onto its first factor.
Note that the specific choice of r, s, or h above determines the other two maps
precisely.
CHAPTER 7. RINGS AND MODULES
265
Definition 7.7.50. If a short exact sequence
g
f
0 → M −
→M −
→ M → 0
satisfies the three properties listed in Proposition 7.7.49, then we say the sequence splits.
A specification of a specific retraction for f , a specific section of g, or a specific isomorphism h as in the third condition is called a splitting for the sequence.
Note that the situation with regard to splitting extensions of A-modules is cleaner
than that for nonabelian groups. Here, there is no analogue of a semidirect product, and
every split extension is a direct sum.
If the third term of a short exact sequence is free, then it always splits.
Proposition 7.7.51. Suppose given a short exact sequence
g
f
0 → M −
→M −
→ M → 0
with M free. Then the sequence splits.
Proof Let {mi | i ∈ I} be a basis for M , and, for each i ∈ I, choose mi ∈ M such
that π(mi ) = mi . We claim that there is an A-module map s : M → M such that
s(mi ) = mi for all i ∈ I. Once we show this we’ll be done, as then g ◦ s is the identity
on {mi | i ∈ I}. But any two A-module homomorphisms that agree on a generating set
for a module must be equal, so g ◦ s = 1M , and hence s is a section for g.
We now show that such an s exists. Recall from Lemma 7.7.32 that if N is any
A-module and if*
{ni | i ∈ I} are elements of N , then there’s a unique A-module homomorphism from i∈I A to N which carries the canonical basis vector ei to ni for each
i ∈ I.
*
Let f : i∈I A → M be the unique A-module homomorphism that carries
* ei to mi
for each i ∈ I. Then f is an isomorphism by the definition of a basis. Let s : i∈I A → M
be the unique A-module homomorphism that carries ei to mi for each i ∈ I. Then the
desired homomorphism s is the composite s ◦ f −1 .
Exercises 7.7.52.
1. Let A be any ring and let
0 → M → M → M → 0
be an exact sequence of A-modules. Show that if M and M are finitely generated,
then so is M .
† 2. Snake Lemma Suppose given a commutative diagram with exact rows:
M
i
/ M
f
0
/ N
p
/ N
/ 0
f f
j
/ M q
/ N Show that there is an exact sequence
ker f → ker f → ker f −
→ coker f → coker f → coker f ,
δ
CHAPTER 7. RINGS AND MODULES
266
where “coker” stands for cokernel, and δ is defined as follows. For m ∈ ker f , let
m = p(m). Then f (m) = j(n ) for some n ∈ N , and we let δ(m ) be the element
of coker f represented by n . Show also that if i is injective, so is ker f → ker f ,
and if q is surjective, so is coker f → coker f . (Hint: The key here is showing
that δ, as described, is a well defined homomorphism.)
‡ 3. Let A be an integral domain and let (x) be the principal ideal generated by x ∈ A.
Show that there is a short exact sequence
fx
π
→ A/(x) → 0
0 → A −→ A −
of A-modules, where fx (a) = ax.
4. Let Zn = T and let A be any commutative ring. Show that there is an exact
sequence of A[Zn ]-modules
· · · → Ck → · · · → C0 → A → 0,
where Ck = A[Zn ] for all k ≥ 0, and the maps are given as follows. The map
C0 → A is the standard augmentation. Each map C2k+1 → C2k with k ≥ 0 is
multiplication by T − 1, and each map C2k → C2k−1 with k ≥ 1 is multiplication
by Σ = 1 + T + · · · + T n−1 . (In the parlance of homological algebra, the sequence
· · · → Ck → · · · → C0 is called a resolution of A over A[Zn ].
Rings Without Identity
Not all algebraists require rings to have multiplicative identity elements. Here, we discuss
the relationship between such a theory and that of rings as we have defined them.
Definition 7.7.53. A ring without identity is an abelian group R together with an associative binary operation, to be thought of as multiplication, that satisfies the distributive
laws:
a(b + c) = ab + ac
and
(a + b)c = ac + bc.
When we speak of a ring, we will continue to assume that it has a multiplicative
identity. The new notion must be named in full: rings without identity.
Example 7.7.54. Let a be a left or right ideal of a ring. Then, under the multiplication
inherited from A, a is a ring without identity.
We shall see presently that every ring without identity may be embedded as an ideal
in a ring. But it may be possible to embed it as an ideal in many different rings.
We define homomorphisms, modules, and ideals for rings without identity in the
obvious ways. Since rings with identity may also be considered in this broader context,
we shall, at the risk of redundancy, refer to the modules as “nonunital modules.”
Definitions 7.7.55. A homomorphism f : R → S of rings without identity is a homomorphism of additive groups with the additional property that f (rs) = f (r)f (s) for all
r, s ∈ R.
CHAPTER 7. RINGS AND MODULES
267
A nonunital module M over a ring R without identity is an abelian group together
with a multiplication R × M → M such that the following conditions hold:
r(m + m )
= rm + rm
(r + r )m = rm + r m
r(r m) = (rr )m,
for all r, r ∈ R and all m, m ∈ M .
A homomorphism f : M → N of nonunital modules over a ring R without identity is
a group homomorphism such that f (rm) = rf (m) for all r ∈ R and m ∈ M .
The basics of module theory, such as direct sums, exact sequences, quotient modules,
etc. go over to this context without change. Free modules, on the other hand, are
problematic, as R itself is not necessarily free in the categorical sense. There is no good
notion of basis here.
Left, right, and two-sided nonunital ideals may now be defined in the obvious way,
and the quotient by a two-sided nonunital ideal satisfies the analogue of Proposition 7.2.9.
There is a natural way to adjoin an identity to a ring without identity.
) of R is
Definition 7.7.56. Let R be a ring without identity. The associated ring, R,
)
given by R = Z ⊕ R, as an abelian group, with the multiplication
(m, r) · (n, s) = (mn, ms + nr + rs),
where ms and nr denote the m-th power of s and the n-th power of r, respectively, in
the additive group of R.
) for the map ι(r) = (0, r) for r ∈ R.
Write ι : R → R
The reader may check the details of the following proposition.
) is
Proposition 7.7.57. Let R be a ring without identity. Then the associated ring R
)
a ring with identity element (1, 0), and ι : R → R is a homomorphism of rings without
)
identity, whose image is a two-sided ideal of R.
Moreover, if A is a ring, then for each homomorphism f : R → A of rings without
) → A such that f) ◦ ι = f .
identity, there is a unique ring homomorphism f) : R
) is a left adjoint to
In categorical language, this says that the passage from R to R
the forgetful functor that forgets that a ring has an identity element.
Given the relationship between module structures and homomorphisms into endomorphism rings, the following proposition should not be a surprise.
Proposition 7.7.58. Let M be a nonunital module over the ring R without identity.
)
Then there is a unique R-module
structure on M whose restriction to the action by
elements in the image of ι gives the original nonunital R-module structure on M :
(n, r) · m = nm + rm,
where nm is the n-th power of m in the additive group of M .
)
)
The nonunital R-submodules of M and the R-submodules
of this induced R-module
structure coincide.
This correspondence is natural in M in that it gives a functor from the category of
)
nonunital R-modules to the category of R-modules.
CHAPTER 7. RINGS AND MODULES
268
Exercises 7.7.59.
1. Suppose the ring R without identity has exponent n as an additive group. Show
that there is a ring structure on Zn ⊕R that extends the multiplication on R = 0⊕R.
Show that the analogue of Proposition 7.7.57 holds for homomorphisms into rings
of characteristic n.
2. Let A be a commutative ring and let R be a ring without identity which is also an
A-module, such that
r · as = ar · s = a(r · s)
for r, s ∈ R, and a ∈ A. Construct an A-algebra associated to R and state and prove
an analogue of Proposition 7.7.57 for appropriate homomorphisms into A-algebras.
7.8
Chain Conditions
We return to the discussion of chain conditions that was begun in Section 7.6. The
module theory we’ve now developed will be very useful in this regard. First, let us
generalize the chain conditions to the study of modules and of left or right ideals.
Definitions 7.8.1. A ring is left Noetherian if it has the ascending chain condition for
left ideals. In other words, any ascending sequence
a1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · ·
of inclusions of left ideals is eventually constant in the sense that there’s an n such that
an = ak for all k ≥ n.
Similarly, left Artinian rings are those with the descending chain condition for left
ideals. Here, any descending sequence
a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · ·
of inclusions of left ideals is eventually constant in the sense that there’s an n such that
an = ak for all k ≥ n.
Right Noetherian and right Artinian rings are defined analogously via chain conditions
on right ideals.
An A-module M is a Noetherian A-module if it has the ascending chain condition for
its submodules. It is an Artinian module if it has the descending chain condition for its
submodules.
Clearly, A is left Noetherian (resp. left Artinian) if and only if it is Noetherian
(resp. Artinian) as a left A-module. We shall not make use of chain conditions on
two-sided ideals. When we speak of a Noetherian or Artinian ring, without specifying
left or right, we shall mean a commutative ring with the appropriate chain condition on
its ideals.
Examples 7.8.2.
1. Recall that a division ring has no left (or right) ideals other than 0 and the ring
itself. Thus, division rings are left and right Noetherian and Artinian.
2. Recall that the left ideals of a product A × B all have the form a × b, with a a
left ideal of A and b a left ideal of B. Thus, if A and B are both left Noetherian
(resp. left Artinian), so is A × B.
CHAPTER 7. RINGS AND MODULES
269
3. Regardless
*∞ of any chain conditions that hold for a nonzero ring A, the infinite direct
sum i=1 A is neither Noetherian nor Artinian as an A-module.
4. The integers, Z, are not Artinian, but Zn is an Artinian Z-module for all n > 0.
!∞
5. The infinite product i=1 Zp satisfies neither chain condition on its ideals, but the
module structure on Zp obtained from the projection onto the first factor satisfies
both chain conditions.
6. Let
A=
a b
0 c
a ∈ Q, b, c ∈ R .
Let a ⊂ A be the set of all elements of the form ( 00 0b ) with b ∈ R. Then a is a
two-sided ideal of A. Note that
a b
0 d
0 ad
=
.
0 c
0 0
0 0
Thus, the left submodules of a may be identified with the Q-submodules of R.
Since R is infinitely generated as a vector space over Q, it satisfies neither chain
condition by Example 3. (See Corollary 7.9.14 for a demonstration that R is free
as a Q-module.) Thus, A itself is neither left Artinian nor left Noetherian. But
0 d
a b
0 dc
=
,
0 0
0 c
0 0
so the right submodules of a may be identified with the R-submodules of R, which
consist only of 0 and R itself. Thus, a is both right Noetherian and right Artinian
as an A-module.
There is a surjective ring homomorphism f : A → Q × R, given by
a b
f
= (a, c).
0 c
The kernel of f is a. Since Q × R is commutative, the left and right A-module
structures on it coincide. Note that the only proper A-submodules of Q × R are
Q × 0, 0 × R, and 0. Thus, Q × R satisfies both chain conditions as an A-module.
Moreover, we have a short exact sequence
0→a→A→Q×R→0
of two-sided A-modules. As right modules, both a and Q × R are both Noetherian
and Artinian. As we shall see in Lemma 7.8.7, this implies that A itself is both
right Noetherian and right Artinian. Thus, A satisfies both chain conditions on
one side, and satisfies neither chain condition on the other.
The Noetherian property can be characterized in terms of finite generation.
Proposition 7.8.3. An A-module M is a Noetherian module if and only if each of its
submodules is finitely generated.
CHAPTER 7. RINGS AND MODULES
270
Proof Suppose that every submodule is finitely generated and that we are given an
ascending chain
N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · ·
of submodules of M . Let N = k≥0 Nk . Then it is easy to see that N is a submodule
of M . Suppose that N is generated by n1 , . . . , nk . Then we may choose m to be large
enough that n1 , . . . , nk ∈ Nm . But then the submodule of M generated by n1 , . . . , nk
must be contained in Nm . Since n1 , . . . , nk generate N , we must have Nm = Nm+l = N
for all l ≥ 0.
Conversely, suppose that M is a Noetherian module and let N be one of its submodules. Suppose by contradiction that N is not finitely generated. Then by induction we
may choose a sequence of elements ni ∈ N such that n1 = 0 and for each i > 1, ni is not
in the submodule generated by n1 , . . . , ni−1 .
Write Nk for the submodule generated by n1 , . . . , nk . Then each inclusion Nk−1 ⊂ Nk
is proper, and hence the sequence
N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · ·
contradicts the hypothesis that M is a Noetherian module.
Since a ring is left Noetherian if and only if it is Noetherian as a left module over
itself, we obtain the following corollary.
Corollary 7.8.4. A ring A is left Noetherian if and only if every left ideal is finitely
generated.
Note that every ideal in a principal ideal ring is a cyclic module, and hence is finitely
generated.
Corollary 7.8.5. Principal ideal rings are Noetherian.
The next very useful lemma shows that chain conditions are inherited by submodules
and quotient modules.
Lemma 7.8.6. Let A be a ring and let M be a Noetherian (resp. Artinian) A-module.
Then any submodule or quotient module of M is Noetherian (resp. Artinian) as well.
Proof An ascending (resp. descending) chain in a submodule of M is a chain in M .
But then an application of the appropriate chain condition in M shows that it must hold
in N as well.
For quotient modules M/N , the result follows from the usual one-to-one correspondence between submodules of the quotient and the submodules of M containing N .
We wish to show that the free modules An share whatever chain conditions A has.
The next lemma will help us to do so.
Lemma 7.8.7. Suppose given a short exact sequence
f
g
0 → M −
→M −
→ M → 0
of A-modules. Then M is Noetherian (resp. Artinian) if and only if both M and M are.
CHAPTER 7. RINGS AND MODULES
271
Proof If M satisfies the stated chain condition, then so do M and M by Lemma 7.8.6.
For the converse, we restrict attention to the Noetherian case. The Artinian case is
similar. Thus, suppose that M and M are Noetherian modules, and suppose given an
ascending chain
N1 ⊂ N2 ⊂ · · · ⊂ N k ⊂ · · ·
of submodules of M . Then we obtain ascending chains
g(N1 ) ⊂ g(N2 ) ⊂ · · · ⊂ g(Nk ) ⊂ · · ·
in M and
f −1 (N1 ) ⊂ f −1 (N2 ) ⊂ · · · ⊂ f −1 (Nk ) ⊂ · · ·
in M . Thus, since M and M are Noetherian, for k large enough, we have g(Nk ) =
g(Nk+l ) and f −1 (Nk ) = f −1 (Nk+l ) for all l ≥ 0. But we claim then that Nk = Nk+l .
To see this, consider the following diagram, in which the vertical maps are the inclusion maps from the sequences above. Since f (f −1 (Ni )) = Ni ∩ ker g, the rows are
exact.
f
g
/ f −1 (Nk )
/ Nk
/ g(Nk )
/ 0
0
0
/ f −1 (Nk+1 )
f
/ Nk+1
g
/ g(Nk+1 )
/ 0
Since the inclusions f −1 Nk ⊂ f −1 Nk+l and g(Nk ) ⊂ f (Nk+l ) are isomorphisms, the
inclusion of Nk in Nk+l is also by the Five Lemma.
Since sequences of the form
0→M →M ⊕N →N →0
are exact, a quick induction on n gives the next corollary.
Corollary 7.8.8. If A is left Noetherian (resp. left Artinian), then An is Noetherian
(resp. Artinian) as a left A-module for all n ≥ 0.
Now, a finitely generated module is a quotient module of An for some n. Since passage
to quotient modules preserves chain conditions, the next proposition is immediate.
Proposition 7.8.9. Let A be a left Noetherian (resp. left Artinian) ring. Then any
finitely generated left A-module is a Noetherian (resp. Artinian) module over A.
This now allows us to characterize the Noetherian property.
Corollary 7.8.10. A ring A is left Noetherian if and only if it has the property that
every submodule of every finitely generated left A-module is finitely generated.
Proof If every left submodule of A itself is finitely generated, then A is left Noetherian
by Corollary 7.8.4.
Conversely, if A is left Noetherian and M is a finitely generated A-module, then M is
Noetherian, and hence each of its submodules is finitely generated by Proposition 7.8.3.
CHAPTER 7. RINGS AND MODULES
272
Proposition 7.8.9 also has immediate applications to establishing chain conditions in
some of the examples we’ve seen of rings.
Corollary 7.8.11. Let f : A → B be a ring homomorphism and suppose that B is
finitely generated as a left A-module. Then if A is left Noetherian or left Artinian, so is
B.
Proof Every B-module is an A-module via f . So a chain of ideals in B is a chain of
A-submodules of the finitely generated A-module B.
We obtain two immediate corollaries.
Corollary 7.8.12. Let G be a finite group. Then if A is left or right Noetherian or
Artinian, then so is the group ring A[G]. In particular the group ring D[G] over a
division ring D satisfies all four chain conditions.
Corollary 7.8.13. If A is left or right Noetherian or Artinian, then so are the matrix
rings Mn (A) for n ≥ 1. In particular the matrix rings Mn (D) over a division ring D
satisfy all four chain conditions.
Of course, A[X] is infinitely generated as an A-module, so we shall have to work
harder to show that it inherits the Noetherian property from A. As a result, however,
we shall obtain a considerable strengthening of the Noetherian case of Corollary 7.8.11
when A is commutative.
Theorem 7.8.14. (Hilbert Basis Theorem) Let A be a left Noetherian ring. Then the
polynomial ring A[X] is also left Noetherian.
Proof We show that every left ideal of A[X] is finitely generated. Let a be a left ideal
of A[X] and define b ⊂ A to be the set of all leading coefficients of elements of a. Note
that if a and b are the leading coefficients of f and g, respectively, and if f and g have
degrees m and n, respectively, with m ≥ n, then the leading coefficient of f + X m−n g is
a + b. Thus, b is a left ideal of A.
Since A is left Noetherian, there is a finite generating set x1 , . . . , xk for b. But then
we can find polynomials f1 , . . . , fk whose leading coefficients are x1 , . . . , xk , respectively.
After stabilizing by multiplying by powers of X, if necessary, we may assume that the
polynomials f1 , . . . , fk all have the same degree, say m.
Let M be the A-submodule of A[X] generated by 1, X, X 2 , . . . , X m−1 . Thus, the
elements of M are the polynomials of degree less than m. Now a ∩ M is a submodule
of a finitely generated A-module. Since A is Noetherian, a ∩ M is a finitely generated
A-module. We claim that every element of a may be written as the sum of an element of
A[X]f1 + · · · + A[X]fk (i.e., the ideal generated by f1 , . . . , fk ) with an element of a ∩ M .
This will be sufficient, as then a is generated as an ideal by f1 , . . . , fk together with any
set of A-module generators for a ∩ M .
But the claim is an easy induction on degree. Let f ∈ a have degree n. If n <
m, then f ∈ a ∩ M , and there’s nothing to show. Otherwise, let a be the leading
coefficient of f and write a = a1 x1 + · · · + ak xk with ai ∈ A for all i. Then a is
the leading coefficient of a1 X n−m f1 + · · · + ak X n−m fk , which has degree n. But then
f − (a1 X n−m f1 + · · · + ak X n−m fk ) has degree less than n, and hence the claim follows
by induction.
Recall that an A-algebra of finite type is one which is finitely generated as an Aalgebra.
CHAPTER 7. RINGS AND MODULES
273
Corollary 7.8.15. Let A be a commutative Noetherian ring and let B be a commutative
A-algebra of finite type. Then B is Noetherian.
Proof The hypothesis on B is that there is a surjective ring homomorphism f :
A[X1 , . . . , Xn ] → B for some n ≥ 0. By Corollary 7.8.11, it suffices to show that
A[X1 , . . . , Xn ] is Noetherian. But Proposition 7.3.23 shows that A[X1 , . . . , Xn ] is a polynomial ring in one variable on A[X1 , . . . , Xn−1 ], so the result follows from the Hilbert
Basis Theorem and induction.
Exercises 7.8.16.
1. Let A be a nonzero ring. Show that A[X] is not left Artinian.
2. In multiplicative notation, we can write the abelian group Z as {X n | n ∈ Z},
with 1 = X 0 as the identity element. For a ring A, the group ring A[Z] is sometimes called the ring of Laurent polynomials in A, and is sometimes denoted by
A[X, X −1 ]. (This is absolutely not to be confused with a polynomial ring in two
variables.) Note that A[X, X −1 ] contains the polynomial ring A[X] as a subring.
Show that for any f ∈ A[X, X −1 ] there is a unit u such that uf ∈ A[X]. Use this,
together with either the statement or the proof of the Hilbert Basis Theorem, to
show that if A is left Noetherian, so is A[X, X −1 ].
3. Since an abelian group is precisely a Z-module, any finitely generated abelian group
is a quotient group of Zn for some n. Show that if G is a finitely generated abelian
group and if A is a left Noetherian ring, then A[G] is also left Noetherian.
4. Suppose there is a finitely generated group, say with n generators, such that Z[G]
fails to be left Noetherian. Show, then, that if F is a free group on n or more
generators, then Z[F ] must also fail to be left Noetherian.
7.9
Vector Spaces
We give some of the basic properties of vector spaces, which we shall use in studying
modules over more general rings.
In particular, we shall show that if A is a commutative ring and if we have an Amodule isomorphism Am ∼
= An , then m = n. Thus, using vector spaces, we shall show
that the rank of a finitely generated free module over a commutative ring is well defined,
independently of the choice of an isomorphism to one of the canonical free modules An
(i.e., independently of the choice of a basis).
For the sake of generality, we will work over a division ring, rather than a field. (In
particular, we shall not treat determinants here.) Thus, “vector space” here will mean a
left module over a division ring D, and a D-linear map is a D-module homomorphism.
The results, of course, will all extend to right modules as well.
Definition 7.9.1. We say that a vector space V over D is finite dimensional if it is
finitely generated as a D-module.
Recall that a finite sequence, v1 , . . . , vk , of elements of a D-module V is a basis
for V if the unique D-module homomorphism f : Dk → V for which f (ei ) = vi for
i = 1, . . . , k is an isomorphism, where ei is the i-th canonical basis vector of Dk . Recall
also (from Proposition 7.7.40) that v1 , . . . , vk is a basis if and only if v1 , . . . , vk are linearly
independent and generate V .
CHAPTER 7. RINGS AND MODULES
274
Proposition 7.9.2. Let V be a finite dimensional vector space over D, with generators
v1 , . . . , vn . Then there is a subset vi1 , . . . , vik of v1 , . . . , vn which is a basis for V . In
addition, if v1 , . . . , vm are linearly independent for some m ≤ n, then we may choose the
subset vi1 , . . . , vik so that it contains v1 , . . . , vm .
In particular, every finite dimensional D-vector space is free as a D-module.
Proof If V = 0, then the null set is a basis. Otherwise, we argue by induction on
n − m. If v1 , . . . , vn is linearly independent, there is nothing to prove. Otherwise, we
can find x1 , . . . , xn ∈ D, not all 0, such that x1 v1 + · · · + xn vn = 0. Now at least
one of xm+1 , . . . , xn must be nonzero, as otherwise the elements v1 , . . . , vm would be
linearly dependent. For simplicity, suppose that xn = 0. Then vn = (−x−1
n x1 )v1 +
x
)v
.
Thus,
v
is
in
the
submodule
generated
by
v
,
.
.
.
,
v
· · · + (−x−1
n−1 n−1
n
1
n−1 . Since
n
v1 , . . . , vn generate V , V must in fact be generated by v1 , . . . , vn−1 . But the result now
follows by induction.
Corollary 7.9.3. Let V be a finite dimensional vector space and let v1 , . . . , vm be linearly
independent in V . Then there is a basis of V containing v1 , . . . , vm .
Proof Let v1 , . . . , vn be obtained from v1 , . . . , vm by adjoining any finite generating set
for V , and apply Proposition 7.9.2.
Proposition 7.9.4. Let v1 , . . . , vn be a basis for the vector space V and let w1 , . . . , wk
be linearly independent in V . Then k ≤ n.
Proof We shall show by induction on i ≥ 0 that we may relabel the v1 , . . . , vn in
such a way that w1 , . . . , wi , v1 , . . . , vn−i generate V . If k were greater than n, this
would say that w1 , . . . , wn is a generating set for V , and hence that wn+1 is a linear
combination of w1 , . . . , wn , say wn+1 = x1 w1 +· · ·+xn wn , with x1 , . . . , xn ∈ D. But then
x1 w1 + · · · + xn wn + (−1)wn+1 = 0, contradicting the linear independence of w1 , . . . , wk .
Thus, for i ≥ 1, we suppose inductively that for some relabelling of the v’s, the set
w1 , . . . , wi−1 , v1 , . . . , vn−i+1 generates V . Thus, we may write wi as a linear combination
of these elements, say wi = x1 w1 + · · · + xi−1 wi−1 + xi v1 + · · · + xn vn−i+1 . Now it cannot
be the case that xj = 0 for all j ≥ i, as then the elements w1 , . . . , wi would be linearly
dependent. After relabelling, if necessary, we may assume that xn = 0, in which case
vn−i+1 may be written as a linear combination of w1 , . . . , wi , v1 , . . . , vn−i , which then
must generate V .
Corollary 7.9.5. Let V be a finite dimensional vector space. Then any two bases of V
have the same number of elements.
Proof Since any basis of V is linearly independent, the number of elements in each
basis must be less than or equal to the number in the other, by Proposition 7.9.4.
Thus, we can make the following definition.
Definition 7.9.6. Let V be a finite dimensional vector space. Then the dimension of
V , written dim V , is the number of elements in any basis of V .
By the definition of a basis, V has a basis with n elements (and hence has dimension
n) if and only if V is isomorphic to Dn .
Corollary 7.9.7. Two finite dimensional vector spaces over D have the same dimension
if and only if they are isomorphic. In particular, Dn ∼
= Dm if and only if n = m.
CHAPTER 7. RINGS AND MODULES
275
We can now deduce an important fact about free modules over commutative rings.
Theorem 7.9.8. Let A be a commutative ring. Then An is isomorphic to Am if and
only if m = n. Thus, finitely generated free modules over a commutative ring have a well
defined rank, independent of the choice of an isomorphism to some An .
Proof Suppose that An is isomorphic to Am and let m be a maximal ideal of A. Then
An /mAn is isomorphic as a vector space over the field A/m to Am /mAm . But for k arbitrary, Corollary 7.7.37 provides an isomorphism of A/m-vector spaces between Ak /mAk
and (A/m)k . Thus, An /mAn and Am /mAm have dimensions n and m, respectively, as
A/m-vector spaces, and hence n = m.
The following lemma is useful in detecting the difference between finite and infinite
dimensional vector spaces.
Lemma 7.9.9. Let V be an infinite dimensional vector space over D. Then there are
linearly independent subsets of V of arbitrary length. By passage to the vector subspaces
that they generate, we see that V has subspaces of every finite dimension.
Proof We argue by contradiction. Suppose that the maximal length of a linearly
independent subset of V is k, and let v1 , . . . , vk be linearly independent. Let W be the
subspace of V generated by v1 , . . . , vk . Then W is finite dimensional, so there must be an
element vk+1 of V that is not in W . We claim that v1 , . . . , vk+1 is linearly independent,
contradicting the maximality of k.
To see this, suppose that x1 v1 + · · · + xk+1 vk+1 = 0, with x1 , . . . , xk+1 ∈ D. If
−1
xk+1 = 0, then vk+1 = (−x−1
k+1 x1 )v1 + · · · + (−xk+1 xk )vk . But this says that vk+1 ∈ W ,
contradicting the choice of vk+1 , so we must have xk+1 = 0. But this gives x1 v1 + · · · +
xk vk = 0. Since v1 , . . . , vk are linearly independent, we must have x1 = · · · = xk = 0,
and hence v1 , . . . , vk+1 are indeed linearly independent.
This now gives a quick proof of the following fact, which we already knew from the
fact that D is left Noetherian.
Corollary 7.9.10. Let V be a finite dimensional vector space. Then any subspace of V
is finite dimensional as well.
Proof If V ⊂ V , then any linearly independent subset of V is a linearly independent
subset of V . By Proposition 7.9.4, the maximal length of a linearly independent subset
of V is dim V . So V must be finite dimensional by Lemma 7.9.9.
We can now analyze the behavior of extensions with respect to dimension.
Proposition 7.9.11. Suppose given a short exact sequence
0→V −
→V −
→ V → 0
i
π
of vector spaces over D. Then V is finite dimensional if and only if both V and V are
finite dimensional, in which case dim V = dim V + dim V .
Proof Suppose V is finite dimensional. Since π : V → V is surjective, any set of
generators of V is carried onto a set of generators of V by π. Since i(V ) is a subspace
of V that is isomorphic to V , finite generation of V was just shown in Corollary 7.9.10.
Now suppose that V and V are finite dimensional. We identify V with the image
of i. Let v1 , . . . , vk be a basis for V and let z1 , . . . , zl be a basis for V . We claim that
CHAPTER 7. RINGS AND MODULES
276
if we’re given any choice of w1 , . . . , wl ∈ V such that π(wi ) = zi for 1 ≤ i ≤ k, then
v1 , . . . , vk , w1 , . . . , wl is a basis for V .
First, we show that v1 , . . . , vk , w1 , . . . , wl is linearly independent. Suppose given
x1 , . . . , xk+l ∈ D with x1 v1 + · · · + xk vk + xk+1 w1 + · · · + xk+l wl = 0. Then applying π, we see that xk+1 z1 + · · · + xk+l zl = 0. Since z1 , . . . , zl are linearly independent,
we must have xk+1 = · · · = xk+l = 0.
But then the original equation gives x1 v1 +· · ·+xk vk = 0. Since v1 , . . . , vk are linearly
independent, we must have x1 = · · · = xk = 0 as well.
The proof that v1 , . . . , vk , w1 , . . . , wl generate V works in an exact sequence of finitely
generated modules over any ring. Let v ∈ V . Then π(v) = xk+1 z1 + · · · + xk+l zl for some
l
l
xk+1 , . . . , xk+l ∈ D. But then π(v − i=1 xk+i wi ) = 0, and hence (v − i=1 xk+i wi ) ∈
ker π. The result now follows, since ker π = V is generated by v1 , . . . , vk .
We obtain a characterization of isomorphisms of finite dimensional vector spaces.
Corollary 7.9.12. Let V and W be finite dimensional vector spaces over D with the
same dimension. Let f : V → W be a D-linear map. Then the following conditions are
equivalent.
1. f is an isomorphism.
2. f is injective.
3. f is surjective.
Proof Clearly, an isomorphism is both injective and surjective. We shall show that if
f is either injective or surjective, then it is an isomorphism.
If f is injective, then we get an exact sequence
f
π
0→V −
→W −
→ C → 0,
where C is the cokernel of f . But then dim C = dim W − dim V = 0. But then C = 0,
as otherwise, C would have at least one linearly independent element in it.
If f is surjective, then a similar argument shows that the kernel of f is 0.
For infinite dimensional vector spaces, we can give an analogue of Proposition 7.9.2.
Proposition 7.9.13. Let V be an infinite dimensional vector space over D. Let {vi | i ∈
I} be linearly independent in V , and suppose given a family {vi | i ∈ J} such that V is
generated by {vi | i ∈ I ∪ J}. Then there is a subset, K, of I ∪ J containing I such that
{vi | i ∈ K} is a basis for V .
Proof Let
S = {K ⊂ I ∪ J | I ⊂ K and {vi | i ∈ K} is linearly independent } .
Then S is nonempty, and has a partial ordering given by inclusion of subsets. We claim
that S satisfies the hypothesis of Zorn’s Lemma.
=
Thus, let T = {Kα | α ∈ A} be a totally ordered subset of S and let K
α∈A Kα .
We claim that K is linearly independent, and hence K is an upper bound for T in S.
To see this, suppose that xi1 vi1 + · · · + xik vik = 0, where vi1 , . . . , vik are distinct
Then for each j, vi ∈ Kα for some αj ∈ A. But since T is totally
elements of K.
j
j
ordered, there is an r ∈ {1, . . . , k} such that Kαj ⊂ Kαr for j = 1, . . . , k. But then
CHAPTER 7. RINGS AND MODULES
277
vi1 , . . . , vik all lie in Kαr . Since Kαr is linearly independent, this forces xij = 0 for
is linearly independent as well.
j = 1, . . . , k. Thus, K
Thus, by Zorn’s Lemma, there is a maximal element, K, in S. We claim that {vi | i ∈
K} is a basis for V . To see this, it suffices to show that {vi | i ∈ K} generates V , which
in turn will follow if we show that each vi for i ∈ J is contained in the vector subspace
generated by {vi | i ∈ K}.
Thus, suppose that r is contained in the complement of K in I ∪ J, and let K =
K ∪{r}. Then {vi | i ∈ K } must be linearly dependent by the maximality of K. Suppose
given a relation of the form xi1 vi1 +· · ·+xik vik = 0, where vi1 , . . . , vik are distinct elements
of K , and xij is nonzero for j = 1, . . . , k.
Since {vi | i ∈ K} is linearly independent, then at least one of the ij must equal r,
−1
say r = ik . But then vr = −x−1
r xi1 vi1 − · · · − xr xik−1 vik−1 lies in the vector subspace
of V generated by {vi | i ∈ K}.
Taking the set {vj | j ∈ J} to be the set of nonzero elements of V , if necessary, we
see that every vector space over D has a basis.
Corollary 7.9.14. Every module over a division ring is free.
The following corollary is now immediate from Proposition 7.7.51.
Corollary 7.9.15. Every short exact sequence
0 → V → V → V → 0
of modules over a division ring splits.
7.10
Matrices and Transformations
Recall that an m × n matrix is one with m rows and n columns. We review the basic
definitions and properties of nonsquare matrices.
Definitions 7.10.1. Addition of nonsquare matrices is defined coordinate-wise: If M1 =
(aij ) and M2 = (bij ) are both m × n matrices with coefficients in A, then M1 + M2 is
the m × n matrix whose ij-th coefficient is aij + bij .
Multiplication of non-square matrices is defined analogously to that of square matrices, except that we may form the product M1 M2 whenever the number of columns of
M1 is equal to the number of rows of M2 . Thus, if M1 = (aij ) and M2 = (bij ) are m × n
and n × l
matrices, respectively, then we define M1 M2 to be the m × l matrix whose ij-th
n
entry is k=1 aik bkj for 1 ≤ i ≤ m and 1 ≤ j ≤ l.
Nonsquare matrices behave analogously to square ones. The proof of the next lemma
is left to the reader.
Lemma 7.10.2. Let A be a ring and let M1 , M2 , and M3 be m × n, n × k, and k × l
matrices over A, respectively. Then (M1 M2 )M3 = M1 (M2 M3 ).
In addition, the distributive laws hold: with the Mi as above, if M1 and M2 are
m × n and n × k matrices, respectively, then (M1 + M1 )M2 = M1 M2 + M1 M2 and
M1 (M2 + M2 ) = M1 M2 + M1 M2 .
Finally, for a ∈ A, let aIn ∈ Mn (A) be the matrix whose diagonal entries are all a
and whose off-diagonal entries are all 0. Then if M1 = (bij ) and M2 = (cij ) for M1
and M2 above, then M1 · aIn is the matrix whose ij-th entry is bij a, and aIn · M2 is the
matrix whose ij-th entry is acij .
CHAPTER 7. RINGS AND MODULES
278
The following corollary is now immediate.
Corollary 7.10.3. Right multiplication by the matrices aIn gives a right A-module structure to the set of m × n matrices over A, and left multiplication by aIm gives it a left
A-module structure.
In both cases the underlying abelian group structure of the module is given by matrix
addition, but the two module structures have no direct relationship to each other if A is
not commutative. (What we have instead is an example of an A-A-bimodule, as we shall
encounter in our discussions of tensor products in Section 9.4. We shall not make use of
bimodules here.)
Let us first establish notation:
Definitions 7.10.4. For m, n ≥ 1, we write Mm,n (A) for the set of m × n matrices over
A. We write eij ∈ Mm,n (A) for the matrix whose ij-th coordinate is 1 and whose other
coordinates are all 0, for 1 ≤ i ≤ m and 1 ≤ j ≤ n.
The next lemma generalizes a result we’ve seen for square matrices.
Lemma 7.10.5. Let A be a ring. Then as either a right or a left A-module, Mm,n (A)
is a free module with basis given by any ordering of {eij | 1 ≤ i ≤ m, 1 ≤ j ≤ n}.
In the study of classical linear algebra over a field, it is shown that if K is a field, then
the K-linear maps from K n to K m are in one-to-one correspondence with Mm,n (K) We
shall generalize this here to show that if A is a commutative ring, then the A-module
homomorphisms from An to Am are in one-to-one correspondence with Mm,n (A).
Thus, matrices correspond to transformations of free modules.
The situation for noncommutative rings is more complicated, but only by a very little.
The (usually) standard situation, where an m × n matrix acts from the left on an n × 1
column vector, gives the right A-module homomorphisms from An to Am . Left module
homomorphisms, on the other hand, are given by the right action of the n × m matrices
on the 1 × n row vectors. We now give formal treatments of these correspondences,
beginning with the case of right modules.
Thus, for r ≥ 1, identify Ar with the right A-module of r × 1 column matrices (i.e.,
column vectors) for all r ≥ 1. A typical element is
⎞
⎛
a1
⎟
⎜
x = ⎝ ... ⎠ .
ar
As shown above, the standard right module structure on Ar is given by setting xa equal
to the matrix product x · aI1 for a ∈ A. We write ei for the basis element ei1 of
Lemma 7.10.5, so that
⎛ ⎞
0
⎜ .. ⎟
⎜ . ⎟
⎜ ⎟
⎜ 0 ⎟
⎜ ⎟
⎟
ei = ⎜
⎜ 1 ⎟.
⎜ 0 ⎟
⎜ ⎟
⎜ . ⎟
⎝ .. ⎠
0
CHAPTER 7. RINGS AND MODULES
279
where
nthe 1 is in the i-th place, with 1 ≤ i ≤ n. Thus, with x as above, we have
x = i=1 ei ai .
Regarding An and Am are right A-modules, recall that HomA (An , Am ) denotes the
set of right A-module homomorphisms from An to Am . Recall also that HomA (An , Am )
is an abelian group under the operation that sets (f + g)(x) = f (x) + g(x) for f, g ∈
HomA (An , Am ) and x ∈ An .
Proposition 7.10.6. Let m, n ≥ 1. Then there is an isomorphism of abelian groups
∼
=
Φm,n : Mm,n (A) −→ HomA (An , Am ) defined as follows. For an m × n matrix M , we
define Φm,n (M ) : An → Am by setting (Φm,n (M ))(x) to be equal to the matrix product
M x for all x ∈ An .
Proof Φm,n (M ) is a group homomorphism from An to Am by the distributive law of
matrix multiplication, and is an A-module homomorphism by associativity: M · (xa) =
M (x · aI1 ) = (M x)aI1 = (M x)a. Thus, Φm,n : Mm,n (A) → HomA (An , Am ) is a well
defined function, and by the distributive law, it is a group homomorphism.
It is easy to see, by direct calculation of matrix products, that M ei is the i-th column
of M . Thus, if Φm,n (M ) and Φm,n (M ) have the same effect on the canonical basis
vectors, the matrices M and M must be equal. Thus, Φm,n is injective.
Let f ∈ HomA (An , Am ), and let M be the matrix whose i-th column is f (ei ) for
i = 1, . . . , n. Then Φm,n (M ) and f have the same effect on the canonical basis elements.
Since the canonical basis generates An as an A-module, Φm,n (M ) and f must have the
same effect on every element of An , and hence Φm,n (M ) = f . Thus, Φm,n is surjective.
As we shall see in Section 9.7, there is a natural left A-module structure on HomA (An , Am )
coming from the left action of A on Am . Similarly, there is a natural right A-module
structure on HomA (An , Am ) coming from the left action of A on An . As the interested
reader may check upon reading Section 9.7, Φm,n gives an A-module isomorphism from
the standard left module structure on matrices to the standard left module structure on
homomorphisms, as well as from the standard right module structure on matrices to the
standard right module structure on homomorphisms.
Recall from Problem 24 of Exercises 7.7.27 that HomA (An , An ) forms a ring, otherwise
denoted EndA (An ), called the endomorphism ring of An over A. The addition is the
same one we’ve just considered: (f + g)(x) = f (x) + g(x). Multiplication is given by
composition of functions: (f · g)(x) = f (g(x)).
Proposition 7.10.7. For n ≥ 1, Φn,n induces an isomorphism of rings from the matrix
ring Mn (A) to the endomorphism ring EndA (An ). (Here, An is considered as a free right
A-module.)
More generally, if M is an m × n matrix and M is an n × k matrix, then the
A-module homomorphism induced by the matrix product M M is the composite of the
homomorphism induced by M and the homomorphism induced by M :
Φm,k (M M ) = (Φm,n (M )) ◦ (Φn,k (M )) .
Proof We’ve seen that Φn,n is a group homomorphism. Moreover, the multiplicative
identity element In of Mn (A) is known to induce the identity transformation of An .
Thus, Φn,n is a ring homomorphism if and only if it respects multiplication, and hence
the first statement follows from the second. But the second statement is immediate from
the associativity of matrix multiplication:
(Φm,n (M ) ◦ Φn,k (M )) (x) = M (M x) = (M M )x = Φm,k (M M )(x).
CHAPTER 7. RINGS AND MODULES
280
Recall that a matrix is invertible if it is a unit in Mn (A), and that the group of units
there is Gln (A), the n-th general linear group of A. Note that M ∈ Gln (A) if and only
if there is a matrix M ∈ Mn (A) with M M = M M = In .
There is a module theoretic analogue of the general linear group.
Definition 7.10.8. If N is an A-module, we shall write AutA (N ), the group of A-module
automorphisms of N , for the unit group of EndA (N ).
Since composition is the ring product in EndA (An ) and the identity function is the
multiplicative identity element, an A-module automorphism is precisely an isomorphism
f : N → N of A-modules. Since Φn,n is an isomorphism of rings, we obtain the following
corollary.
Corollary 7.10.9. An n × n matrix M over A is invertible if and only if the right Amodule homomorphism that it induces (i.e., Φn,n (M ) : An → An ) is an isomorphism.
In particular, Φn,n restricts to an isomorphism
∼
=
Φn,n : Gln (A) −→ AutA (An ).
Recall from Lemma 7.7.36 that if f : N → N is an A-module homomorphism between
free modules and if x1 , . . . , xn is a basis for N , then f is an isomorphism if and only if
f (x1 ), . . . , f (xn ) is a basis for N . Since M ei is the i-th column of M , we obtain the
following corollary.
Corollary 7.10.10. An n × n matrix M over A is invertible if and only if its columns
form a basis of An .
It is useful to be able to do manipulations on the matrix level that match standard
constructions in module theory. One such construction is the block sum (sometimes
called Whitney sum13 ) of matrices: given an m × n matrix M and an m × n matrix
M , the block sum M ⊕ M of M and M is the matrix
M
0
,
0 M
where the 0’s are 0-matrices of the appropriate size. In coordinates, the ij-th coordinate
of M ⊕ M is given as follows. If 1 ≤ i ≤ m and 1 ≤ j ≤ n, then the ij-th coordinate
of M ⊕ M is equal to the ij-th coordinate of M . If 1 ≤ i ≤ m and j > n, or if i > m
and j ≤ n, then the ij-th coordinate of M ⊕ M is 0. If m + 1 ≤ i ≤ m + m and
n + 1 ≤ j ≤ n + n , then the ij-th coordinate of M ⊕ M is equal to the (i − m)(j − n)-th
coordinate of M .
The use of the symbol “⊕” for block sum is meant to be suggestive. In moduletheoretic terms, we use it as follows: If f1 : N1 → N1 and g : N2 → N2 are A-module
homomorphisms, we write f ⊕ g : N1 ⊕ N2 → N1 ⊕ N2 for the cartesian product of f and
g. Thus, (f ⊕ g)(n1 , n2 ) = (f (n1 ), g(n2 )).
13 After
the topologist Hassler Whitney.
CHAPTER 7. RINGS AND MODULES
281
Proposition 7.10.11. Let M be an m × n matrix and let M be an m × n matrix
over A. Then if we identify An ⊕ An with An+n in the standard way (i.e., identifying
n
n
n
n
( i=1 ei ai , i=1 ei ai ) with i=1 ei ai + i=1 ei+n ai ) and identify Am ⊕Am with Am+m
similarly, then the A-module homomorphism represented by the block sum M ⊕ M is
equal to the direct sum of the A-module homomorphisms represented by M and by M .
Symbolically, we have
Φm+m ,n+n (M ⊕ M ) = Φm,n (M ) ⊕ Φm n (M ).
Proof Both sides agree on the canonical basis elements of An+n .
So far we have only treated the module homomorphisms from An to Am , rather than
treating the module homomorphisms between arbitrary finitely generated free modules.
The reason is that there is no way to get from module homomorphisms to matrices that
doesn’t depend on making explicit choices of bases for the two modules: a different choice
of basis results in a different matrix to represent it.
Thus, we should think of An or Am in this discussion as a finitely generated free
module with a specific preferred basis. Of course, the point is that if N is a finitely
generated free module, then an ordered subset B = n1 , . . . , nk of N is a basis if and
only if the unique homomorphism fB : Ak → N with fB (ei ) = ni for 1 ≤ i ≤ k is an
isomorphism. Thus, the bases of N with k elements are in one-to-one correspondence
with the isomorphisms from Ak to N .
Definition 7.10.12. Suppose given finitely generated free right A-modules N and N ,
and let B = n1 , . . . , nk and B = n1 , . . . , nm be bases for N and N , respectively. Write
fB : Ak → N and fB : Am → N for the isomorphisms induced by B and B . Then
if g : N → N is an A-module homomorphism, we define the matrix of g with respect
to the bases
B to be the m × k matrix defined as follows. For 1 ≤ j ≤ k, write
m B and
g(nj ) = i=1 ni aij . Then the ij-th entry of the matrix of g is defined to be aij .
Note that the j-th column of the matrix of g with respect to the bases B and B is
precisely fB−1 g(nj ). Thus, the following lemma is immediate.
Lemma 7.10.13. The matrix of g : N → N with respect to the bases B and B is equal
−1
to the matrix of (fB−1 ◦ g ◦ fB ) : Ak → Am , i.e., it is the matrix Φ−1
m,k (fB ◦ g ◦ fB ).
Of particular interest is the use of matrices to study the A-module homomorphisms
from a free right A-module N to itself (i.e., to study elements of EndA (N )). Since the
source and target of these homomorphisms are the same, we need only one basis to define
a matrix.
Definition 7.10.14. Let B = x1 , . . . , xn be a basis for the free right A-module N and
let fB : An → N be the A-module homomorphism that takes ei to xi for each i. Let
g : N → N be an A-module homomorphism. Then the matrix of g with respect to the
−1
−1
basis B is defined to be Φ−1
n,n (fB ◦ g ◦ fB ), the matrix whose i-th column is fB (g(xi )).
The next result makes it clearer what is going on.
Lemma 7.10.15. Let B = x1 , . . . , xn be a basis for the free right A-module N and let
fB : An → N be the A-module homomorphism that takes ei to xi for each i. Then
there is a ring isomorphism fB∗ : EndA (N ) → EndA (An ) defined by fB∗ (g) = fB−1 ◦ g ◦ fB .
CHAPTER 7. RINGS AND MODULES
282
Moreover, the matrix of g with respect to the basis B is the image of g under the composite
ring isomorphism
f∗
Φ−1
n,n
B
EndA (An ) −−−→ Mn (A).
EndA (N ) −→
In particular, the passage from a transformation to its matrix with respect to B gives an
isomorphism of rings from EndA (N ) to Mn (A).
Proof The function fB∗ is easily seen to be a ring homomorphism. It is an isomorphism
because the correspondence that takes g ∈ EndA (An ) to fB ◦ g ◦ fB−1 gives an inverse for
it. The rest follows from the definition of the matrix of g with respect to B.
It is valuable to discern which properties of a transformation may be detected by its
matrix regardless of the basis used. Thus, it is important to understand what happens
to the matrix of g when we change the basis used to define it. Of course, as we’ve stated
above, the rank of a free module is not well defined over every ring A (though it is for
the ones that we care about, as we’ll see below). We shall not consider the base-change
problem between bases with a different number of elements, but shall restrict attention
to understanding what happens when we change between bases with the same number
of elements.
Definition 7.10.16. Suppose given bases B = x1 , . . . , xn and B = x1 , . . . , xn of the
free right A-module N and let fB : An → N and fB : An → N be the A-module
homomorphisms that take ei to xi and xi , respectively, for 1 ≤ i ≤ n. Then the base
change matrix MBB is defined to be the matrix of the identity map 1N : N → N , where
B is used as the basis for the domain of 1N , and B is used as the basis for the target
−1
copy of N . Notationally, we have MBB = Φ−1
n,n (fB fB ).
Lemma 7.10.17. Suppose given bases B, B , and B for the free right A-module N , all
of which have n elements. Then the base change matrices satisfy
MB B · MBB = MBB .
Additionally, MBB is the identity matrix, and hence MBB is invertible, with inverse
given by MB B .
Proof We first show that the displayed formula holds. By definition, it suffices to
show that Φn,n (MB B · MBB ) = fB−1 fB . But Φn,n is a ring homomorphism, so that
Φn,n (MB B · MBB ) = Φn,n (MB B ) ◦ Φn,n (MBB ) = fB−1 ◦ fB ◦ fB−1 ◦ fB , and hence
the formula holds.
Now MBB = Φ−1
n,n (1) = In , and hence MB B MBB = In by setting B = B in the
displayed equation. Exchanging the roles of B and B now gives the desired result.
We may now express the difference between the matrices of a given homomorphism
with respect to different bases.
Corollary 7.10.18. Suppose given bases B = x1 , . . . , xn and B = x1 , . . . , xn of the
free right A-module N and let MBB be the base change matrix defined above. Then if
g ∈ EndA (N ) and if M is the matrix of g with respect to the basis B, then the matrix of
−1
g with respect to B is the matrix product MBB M MBB
.
CHAPTER 7. RINGS AND MODULES
283
Proof Since Φn,n is a ring homomorphism, we have
−1
−1
−1
Φn,n (MBB M MBB
◦ g ◦ fB ) ◦ (fB−1 ◦ fB ).
) = (fB ◦ fB ) ◦ (fB
The result now follows.
Definition 7.10.19. We say that two n × n matrices, M, M ∈ Mn (A) are similar if
there is an invertible n × n matrix M such that M = M M (M )−1 .
Corollary 7.10.20. Let N be a finitely generated free right A-module and let g ∈ EndA (N ).
Suppose we’re given two bases of N with the same number of elements. Then the matrices
of g with respect to these two bases are similar.
Of course, we’d like to treat left modules as well. When A is commutative, the usual
thing is to identify left modules with right modules, and to study homomorphisms of
free left modules by the methods above. For noncommutative rings, this doesn’t work,
and hence further argumentation is needed. In this case, we identify the free left module
An with the 1 × n row matrices (i.e., row vectors). This has the advantage that the
notation agrees with the notation we’re used to for free modules: A typical element is
x = (a1 , . . . , an ).
Again we write ei = (0, . . . , 0, 1, 0, . . . , 0), where the 1 is in the i-th place. Here, the
key is that if M is an n × m matrix, then the matrix product ei M is precisely the i-th
row of M . Given this, and the fact that the left A-module structure on An is obtained
by setting ax equal to the matrix product (aI1 )x, the proof of the next proposition is
entirely analogous to that of Proposition 7.10.6, and is left to the reader.
Proposition 7.10.21. Write HomA (An , Am ) for the set of left A-module homomorphisms from An to Am . Then there is a group isomorphism
Ψn,m : Mn,m (A) → HomA (An , Am )
defined by setting (Ψn,m (M ))(x) equal to the matrix product xM .
In this case, there is a reversal between composition of functions and matrix multiplication: if M is n × m and M is k × n, then
(Ψn,m (M ) ◦ Ψk,n (M )) (x) = (xM )M = x(M M ) = Ψk,m (M M )(x).
We obtain the following proposition.
Proposition 7.10.22. For n ≥ 1, Ψn,n induces an isomorphism from the matrix ring
Mn (A) to the opposite ring of the endomorphism ring of the free left A-module An .
As in the case of right modules, the block sum of matrices represents the direct sum
of the associated A-module homomorphisms.
The treatment of matrices associated to different bases is analogous to that given for
right modules. We leave the details to the reader.
Exercises 7.10.23.
1. Let N be a free right A-module with basis B = x1 , . . . , xn . Let g ∈ EndA (N ) and
let M be the matrix of g with respect to the basis B. Let M be an n × n matrix
that is similar to M . Show that M is the matrix of g with respect to some other
basis of N .
CHAPTER 7. RINGS AND MODULES
2. Suppose given matrices
M1
M=
M3
M2
M4
284
and
M =
M1
M3
M2
M4
,
where M1 is m × n, M2 is m × s, M3 is r × n, and M4 is r × s, while M1 is n × k,
M2 is n × t, M3 is s × k, and M4 is s × t. Show that
M1 M1 + M2 M3 M1 M2 + M2 M4
.
MM =
M3 M1 + M4 M3 M3 M2 + M4 M4
3. Let M be an (n + k) × (n + k) matrix over A. Show that M has the form
M1 M 2
M=
0 M3
with M1 n × n, M2 n × k, and M3 k × k if and only if (Φn,n (M ))(An ) ⊂ An (i.e.,
the standard right A-module homomorphism induced by M carries An into itself).
Here An is the submodule of An+k generated by e1 , . . . , en . Show also that the
matrices of the above form (where n is fixed) give a subring of Mn+k (A).
4. Let B = Mn (A). Show that the matrix ring Mk (B) is isomorphic as a ring to
Mnk (A).
‡ 5. If A is commutative, then, since left and right A-modules may be identified with
each other, Ψ−1
n,n ◦ Φn,n provides an isomorphism from Mn (A) to its opposite ring
for each n ≥ 1. Explicitly write down the effect of Ψ−1
n,n ◦ Φn,n on a matrix (aij ).
7.11
Rings of Fractions
Think for a moment about how our number system evolved. First, there were the nonnegative integers. Then negative numbers were adjoined, producing the integers. This
process of adjoining inverses has since been generalized to a construction that produces
an abelian group from any abelian monoid. It is called the Grothendieck construction,
and will be studied below.
The next stage in the development of our number system was the creation of the
rational numbers, by adjoining multiplicative inverses to the nonzero integers. We shall
generalize this last process in this section. One of the consequences will be that we can
then embed any integral domain in a field.
Definition 7.11.1. Let A be a commutative ring. A multiplicative subset of A is a
subset S ⊂ A such that 0 ∈ S and such that S is closed under multiplication: For
s, t ∈ S, the product st is also in S.14
Note: A ring called “A” in this section will be implicitly assumed to be commutative.
Other rings, however, will not.
Let S be a multiplicative subset of the commutative ring A. We shall construct a
new ring S −1 A, in which the elements of S are invertible.
14 Some people permit 0 to be an element of a multiplicative subset. Indeed, the entire theory of rings
of fractions generalizes to that context, but, if 0 ∈ S, then the ring of fractions S −1 A is the 0 ring. So
all we’re doing is excluding this rather boring case.
CHAPTER 7. RINGS AND MODULES
285
The elements of S −1 A will be equivalence classes of ordered pairs (a, s) with a ∈ A
and s ∈ S. We say that (a, s) is equivalent to (b, t) (written (a, s) ∼ (b, t)) if there is an
s ∈ S such that s (ta − sb) = 0. Notice that if A is an integral domain, then this is just
the familiar condition we use to test when two fractions a/s and b/t are equal in Q.
Lemma 7.11.2. Let S be a multiplicative subset of the commutative ring A. Then the
above relation ∼ is an equivalence relation.
Proof The relation is obviously symmetric and reflexive, so it suffices to show transitivity. Suppose that (a, s) ∼ (b, t) and that (b, t) ∼ (c, u). Then there are elements
s , s ∈ S with s (ta − sb) = 0 and s (ub − tc) = 0. We claim now that s s t(ua − sc) = 0,
and hence (a, s) ∼ (c, u).
Now s s tua = s s usb, because s (ta − sb) = 0. Similarly, s s usb = s s stc, because
s (ub − tc) = 0. Putting this together, we see that s s t(ua − sc) = 0 as desired.
Definition 7.11.3. Let S be a multiplicative subset of the commutative ring A. We
define the ring of fractions S −1 A as the set of equivalence classes of pairs (a, s), and,
to emphasize the analogy with fractions, we write a/s, or as , for the element of S −1 A
represented by (a, s).
The addition and multiplication operations in S −1 A are given by
ta + sb
a b
+ =
s
t
st
and
a b
ab
· = .
s t
st
We define the canonical map η : A → S −1 A by setting η(a) = as/s for any element
s ∈ S.
The proof of the following proposition is tedious but straightforward. We leave it to
the reader.
Proposition 7.11.4. Let S be a multiplicative subset of A. Then S −1 A is a commutative ring under the above operations. The additive identity is 0/s for any s ∈ S, and
multiplicative identity is s/s for any s ∈ S.
The canonical map η : A → S −1 A is a ring homomorphism whose kernel is the set of
all a ∈ A that are annihilated by some s ∈ S. For s ∈ S, the element η(s) is a unit in
S −1 A, with inverse 1/s. The generic element of S −1 A may then be characterized by
a
= η(a) · (η(s))−1 .
s
The reader may very well ask: Why don’t we insist that 1 ∈ S? The answer, it turns
out, is that there isn’t a good reason other than the fact that it isn’t necessary. We’ll see
in Exercises 7.11.27 that if 1 ∈ S, and if we take S = S ∪ {1}, then the natural inclusion
−1
of S −1 A in S A is an isomorphism.
The principal examples of rings of fractions are important enough to warrant names.
Definitions 7.11.5.
1. Let A be an integral domain and let S be the set of all nonzero
elements in A. The resulting ring S −1 A is called the field of fractions, A(0) , of A.
2. Let A be any commutative ring and let p be a prime ideal of A. For
S = {a ∈ A | a ∈ p}
and call it the localization of A at p.
we write
S −1 A = Ap ,
CHAPTER 7. RINGS AND MODULES
286
3. Let A be any commutative ring and let a ∈ A be any element that is not nilpotent.
Let S = {ak | k ≥ 0}. We write
S −1 A = A [1/a] ,
and call it the ring obtained by adjoining an inverse to a.
It is easy to see that Q is the field of fractions of Z, so that we have, as promised,
generalized the construction of the rationals from the integers. Note that the field of
fractions of an integral domain A really is a field, as η(a)(η(s))−1 is a unit in A(0)
whenever a = 0. We obtain the following corollary.
Corollary 7.11.6. The field of fractions, A(0) , of an integral domain A is a field, and
η : A → A(0) an embedding. Thus, a ring is an integral domain if and only if it is a
subring of a field.
The following examples are important in field theory and number theory, respectively.
Examples 7.11.7.
1. Let K be a field and let X1 , . . . , Xn be variables. We write K(X1 , . . . , Xn ) for
the field of fractions of the polynomial ring K[X1 , . . . , Xn ]. We call it the field of
rational functions over K in the variables X1 , . . . , Xn .
p form an
2. Recall from Problem 1 of Exercises 7.1.41 that the p-adic integers Z
integral domain. We write Qp for its field of fractions, known as the field of p-adic
rationals.
Note that a field of fractions is a special case of localization, as the ideal (0) is prime
in an integral domain.
If p ∈ Z is prime, then the ring Z [1/p] is sometimes called the “localization of Z away
from p.” As we shall see below, this is a definite abuse of the word localization.
Lemma 7.11.8. Let B be a ring and let a and b be elements of B that commute with
each other, where b ∈ B × . Then a commutes with b−1 .
Proof Clearly, b annihilates ab−1 − b−1 a. Since b is a unit, the result follows.
We return to the general case of a multiplicative subset of a commutative ring A.
The ring of fractions S −1 A satisfies a useful universal property.
Proposition 7.11.9. Let f : A → B be a ring homomorphism such that f (s) is a unit
for all s ∈ S. Then there is a unique ring homomorphism f : S −1 A → B such that the
following diagram commutes.
f
/ B
A?
?
??

??



η ??
 f
S −1 A
Moreover, if f gives an A-algebra structure on B, then f , if it exists, induces an
S −1 A-algebra structure on B. We obtain a one-to-one correspondence between the S −1 Aalgebra structures on B and the A-algebra structures f : A → B with the property that
f (S) ⊂ B × . And if g : B → C is an A-algebra homomorphism between S −1 A-algebras,
then it is also an S −1 A-algebra homomorphism.
CHAPTER 7. RINGS AND MODULES
287
Proof Suppose there is a ring homomorphism f that makes the diagram commute.
Since η(s) is a unit in S −1 A for each s ∈ S, f (s) = f (η(s)) must be a unit in B.
Moreover, since a/s = η(a)(η(s))−1 , f must satisfy the formula f (a/s) = f (a)f (s)−1 .
Thus, f is uniquely defined by the commutativity of the diagram.
Conversely, if f (s) is a unit for each s ∈ S, we define f by the same formula: f (a/s) =
f (a)f (s)−1 . This is well defined, since if s (ta − sb) = 0, we have f (ta − sb) = 0 by the
invertibility of f (s ). Making use of Lemma 7.11.8, if B is noncommutative, it is easy to
see that f is a ring homomorphism.
The statements about existence and uniqueness of algebra structures are straightforward. To see that every A-algebra homomorphism f : B → C between S −1 A-algebras is
an S −1 A-algebra homomorphism, write ν and μ for the S −1 A-algebra structures on B
and C, respectively. We want to show that f ◦ ν = μ.
Our assumption is that f is an A-algebra homomorphism, which means that f ◦ ν
agrees with μ if we precompose with η : A → S −1 A. But two ring homomorphisms out
of S −1 A that agree when precomposed with η must be equal.
Now Q is the field of fractions of Z. Recall that for any ring B, there is a unique ring
homomorphism f : Z → B, where f (n) = n · 1, the n-th “power” of 1 in the additive
group structure on B. The homomorphism f gives B a unique Z-algebra structure, and
B has characteristic 0 if and only if f is injective. We obtain the following corollary.
Corollary 7.11.10. A ring B is a Q-algebra if and only if n · 1 is a unit in B for
each 0 = n ∈ Z. The Q-algebra structure on B, if it exists, is unique, and any ring
homomorphism between Q-algebras is a Q-algebra homomorphism.
In particular, there is a unique embedding (as a ring) of Q in any field of characteristic
0.
We also have the following.
Corollary 7.11.11. Let f : A → K be an embedding of an integral domain A into a
field K. Then there is a unique extension of f to an embedding f : A(0) → K, where A(0)
is the field of fractions of A. The image of f is the smallest subfield of K containing the
image of f .
We now have a framework to classify the subrings of Q.
Proposition 7.11.12. Let A be a subring of Q and let S = A× ∩ Z. Then A is isomorphic as a ring to S −1 Z.
Proof The elements of S become units in A. Thus, Proposition 7.11.9 provides a unique
ring homomorphism f : S −1 Z → A. We claim that f is an isomorphism.
f
First, note that the composite S −1 Z −
→ A ⊂ Q takes m/s to m/s. Since the equivalence relations defining S −1 Z and Q are the same (because Z is an integral domain), the
above composite is injective, and hence so is f .
Now suppose given an element of A. Since A ⊂ Q, we may write it as m/n for
m, n ∈ Z with n = 0. By reducing the fraction, we may assume that m and n are
relatively prime. It suffices to show that n is a unit in A, i.e., that the element 1/n of Q
lies in A.
Since m and n are relatively prime, we can find a, b ∈ Z with am + bn = 1. Dividing
both sides by n, we see that a(m/n) + b = 1/n. Since a, m/n and b are in A, so is 1/n.
CHAPTER 7. RINGS AND MODULES
288
The universal property for homomorphisms out of rings of fractions gives the simplest
proof of the following characterization of modules over these rings. The situation is
similar to that given by Proposition 7.7.15.
Because S −1 A is an A-algebra, every S −1 A-module has an underlying A-module
structure. So an immediate question is: Which A-modules admit a compatible S −1 Amodule structure? And if a compatible S −1 A-module structure exists, is it unique?
(Compatibility here means that the A-module structure induced by the S −1 A-module
structure is the same as the one we started with.) A good answer will give us a complete
understanding of S −1 A-modules.
Proposition 7.11.13. Let M be an A-module. Then M admits a compatible S −1 A module structure if and only if multiplication by s induces an isomorphism (of abelian groups)
from M to M for each s ∈ S. Moreover, there is at most one S −1 A-module structure
compatible with the original A-module structure on M . Finally, if the A-modules M and
N admit compatible S −1 A-module structures, then every A-module homomorphism from
M to N is an S −1 A-module homomorphism as well.
Proof We use the correspondence given by Problem 21 of Exercises 7.7.27 between
A-module structures on M and ring homomorphisms from A to EndZ (M ). Note that
if f : A → EndZ (M ) and f : S −1 A → EndZ (M ) are ring homomorphisms, then the
S −1 A-module structure on M induced by f is compatible with the A-module structure
coming from f if and only if the diagram
f
/ EndZ (M )
A JJ
JJ
t:
JJ
ttt
J
t
J
t
η
J$
tt f
−1
S A
commutes.
For a given homomorphism f : A → EndZ (M ), the homomorphisms f : S −1 A →
EndZ (M ) that make the diagram commute are characterized by Proposition 7.11.9: For
a given f there is at most one f with this property, and such an f exists if and only if
f (s) is a unit in EndZ (M ) (i.e., a group isomorphism of M ) for each s ∈ S.
It remains to show that if f : M → N is an A-module homomorphism between
the S −1 A-modules M and N , then f is in fact an S −1 A-module homomorphism. In
particular, it suffices to show that f ((1/s)m) = (1/s)f (m) for s ∈ S and m ∈ M .
Write s : M → M and s : N → N for the homomorphisms induced by multiplication
by s. Then the statement that f ((1/s)m) = (1/s)f (m) for all m ∈ M amounts to the
−1
statement that f ◦ −1
s = s ◦ f . Now, since f is an A-module homomorphism, we have
f ◦ s = s ◦ f . So the result now follows as in the proof of Lemma 7.11.8.
We’d like now to study ideals in rings of fractions. Indeed, the properties of the ideals
in a localization form one of the primary motivations for studying rings of fractions. Given
the relationship between ideals and modules, it seems reasonable to discuss modules of
fractions first.
Let M be an A-module and let S be a multiplicative subset of A. We shall define
an S −1 A-module S −1 M as follows. The elements of S −1 M are equivalence classes of
ordered pairs (m, s) with m ∈ M and s ∈ S, where (m, s) is equivalent to (n, t) (written
(m, s) ∼ (n, t)) if there is an s ∈ S with s (tm − sn) = 0. As expected, we write m/s for
the element of S −1 M represented by the ordered pair (m, s). The proof of the following
lemma is straightforward and is left to the reader.
CHAPTER 7. RINGS AND MODULES
289
Lemma 7.11.14. The relation ∼ above is an equivalence relation. The resulting set
S −1 M of equivalence classes is an abelian group and an S −1 A-module via
m n
tm + sn
+ =
s
t
st
a m
am
·
=
,
s t
st
and
respectively. There is a canonical A-module homomorphism η : M → S −1 M given by
η(m) = ms/s for any s ∈ S.
If N is an S −1 A-module and f : M → N is an A-module homomorphism, then
there is a unique S −1 A-module homomorphism f : S −1 M → N that makes the following
diagram commute.
f
/ N
M?
?
??

??



η ???
 f
S −1 M
We obtain an isomorphism
η∗
HomS −1 A (S −1 M , N ) −→ HomA (M, N )
∼
=
via η ∗ (f ) = f ◦ η.
Localization forms an important special case of modules of fractions:
Definition 7.11.15. Let p be a prime ideal of A and let S be the complement of p in
A. Then we write Mp for S −1 M . We call it the localization of M at p.
Let S be an arbitrary multiplicative subset of A and let f : M → N be an A-module
homomorphism. Let S −1 f : S −1 M → S −1 N be defined by (S −1 f )(m/s) = f (m)/s.
Then S −1 f is clearly the unique S −1 A-module homomorphism that makes the following
diagram commute.
f
M
η
/ N
η
S
S −1 M
−1
f/
S −1 N
Clearly, passage from M to S −1 M and from f to S −1 f provides a functor from Amodules to S −1 A-modules (i.e., S −1 f ◦S −1 g = S −1 (f ◦g), and S −1 (1M ) = 1S −1 M , where
1N denotes the identity map of the module N ).
Proposition 7.11.16. Suppose given an exact sequence
f
g
· · · → M −
→M −
→ M → · · ·
of A-modules. Then the induced sequence
S −1 f
S −1 g
· · · → S −1 M −−−→ S −1 M −−−→ S −1 M → · · ·
of S −1 A-modules is also exact.
CHAPTER 7. RINGS AND MODULES
290
Proof A sequence is exact if it is exact at every module in it. We show exactness at M .
First, note that S −1 g ◦ S −1 f = S −1 (g ◦ f ) = S −1 0, since g ◦ f is the 0-homomorphism 0
(since im f ⊂ ker g). But S −1 0 = 0, so im(S −1 f ) ⊂ ker(S −1 g).
Now let m/s ∈ ker S −1 g. Then a glance at the equivalence relation defining S −1 M shows that tg(m) = 0 for some t ∈ S. But then g(tm) = tg(m) = 0, so tm ∈ ker g = im f .
−1
f is onto.
Let tm = f (m ). Then m/s = (S −1 f )( m
st ). Thus, S
This behavior may be abstracted as follows.
Definition 7.11.17. A functor which takes exact sequences to exact sequences is called
an exact functor.
In particular, the passage from M to S −1 M is an exact functor. This has a number
of uses, since certain kinds of relationships can be described by exactness properties. For
⊂
instance, N is a submodule of M if and only if the sequence 0 → N −→ M is exact. We
immediately obtain the following corollary.
Corollary 7.11.18. Let S be a multiplicative subset of A and let M be an A-module.
Then if N is a submodule of M , the natural map from S −1 N to S −1 M is an inclusion
of S −1 A-modules.
Thus, if a is an ideal of A, then S −1 a is an ideal of S −1 A. We shall show that every
ideal of S −1 A arises this way:
Proposition 7.11.19. Let S be a multiplicative subset of A and let b be an ideal of
S −1 A. Let η : A → S −1 A be the canonical homomorphism and let a = η −1 b. Then
b = S −1 a.
Proof Let x ∈ a. Then xs/s ∈ b for any s ∈ S, and hence so is x/t = (xs)/(ts) for any
t ∈ S, since b is an ideal of S −1 A. Thus, S −1 a ⊂ b.
Suppose given an arbitrary element x/s ∈ b with x ∈ A and s ∈ S. It suffices to show
that x ∈ a. But this follows, since η(s) · (x/s) = η(x) (i.e., η(x) ∈ b).
Corollary 7.11.20. Let S be a multiplicative subset of A. Then the prime ideals of
S −1 A are those of the form S −1 p, where p is a prime ideal of A such that p ∩ S = ∅.
Proof Since the preimage under a ring homomorphism of a prime ideal is prime, every
prime ideal of S −1 A has the form S −1 p for some prime ideal p of A. Also, since S −1 p
must be a proper ideal of S −1 A in order to be prime, S −1 p must not contain a unit of
S −1 A, and hence p ∩ S must be empty. Thus, it suffices to show that if p is any prime
ideal of A that doesn’t meet S, then S −1 p is prime in S −1 A.
Suppose given such a p, and suppose that (a/s) · (b/t) ∈ S −1 p. Say (ab)/(st) = x/s ,
with x ∈ p and s ∈ S. This says there is an s ∈ S with s (s ab − stx) = 0. But then
s s ab ∈ p. Since s and s are not in p, one of a and b must be. Thus, one of a/s and
b/t is in S −1 p.
Thus, it suffices to show that S −1 p is proper. Suppose that x/s = 1 in S −1 A,
with x ∈ p and s ∈ S. Since 1 = s /s for any s ∈ S, we can find s , s ∈ S with
s (s x − s s) = 0. But this places s s s in p, which is impossible, and hence S −1 p cannot
be all of S −1 A.
Of course, if S −1 p and S −1 p are prime ideals of S −1 A, then S −1 p ⊂ S −1 p if and
only if p ⊂ p, by Proposition 7.11.19.
We now are in a position to explain the term “localization.”
CHAPTER 7. RINGS AND MODULES
291
Definition 7.11.21. A local ring is a commutative ring that has only one maximal ideal.
Fields, of course, are local rings, as are localizations, as we shall see shortly. Another
p.
example of a local ring that we’ve seen is Z
Example 7.11.22. Recall from Proposition 7.1.40 that the ring homomorphism
p → Zp
π1 : Z
p is a unit if and only if π1 (a) is nonzero in Zp . Thus, any
has the property that a ∈ Z
p
p which doesn’t lie in ker π1 is a unit. Therefore, any proper ideal of Z
element of Z
must be contained in ker π1 , and hence ker π1 is the unique maximal ideal of Zp .
Note that Problem 4 of Exercises 7.1.41 shows that ker π1 = (p), the principal ideal
p is a local ring with maximal ideal (p).
generated by p. Thus, Z
Proposition 7.11.23. Let p be a prime ideal in the commutative ring A. Then the
localization Ap is a local ring whose maximal ideal is pp .
Proof Let S be the complement of p in A. Then p ∩ S = ∅, so that pp = S −1 p is a
prime ideal of Ap . Since maximal ideals are prime, it suffices to show that any other
prime ideal of Ap must be contained in pp .
But the prime ideals of Ap all have the form pp , where p ∩ S = ∅. By the definition
of S, this forces p ⊂ p, and the result follows.
This, of course, begs the question of the relationship of the field Ap /pp to A itself.
Proposition 7.11.24. Let p be a prime ideal of A. Then Ap /pp is isomorphic to the
field of fractions of A/p.
In particular, if m is a maximal ideal of A, then Am /mm is just A/m.
Proof We have a commutative diagram
A
π A/p
η
η
/ Ap
π / Ap /pp
where π and π are the canonical maps. It is easy to check that p = η −1 (pp ), and hence
η is an embedding.
By Corollary 7.11.11, η extends uniquely to an embedding, η : K → Ap /pp of the
field of fractions, K, of A/p. But η is surjective, as π (a/s) = η(π(a)/π(s)).
If p is maximal, then A/p is a field, and hence A/p = K by Corollary 7.11.11.
Localization turns out to be a very important technique in understanding A-modules.
The point is that modules over a local ring are easier to classify than those over a non-local
ring. This will become especially clear in our study of projective modules in Section 9.8.
So it is valuable to be able to deduce information about a module M from information
about the localizations Mp .
Proposition 7.11.25. Let M be a module over the commutative ring A. Then the
following conditions are equivalent.
CHAPTER 7. RINGS AND MODULES
292
1. M = 0.
2. Mp = 0 for each prime ideal p of A.
3. Mm = 0 for each maximal ideal m of A.
Proof Clearly, the first condition implies the second, which implies the third. We shall
show that the third implies the first. Let M be an A-module such that Mm = 0 for each
maximal ideal m of A. Let m ∈ M . It suffices to show that the annihilator of m is A.
Thus, let m be a maximal ideal of A. Then m is in the kernel of the canonical map
from M to Mm . Thus, m/1 = 0/1 in Mm , so there exists s in A which is not in m such
that sm = 0. In other words, Ann(m) is not contained in m. But since Mm = 0 for
every maximal ideal m of A, we see that Ann(m) is not contained in any of the maximal
ideals of A. Since Ann(m) is an ideal, it must be all of A.
Corollary 7.11.26. Let f : M → N be an A-module homomorphism. Then the following conditions are equivalent.
1. f is injective.
2. fp is injective for each prime ideal p of A.
3. fm is injective for each maximal ideal m of A.
The same holds if we replace injective by surjective.
Proof Consider the exact sequence
i
f
π
→M −
→C→0
→N −
0→K−
of Lemma 7.7.47. Since localization is an exact functor, we see that Kp is the kernel
of fp and Cp is the cokernel of fp for each prime ideal p of A. Thus, the result follows
immediately from Proposition 7.11.25.
Exercises 7.11.27.
1. Show that Zn is a local ring if and only if n is a prime power.
2. Show that the localization, Z(p) , of Z at (p) embeds uniquely as a subring of the
p.
p-adic integers Z
3. Let a be any non-nilpotent element of the commutative ring A. Show that A[1/a]
is isomorphic as an A-algebra to A[X]/(aX − 1), where (aX − 1) is the principal
ideal generated by aX − 1 in the polynomial ring A[X].
4. Let A be an integral domain and let K be its field of fractions. Show that if
0 = a ∈ A, then the ring of fractions A[1/a] may be identified with the image of
the evaluation map ε1/a : A[X] → K obtained by evaluating X at 1/a. Thus, the
notation A[1/a] is consistent with that for adjoining an element to A.
5. Let A be a P.I.D. and let S be a multiplicative subset of A. Show that the ring of
fractions S −1 A is also a P.I.D.
CHAPTER 7. RINGS AND MODULES
293
‡ 6. Let S be a multiplicative subset of A with 1 ∈ S. Suppose that S is generated
as a monoid by a finite set s1 , . . . , sk , i.e., an element is in S if and only if it has
the form sr11 . . . srkk where ri ≥ 0 for 1 ≤ i ≤ k. Show that S −1 A is isomorphic
as an A-algebra to A[X1 , . . . , Xk ]/(s1 X1 − 1, . . . , sk Xk − 1). Here, X1 , . . . , Xk are
indeterminates over A, and (s1 X1 − 1, . . . , sk Xk − 1) is the ideal generated by
s1 X1 − 1, . . . , sk Xk − 1. It is common to write A[1/s1 , . . . , 1/sk ] for S −1 A in this
context, a usage which may be of assistance in designing the proof.
7. Let A be a commutative ring and let S ⊂ A[X] be given by S = {X k | k ≥ 0}.
Show that S −1 A[X] is isomorphic to the group ring A[Z] of Z over A.
8. We say that a multiplicative subset S ⊂ A is saturated if for any s ∈ S and any
a ∈ A dividing s, we must have a ∈ S. Show that if S is any multiplicative subset,
then T = {a ∈ A | a divides s for some s ∈ S} is a multiplicative subset. Show also
that T is saturated. We call T the saturation of S.
† 9. Let S be a multiplicative subset of A and let T be its saturation. Show that for
−1
any multiplicative subset S with S ⊂ S ⊂ T , the natural maps S −1 A → S A →
−1
T A are isomorphisms of rings. Show also that if M is an A-module, then the
−1
natural maps S −1 M → S M → T −1 M are isomorphisms of S −1 A-modules.
10. To complete the classification of the subrings of Q, it suffices to find all of the saturated multiplicative subsets of Z. Show that there is a one-to-one correspondence
between the saturated multiplicative subsets of Z and the subsets of the set of all
primes in Z. Here, if T is a set of primes in Z, then the associated multiplicative
subset is given by
S = {±pr11 . . . prkk | k ≥ 0, p1 , . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}.
11. Let A be an integral domain and let B be any commutative ring. Show that the
ideal p = (0 × B) of A × B is prime and that the localization (A × B)p is isomorphic
to the field of fractions of A.
12. Let p be a prime number and let M be a finite abelian group. Show that the localization M(p) is isomorphic to the subgroup Mp of M consisting of those elements
whose order is a power of p. (Hint: Consider the restriction of the natural map
η : M → M(p) to Mp .)
13. Let S be a multiplicative subset of Z and let M be a finite abelian group. Show
that S −1 M is isomorphic to the subgroup of M consisting of those elements whose
order is relatively prime to every element in S.
14. Let S be a multiplicative subset of A. Show that if S ⊂ A× , then the natural map
η : A → S −1 A is an isomorphism.
15. Let S be a multiplicative subset of A and let M be an A-module that admits a
compatible S −1 A-module structure. Show that the natural map η : M → S −1 M
is an isomorphism.
16. Let F be a functor from A-modules to B-modules. Show that F is an exact functor
if and only if for any short exact sequence
f
g
→M −
→ M → 0
0 → M −
CHAPTER 7. RINGS AND MODULES
294
of A-modules, the induced sequence
F (f )
F (g)
0 → F (M ) −−−→ F (M ) −−−→ F (M ) → 0
of B-modules is also exact.
Chapter 8
P.I.D.s and Field Extensions
The theories of fields and of P.I.D.s are inextricably intertwined. If K is a field, then
the polynomial ring K[X] is a P.I.D. We shall show in Section 8.1 that the elements of
a P.I.D. have prime decompositions similar to the ones in the integers. This is used to
study the extension fields L of K, via the evaluation maps εα : K[X] → L obtained by
evaluating X at an element α of L.
Also, we give a classification of the finitely generated modules over a P.I.D. in Section 8.9. We shall use this to classify the n × n matrices over a field K up to similarity
in Section 10.6. The point is that an n × n matrix induces a K[X]-module structure on
K n in which X acts on K n by multiplication by the matrix in question.
The theory of prime factorization in P.I.D.s also gives some elementary examples of
the behavior of primes under finite extensions of the rational numbers, Q. The point is
that if K is a finite extension field of Q, then there is a subring O(K), called the ring
of integers of K, that plays a role in K similar to the role of the integers in Q. An
important topic in number theory is the way in which the prime numbers in Z factor in
O(K). In the general case, O(K) is a Dedekind domain, but not a P.I.D. In that case,
we study the factorizations of the principal ideal (p) of O(K) as a product of ideals.
But in some nice cases, which we shall illustrate in the exercises to Section 8.1, the ring
of integers is a P.I.D., and we may study this factorization in terms of factorizations of
prime elements.
The theory of prime factorization is developed in Section 8.1 in a generalization of
P.I.D.s called unique factorization domains, or U.F.D.s.
In Section 8.2, we study the algebraic extensions of a field K. These are the extension
fields with the property that the evaluation maps εα : K[X] → L have nontrivial kernels
for each α ∈ L. They will be the primary objects of study in Chapter 11. We begin to
analyze them here.
Section 8.3 studies extension fields which are not algebraic. There is a notion of
transcendence degree, which measures how far a given extension is from being algebraic.
There is a theory of transcendence bases, which behave toward the transcendence degree
in a manner analogous to the relationship between a basis for a vector space and its
dimension.
In Section 8.4, we construct an algebraic closure for a field K. Uniqueness statements
regarding algebraic closures are deferred to Section 11.1. We shall see that every algebraic
extension of a field K may be embedded in its algebraic closure. So algebraic closures
will be an important tool in the material on Galois theory.
Algebraic closures are also important in matrix theory, as the Jordan canonical form
295
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
296
may be defined for n × n matrices over an algebraically closed field. Thus, we may study
matrices over K by passage to the matrix ring of its algebraic closure.
The various applications of prime factorization in the P.I.D. K[X] depend strongly
on our ability to actually compute these factorizations. In particular, we need to be able
to test polynomials for primality. In Section 8.5, we give criteria for primality in Q[X],
or, more generally, in K[X], where K is the field of fractions of a unique factorization
domain. In the process, we show that if A is a U.F.D., then so is A[X]. In particular,
the polynomial ring K[X1 , . . . , Xn ] is a U.F.D. for any field K.
In Section 9.3, we make use of unique factorization in K[X1 , . . . , Xn ], the study of
transcendence bases, and the theory of Noetherian rings to prove Hilbert’s Nullstellensatz, which states that if the extension field L of K is finitely generated as a K-algebra,
then L is a finite extension of K. In consequence, we determine all of the maximal
ideals of K[X1 , . . . , Xn ] when K is an algebraically closed field. This is the starting
point for the study of algebraic varieties. We proceed to define algebraic varieties and
to describe the Zariski topology on affine n-space, K n , as well as on the prime spectrum
of a commutative ring. This material forms an introduction to the study of algebraic
geometry.
The next three sections study field extensions, culminating in the determination of
the degree of the cyclotomic extension Q[ζn ] over Q, as well as the determination of the
minimal polynomial of ζn over Q. This establishes the most basic facts about a family
of examples that has vital connections to a number of branches of mathematics. For
instance, we shall see that the cyclotomic rationals play a vital role in computations
of Galois theory. And their rings of integers, Z[ζn ], are of crucial importance in issues
related to Fermat’s Last Theorem, as well as to calculations of the K-theory of integral
group rings. The latter plays an important role in topology.
Section 8.6 introduces the Frobenius homomorphism and discusses perfect fields. Section 8.7 introduces the notion of repeated roots, which forms the basis for the study of
separable extensions. We shall return to these ideas in Chapter 11.
8.1
Euclidean Rings, P.I.D.s, and U.F.D.s
Recall that the Euclidean algorithm was the main tool we used to analyze the structure of
the integers. There is an analogue of it which holds in a variety of other rings, including
the polynomial ring K[X] over a field K. The consequences, for rings that satisfy this
property, are very similar to the consequences that hold in Z.
Definition 8.1.1. A Euclidean domain is an integral domain A that admits a function
ϕ : A → Z with the following properties.
1. If b divides1 a and a = 0, then ϕ(b) ≤ ϕ(a).
2. For any a, b ∈ A with b = 0, there are elements q, r ∈ A such that a = qb + r and
ϕ(r) < ϕ(b).
We call the function ϕ : A → Z a Euclidean structure map for A.
1 I.e.,
b.
a = bc for some c ∈ A. Once again, this just says that a ∈ (b), the principal ideal generated by
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
297
Examples 8.1.2.
1. Z is Euclidean, with the structure map given by the absolute value function | | :
Z → Z.
2. Let K be a field, and define ϕ : K[X] → Z by setting ϕ(f (X)) equal to the degree
of f (X) if f (X) = 0, and setting ϕ(0) = −1. Then Proposition 7.3.10 shows that
ϕ puts a Euclidean structure on K[X].
3. In Problem 5 of Exercises 7.1.41, it is shown that every nonzero element of the
p may be written as a product pk u with k ≥ 0 and u ∈ Z
× . Define
p-adic integers Z
p
p → Z by setting ϕ(pk u) = k and ϕ(0) = −1. Since p is not a unit in the
ϕ:Z
p , ϕ is a well defined function. Note that if a and b are nonzero elements
domain Z
of Zp , then ϕ(b) ≤ ϕ(a) if and only if b divides a. It follows that ϕ is a Euclidean
p.
structure map for Z
p is also a local ring, meaning
Recall from the discussion in Example 7.11.22 that Z
p is an
that it has only one maximal ideal, in this case the principal ideal (p). Thus, Z
example of the following notion.
Definition 8.1.3. A discrete valuation ring, or D.V.R., is a Euclidean domain that is
also a local ring.
The Euclidean structure on Z was crucial in our analysis of the subgroups of Z. It
will have similar value for the study of the ideals in other Euclidean domains.
There is an additional property one could ask for from the function ϕ:
Definition 8.1.4. We say that a Euclidean structure map ϕ : A → Z is positive definite
if ϕ(a) > 0 for all 0 = a ∈ A and ϕ(0) = 0.
Lemma 8.1.5. Let A be a Euclidean domain. Then there is a Euclidean structure map
for A which is positive definite.
Proof We claim that if ϕ : A → Z is any Euclidean structure map for A, then ϕ(0) <
ϕ(a) for all a = 0. Given this, it will suffice to replace ϕ by a function ϕ given by
ϕ (a) = ϕ(a) − ϕ(0).
To prove the claim, let a = 0 and write 0 = qa + r with ϕ(r) < ϕ(a). But then
r = (−q)a is divisible by a, which would force ϕ(a) ≤ ϕ(r) unless r = 0. But then r
must equal 0, displaying ϕ(0) < ϕ(a) as desired.
Recall that a principal ideal domain, or P.I.D., is an integral domain in which every
ideal is principal.
Proposition 8.1.6. A Euclidean domain is a principal ideal domain. Indeed, let ϕ :
A → Z be a positive definite Euclidean structure map for A. Let a be a nonzero ideal of
A and let a ∈ a be such that ϕ(a) is the smallest positive value taken on by the restriction
of ϕ to a. Then a = (a).
Proof Let 0 = b ∈ a and write b = qa + r with ϕ(r) < ϕ(a). But r = b − qa is in a, so
by the minimality of ϕ(a), we must have ϕ(r) = 0. But then r = 0 and b ∈ (a).
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
298
Lemma 8.1.7. Let A be an integral domain and let a, b ∈ A. Then (a) = (b) if and only
if a = ub for some unit u of A. Indeed, if a = 0 and (a) = (b), and if a = bc, then c
must be a unit.
Thus, if b = 0 and c is not a unit, then the inclusion
(bc) ⊂ (b)
is proper.
Proof The case a = 0 is immediate, so we assume a = 0. Here, if (a) = (b), we can
find c, d such that a = bc and b = ad. Then a = adc. By the cancellation property in a
domain, dc = 1.
Since the units in K[X] (K a field) are the nonzero elements of K, we obtain an
immediate corollary.
Corollary 8.1.8. Let K be a field and let a be a nonzero ideal in K[X]. Let n be the
smallest non-negative integer such that a contains an element of degree n. Then there is
a unique monic polynomial, f (X), in a of degree n, and a is the principal ideal generated
by f (X). Moreover, f (X) is the unique monic polynomial generating a.
Definition 8.1.9. Let B be an algebra over the field K and let b ∈ B. Let εb : K[X] →
B be the K-algebra map obtained by evaluating X at b. If εb is injective, we say that
the minimal polynomial of b over K is 0. Otherwise, the minimal polynomial of b over
K is the monic polynomial of smallest degree in ker εb . We denote it by minb (X), if the
context is obvious, or more formally by minb/K (X).
Corollary 8.1.10. Let B be an algebra over the field K and let b ∈ B. Then K[b] is
isomorphic as a K-algebra to K[X]/(minb (X)). In particular, K[b] is finite dimensional
as a K-vector space if and only if minb (X) = 0, and in this case, the dimension of K[b]
over K is equal to the degree of minb (X).
Proof If minb (X) = 0, then the determination of the dimension of K[b] follows from
Proposition 7.7.35. Otherwise, K[b] is isomorphic to K[X], which is easily seen to be an
infinite dimensional vector space over K, with basis {X i | i ≥ 0}.
It turns out that P.I.D.s have almost all the nice properties exhibited by the integers,
including those shown in Sections 2.3 and 4.4. We shall treat the material from the
former in this section and the latter in Section 8.9.
Definitions 8.1.11. Let A be a commutative ring. Then p ∈ A is irreducible if p is not
a unit, but whenever p = ab, then either a or b must be a unit.
We say that p ∈ A is a prime element if p = 0 and the principal ideal (p) is a prime
ideal.
Note that if A is a domain, then (0) is a prime ideal, but 0 is neither a prime element
nor an irreducible element.
There is a basic ideal theoretic consequence of irreducibility.
Lemma 8.1.12. Let p be an irreducible element in the commutative ring A and suppose
that a divides p. Then either a is a unit, or p divides a.
In ideal theoretic terms, if (p) ⊂ (a), then either (a) = (p) or (a) = A. Thus, the
ideal generated by an irreducible element is a maximal element in the partially ordered
set of proper principal ideals, ordered by inclusion.
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Proof If a divides p, then p = ab for some b ∈ A. Since p is irreducible, either a or b is
a unit. If b is a unit, then a = pb−1 , and hence p divides a.
In terms of ideals, (p) ⊂ (a) if and only if p ∈ (a), which is equivalent to a dividing
p. If a is a unit, (a) = A. Otherwise, p divides a, and hence (a) ⊂ (p).
In a domain, Lemma 8.1.12 has a converse.
Lemma 8.1.13. Let A be an integral domain. Then a nonzero nonunit p ∈ A is irreducible if and only if the principal ideal (p) satisfies the property that (p) ⊂ (a) implies
that either (a) = (p) or (a) = A.
Proof Suppose that (p) satisfies the stated condition and suppose that p = ab. Then
(p) ⊂ (a). If (a) = A, then a is a unit. If (p) = (a), then Lemma 8.1.7 shows b to be a
unit. Thus, p is irreducible.
If every ideal is principal, then Lemma 8.1.13 may be restated as follows.
Corollary 8.1.14. Let A be a P.I.D. Then a nonzero element p ∈ A is irreducible if
and only if the principal ideal (p) is a maximal ideal of A.
Thus, since maximal ideals are prime, every irreducible element in a P.I.D. is prime.
But this does not hold in a general domain. Surprisingly, the reverse implication does
hold.
Lemma 8.1.15. Let p be a prime element in the integral domain A. Then p is irreducible.
Proof Suppose that p = ab. Then ab ∈ (p), a prime ideal, so either a or b must lie in
(p). Say a ∈ (p). But p is also in (a), since p = ab, so (a) = (p), and hence b is a unit by
Lemma 8.1.7.
Since maximal ideals are prime, the next corollary is immediate from Lemma 8.1.15
and Corollary 8.1.14.
Corollary 8.1.16. Let A be a P.I.D. Then an element p ∈ A is irreducible if and only
if it is a prime element. In particular, if p ∈ A is irreducible and if p divides ab, then it
must divide either a or b.
Moreover, since every nonzero prime ideal of A is generated by a prime element, every
nonzero prime ideal of A is maximal, and has the form (p), where p is an irreducible
element of A.
Thus, the longest possible chain of proper inclusions of prime ideals in a P.I.D. has
the form
0 ⊂ (p)
where p is a prime element of A. This may only occur if A is not a field.
Corollary 8.1.17. A P.I.D. that is not a field has Krull dimension 1.
The statement in the first paragraph of Corollary 8.1.16 has a more classical proof,
which is modelled on the one that we used for the integers. We shall give it to illustrate
that it really uses exactly the same ideas as the proof we just gave.
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Definition 8.1.18. Let A be a P.I.D. Then the greatest common divisor of a and b is
the ideal (c) = (a) + (b). We write (a, b) = (c). If (a) + (b) = A, then we say that a and
b are relatively prime, and write (a, b) = 1.
Note that the notation of (a, b) for the ideal (a) + (b) is consistent with the notation
that (a, b) is the ideal generated by a and b.
We will occasionally refer to c as the greatest common divisor of a and b when
(a, b) = (c). But it is really more correct for g.c.d. to refer to the ideal (c), rather than
the generator c.
Lemma 8.1.19. Let A be a P.I.D. and let (a, b) = (c). Then x divides both a and b if
and only if x divides c.
Proof Since a and b are both in (c), c divides a and b. Thus, it suffices to show that
an element dividing both a and b must divide c. But since (c) = (a) + (b), c = ra + sb
for some r, s ∈ A. Thus, any element that divides a and b will divide c.
The entire argument above reduces to the statement that (a) + (b) is the smallest
ideal containing both a and b, in the context that every ideal is principal.
Second Proof of Corollary 8.1.16 Let p be an irreducible element of A. We shall
show that (p) is a prime ideal.
Suppose that p divides ab but does not divide a. Since p is irreducible, Lemma 8.1.12
shows that any divisor of p is either a unit or is divisible by p. Thus, a and p cannot
have any common divisors other than units. So a and p are relatively prime, and hence
there are elements r, s ∈ A with ra + sp = 1. But then rab + spb = b. Since p divides ab,
it divides the left-hand side, so it must divide b.
Note the role that the maximality of (p) plays in the above argument.
Now we would like to show that every nonzero element of a P.I.D. may be factored
in some sense uniquely as a product of primes. This may be read as a statement that
every P.I.D. is an example of a more general type of ring called a unique factorization
domain, or U.F.D.
Definition 8.1.20. A unique factorization domain, or U.F.D., is an integral domain in
which every nonzero nonunit may be written as a product of prime elements.
The next lemma has already been seen to hold in P.I.D.s.
Lemma 8.1.21. Irreducible elements in a U.F.D. are prime. Thus, in a U.F.D., the
prime elements and the irreducible elements coincide.
Proof Let p be an irreducible element in the U.F.D. A. Then p is a product of prime
elements, and hence is divisible by a prime element, q. Say p = qr. Since q is prime, it
is not a unit. But p is irreducible, so r must be a unit. Thus, (p) = (q), a prime ideal,
and hence p is prime.
We now wish to show that every P.I.D. is a U.F.D. Since we’ve shown that every
irreducible element in a P.I.D. is prime, it suffices to show that every nonzero nonunit is
a product of irreducible elements.
If our P.I.D. is actually a Euclidean domain, this may be shown by a quick induction
on ϕ(a), much as the proof for the integers worked. For the general P.I.D., we need to
work a little bit harder. The main tool comes from the Noetherian property.
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301
Proposition 8.1.22. Every P.I.D. is a U.F.D.
Proof Let A be a P.I.D. Since every ideal is principal, the ideals are certainly finitely
generated, and hence A is Noetherian by Corollary 7.8.4. Let 0 = a ∈ A be a nonunit,
and suppose by contradiction that a is not a product of primes. Since a is not a unit, (a)
is proper, and hence (a) is contained in a maximal ideal (p1 ). Thus, a = p1 a1 for some
a1 ∈ A.
Now p1 is irreducible by Corollary 8.1.14, and hence prime, by Corollary 8.1.16. Since
p1 is prime and a = p1 a1 is not a product of primes, a1 cannot be a product of primes.
Also, the inclusion (a) ⊂ (a1 ) must be proper by Lemma 8.1.7, as p1 and a1 are nonzero
nonunits.
We may continue by induction, obtaining elements ak ∈ A for k ≥ 1, such that
ak−1 = pk ak with pk prime, and hence ak cannot be a product of primes. Thus, we
obtain an infinite sequence of proper inclusions
(a) ⊂ (a1 ) ⊂ · · · ⊂ (ak ) ⊂ · · ·
contradicting the fact that A is Noetherian.
We shall show in Section 8.5 that if A is a U.F.D., so is A[X]. Taking A to be a
P.I.D. that is not a field, this will give a family of examples of U.F.D.s that are not
P.I.D.s.
We shall now discuss the uniqueness properties of the factorizations of the elements
in a U.F.D. as products of primes.
Definition 8.1.23. Let p and q be prime elements in the U.F.D. A. We say that p and
q are equivalent if (p) = (q).
In other words, p and q are equivalent if p = uq for some unit u. In the integers, this
doesn’t say much, since the only units are ±1. Thus, it is easy to choose a representative
for each equivalence class of primes in Z: Just pick the positive one. That’s what we did
in Section 2.3.
In the generic P.I.D. there can be infinitely many units, so choosing a representative
for each equivalence class of primes could get cumbersome. For this reason, we shall give
a uniqueness statement first that uses equivalence classes, and then follow it up with
another in which actual representatives are chosen.
We say that a collection p1 , . . . , pk of primes is pairwise inequivalent if pi and pj are
inequivalent for i = j.
Proposition 8.1.24. Let A be a U.F.D., and suppose given an equality
upr11 . . . prkk = vq1s1 . . . qlsl
where u and v are units, p1 , . . . , pk are pairwise inequivalent primes, q1 , . . . , ql are pairwise inequivalent primes, and the exponents ri and si are all positive. Then k = l, and
after reordering (and relabelling) the q1 , . . . , ql if necessary, pi is equivalent to qi and
ri = si for 1 ≤ i ≤ k.
Proof We argue by induction on r = r1 + · · · + rk . Here, we take the empty product
to be 1, so if r = 0, then the left side is just u. But then vq1s1 . . . qlsl is a unit, and hence
q1s1 . . . qlsl must be the empty product, too.
If r ≥ 1, then p1 divides the left-hand side and hence divides the right as well. In
particular, it must divide qi for some i, since the ideal (p1 ) is prime. But since p1 is not
a unit and qi is irreducible, p1 and qi must be equivalent, say qi = wp1 , with w a unit.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
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Reorder the q’s so that i = 1. Then
p1 (up1r1 −1 . . . prkk − vwq1s1 −1 . . . qlsl ) = 0.
Since A is an integral domain, we have
upr11 −1 . . . prkk = vwq1s1 −1 . . . qlsl ,
and the result follows by induction.
By a choice of representatives for the primes in A, we mean a collection {pi | i ∈ I}
of primes in A such that for each prime p of A there is a unique i ∈ I such that p is
equivalent to pi .
Thus, for example, the positive primes form a choice of representatives for the primes
in Z.
Corollary 8.1.25. Let A be a U.F.D. and let {pi | i ∈ I} be a choice of representatives
for the primes in A. Then any nonzero nonunit in A may be written uniquely in the form
upri11 . . . prikk , where k > 0, u is a unit, i1 , . . . , ik are distinct elements of I, and ri > 0 for
1 ≤ i ≤ k.
Proof Every element may be written in the above form because equivalent primes differ
by multiplication by a unit. Thus, the u term can absorb the differences between a given
product of primes and a product of pi ’s. The uniqueness statement now follows from that
in Proposition 8.1.24, since pi is not equivalent to pj unless i = j. Thus, if two elements
of the form upri11 . . . prikk are equal, the primes in them, together with their exponents,
must be identical. Thus, we’re reduced to the case where upri11 . . . prikk = vpri11 . . . prikk , with
u and v units. But since we’re in an integral domain, we can cancel off the primes in the
above equation, leaving u = v.
Corollary 8.1.26. Let A be a U.F.D. and let a ∈ A. Suppose that a = upr11 . . . prkk ,
where u is a unit, p1 , . . . , pk are pairwise inequivalent primes, and ri ≥ 0 for i = 1, . . . , k.
Then an element b divides a if and only if b = vps11 . . . pksk , where v is a unit and 0 ≤
si ≤ ri for 1 ≤ i ≤ k.
Proof If b divides a, then any prime that divides b will also divide a. If a = bc,
then c also divides a, and the same reasoning applies. Thus, b = vps11 . . . pskk , and
c = wpt11 . . . ptkk , where v and w are units and the exponents are all non-negative.
But then a = bc = vwps11 +t1 . . . pskk +tk . By the uniqueness of factorization, si + ti = ri
for i = 1, . . . , k, and hence si ≤ ri as claimed.
Conversely, if si ≤ ri for i = 1, . . . , k and if v is a unit, then vps11 . . . pskk is easily seen
to divide a.
We’d like to be able to talk about greatest common divisors in a U.F.D., but the
definition that we used in P.I.D.s fails for U.F.D.s that are not P.I.D.s. The next corollary
will help us.
Corollary 8.1.27. Let A be a U.F.D. and let 0 = a, b ∈ A. Let p1 , . . . , pk be a set of
representatives for the set of primes dividing either a or b. Thus, p1 , . . . , pk are pairwise
inequivalent, and we may write a = upr11 . . . prkk and b = vps11 . . . pskk , where u and v are
units, and the exponents are non-negative.
Let ti = min(ri , si ) for 1 ≤ i ≤ k and let c = pt11 . . . ptkk . (Note that c = 1 if no prime
element divides both a and b.) Then c divides both a and b. In addition, if d is any
element that divides both a and b, then d divides c.
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303
In the language of ideals, this says that (a) + (b) ⊂ (c), and if d is any other element
such that (a) + (b) ⊂ (d), then (c) ⊂ (d). In particular, (c) is the smallest principal ideal
that contains a and b.
Proof This is almost immediate from Corollary 8.1.26: If d divides both a and b, then
d = wpl11 . . . plkk , where, if 1 ≤ i ≤ k, then 0 ≤ li ≤ ri , because d divides a, and 0 ≤ li ≤ si ,
because d divides b. But then li ≤ ti for all i, so d divides c. Clearly, c divides a and b,
so the result follows.
Definitions 8.1.28. Let A be a U.F.D. and let a, b ∈ A. Then the greatest common
divisor of a and b is the smallest principal ideal that contains (a) + (b). (Such an ideal
exists by Corollary 8.1.27.) If the greatest common divisor of a and b is c, we write
gcd(a, b) = (c),
or sometimes, by abuse of notation, gcd(a, b) = c.
We say that a and b are relatively prime if their greatest common divisor is (1).
Notice that if A is a P.I.D., then these definitions agree with the old ones. Of course,
Corollary 8.1.27 tells us exactly how to calculate greatest common divisors in a U.F.D.
We state the result formally here:
Corollary 8.1.29. Let A be a U.F.D. and let 0 = a, b ∈ A. Let p1 , . . . , pk be a set of
representatives for the set of primes dividing either a or b. Thus, p1 , . . . , pk are pairwise
inequivalent, and we may write a = upr11 . . . prkk and b = vps11 . . . pskk , where u and v are
units, and the exponents are non-negative.
Let ti = min(ri , si ) for 1 ≤ i ≤ k. Then the greatest common divisor of a and b is
(c), where c = pt11 . . . ptkk . In particular, a and b are relatively prime if and only if no
prime element of A divides both a and b.
More generally, we can talk about the greatest common divisor of a set of elements
in a U.F.D.
Definition 8.1.30. Let a1 , . . . , an be elements of the U.F.D. A. Then the greatest common divisor, gcd(a1 , . . . , an ) of these elements is the smallest principal ideal containing
(a1 ) + · · · + (an ). By abuse of notation, we also write gcd(a1 , . . . , an ) for any generator
of this ideal.
We say that a1 , . . . , an are relatively prime if gcd(a1 , . . . , an ) = 1.
Notice the difference between saying that a1 , . . . , an are relatively prime and saying
that they are pairwise relatively prime. The reader should supply the proof of the next
corollary.
Corollary 8.1.31. Let a1 , . . . , an be nonzero elements of the U.F.D. A. Let p1 , . . . , pk
be a set of representatives for the set of all primes that divide the elements a1 , . . . , an ,
and write
ai = ui pr1i1 . . . prkik
for 1 ≤ i ≤ n, with ui a unit in A.
Let tj = min(r1j , . . . , rnj ) for 1 ≤ j ≤ k. Then
gcd(a1 , . . . , an ) = pt11 . . . ptkk .
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Another construction which makes sense in a U.F.D. is the least common multiple.
Once again we shall need a computation to set it up.
Proposition 8.1.32. Let A be a U.F.D. and let a1 , . . . , an ∈ A. Define an element
a ∈ A as follows. If any of the ai = 0, then set a = 0. Otherwise, let p1 , . . . , pk be a set
of representatives for the primes that divide at least one of the ai , and let
ai = ui pr1i1 . . . prkik
for 1 ≤ i ≤ n, where ui is a unit and the exponents rij are all non-negative. Then define
tj = max(r1j , . . . , rnj ), and set
a = pt11 . . . pktk .
Then the intersection of the principal ideals generated by the ai is the principal ideal
generated by a:
n
(ai ) = (a).
i=1
n
Proof Here, i=1 (ai ) is the collection of elements divisible by each of the ai , so the
result follows immediately from Corollary 8.1.26.
Definition 8.1.33. Let A be a U.F.D. and let a
1 , . . . , an ∈ A. By the least common
n
multiple of a1 , . . . , an we mean the intersection, i=1
n(ai ), of the principal ideals they
generate, or, by abuse of notation, any generator of i=1 (ai ).
We also obtain a sharpening of the Chinese Remainder Theorem.
Corollary 8.1.34. (Chinese Remainder Theorem, second form) Let A be a P.I.D. and
let a1 , . . . , ak ∈ A, such that ai and aj are relatively prime for i = j. Let a = a1 . . . ak .
Then there is an A-algebra isomorphism
∼
=
f : A/(a) −→ A/(a1 ) × · · · × A/(ak )
given by f (x) = (x, . . . , x) for all x ∈ A.
Proof Since ai and aj are relatively prime for i = j, (ai )+(aj ) = A for i = j. Thus, the
first form of the Chinese Remainder Theorem (Proposition 7.2.23) applies, showing that
the A-algebra homomorphism f : A → A/(a1 ) × · · · × A/(ak ) given by f (a) = (a, . . . , a)
k
k
is a surjection whose kernel is i=1 (ai ). But i=1 (ai ) = (a) by Proposition 8.1.32.
If a U.F.D. A is not a P.I.D., then its prime elements generate prime ideals which will
not, in general, be maximal ideals. In fact, they turn out to be minimal nonzero prime
ideals of A.
Definition 8.1.35. Let A be a domain. A minimal nonzero prime ideal of A is a nonzero
prime ideal with the property that the only prime ideal which is properly contained in
it is 0.
Proposition 8.1.36. Let A be a U.F.D. Then a prime ideal p of A is a minimal nonzero
prime ideal if and only if p = (p) for a prime element p of A.
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305
Proof To show that every minimal nonzero prime ideal of A is principal, it suffices to
show that every nonzero prime ideal q contains a prime element. But if 0 = a ∈ q, let
a = p1 . . . pk , a product of (not necessarily distinct) prime elements of A. Because a ∈ q
and q is prime, at least one of the pi must lie in q.
Conversely, let p be a prime element and suppose that q is a prime ideal that’s properly
contained in (p). Let K be the fraction field of A and let a = (1/p)q = {(1/p)a | a ∈ q} ⊂
K. Since every element of q is divisible by p, a ⊂ A, and is easily seen to be an ideal of
A.
But now q = pa = (p)a. Since q is prime, either (p) ⊂ q or a ⊂ q by Lemma 7.6.9.
Since q is properly contained in (p), we must have a ⊂ q.
But the fact that q = (p)a shows that q ⊂ a, so q = a. In other words, q = (p)q = pq.
But by induction, we see that q = pn q for all n, and hence each a ∈ q is divisible by pn
for all n. By the uniqueness of factorization in A, no element but 0 can be divisible by
all powers of a prime element, so q must be the 0 ideal. Thus, the only prime ideal that
can be properly contained in a principal prime ideal is 0.
Exercises 8.1.37.
1. Let A be a U.F.D. Suppose that a divides bc in A, where a and b are relatively
prime. Show that a divides c.
2. Let p1 , . . . , pk be pairwise inequivalent primes in the U.F.D. A and let a = upr11 . . . prkk
for some unit u. Show that a divides b if and only if pri i divides b for 1 ≤ i ≤ k.
3. Let K be a field. Show that every degree 1 polynomial of K is irreducible in K[X].
If f (X) has degree 1, what is K[X]/(f (X))?
4. Let K be a field and let f (X) ∈ K[X] be a polynomial whose degree is either 2 or
3. Show that f (X) is irreducible if and only if it has no roots.
5. Let K be the field of fractions of the U.F.D. A. Let a/b ∈ K be a root of f (X) =
am X m + · · · + a0 ∈ A[X], where am = 0 and where a, b ∈ A are relatively prime.
Show that b must divide am and a must divide a0 .
6. The simplest procedure for finding primes in Z is to proceed inductively, testing
each number by dividing it by all primes smaller than its square root. If it has no
prime divisor less than or equal to its square root, it is prime. One could adopt a
similar strategy in F [X] for a finite field F . Here, a polynomial f (X) is irreducible
if it has no irreducible factors of degree ≤ 12 deg f . Carry out this procedure for
F = Z2 to find all irreducible polynomials of degree ≤ 4.
7. Let A be a P.I.D. and let B be a subring of the field of fractions of A with A ⊂ B.
Show that B = S −1 A for a suitable multiplicative subset S of A.
8. Let A be a U.F.D. and let S be a saturated multiplicative subset of A. Let T be
the set of prime elements of A that lie in S. Show that
S = {up1r1 . . . prkk | k ≥ 0, u ∈ A× , p1 , . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}.
9. Let A be a U.F.D. with only one prime element, p, up to equivalence of primes.
Show that there is a Euclidean structure map ϕ : A → Z obtained by setting
ϕ(0) = −1 and setting ϕ(a) = k if pk is the highest power of p that divides the
nonzero element a. Deduce that a ring is a discrete valuation ring if and only if it
is a U.F.D. with at most one prime element up to equivalence of primes.
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306
10. Let A be a U.F.D. and let S be a multiplicative subset of A. Show that S −1 A is a
U.F.D.
11. Let A be a U.F.D. and let p be a prime element of A. Show that the localization
A(p) of A at the prime ideal (p) is a discrete valuation ring.
12. Let A be a U.F.D. that has infinitely many equivalence classes of prime elements.
Show that the field of fractions of A is infinitely generated as an A-algebra.
13. Show that if A is not a field, then A[X] is not a P.I.D.
14. Recall the ring of Gaussian integers, Z[i], introduced in Example 7.3.6: Z[i] is the
subring of C consisting of the elements a + bi with a, b ∈ Z.
(a) Show that Z[i] is Euclidean, via the function ϕ(a + bi) = a2 + b2 .
(b) Show that complex conjugation restricts to an isomorphism of rings from Z[i]
to itself. Show also that ϕ(z) = zz for z ∈ Z[i], where z is the complex
conjugate of z. Deduce that ϕ(zw) = ϕ(z)ϕ(w) for z, w ∈ Z[i].
(c) Show that the only units in Z[i] are ±i and ±1.
(d) Show that anelement z ∈ Z[i] is prime if and only if it has no divisors w with
1 < ϕ(w) ≤ ϕ(z).
(e) Show that 1 + i is a prime and that 1 + i = −i · (1 + i). Show also that
ϕ(1 + i) = 2, so that
2 = (1 + i) · (1 + i) = −i · (1 + i)2 .
(In the language of algebraic number theory, this implies that 2 ramifies in
Z[i].) Deduce that 2 is not prime in Z[i].
(f) Show that if z = a+bi ∈ Z[i] with a, b ∈ Z is a prime element such that neither
a nor b is zero and (z) = (1 + i), then z is a prime that is not equivalent to z.
Deduce that every nonzero element n ∈ Z has a prime decomposition in Z[i]
of the form
n = u · (1 + i)2r · pr11 . . . prkk · z1s1 z s11 . . . zlsl z sl l ,
where u is a unit, the pi are primes in Z that remain prime in Z[i], and the zi
are primes in Z[i] of the form a + bi, where a and b are positive integers with
b < a.
(g) Let z = a + bi where a and b are nonzero integers. Show that z is prime in
Z[i] if and only if ϕ(z) is prime in Z.
(h) Let p be a prime in Z that is not prime in Z[i]. Show that p = ϕ(z) for some
z ∈ Z[i]. Deduce that there are integers a and b with p = a2 + b2 .
(i) Show that a prime in Z that is congruent to 3 mod 4 remains prime in Z[i].
(j) Suppose that a prime p in Z remains prime in Z[i]. Deduce that Z[i]/(p) is
a field of order p2 , which is additively a vector space over Zp of dimension
2, with a basis given by the images of 1 and i under the canonical map from
Z[i]. Deduce that the only elements of order 4 in the unit group (Z[i]/(p))×
must lie outside the image of the unique ring homomorphism from Zp to
Z[i]/(p). Deduce that Zp × has no elements of order 4, and hence that p must
be congruent to 3 mod 4.
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307
(k) Let p be a prime in Z that is congruent to 1 mod 4. Deduce from the results
above that p = zz for some prime z of Z[i]. Show that the field Z[i]/(z)
is of characteristic p, and is generated additively by the images of 1 and i
under the canonical map from Z[i]. Deduce from the fact that Zp has units
of order 4 that the image of i under the canonical map must lie in the image
of the unique ring homomorphism from Zp to Z[i]/(z). Deduce that Z[i]/(z)
is isomorphic to Zp .
15. Characterize the collection of integers that may be written as the sum of two
squares.
16. We study the subring Z[ζ3 ] of C.
(a) Show that the kernel of the evaluation map εζ3 : Z[X] → Z[ζ3 ] is the principal
ideal of Z[X] generated by X 2 + X + 1. Deduce from Proposition 7.7.35
that the elements of Z[ζ3 ] may be written uniquely in the form m + nζ3 with
m, n ∈ Z. Write down the multiplication formula for the product of two such
elements.
(b) Using the fact that ζ 3 = ζ3−1 = ζ32 , where ζ 3 is the complex conjugate of ζ3 ,
show that setting ϕ(α) = αα defines a Euclidean structure map ϕ : Z[ζ3 ] → Z.
(c) Show that Z[ζ3 ]× = {±1, ±ζ3 , ±ζ32 }, where ζ32 = −1−ζ3 and −ζ32 = ζ6 = 1+ζ3 .
(d) Show that 3 ramifies in Z[ζ3 ] in the sense that 3 = up2 , where u ∈ Z[ζ3 ]× and
p is a prime of Z[ζ3 ].
(e) Show that an element of the form m + nζ3 , where m and n are both nonzero
elements of Z, is prime in Z[ζ3 ] if and only if its image under ϕ is a prime
in Z.
(f) Show that a prime in Z remains prime in Z[ζ3 ] if and only if it is not in the
image of ϕ : Z[ζ3 ] → Z.
(g) Show that a prime p = 3 in Z remains prime in Z[ζ3 ] if and only if p is not
congruent to 1 mod 3.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
308
√
17. Show that Z[i 2] is a Euclidean√domain via the square of the complex norm, as
above. What are the units of Z[i√ 2]? Find congruences that will guarantee that a
prime in Z remains prime in Z[i 2].
√
18. Show that Z[ 2] admits a Euclidean structure map.
8.2
Algebraic Extensions
Definitions 8.2.1. Let K be a field. By an extension of K, or, more formally, an
extension field of K, we mean a field L that contains K as a subfield.
We say that L is a finite extension of K if it is finite dimensional as a vector space
over K.
If L is a finite extension of K, then the degree of L over K, written [L : K], is
the dimension of L as a K-vector space. If L is an infinite extension of K, we write
[L : K] = ∞.
Throughout this section, K is a field.
Lemma 8.2.2. Let L be an extension of K and let L be an extension of L. Then L is
finite over K if and only if both L is finite over L and L is finite over K, in which case
[L : K] = [L : L] · [L : K].
Proof If L is finite over K, then the other two extensions are clearly finite as well.
Conversely, if y1 , . . . , yn is a basis for L as a vector space over L, and x1 , . . . , xm is a
basis for L as a vector space over K, then any ordering of {xi yj | 1 ≤ i ≤ m, 1 ≤ j ≤ n}
will give a basis for L over K.
We now consider the elements of an extension field L of K. For each α ∈ L, there is a
unique K-algebra homomorphism εα : K[X] → L that takes X to α. Here, εα evaluates
polynomials at α:
" n
#
n
i
εα
ai X =
ai αi .
i=0
i=0
Thus, f (X) ∈ ker εα if and only if α is a root of f . In particular, ker εα = 0 if and only
if α is a root of some nonzero polynomial over K.
Recall from Corollary 8.1.10 that the kernel of εα is the principal ideal generated by
minα (X), the minimal polynomial of α. Here, minα (X) = 0 if εα is injective, and is the
monic polynomial of lowest degree in ker εα otherwise. In particular, minα (X) divides
any polynomial that has α as a root.
The image of εα , of course, is K[α], the K-subalgebra of L generated by α.
Definition 8.2.3. Let L be an extension field of K and let α ∈ L. We say that α is
algebraic over K if α is a root of a nonzero polynomial over K, and hence minα (X) = 0.
If α is not a root for any nonzero polynomial over K, we say that α is transcendental
over K.
Examples 8.2.4.
2π
1. Let ζn = ei· n ∈ C be the standard primitive n-th root of unity in C. Since ζn has
order n in C× , it is a root of the polynomial X n − 1 ∈ Z[X], and hence is algebraic
over Q.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
309
√
n
a is a root of the integer polynomial
2. For positive integers n and a, the real number
X n − a, and hence is algebraic over Q.
3. The real numbers π and e are known to be transcendental over Q. The standard
proofs of this use analytic number theory.
Let α be an element of the extension field L of K. Then K[X]/(minα (X)) ∼
= K[α],
which, as a subring of L, is an integral domain. Thus, (minα (X)) is a prime ideal in
K[X]. But by Corollary 8.1.16, the nonzero prime ideals of the P.I.D. K[X] are all
maximal, generated by irreducible elements of K[X]. Thus, if α is algebraic over K,
then K[α] ∼
= K[X]/(minα (X)) is a field. The following proposition is now immediate
from Corollary 8.1.10.
Proposition 8.2.5. Let L be an extension field of K and let α ∈ L be algebraic over K.
Then K[α] is a finite extension field whose degree over K is equal to the degree of the
minimal polynomial of α over K:
[K[α] : K] = deg minα (X).
Recall that K[α] is the smallest K-subalgebra of L that contains α.
Definition 8.2.6. Let L be an extension field of K and let α ∈ L. We write K(α) for
the smallest subfield of L that contains both K and α.
More generally, if {αi | i ∈ I} is any family of elements in L, we write K(αi | i ∈ I) for
the smallest subfield of L containing both K and {αi | i ∈ I}. For a finite set α1 , . . . , αn
of elements of L, the notation for this subfield, of course, becomes K(α1 , . . . , αn ).
In particular, Proposition 8.2.5, when α is algebraic over K, or Corollary 7.11.11,
when α is transcendental, now gives the following corollary.
Corollary 8.2.7. Let L be an extension field of K and let α ∈ L.
If α is algebraic over K, then K(α) = K[α].
If α is transcendental over K, then K(α) is isomorphic to the field of fractions of
K[α], and consists of the elements of the form f (α)/g(α), where f (X), g(X) ∈ K[X],
with g(α) = 0. Since α is transcendental over K, this just says that g(X) = 0, and hence
K(α) is isomorphic to K(X), the field of rational functions over K.
The extensions K(α) with α algebraic over K are of sufficient interest to warrant a
name.
Definitions 8.2.8. An extension field L of K is simple if L = K(α) for some α ∈ L
that is algebraic over K.
An element α that is algebraic over K and satisfies L = K(α) is called a primitive
element for L over K.
An alternative name for a simple extension is a primitive extension.
We shall see in Chapter 11, using the Fundamental Theorem of Galois Theory, that
if K has characteristic 0, then every finite extension of K is simple. For fields of nonzero
characteristic, we shall see that every finite separable extension is simple.
The degree of an extension is a useful number to calculate. Thus, to study a simple
extension K(α) of K, it is useful to be able to determine the minimal polynomial of α
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
310
over K. In practice, what we’re often given is some polynomial f (X) ∈ K[X] of which
α is a root (e.g., ζn is a root of X n − 1, but we don’t yet know the minimal polynomial
of ζn over Q, nor, indeed, the degree [Q(ζn ) : Q]). All we know is that minα (X) is one
of the prime divisors of f (X).
Thus, it is quite valuable to be able to compute prime decompositions in K[X],
and, in particular, to test the elements of K[X] to see if they are irreducible. Tests of
irreducibility will be a topic in Section 8.5.
Even in the case of a simple extension, it is sometimes difficult to find a primitive
element. Thus, it is valuable to be able to handle extensions that are generated as
algebras by more than one element. Recall that if B is a commutative K-algebra and if
b1 , . . . , bn ∈ B, then we write K[b1 , . . . , bn ] for the image of the K-algebra homomorphism
εb1 ,...,bn : K[X1 , . . . , Xn ] → B
obtained by evaluating Xi at bi for i = 1, . . . , n.
It is easy to see that K[b1 , . . . , bn ] is the smallest K-subalgebra of B containing
b1 , . . . , bn . The next lemma gives a partial generalization of Proposition 8.2.5.
Lemma 8.2.9. Let L be an extension field of K and let α1 , . . . , αn ∈ L be algebraic over
K. Then K[α1 , . . . , αn ] is a field. Indeed, K[α1 , . . . , αn ] is a finite extension of K.
Proof We argue by induction on n, with the case n = 1 being given by Proposition 8.2.5.
Suppose then by induction that K[α1 , . . . , αn−1 ] is a finite extension field of K.
Since K[α1 , . . . , αn ] is the smallest K-subalgebra of L that contains the elements
α1 , . . . , αn , it must be the case that
K[α1 , . . . , αn ] = K[α1 , . . . , αn−1 ][αn ].
Now αn is algebraic over K, and hence is a root of a nonzero polynomial f (X) ∈ K[X].
But f may be regarded as a polynomial over K[α1 , . . . , αn−1 ], so αn is algebraic over
K[α1 , . . . , αn−1 ] as well. But Proposition 8.2.5 now shows that K[α1 , . . . , αn ] is a finite
extension of K[α1 , . . . , αn−1 ], and hence also of K by Lemma 8.2.2.
Corollary 8.2.10. Let L be an extension of K and let α1 , . . . , αn ∈ L be algebraic over
K. Then
K(α1 , . . . , αn ) = K[α1 , . . . , αn ]
is a finite extension of K.
We would like to study the situation of adjoining infinitely many algebraic elements
to a field K.
Definition 8.2.11. Let B be a commutative A-algebra and let {bi | i ∈ I} be a family
of elements of B. We write A[bi | i ∈ I] for the set of all elements in B lying in one of
the subalgebras A[bi1 , . . . , bik ] for a finite subset {i1 , . . . , ik } ⊂ I.
Clearly, A[bi | i ∈ I] is the smallest A-subalgebra of B that contains {bi | i ∈ I}.
Proposition 8.2.12. Let L be an extension field of K and let {αi | i ∈ I} be elements
of L that are algebraic over K. Then K[αi | i ∈ I] is a field, and hence
K(αi | i ∈ I) = K[αi | i ∈ I]
is the smallest K-subalgebra of L containing {αi | i ∈ I}.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
311
Proof Every nonzero element β ∈ K[αi | i ∈ I] lies in an extension K[αi1 , . . . , αik ],
which is a field, since the α’s are algebraic over K. So β is a unit in K[αi | i ∈ I].
We shall be especially interested in extension fields that are algebraic over K:
Definitions 8.2.13. An extension L of K is algebraic if every element of L is algebraic
over K.
An extension that is not algebraic is said to be transcendental.
The next proposition characterizes the algebraic elements in an extension field L.
Proposition 8.2.14. Let L be an extension field of K and let α ∈ L. Then the following
conditions are equivalent.
1. α is algebraic over K.
2. K[α] is a finite extension of K.
3. There is a field L ⊂ L, containing both K and α, such that L is a finite extension
of K.
Proof The first condition implies the second by Proposition 8.2.5, while the second
implies the first by Corollary 8.1.10. The second implies the third by taking L = K[α],
so it suffices to show that the third implies the second.
But if L is a finite extension of K containing α, then K[α] ⊂ L , and hence K[α] is
also finite over K.
Corollary 8.2.15. An extension L of K is algebraic if and only if every element of L
is contained in a finite extension of K. In particular, every finite extension of K is
algebraic.
Corollary 8.2.16. Let L be an extension field of K. Then L is a finite extension of K
if and only if L = K(α1 , . . . , αn ), where α1 , . . . , αn are algebraic over K.
Proof If L = K(α1 , . . . , αn ), where α1 , . . . , αn are algebraic over K, then L is finite
over K by Corollary 8.2.10.
Conversely, if L is finite over K, then every element of L is algebraic over K by
Proposition 8.2.14. If α1 , . . . , αn is a basis for L over K, then clearly, L = K[α1 , . . . , αn ],
so the result follows from Corollary 8.2.10.
Thus, if α1 , . . . , αn are algebraic over K, then K(α1 , . . . , αn ) is an algebraic extension
of K. Indeed, we may extend this to infinitely generated extensions.
Corollary 8.2.17. An extension L of K is algebraic if and only if L = K(αi | i ∈ I) for
some collection of elements {αi | i ∈ I} that are algebraic over K.
Proof Let L = K(αi | i ∈ I), where each αi is algebraic over K. Then L = K[αi | i ∈ I]
by Proposition 8.2.12. In particular, every element of L lies in a finitely generated
K-subalgebra of the form K[αi1 , . . . , αik ]. Since the αi are algebraic over K, these
subalgebras are finite extension fields of K, and hence L is algebraic over K by Proposition 8.2.14.
Conversely, if L is algebraic over K and if {αi | i ∈ I} is any basis for L as a vector
space over K, then L = K(αi | i ∈ I).
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
312
We can also use Proposition 8.2.14 to obtain an important closure property for algebraic extensions.
Corollary 8.2.18. Let L be an algebraic extension of K and let L1 be any extension of
L. Suppose that α ∈ L1 is algebraic over L. Then α is algebraic over K as well.
In particular, if L is an algebraic extension of K, then any algebraic extension of L
is an algebraic extension of K.
Proof Let α ∈ L1 be algebraic over L. Then α is a root of a nonzero polynomial f (X)
over L. Say f (X) = αn X n + · · · + α0 , with αi ∈ L for 0 ≤ i ≤ n.
Since L is algebraic over K, so is αi for 0 ≤ i ≤ n, and hence K(α0 , . . . , αn ) is a finite
extension of K by Corollary 8.2.10.
But α is algebraic over K(α0 , . . . , αn ). Thus, K(α0 , . . . , αn )(α) is a finite extension
of K(α0 , . . . , αn ), and hence of K as well. So α is contained in a finite extension of K,
and hence is algebraic over K by Proposition 8.2.14.
We shall be primarily interested in finite extensions, but some results regarding algebraic closures will require information about infinite algebraic extensions. The next result
is trivial for finite extensions and false for transcendental extensions (see Corollary 8.6.7).
Thus, it shows the strength of the condition that an extension is algebraic.
Proposition 8.2.19. Let L be an algebraic extension of K and let ν : L → L be a homomorphism of fields that restricts to the identity map on K. Then ν is an automorphism
of L.
Proof Every homomorphism of fields is injective. So it suffices to show ν is onto. If L
is a finite extension, this is immediate from a dimension argument. Otherwise, let α ∈ L.
Let f (X) ∈ K[X] be the minimal polynomial of α over K, and let α1 , . . . , αn be the
set of all roots of f in L, with α = α1 . Then K(α1 , . . . , αn ) is a finite extension of K,
and it suffices to show that ν carries K(α1 , . . . , αn ) into itself.
Since K(α1 , . . . , αn ) is the smallest K-subalgebra of L containing α1 , . . . , αn , it suffices to show that ν carries each αi into {α1 , . . . , αn }.
Write f (X) = am X m + · · · + a0 , with a0 , . . . , am ∈ K. Then f (αi ) = 0, so
0 = ν(f (αi )) = am (ν(αi ))m + · · · + a0 = f (ν(αi )),
since ν acts as the identity on K. In particular, ν(αi ) is a root of f in L. And the set of
all such roots is precisely {α1 , . . . , αn }.
We shall next show that in any extension L of K, there is a largest subfield K1 of L
such that K1 is an algebraic extension of K.
Definition 8.2.20. Let L be an extension of K. Then the algebraic closure of K in L
is the set of all elements of L that are algebraic over K.
Proposition 8.2.21. Let L be an extension of K. Then the algebraic closure of K in
L is a subfield of L containing K. In particular, it is an algebraic extension of K.
Proof If α is algebraic over K, then K(α) is a finite extension of K, and hence every
element of K(α) must be algebraic over K by Proposition 8.2.14. In particular, −α and
α−1 are algebraic over K.
If α and β are both algebraic over K, then β is a root of a nonzero polynomial over
K, and hence also over K(α). Thus, β is contained in a finite extension, L , of K(α),
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
313
which is then finite over K by Lemma 8.2.2. But α + β and αβ lie in L , and hence α + β
and αβ are algebraic over K.
Thus, the elements of L that are algebraic over K form a subfield of L. This subfield
includes K, as each a ∈ K is a root of X − a.
Definition 8.2.22. Let L be an extension of K. We say that K is algebraically closed
in L if the only elements of L that are algebraic over K are those of K itself.
Proposition 8.2.23. Let L be an extension of K and let K1 be the algebraic closure of
K in L. Then K1 is algebraically closed in L.
Proof Let α ∈ L be algebraic over K1 . Then α is algebraic over K by Corollary 8.2.18,
and hence an element of K1 , as K1 is the algebraic closure of K in L.
Exercises 8.2.24.
1. Let α ∈ C with α ∈ R. Show that both α + α and α · α are real. Deduce that
minα/R (X) = (X − α)(X − α) = X 2 − (α + α)X + αα.
2. Find the minimal polynomial of i over Q.
3. Find the minimal polynomial of ζ3 over Q.
4. Show that minζ8 /Q (X) = X 4 + 1. (Hint: Show that Q(ζ8 ) is a degree 2 extension
of Q(i).)
5. Let n be any √
integer that is not a perfect cube (i.e., not the cube of an integer).
Show that Q( 3 n) is a degree 3 extension of Q.
6. Let L be an extension of K of degree p, with p prime. Show that every element α
of L that is not in K is a primitive element for L over K (i.e., that L = K(α)).
7. Let K be a field of characteristic = 2 and let L be a quadratic extension of K (i.e.,
[L : K] = 2). Show that there is a primitive element, α, for L over K such that the
minimal polynomial
of α over K has the form X 2 − d with d ∈ K. Thus, we may
√
write L = K( d). (Hint: Suppose that L has a primitive element whose minimal
polynomial is aX 2 + bX + c. Show that d = b2 − 4ac has a square root, α, in L and
that L = K(α).)
8. Let K be the field of fractions of a domain A of √
characteristic =
2, and let L be
a quadratic
extension
of
K.
Show
that
L
=
K(
d)
for
some
d ∈ A. (Hint: If
L = K( a/b), with a, b ∈ A, take d = ab.)
If A is a U.F.D., show that we may additionally assume that the element d ∈ A is
square-free, in the sense that it is not divisible by the square of any prime element
of A.
9. Let K be any field and let K(X) be the field of fractions of K[X]. Show that K is
algebraically closed in K(X).
10. Deduce from the preceding problem that if K is finite, then the only finite subfields
of K(X) are those contained in K itself.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
8.3
314
Transcendence Degree
Some extensions are more transcendental than others. We shall discuss this measurement
here. There’s a notion of transcendence basis analogous to that of a basis for a vector
space. We obtain a notion of finite transcendence degree, measured by the number of
elements in a transcendence basis.
Definitions 8.3.1. Let A be a subring of the commutative ring B. We say that the
elements b1 , . . . , bn of B are algebraically independent over A if the evaluation map
εb1 ,...,bn : A[X1 , . . . , Xn ] → B that evaluates each Xi at bi is injective.
Let L be an extension field of K We say that α1 , . . . , αn is a transcendence basis for
L if they are algebraically independent and L is algebraic over K(α1 , . . . , αn ). (Infinite
transcendence bases may be defined analogously.)
We say that an extension L of K has finite transcendence degree over K if there exists
a finite transcendence basis for L over K.
Examples 8.3.2.
1. A single element α ∈ L is algebraically independent if and only if it is transcendental
over K.
2. The null set forms a transcendence basis for an algebraic extension of K. In particular, algebraic extensions have finite transcendence degree.
3. Let α = f (X)/g(X) be any element of K(X) that’s not in K. Then α is a transcendence basis for K(X) over K. The point is that X is algebraic over K(α), as
it is a root of the polynomial αg(T ) − f (T ) in K(α)[T ]. Thus, α is transcendental
over K, as otherwise, X would be algebraic over K.
4. Of course, X1 , . . . , Xn is a transcendence basis for the ring of rational functions
K(X1 , . . . , Xn ).
Recall the isomorphism A[X1 , . . . , Xn ] ∼
= A[X1 , . . . , Xn−1 ][Xn ]. Under this isomorphism, we may write any polynomial f (X1 , . . . , Xn ) uniquely as a polynomial in Xn with
coefficients in A[X1 , . . . , Xn−1 ]:
f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ).
Definitions 8.3.3. Suppose given a polynomial
f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 )
in A[X1 , . . . , Xn ]. We say that f has degree k in Xn if fk (X1 , . . . , Xn−1 ) = 0 in the
equation above.
Let A be a subring of the commutative ring B and b1 , . . . , bn ∈ B. Suppose given a
polynomial
f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 )
in A[X1 , . . . , Xn ]. We say that f has degree m in bn when evaluated at b1 , . . . , bn if the
polynomial
fk (b1 , . . . , bn−1 )Xnk + · · · + f0 (b1 , . . . , bn−1 )
has degree m as a polynomial in Xn over A[b1 , . . . , bn−1 ]. Here, by abuse of notation, we
shall say that f (b1 , . . . , bn ) has degree m in bn .
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
315
Lemma 8.3.4. Let L be an extension of K and let α1 , . . . , αn ∈ L be algebraically
independent over K. Then an element β ∈ L is algebraic over K(α1 , . . . , αn ) if and only
if α1 , . . . , αn , β are algebraically dependent over K.
Proof Suppose that β is algebraic over K(α1 , . . . , αn ), so it satisfies an equation of the
form
fm (α1 , . . . , αn ) m
f0 (α1 , . . . , αn )
β + ··· +
= 0,
gm (α1 , . . . , αn )
g0 (α1 , . . . , αn )
with positive degree in β. Now, multiplying both sides by the product
g0 (α1 , . . . , αn ) . . . gm (α1 , . . . , αn ),
we obtain an algebraic dependence relation between α1 , . . . , αn , β.
Conversely, if α1 , . . . , αn , β are algebraically dependent, let f (X1 , . . . , Xn+1 ) be a
nonzero polynomial in A[X1 , . . . , Xn+1 ] that vanishes when evaluated at α1 , . . . , αn , β.
Write
k
f (X1 , . . . , Xn+1 ) = fk (X1 , . . . , Xn )Xn+1
+ · · · + f0 (X1 , . . . , Xn ).
Since α1 , . . . , αn are algebraically independent, fi (α1 , . . . , αn ) is nonzero whenever fi is
a nonzero polynomial. Thus, f has positive degree in β, and hence β is algebraic over
K(α1 , . . . , αn ).
A new characterization of transcendence bases is now immediate.
Corollary 8.3.5. Let L be an extension of K and let α1 , . . . , αn ∈ L be algebraically
independent over K. Then α1 , . . . , αn is a transcendence basis for L over K if and only
if there is no element β in L such that α1 , . . . , αn , β are algebraically independent over
K.
The theory of transcendence bases now proceeds much like the theory of bases in
ordinary linear algebra.
Proposition 8.3.6. Let L be an extension of K and let α1 , . . . , αn in L such that L
is algebraic over K(α1 , . . . , αn ). Suppose in addition that α1 , . . . , αm are algebraically
independent for some m < n. Then we can find i1 , . . . , ik ∈ {m + 1, . . . , n} such that
α1 , . . . , αm , αi1 , . . . , αik is a transcendence basis for L over K.
Proof By induction on n − m, we may choose i1 , . . . , ik so that α1 , . . . , αm , αi1 , . . . , αik
is algebraically independent, but is not properly contained in any other algebraically
independent subset of α1 , . . . , αn . For simplicity, we reorder the α’s if necessary so that
α1 , . . . , αm+k is our maximal algebraically independent subset containing α1 , . . . , αm .
Now, for i > m+k, α1 , . . . , αm+k , αi is algebraically dependent, so Lemma 8.3.4 shows
αi to be algebraic over K(α1 , . . . , αm+k ). Corollary 8.2.10 now shows that K(α1 , . . . , αn )
is algebraic over K(α1 , . . . , αm+k ), and hence L is also.
Recall that the total degree of a monomial aX1i1 . . . Xnin is i1 + · · · + in and that
the total degree of a polynomial f (X1 , . . . , Xn ) is the largest of the total degrees of the
nonzero monomials in it.
Definition 8.3.7. Let A be a subring of a commutative ring B and let b1 , . . . , bn ∈
B. We say that a nonzero polynomial f (X1 , . . . , Xn ) ∈ A[X1 , . . . , Xn ] gives a minimal
algebraic dependence relation between b1 , . . . , bn if f (b1 , . . . , bn ) = 0 and if no nonzero
polynomial whose total degree is less than that of f vanishes when evaluated on b1 , . . . , bn .
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
316
Lemma 8.3.8. Let A be a subring of a commutative ring B and let b1 , . . . , bn ∈ B. Suppose that f (X1 , . . . , Xn ) ∈ A[X1 , . . . , Xn ] gives a minimal algebraic dependence relation
between b1 , . . . , bn . Suppose also that f has degree k > 0 in Xn . Then f (b1 , . . . , bn ) has
degree k in bn .
Proof Suppose that
f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ).
Then fk (X1 , . . . , Xn−1 ) is a nonzero polynomial whose total degree is k less than that of
f . If fk (b1 , . . . , bn−1 ) = 0, then fk may be regarded as an algebraic dependence relation
for b1 , . . . , bn , contradicting the minimality of f . Thus, fk (b1 , . . . , bn−1 ) = 0.
Proposition 8.3.9. Let α1 , . . . , αn be a transcendence basis for L over K and let β1 , . . . , βk ∈
L be algebraically independent over K. Then k ≤ n, and if k = n, then β1 , . . . , βn is a
transcendence basis for L over K.
Proof Suppose that k ≥ n. We shall show by induction on i that we may reorder the
α’s in such a way that L is algebraic over K(β1 , . . . , βi , α1 , . . . , αn−i ) for each i ≤ n.
When i = n, this says that L is algebraic over K(β1 , . . . , βn ), so that β1 , . . . , βn is a
transcendence basis for L over K. But if k > n, Lemma 8.3.5 then shows that β1 , . . . , βk
cannot be algebraically independent over K, so the result will follow from our induction
on i.
Suppose, inductively, that L is algebraic over K(β1 , . . . , βi−1 , α1 , . . . , αn−i+1 ). Then
Proposition 8.3.6 shows that we may reorder the α’s if necessary so that β1 , . . . , βi−1 , α1 , . . . , αr
is a transcendence basis for L over K for some r ≤ n − i + 1.
Thus, Lemma 8.3.4 shows that β1 , . . . , βi , α1 , . . . , αr are algebraically dependent. Let
f (X1 , . . . , Xr+i ) be a minimal algebraic dependence relation between β1 , . . . , βi , α1 , . . . , αr .
By Lemma 8.3.8, f must have positive degree in at least one of the α’s, as otherwise, f
only involves the variables X1 , . . . , Xi , and hence gives an algebraic dependence relation
for the β’s.
Renumbering the α’s, if necessary, we may assume that f has positive degree in
αr . But then αr is algebraic over K(β1 , . . . , βi , α1 , . . . , αr−1 ). Since L is algebraic over
K(β1 , . . . , βi , α1 , . . . , αr ), we have established the inductive step.
Corollary 8.3.10. Let L be an extension of finite transcendence degree over K. Then
any two transcendence bases of L over K have the same number of elements.
Thus, we may define transcendence degree.
Definition 8.3.11. Let L be an extension of finite transcendence degree over K. Then
the transcendence degree of L over K is the number of elements in any transcendence
basis of L over K.
Exercises 8.3.12.
1. Suppose that L has transcendence degree n over K and that L is algebraic over
K(α1 , . . . , αn ). Show that α1 , . . . , αn is a transcendence basis for L over K.
2. Suppose given extensions K ⊂ L ⊂ L1 such that L has finite transcendence degree
over K and L1 has finite transcendence degree over L. Show that L1 has finite
transcendence degree over K, and that the transcendence degree of L1 over K is
the sum of the transcendence degree of L over K and the transcendence degree of
L1 over L.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
8.4
317
Algebraic Closures
Definition 8.4.1. A field K is algebraically closed if every polynomial in K[X] of degree
≥ 1 has a root in K.
We shall show in Section 11.6, in what’s known as the Fundamental Theorem of Algebra, that the complex numbers, C, are an algebraically closed field. Every argument
for this relies at some point on topology. Perhaps the simplest proof makes use of fundamental groups, but requires no ring or field theory beyond the definitions of complex
multiplication and of polynomials. However, the topology would require too much development to present here. So we shall give a proof that is more algebraic, and relies on
the Fundamental Theorem of Galois Theory, in Chapter 11.
Lemma 8.4.2. Let K be an algebraically closed field. Then every polynomial f (X) of
degree ≥ 1 in K[X] breaks up as a product
f (X) = a(X − a1 ) . . . (X − an )
in K[X], where a, a1 , . . . , an ∈ K.
Proof Since every polynomial of degree ≥ 1 has a root in K, it has a degree 1 factor.
So no polynomial of degree > 1 is irreducible. So the irreducibles are precisely the degree
1 polynomials, and the result follows from unique factorization in K[X].
A simpler argument, by induction on degree, would have sufficed, of course. We shall
make use of the next lemma in Section 9.3.
Lemma 8.4.3. Every algebraically closed field is infinite.
Proof First note that Z2 is not algebraically closed, as X 2 + X + 1 has no roots in Z2 .
Let K be a finite field with more than two elements and let a be a non-identity
element of K × . Let n be the order of K × . Then every element of K × has exponent n,
and hence X n − a has no roots in K.
Often, problems over R are most easily solved by passing to C and first solving the
problem there. In general, it can be very useful to study a given field by making use of
an algebraically closed field in which it embeds.
Definition 8.4.4. Let K be a field. An algebraic closure of K is an algebraic extension,
L, of K, such that L is algebraically closed.
Notice that this is a quite different notion from the algebraic closure of K in some
extension, L, of K (Definition 8.2.20), though that concept can be useful in constructing
some algebraic closures.
Proposition 8.4.5. Suppose L is an extension of K that is algebraically closed. Let
K1 ⊂ L be the algebraic closure of K in L (i.e., the set of elements of L that are
algebraic over K.) Then K1 is an algebraic closure of K.
Proof We’ve seen in Proposition 8.2.21 that K1 is an algebraic extension of K. So it
suffices to show that K1 is algebraically closed.
Thus, suppose that f (X) ∈ K1 [X] has positive degree. Since L is algebraically closed,
f has a root, α, in L. It suffices to show that α is in K1 .
But α is algebraic over K1 , and by Proposition 8.2.23, the only elements of L that
are algebraic over K1 are those of K1 .
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
318
Thus, once we’ve established the Fundamental Theorem of Algebra, we’ll know that
the algebraic closure of Q in C is an algebraic closure of Q. In particular, it will be
our preferred model for the algebraic closure of Q, as we’ve become accustomed to
visualizing the elements of C. It also allows the use of complex analysis to study the
algebraic extensions of Q, and forms the basis for the field of analytic number theory.
Here, we shall give an abstract proof that algebraic closures exist. We shall treat the
uniqueness of algebraic closures in Chapter 11.
Lemma 8.4.6. Let f1 (X), . . . , fk (X) be polynomials over K. Then there exists an extension of K in which each of the polynomials fi has a root.
Proof By induction, it suffices to treat the case of a single polynomial f (X). And
by factoring f , if necessary, we may assume that f (X) is irreducible in K[X]. Thus,
L = K[X]/(f (X)) is a field, containing K as a subfield. Moreover, if α ∈ L is the image
of X under the canonical map from K[X], then f (α) = 0, and hence α is a root of f in
L.
The construction we shall use relies on polynomial rings in infinitely many variables.
Here, if A is a ring and S is a set of variables, the elements of the polynomial ring A[S]
are simply polynomials in finitely many of the variables of S with coefficients in A. We
add and multiply these as we would in a polynomial ring in finitely many variables, and
may do so because for any finite collection of elements of A[S], there is a finite subset
{X1 , . . . , Xn } such that each of the polynomials in this collection lies in A[X1 , . . . , Xn ] ⊂
A[S]. (In categorical language, A[S] is the direct limit of the polynomial rings on the
finite subsets of S.)
Polynomials in infinitely many variables behave analogously to polynomials in finitely
many variables. In particular, there is an inclusion map i : S → A[S] which, as the reader
may verify, is universal in the following sense.
Proposition 8.4.7. Let A be a commutative ring and let B be a commutative A-algebra.
Let S be a set of variables and let f : S → B be any function. Then there is a unique
A-algebra homomorphism εf : A[S] → B such that εf ◦ i = f . The image of εf is
A[f (Y ) | Y ∈ S], the smallest A-subalgebra of B containing the image of f .
We shall refer to εf as the A-algebra homomorphism obtained by evaluating each
Y ∈ S at f (Y ). The next lemma is the key step in constructing algebraic closures.
Lemma 8.4.8. Let K be a field. Then there is an algebraic extension K1 of K such that
every polynomial in K[X] has a root in K1 .
Proof It suffices to construct K1 so that each monic irreducible polynomial of degree
> 1 in K[X] has a root in K1 . Thus, for each monic irreducible polynomial f (X) of
degree > 1 in K[X], we define a variable Xf , and we let S be the set of all such Xf .
Then substituting Xf for the variable X, we obtain the polynomial f (Xf ) ∈ K[S].
Let a be the ideal of K[S] generated by the set {f (Xf ) | Xf ∈ S}. We claim that a is
a proper ideal of K[S].
To see this, we argue by contradiction. Suppose that 1 ∈ a. Then there are
monic irreducible polynomials f1 (X), . . . , fk (X) of degree > 1 in K[X] and polynomials
g1 , . . . , gk ∈ K[S] such that
1 = g1 f1 (Xf1 ) + · · · + gk fk (Xfk ).
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
319
By Lemma 8.4.6, there is an extension L of K in which each of f1 (X), . . . , fk (X) has a
root. Say αi ∈ L is a root of fi (X) for 1 ≤ i ≤ k. Let h : S → L be any function such
that h(Xfi ) = αi for 1 ≤ i ≤ k and let εh : K[S] → L be the K-algebra map obtained
by evaluating each Xf in S at h(Xf ).
Applying εh to the displayed equation above, we get
1
= εh (g1 )f1 (α1 ) + · · · + εh (gk )fk (αk )
=
0,
as αi is a root of fi for 1 ≤ i ≤ k.
Thus, a is a proper ideal of K[S], and hence there is a maximal ideal m of K[S] that
contains it. So let K1 = K[S]/m. Now write βf for the image in K1 of Xf under the
canonical map π : K[S] → K1 for each Xf ∈ S. Then f (βf ) = π(f (Xf )) = 0, and
hence each monic irreducible polynomial f (X) ∈ K[X] of degree > 1 has a root in K1 ,
as claimed. Also, each βf is algebraic over K. Since K1 = K[βf | Xf ∈ S], it is algebraic
over K by Proposition 8.2.12 and Corollary 8.2.17.
Now, we can construct algebraic closures.
Proposition 8.4.9. Let K be a field. Then K has an algebraic closure.
Proof By induction, using Lemma 8.4.8, construct a sequence
K = K 0 ⊂ K1 ⊂ · · · ⊂ K n ⊂ · · ·
of field extensions such that for i ≥ 0, Ki+1 is an algebraic extension of Ki with the
property that each polynomial in Ki [X] has a root in Ki+1 . Note that an inductive
application of Corollary 8.2.18 now shows that Ki is an algebraic extension of K for all
i.
Using set theory, we may construct a set in which the Ki all embed, compatibly
with their embeddings in each other. For instance, the direct limit −
lim
→ Ki constructed in
Section 6.8 has that property. We set L equal to the union of the Ki in such a set.
Thus, each element of L lies in Ki for some i, and for any finite set of elements in L,
we can find a Ki that contains them all. Thus, the addition and multiplication operations
on the Ki induce addition and multiplication operations in L, and the fact that each Ki
is a field implies that L is a field under these operations.
Since every element of L lies in some Ki , each element of L is algebraic over K.
Also, if f (X) ∈ L[X], then there is some Ki that contains all of the coefficients of f . In
particular, f (X) ∈ Ki [X] ⊂ L[X], and hence has a root in Ki+1 , and hence in L. Thus,
L is algebraically closed, and hence an algebraic closure for K.
Exercises 8.4.10.
1. Show that the algebraic closure of Q in C (and hence any algebraic closure of Q,
once we have the uniqueness statement) is countable.
2. Show that if K is an algebraically closed field and if L is any extension of K, then
K is algebraically closed in L.
3. Let K be an algebraically closed field and let D be a division ring that is also a Kalgebra. Suppose that D is finite dimensional as a vector space over K, under the
vector space structure induced by its K-algebra structure. Show that the inclusion
K ⊂ D given by the K-algebra structure map is a bijection, so that K = D.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
8.5
320
Criteria for Irreducibility
We wish to study the irreducibility of polynomials over Q. For this purpose, it is useful
to note that if f (X) = αn X n + · · · + α0 with α0 , . . . , αn ∈ Q, and if m is divisible by the
least common multiple of the denominators (when placed in lowest terms) of α0 , . . . , αn ,
then mf (X) has coefficients in Z.
We can use this technique to reduce questions about Q[X] to questions about Z[X].
Along the way, we shall show that Z[X] is a U.F.D.
Let p be a prime number in Z. Then the divisors of p in Z[X] all lie in Z, and hence
p is irreducible in Z[X]. Thus, the following result is necessary for Z[X] to be a U.F.D.
Lemma 8.5.1. Let p be a prime number and let f (X), g(X) ∈ Z[X] be such that p
divides f (X) · g(X). Then p must divide at least one of f and g.
In other words, the principal ideal generated by p in Z[X] is a prime ideal.
n
m
i
i
Proof Write f (X) =
i=0 ai X and g(X) =
i=0 bi X . Suppose that p does not
divide f . Then there is a smallest index i such that ai is not divisible by p. Similarly, if
g is not divisible by p, there is a smallest index j such that bj is not divisible by p.
i+j
Now the i + j-th coefficient of f (X)g(X) is k=0 ak bi+j−k . If k < i, ak is divisible
by p. If k > i, bi+j−k is divisible by p. Thus, the i + j-th coefficient of f (X)g(X) is
divisible by p if and only if ai bj is divisible by p. Since neither ai nor bj is divisible by
p, this cannot be. Thus, either f or g must have been divisible by p.
The greatest common divisor of a collection of integers is the largest integer that
divides all of them. It can be computed in terms of primes in the usual way (Corollary 8.1.31).
Definitions 8.5.2. The content, c(f ), of a polynomial f (X) ∈ Z[X] is the greatest
common divisor of the coefficients of f .
We say that f (X) is primitive if it has content 1.
The next result is a corollary of Lemma 8.5.1.
Corollary 8.5.3. Let f (X), g(X) ∈ Z[X]. Then c(f (X)g(X)) = c(f ) · c(g).
Proof Dividing, we can write f (X) = c(f )f1 (X) and g(X) = c(g)g1 (X), where f1 (X)
and g1 (X) are primitive. Multiplying these, we see that it suffices to show that the
product of two primitive polynomials is primitive.
But if the product of two polynomials is not primitive, then it must be divisible by
some prime number p, in which case p must divide one of the two polynomials.
This now sets up the very useful Gauss Lemma.
Lemma 8.5.4. (Gauss Lemma) Let f (X) ∈ Z[X] be a polynomial of degree > 1. Then
f (X) is reducible in Q[X] if and only if f factors in Z[X] as the product of two polynomials of degree ≥ 1.
Proof Suppose that f (X) = g(X)h(X) in Q(X), with both g and h of positive degree.
By first clearing the denominators and then factoring out contents, we can write
m
k
and
h(X) = h1 (X),
g1 (X)
n
l
where g1 and h1 are primitive polynomials in Z[X] and k, l, m, n ∈ Z.
Thus, nlf (X) = mkg1 (X)h1 (X). Equating the contents of both sides, we see nlc(f ) =
±mk, and hence nl divides mk. In particular, mk = nlr for some r ∈ Z, and hence
f (X) = rg1 (X)h1 (X) does indeed factor in Z[X] as claimed.
g(X) =
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
321
A variant of the above argument is used in the proof of the next lemma.
Lemma 8.5.5. Let f (X) ∈ Z[X] be a primitive polynomial that is irreducible as an
element of Q[X]. Then f (X) is a prime element in Z[X].
Proof Suppose given g(X), h(X) ∈ Z[X] such that f divides g(X)h(X) in Z[X]. Then
f also divides g(X)h(X) in Q[X]. Since f (X) is a prime element in the P.I.D. Q[X], it
must divide one of g and h there. Let’s say that g(X) = f (X)g1 (X) in Q[X].
Write g1 (X) = m
n g2 (X), where g2 (X) is a primitive polynomial in Z[X] and m, n ∈ Z
with (m, n) = 1. Then
ng(X) = mf (X)g2 (X).
Comparing contents, we see that nc(g) = ±m. Since (m, n) = 1, this gives n = ±1, and
hence g(X) = ±mf (X)g2 (X) is divisible by f (X) in Z[X].
The next lemma adds a little clarity.
Lemma 8.5.6. Let f (X) ∈ Q[X]. Then up to sign, there is a unique primitive polynomial in Z[X] that generates the same principal ideal of Q[X] as f does. In particular, if
f is irreducible, then up to sign there is a unique primitive polynomial in Z[X] equivalent
to f as a prime in Q[X].
Proof It suffices to show that if g(X) and h(X) are primitives in Z[X] that differ by
a unit in Q[X], then g = ±h. But if g and h differ by a unit in Q[X], then there are
integers m, n such that mg(X) = nh(X). Comparing contents, we see that m = ±n, and
hence g = ±h.
To get a unique choice of primitive polynomial in Z[X] equivalent to a given prime
in Q[X], we may choose the one whose leading coefficient is positive.
We now obtain the prime decompositions in Z[X].
Proposition 8.5.7. Let f (X) be a polynomial of degree > 0 over Z and suppose that
the prime decomposition of f (X) in Q[X] is
f (X) =
m
r
r
(p1 (X)) 1 . . . (pk (X)) k ,
n
where p1 (X), . . . , pk (X) are pairwise inequivalent prime elements in Q[X]. Then if qi (X)
is the unique primitive polynomial in Z[X] with positive leading coefficient that is equivalent to pi in Q[X], we have
r
rk
f (X) = ±c(f ) (q1 (X)) 1 . . . (qk (X))
in Z[X].
Proof By uniqueness of prime decomposition in Q[X], we have
f (X) =
k
r
r
(q1 (X)) 1 . . . (qk (X)) k
l
for some k, l ∈ Z, and hence
r
rk
lf (X) = k (q1 (X)) 1 . . . (qk (X))
.
Comparing contents, we see that lc(f ) = ±k, so the result follows by dividing l into k.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
322
This leads directly to the verification that Z[X] is a U.F.D.
Corollary 8.5.8. The polynomial ring Z[X] is a U.F.D. whose prime elements are, up
to sign, the primes in Z, together with the primitive polynomials that are irreducible in
Q[X].
Proof We’ve already seen that the above elements are prime elements of Z[X]. By
factoring c(f ) in Z, Proposition 8.5.7 shows that every nonzero nonunit in Z[X] may be
written as a product of these primes. Thus, Z[X] is a U.F.D.
The uniqueness statement for prime decompositions in a U.F.D. now shows that every
prime element in Z[X] must be equivalent to one of these primes. As the units in Z[X]
are ±1, the primes here are determined up to sign.
Actually, the proofs given apply to a much more general setting. The reader may
verify the following proposition.
Proposition 8.5.9. Let A be a U.F.D. with fraction field K. Then A[X] is a U.F.D.
whose prime elements are the primes in A, together with the primitive polynomials of
A[X] that are irreducible in K[X].
The following criterion for irreducibility shows the value of the Gauss Lemma.
n
Proposition 8.5.10. (Eisenstein’s Criterion) Let f (X) = i=0 ai X i ∈ Z[X] and let p
be a prime. Suppose that the leading coefficient an is relatively prime to p and that p
divides ai for 0 ≤ i < n. Suppose also that p2 does not divide the constant term a0 . Then
f (X) is irreducible in Q[X].
Proof If f (X) is reducible in Q[X], then the Gauss Lemma provides a factorization
f (X) = g(X)h(X) in Z[X], where g and h both have positive degree. Write g(X) =
k
n−k
i
i
i=0 bi X and h(X) =
i=0 ci X .
Since p divides the constant term of f but p2 does not, p divides the constant term
of exactly one of the polynomials g and h. Let’s say that it divides the constant term of
g.
We claim now by induction that p must divide every coefficient of g. Thus, suppose
i
that p divides b0 , . . . , bi−1 for i ≤ k. Then we have ai = j=0 bj ci−j . Now p divides ai ,
since i ≤ k and k < n. Also, for j < i, p divides bj , and hence divides the summand
bj ci−j . Thus, p must divide the remaining summand bi c0 . Since p does not divide c0 , it
must divide bi .
In particular, we obtain that p divides the leading coefficient bk . But an = bk cn−k ,
contradicting the fact that p does not divide an . Thus, f must not be reducible.
We shall see that there are lots of irreducible polynomials that do not satisfy Eisenstein’s criterion, but many do.
Corollary 8.5.11. Let m be a positive integer that’s divisible by a prime p but is not
divisible by p2 . Then X n − m is irreducible in Q[X] for all n ≥ 1. In particular,
√
[Q( n m) : Q] = n.
Proof The polynomial X n − m clearly
√ satisfies Eisenstein’s criterion for the prime p,
and hence is irreducible over Q. Since n m is a root of X n −m, its minimal polynomial is
a prime
divisor of X n − m. But X n − m is prime, so it must be the minimal polynomial
√
n
of m.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
323
Corollary 8.5.12. For each n ≥ 2, there is a simple extension of Q of degree n.
In some cases, we may apply Eisenstein’s criterion via a change of variables.
Corollary 8.5.13. For r ≥ 1, X 2
particular,
r−1
+ 1 is the minimal polynomial of ζ2r over Q. In
[Q(ζ2r ) : Q] = 2r−1 .
Proof Write f (X) = X 2
+ 1. Since ζ2r has order 2r in C× , and since −1 is the
×
unique element of C of order 2, ζ2r is a root of f . Thus, it suffices to show that f (X)
is irreducible over Q.
Let f (X + 1) be the polynomial obtained by evaluating f at X + 1. Note that for
any polynomial g(X), g(X + 1) has the same degree as g(X). Thus, if f (X) is reducible,
f (X + 1) will be reducible also. So it suffices to show that f (X + 1) is irreducible.
r−1
r−1
Now f (X + 1) = (X + 1)2
+ 1, and (X + 1)2
can be expanded by the Binomial
Theorem (Theorem 4.5.16). We have
r−1
2r−1
(X + 1)
=
r−1
2
i=0
2r−1
X i.
i
r−1
In particular, (X + 1)2
is monic, and hence so is f (X + 1), while the constant term of
2r−1
(X + 1)
is 1.
Thus, the constant term of f (X +1) is 2. Thus, the result will follow from Eisenstein’s
r−1 criterion for p = 2, provided that the coefficients 2 i
are even for 0 < i < 2r−1 . But
this follows immediately from Lemma 4.5.17.
Exercises 8.5.14.
1. Show that X n − 1 = (X − 1)(X n−1 + X n−2 + · · · + 1) for all n > 1. Deduce that
the minimal polynomial of ζn divides (X n−1 + X n−2 + · · · + 1).
† 2. Let p be an odd prime and let f (X) = X p−1 + X p−2 + · · · + 1. The preceding
problem gives (X + 1)p − 1 = X · f (X + 1). Deduce from Eisenstein’s criterion that
f (X + 1) is irreducible, and hence f (X) is the minimal polynomial of ζp . Deduce
that
[Q(ζp ) : Q] = p − 1.
3. Suppose that f (X) ∈ Z[X] has a leading coefficient prime to p, and that f (X) is
irreducible in Zp [X]. Show that f (X) must be irreducible in Q[X].
4. Show that X 4 + 1 is not irreducible in Z2 [X]. Deduce that the converse of the
preceding problem is false.
5. Let f (X) ∈ Z[X] be primitive, and write Z[X] · f (X) and Q[X] · f (X) for the
principal ideals of Z[X] and Q[X], respectively, that are generated by f . Show
that
(Q[X] · f (X)) ∩ Z[X] = Z[X] · f (X).
6. Show that Eisenstein’s criterion holds in the field of fractions of any U.F.D.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
8.6
324
The Frobenius
The material here is very basic, but it is important enough to deserve emphasis.
Let p be a prime and let A be any commutative ring of characteristic p. By Proposition 7.2.14, this means that the unique ring homomorphism Z → A factors through
an embedding of Zp in A. But since Zp is a field, any ring homomorphism out of Zp
is injective, and hence the rings of characteristic p are precisely the Zp -algebras. Notice
that the Zp -algebra structure on a ring A of characteristic p is unique, and that A has a
unique subring isomorphic to Zp . We identify this subring with Zp .
Definition 8.6.1. Let p be a prime and let A be a commutative ring of characteristic
p. Then the Frobenius map ϕ : A → A is defined by ϕ(a) = ap for all a ∈ A.
Proposition 8.6.2. Let A be a commutative ring of characteristic p. Then the Frobenius
map ϕ : A → A is a ring homomorphism.
Proof This is an immediate application of the Binomial Theorem. We have
(a + b)p =
p p
i=0
i
ai bp−i .
But for 0 < i < p, Lemma 4.5.17 shows pi to be divisible by p. But every element
in a ring of characteristic p has additive exponent p (Corollary 7.7.16), and hence the
summands with 0 < i < p all vanish. We are left with (a + b)p = ap + bp , so the Frobenius
is a homomorphism of additive groups. But the rest is immediate.
Corollary 8.6.3. Let K be a field of characteristic p. Then the Frobenius map ϕ : K →
K is an embedding. In particular, if K is finite, then ϕ is an automorphism of the field
K.
For K either finite or infinite, the elements fixed by ϕ (i.e., the elements x ∈ K with
ϕ(x) = x are precisely the elements of Zp .
Proof Of course, every homomorphism of fields is an embedding. As far as the fixed
points are concerned, note that x is a fixed point of ϕ if and only if x is a root of X p − X.
As a polynomial of degree p over the field K, X p − X has at most p roots in K. So it
suffices to show that every element of Zp is fixed by ϕ.
p−1
Now Z×
= 1 for all 0 = x ∈ Zp . So xp = x
p is a (cyclic) group of order p − 1, so x
for all x ∈ Zp .
Definition 8.6.4. Let K be a field of characteristic p. We say that K is perfect if the
Frobenius map ϕ : K → K is onto, and hence is an automorphism of the field K.
We also say that every field of characteristic 0 is perfect.
The next corollary is now immediate from Corollary 8.6.3.
Corollary 8.6.5. All finite fields are perfect.
Since Zp [X] has characteristic p, it has a Frobenius homomorphism, ϕ.
Lemma 8.6.6. The Frobenius homomorphism ϕ : Zp [X] → Zp [X] is given by ϕ(f (X)) =
f (X p ), the effect of evaluating f at X p .
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
325
Proof Since ϕ is a homomorphism, we have
#
" n
n
i
=
ai X
ϕ(ai )ϕ(X)i
ϕ
i=0
i=0
=
n
ai (X p )i
i=0
since ϕ is the identity on Zp . But this last expression is f (X p ).
We now show that the field of rational functions, Zp (X), of Zp is imperfect.
Corollary 8.6.7. The Frobenius map on Zp (X) gives an isomorphism from Zp (X) onto
Zp (X p ), the field of fractions of the subring Zp [X p ]. In particular, Zp (X) is not perfect.
Proof The elements of Zp (X) have the form f (X)/g(X), where f (X), g(X) ∈ Zp [X]
with g(X) = 0. Clearly,
ϕ (f (X)/g(X)) = ϕ (f (X)) /ϕ (g(X))
= f (X p )/g(X p ),
so the image of ϕ is Zp (X p ), as claimed. To see that this is not all of Zp (X), note that
if f (X p )/g(X p ) = X/1 in Zp (X), then f (X p ) = Xg(X p ), which is impossible due to the
congruences mod p of the exponents that appear.
Exercises 8.6.8.
1. Show that every algebraically closed field is perfect.
2. Let k be a perfect field and let f (X) ∈ k[X]. Show that there is a polynomial
g(X) ∈ k[X] such that f (X p ) = (g(X))p .
3. Let K be any field of characteristic p = 0. Show that K(X) is not perfect and
calculate the image of ϕ.
4. Let A be a ring of characteristic 2. Show that the Frobenius function ϕ(a) = a2
gives a ring homomorphism from A to A if and only if A is commutative.
8.7
Repeated Roots
The material here is essential in understanding Galois theory for fields of nonzero characteristic. We shall make use of it in studying separable extensions. For now, we look at
the elementary properties involved in the issues of repeated roots.
Definitions 8.7.1. Let L be an extension of K and let α ∈ L be a root of the polynomial
f (X) ∈ K[X]. We say that α is a repeated root, or multiple root, of f if (X − α)2 divides
f in L[X].
We say that α is a root of multiplicity k for f (X) if
f (X) = (X − α)k g(X)
in L[X], where g(α) = 0.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
326
On the surface, this would seem to depend on the field L, but in fact, it only depends
on f and α.
Lemma 8.7.2. Let L be an extension field of K and let f (X) ∈ K[X]. Suppose that
α ∈ L is a root of multiplicity k ≥ 1 of f (X). Then there is a factorization
f (X) = (X − α)k g(X)
in K(α)[X], with g(α) = 0. In particular, α is a root of multiplicity k of f when
considered as an element of any extension field of K(α).
Proof Since f (α) = 0, the Euclidean algorithm in K(α)[X] gives a factorization f (X) =
(X − α)f1 (X) in K(α)[X]. And unique factorization in L[X] says that α is a root of
multiplicity k − 1 of f1 (X). The result follows by induction.
Example 8.7.3. Over Q(i), X 4 + 2X 2 + 1 factors as (X − i)2 (X + i)2 . So i and −i are
both multiple roots of the rational polynomial X 4 + 2X 2 + 1.
It is a little more difficult to produce an example of an irreducible polynomial with
repeated roots.
Example 8.7.4. Let L = Zp (T ) be the field of fractions of the polynomial ring Zp [T ].
Let K = Zp (T p ) be the field of fractions of the subring Zp [T p ] of Zp [T ].
Then T ∈ L is a root of X p −T p ∈ K[X]. Moreover, by consideration of the Frobenius
homomorphism in L[X], we obtain a factorization
X p − T p = (X − T )p
in L[X]. So T is a repeated root of X p − T p .
Since X − T is prime in L[X], an examination of constant terms shows that no proper
factor of (X − T )p can have coefficients in K. So X p − T p must be irreducible in K[X].
We shall see later that this shows L to be an inseparable extension of K.
There is a formal derivative for polynomials, used to detect multiple roots.
n
Definition 8.7.5. Let A be a commutative ring and let f (X) = i=0 ai X i be a polynomial in A. The formal derivative of f is defined as follows:
d
(f (X)) = f (X) =
iai X i−1 .
dX
i=1
n
Lemma 8.7.6. Let A be a commutative ring. Then the formal derivative
A[X] is an A-module homomorphism satisfying the product rule:
d
dX
: A[X] →
d
(f (X)g(X)) = f (X) · g(X) + f (X) · g (X).
dX
A morphism with these properties is often called an A-linear derivation.
The formal derivative detects multiple roots in the following manner.
Lemma 8.7.7. Let L be an extension field of K and let f (X) ∈ K[X]. Then an element
α ∈ L is a repeated root of f if and only if α is a root of both f (X) and f (X).
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
327
Proof If α is a repeated root of f , then we have f (X) = (X − α)2 g(X) in L[X]. The
product rule then gives
f (X) = (X − α) [2g(X) + (X − α)g (X)]
and hence α is a root of f (X) also.
Conversely, if α is a root of both f (X) and f (X), then f (X) = (X − α)g(X) in L[X],
and it suffices to show that α is a root of g(X). Here, the product rule gives
f (X) = g(X) + (X − α)g (X),
and hence f (α) = g(α). Since α is a root of f , the result follows.
This allows us to characterize the irreducible polynomials with multiple roots.
Proposition 8.7.8. Let L be an extension field of K and let f (X) be an irreducible
element of K[X] with a root in L. Then f has a repeated root in L if and only if the
formal derivative f (X) is the zero polynomial, in which case every root of f (X) is a
repeated root.
If K has characteristic 0, this cannot happen at all. If K has characteristic p, then the
derivative f (X) vanishes if and only if f (X) = g(X p ) for some irreducible polynomial
k
g(X) in K[X]. Inductively, if f (X) has a repeated root, we may write f (X) = h(X p )
for some k > 0, where h(X) is irreducible in K[X] and h (X) = 0.
Proof If α ∈ L is a repeated root of f , then α must also be a root of f (X). But f is
irreducible, and hence must be, up to multiplication by a scalar, the minimal polynomial
of α. Since α is a root of f , f must divide f . But the degree of f is less than that of
f , so this can only happen if f is the zero polynomial. Note that if f = 0, then every
root of f is also a root of f and hence is a multiple root of f .
n
Since f is irreducible, it must have degree at least one. Say f (X) = i=0 ai X i with
n > 0 and an = 0. So if K has characteristic 0, the term nan X n−1 in the expansion of
f (X) is nonzero, and hence f (X) is nonzero. Thus, α could not have been a repeated
root of f .
If K has characteristic p, then f (X) vanishes if and only if the only nonzero coeffik
cients of f (X) are those in degrees divisible by p. This then gives f (X) = i=0 bi X pi =
k
g(X p ), where bi = api , k = np and g(X) = i=0 bi X i .
Since any factorization of g would induce a factorization of f (X) = g(X p ), we see
that g is irreducible.
For the final statement, note that if g does not have repeated roots, we may take
h(X) = g(X) and k = 1. Otherwise, the result follows by induction on degree, as the
degree of g is less than that of f .
8.8
Cyclotomic Polynomials
We characterize the minimal polynomial of ζn over Q, for each n.
First, consider the subgroup of Q(ζn )× generated by ζn :
ζn = {1, ζn , . . . , ζnn−1 },
a cyclic group of order n. Thus, the order of ζnk is n/(n, k), and hence ζnk generates ζn if and only if (n, k) = 1.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
328
Definitions 8.8.1. The primitive n-th roots of unity in Q(ζn ) (or in C) are the generators of ζn .
We define the Euler φ function by setting φ(n) equal to the number of primitive n-th
roots of unity for n ≥ 1.
Thus, φ(n) is the number of generators of a cyclic group of order n. As noted above,
m generates Zn if and only if (m, n) = 1. We see (e.g., from Lemma 4.5.3) that φ(n) is
the order of the group of units Z×
n of the ring Zn . As such, a calculation of φ(n) is given
in Section 4.5, where it is shown that φ(n) is the order of the automorphism group of
the cyclic group Zn , which is then calculated in Corollary 4.5.9:
Corollary 8.8.2. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes and the exponents are all positive. Then
φ(n) = p1r1 −1 (p1 − 1) . . . prkk −1 (pk − 1).
Notice that since ζn ⊂ Q(ζn )× has order n, every element in ζn is a root of X n −1.
In particular, this means that there are n distinct roots of X n − 1 in Q(ζn ). Factoring
out the associated degree one polynomials and equating leading coefficients, we obtain a
factorization of X n − 1.
Lemma 8.8.3. In Q(ζn )[X], we have
Xn − 1 =
n−1
.
(X − ζnk ).
k=0
Corollary 8.8.4. In Z[X], there is a decomposition
X n − 1 = p1 (X) . . . pk (X)
where p1 (X), . . . , pk (X) are distinct monic polynomials in Z[X] that are irreducible in
Q[X].
Proof By Proposition 8.5.7, we may write
X n − 1 = ±p1 (X) . . . pk (X)
in Z[X], where p1 (X), . . . , pk (X) are (not necessarily distinct) primitive polynomials in
Z[X] that are irreducible in Q[X]. We may assume that the pi (X) all have positive
leading coefficients. But in this case, since X n − 1 is monic, each pi (X) must be monic
as well, and the sign in the displayed equation must be positive. So it suffices to show
that the pi are all distinct.
Suppose that pi (X) = pj (X) for some i = j. As a factor of X n − 1 of positive degree,
pi (X) must have at least one root, say α, in Q(ζn ) by Lemma 8.8.3. But then α is also
a root of pj , and hence is a repeated root of X n − 1. Since X n − 1 has n distinct roots
in Q(ζn ), this is impossible.
Since each pi (X) is monic and is irreducible in Q[X], it must be the minimal polynomial over Q of any of its roots in Q(ζn ). Since each ζnk is a root of X n − 1, its minimal
polynomial must be one of the pi .
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
329
Definition 8.8.5. We write Φn (X) for the minimal polynomial over Q of ζn . We call
Φn (X) the n-th cyclotomic polynomial over Q.
By the above, we see that Φn (X) is a monic polynomial in Z[X].
Examples 8.8.6.
1. Clearly, Φ1 (X) = X − 1, and Φ2 (X) = X + 1.
2. We’ve seen in Corollary 8.5.13 that
Φ2r (X) = X 2
r−1
+ 1.
3. In Problem 2 of Exercises 8.5.14, it is shown that if p is an odd prime, then
Φp (X) = 1 + X + · · · + X p−1 .
We now obtain our main result.
Proposition 8.8.7. The roots of Φn (X) in Q(ζn ) are precisely the primitive n-th roots
of unity. In particular, [Q(ζn ) : Q] = φ(n), and
.
(X − ζnk )
Φn (X) =
0<k<n
(n,k)=1
in Q(ζn )[X].
Proof Suppose that ζnk is not a primitive n-th root of unity. Then it is a root of X d − 1
for some d < n that divides n, and hence the minimal polynomial of ζnk over Q must
divide X d − 1. Thus, none of the roots of the minimal polynomial of ζnk may be primitive
n-th roots of unity, and, in particular, ζnk cannot be a root of Φn (X).
Thus, it suffices to show that every primitive n-th root of unity is a root of Φn (X).
Since the primitive roots of unity all have the form ζnk with (n, k) = 1, induction on the
sum of the exponents in the prime decomposition of k shows that it suffices to show that
if η is a root of Φn (X) and if p is a prime that does not divide n, then η p is also a root
of Φn (X).
Thus, given a root η of Φn (X) and a prime p that does not divide n, suppose that η p
is not a root of Φn (X). Then let f (X) be the minimal polynomial of η p over Q. By the
discussion above, f (X) is a monic polynomial in Z[X], and we may write
X n − 1 = Φn (X) · f (X) · g(X)
for some g(X) ∈ Z[X].
Since η p is a root of f (X), η is a root of f (X p ). Thus, Φn (X) divides f (X p ) in Q[X],
and hence also in Z[X], by Proposition 8.5.7.
Now let us reduce the coefficients of all these polynomials mod p. In Zp [X], Lemma 8.6.6
shows f (X p ) to be equal to the p-th power (f (X))p . Thus, Φn (X) divides (f (X))p in
Zp [X], and hence Φn (X) and f (X) must have a common prime factor, h(X), in their
prime decompositions in Zp [X].
Since Φn (X)f (X) divides X n − 1, (h(X))2 must divide X n − 1 in Zp [X]. But then
n
X − 1 will have a repeated root in the field Zp [X]/(h(X)). But this is impossible, since
d
(X n − 1) = nX n−1 . Since 0 is not a root of X n − 1, Lemma 8.7.7
0 is the only root of dX
n
shows that X − 1 cannot have repeated roots in any field of characteristic prime to
n.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
330
And in particular, the next corollary will now give an inductive, though tedious, way
to calculate the polynomials Φn (X) explicitly.
Corollary 8.8.8. The prime decomposition of X n − 1 in either Q[X] or Z[X] is given
by
.
Φd (X)
Xn − 1 =
d|n
where the product is taken over all positive integers d that divide n.
Proof Let d divide n, say, n = dr. Then
X n − 1 = (X d − 1)(1 + X d + · · · + X d(r−1) ).
Thus, X d − 1 divides
X n − 1, and hence so does Φd (X). Assembling this over the various
!
d, we see that d|n Φd (X) divides X n − 1.
Thus, it suffices to show that each ζnk is a root of Φd (X) for some d that divides n.
Thus, let r = (n, k) and let d = nr . Then ζnr = ζd , and hence ζnk = ζdl , where l = kr is
relatively prime to d. In particular, ζnk is a primitive d-th root of unity, and hence a root
of Φd (X).
As a dividend, we obtain a determination of the ring structure on the group ring
Q[Zn ].
Corollary 8.8.9. There is an isomorphism of rings
.
∼
=
f : Q[Zn ] −→
Q(ζd )
d|n
where the product ranges over all positive integers d that divide n. Here, if Zn = T ,
then the coordinate of f (T ) corresponding to a given d dividing n is ζd .
Proof We shall construct f to make the following diagram commute.
Q[X]/(X − 1)
n
γ /
∼
=
f
Q[X]/(Φd (X))
d|n
!
εT ∼
=
Q[Zn ]
.
d|n εζd
/
.
∼
=
Q(ζd )
d|n
Here, γ is the isomorphism coming out of the second form of the Chinese Remainder Theorem (Corollary 8.1.34), while the vertical maps are induced by the indicated evaluation
maps.
The right-hand vertical map is an isomorphism by the definition of minimal polynomial. The left-hand vertical map is an isomorphism by Problem 16 of Exercises 7.5.10.
We define f by traversing the diagram. It is easily seen to have the indicated effect on
the generator T .
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
331
The analogous result for Z[Zn ] fails. The maps in the square are all well defined, and
the vertical maps are isomorphisms, but the map analogous to γ fails to be surjective.
The point is that Z[X] is a U.F.D., but not a P.I.D. In a U.F.D., elements a and b may
be relatively prime without (a) + (b) being equal to the whole ring. So the key step in
the proof of the first form of the Chinese Remainder Theorem (Proposition 7.2.23) does
not apply here.
Exercises 8.8.10.
1. Calculate Φ6 (X), Φ9 (X), Φ12 (X), and Φ15 (X).
2. Give a general formula for Φpr (X) for odd primes p.
3. Give a general formula for Φ2p (X) for odd primes p.
4. Let (m, n) = 1. Show that [Q(ζmn ) : Q(ζm )] = φ(n). Deduce that Q(ζn )∩Q(ζm ) =
Q.
5. Show that Q(ζn ) ∩ R = Q(ζn + ζ n ) = Q(ζn + ζn−1 ). Show that if n > 2, then
[Q(ζn ) : Q(ζn + ζn−1 )] = 2.
6. Find the minimal polynomial of ζ8 + ζ8−1 over Q.
7. Show that Φp (1) = p for all primes p.
(a) Deduce that
p = (1 − ζp )(1 − ζp2 ) . . . (1 − ζpp−1 )
in Z[ζp ].
(b) Recall from Problem 20 of Exercises 7.3.16 that if 1 < k < p, then (ζpk −
1)/(ζp −1) = 1+ζp +· · ·+ζpk−1 is a unit in Z[ζp ]. Deduce that p = u(1−ζp )p−1
for some unit u ∈ Z[ζp ]× .
8. Let p be any prime and let r > 0. Show that Φpr (1) = p. Deduce that
.
r−1
p=
(1 − ζpkr ) = u(1 − ζpr )(p−1)p
0<k<pr
(k,p)=1
for some unit u ∈ Z[ζpr ]× .
∼
=
9. Show that Z[X]/(Φn (X)) −→ Z[ζn ]. Deduce that Z[ζn ] is a free abelian group with
φ(n)−1
basis 1, ζn , . . . , ζn
.
10. Let Z2 = T and let
f : Z[Z2 ] → Z[ζ1 ] × Z[ζ2 ] = Z × Z
be the ring homomorphism induced by
f (T ) = (ζ1 , ζ2 ) = (1, −1).
Show that f is an injection whose image has index 2 in Z × Z.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
8.9
332
Modules over P.I.D.s
We give a generalization of the Fundamental Theorem of Finite Abelian Groups. Here,
we shall generalize it to include the case of finitely generated abelian groups as well as
finite ones. We also generalize from the case of abelian groups (Z-modules) to finitely
generated modules over an arbitrary P.I.D.
Let A be an integral domain. The analogue for A-modules of finiteness in Z-modules
is given by finitely generated torsion modules.
Definitions 8.9.1. Let A be any ring and let M be an A-module. An element m ∈ M
is a torsion element if there exists an a ∈ A, a = 0, such that am = 0.
An A-module M is a torsion module if every element in M is a torsion element.
On the other hand, we say that an A-module M is torsion-free if 0 is the only torsion
element in M .
Thus, a nonzero torsion element in A itself is a right zero divisor, so if A is an integral
domain, then A is torsion-free as an A-module.
In general, m ∈ M is a torsion element if and only if the annihilator Ann(m) = 0,
but a module can be a torsion module even if no nonzero element of A annihilates every
element of M . Also, there are examples of finitely generated modules over a commutative
ring that are generated by torsion elements, but are not torsion modules (e.g., consider
Z6 as a Z6 -module). However, over an integral domain, finitely generated modules are
well behaved with respect to torsion phenomena.
Lemma 8.9.2. Let A be an integral domain and let M be a finitely generated module
over A. Suppose that M has a generating set m1 , . . . , mk consisting of torsion elements.
Then M is a torsion module, and
Ann(M ) =
k
Ann(mi )
i=1
is nonzero.
k
Proof Clearly, any element in i=1 Ann(mi ) annihilates M , so it suffices to show that
k
But if 0 = ai ∈ Ann(mi ) for 1 ≤ i ≤ k, then a1 . . . ak is a
i=1 Ann(mi ) is nonzero.
k
nonzero element of i=1 Ann(mi ).
Recall from Proposition 8.1.32 that in a U.F.D., the intersection of a collection of
principal ideals is generated by the least common multiple of their generators.
Corollary 8.9.3. Let M be a finitely generated torsion module over the P.I.D. A, with
generators m1 , . . . , mk . Let Ann(mi ) = (ai ) for 1 ≤ i ≤ k. Then Ann(M ) is the ideal
generated by the least common multiple of a1 , . . . , ak .
In studying the finitely generated modules over a P.I.D., it turns out that we can
study the torsion modules and the torsion-free modules separately and then assemble
the information. The first step in showing this is the next lemma.
Lemma 8.9.4. Let M be a module over an integral domain A, and let Tors(M ) be the
set of torsion elements in M . Then Tors(M ) is a submodule of M , and the quotient
module M/Tors(M ) is torsion-free.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
333
Proof Let m, n ∈ Tors(M ). Then the submodule generated by m and n is a torsion
module by Lemma 8.9.2, and therefore is contained in Tors(M ). In particular, Tors(M )
is a submodule of M .
Let m ∈ M/Tors(M ) be a torsion element, where m represents the image of m ∈
M under the canonical map. Say am = 0 for a = 0. Then am ∈ Tors(M ), since
am = am = 0, while Tors(M ) is the kernel of the canonical map. Since Tors(M ) is the
torsion subgroup of M , we have bam = 0 for some b = 0. But then ba = 0, and hence
m ∈ Tors(M ). So m = 0 in M/Tors(M ). Thus, M/Tors(M ) is torsion-free.
Let M be an A-module. We say that m ∈ M is divisible by a nonunit if we may write
m = am where a is not a unit in A.
Lemma 8.9.5. Let M be a finitely generated torsion-free module over the P.I.D. A and
let 0 = m ∈ M . Then we may write m = am , where m is not divisible by a nonunit.
Proof A is Noetherian by Corollary 7.8.5. Since M is finitely generated, it is a Noetherian A-module by Proposition 7.8.9. We shall argue by contradiction.
Suppose that m cannot be written as a product am where m is not divisible by a
nonunit. Then m itself must be divisible by a nonunit. Say m = am1 , where a is not a
unit in A. We claim that the inclusion Am ⊂ Am1 is proper.
If not, then m1 = bm for some b ∈ A. But then (ab−1)m = 0. Since M is torsion-free
and m = 0, this forces ab = 1, contradicting the assumption that a is not a unit.
But now m1 may not be written as a product a m such that m is not divisible by a
nonunit, as then m would have such a decomposition. By induction, we may construct
an infinite sequence
Am ⊂ Am1 ⊂ · · · ⊂ Amk ⊂ · · ·
of proper inclusions of submodules of M , contradicting the fact that M is a Noetherian
A-module.
Our next result is the key in being able to treat all finitely generated modules, instead
of just the torsion modules.
Proposition 8.9.6. Let A be a P.I.D. and let M be a finitely generated torsion-free
A-module. Then M is isomorphic to An for some n.
Proof We argue by induction on the number of elements needed to generate M . If
M is generated by one element, m, then the unique A-module map fm : A → M with
fm (1) = m is an isomorphism, since Ann(m) = 0 and Am = M .
Thus, suppose that M may be generated by n elements and that any torsion-free
module with fewer than n generators is free. Let m1 , . . . , mn be a generating set for M .
Note that if m1 = am , then m , m2 , . . . , mn generate M . Thus, using Lemma 8.9.5 to
replace m1 if necessary, we may assume that m1 is not divisible by a nonunit.
We claim now that M/Am1 is torsion-free. Thus, we claim that if 0 = a ∈ A and
if am = 0, with m ∈ A/Am1 , then m = 0. But this says that if 0 = a ∈ A, then
multiplication by a gives an injective homomorphism from M/Am1 to itself.
If a is a unit, this is immediate, as multiplication by a gives an isomorphism, whose
inverse is given by multiplication by a−1 . But if a is not a unit, we may write it as a
product of primes, in which case multiplication by a is the composite of the multiplications by these primes. By induction, it suffices to show that multiplication by a prime
element p ∈ A gives an injective homomorphism from A/Am1 to itself.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
334
Thus, suppose that pm = 0 for some m ∈ M/Am1 . But then pm ∈ ker π = Am1 in
M , and hence pm = qm1 for some q ∈ A. Since M is torsion-free, q = 0. Suppose that p
and q are relatively prime, and let rp + sq = 1 for r, s ∈ A. Then m1 = rpm1 + sqm1 =
rpm1 + spm = p(rm1 + sm). But that would imply that m1 is divisible by the nonunit
p, contradicting our assumption about m1 . Thus, p and q cannot be relatively prime.
Since p is prime, this means that q = pr for some r ∈ A. Thus, pm = prm1 , and hence
m = rm1 ∈ Am1 . But then m = 0 as desired, and hence M/Am1 is torsion-free.
Since M/Am1 is generated by m2 , . . . , mn , it is free by induction. Let f : Ak →
M/Am1 be an isomorphism and let ni = f (ei ) for i = 1, . . . , k. Let f : Ak+1 → M be
the A-module homomorphism obtained by setting f(ei ) = ni for i = 1, . . . , k and setting
f(ek+1 ) = m1 . We claim that f is an isomorphism.
Clearly, the elements n1 , . . . , nk , m1 generate m, so f is onto. But if π : M → M/Am1
is the canonical map, then
(π ◦ f)(a1 , . . . , ak+1 ) = f (a1 , . . . , ak ).
Thus, if (a1 , . . . , ak+1 ) is in the kernel of f, then (a1 , . . . , ak ) is in the kernel of f . Since
f is injective, any element of ker f is a multiple of ek+1 . But f(aek+1 ) = am1 , so f is an
isomorphism.
Corollary 8.9.7. Let M be a finitely generated module over the P.I.D. A. Then M is
isomorphic to Tors(M ) ⊕ An for some n.
Proof Since M/Tors(M ) is torsion-free, there is an isomorphism
∼
=
f : An −→ M/Tors(M ).
Let ei be the canonical basis element of An for 1 ≤ i ≤ n and write f (ei ) = mi for
mi ∈ M .
Define g : Tors(M ) ⊕ An → M as follows. The restriction of g to Tors(M ) is just
the inclusion i : Tors(M ) ⊂ M , while g(ei ) = mi for i = 1, . . . , n. Thus, there is a
commutative diagram
0
/ Tors(M )
ι
/ Tors(M ) ⊕ An
/ Tors(M )
i
/ M
/ An
/ 0
f ∼
=
g
0
π
π
/ M/Tors(M )
/ 0
where the maps ι and π in the upper sequence are the natural inclusion and projection
maps of the direct sum. But now g is an isomorphism by the Five Lemma.
To study the torsion modules over A, we need to develop a little information about
cyclic modules. Recall that a cyclic module is a module generated by one element, say,
M = Am. Recall that there is an isomorphism Am ∼
= A/Ann(m).
If A is commutative, Ann(m) = Ann(Am), or, starting out with an ideal rather than
a module, Ann(A/a) = a. Notice, then, that over a commutative ring, the annihilator of
m plays the same role that the order of m plays for a torsion element in an abelian group.
(The annihilator of a module is the analogue of the exponent of an abelian group.)
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
335
This means that we have to find substitutes for some of the counting arguments we
used for abelian groups: We can’t count in order to calculate annihilators. The next
lemma gives the analogue of the calculation of the orders of the elements in a cyclic
group.
Lemma 8.9.8. Let A be a P.I.D. and let x ∈ A/(a) be the image of x ∈ A under
the canonical map. Then Ann(x) may be calculated as follows. Let (c) be the greatest
common divisor of a and x and let a = bc. Then Ann(x) = (b). Additionally, if Ax is the
submodule of A/(a) generated by x, then the quotient module (A/(a))/Ax is isomorphic
to A/(c).
Proof Let fx : A → A be the A-module homomorphism with fx (1) = x. Then Ann(x)
fx
π
is the kernel of the composite A −→ A −
→ A/(a), where π is the canonical map. Note
that the second Noether Theorem gives an isomorphism from (x)/[(x)∩(a)] to Ax. Thus,
Ann(x) = {y ∈ A | yx ∈ (x) ∩ (a)}.
By Proposition 8.1.32, (x) ∩ (a) is the principal ideal generated by the least common
multiple, d, of x and a. But then it is easy to see that (d) = (bx). And unique factorization
now shows that yx ∈ (bx) if and only if y ∈ (b), so Ann(x) = (b).
For the last statement, note that Ax = [(x)+(a)]/(a). But A is a P.I.D., so (x)+(a) =
(c), the greatest common divisor of x and a.
Corollary 8.9.9. Let A be a principal ideal domain and let a, x ∈ A. Let M be the
cyclic module M = A/(a). Then the quotient M/(x)M is isomorphic to A/(c), where
(c) is the greatest common divisor of a and x.
Proof Let x ∈ M = A/(a) be the image of x under the canonical map π : A → A/(a).
Then clearly, Ax = (x)M . The rest now follows from Lemma 8.9.8.
We shall need an analogue of the order of a group, as that was the key for induction
arguments for abelian groups. We shall use something a little cruder, as it depends on a
generating set, rather than a module.
Definitions 8.9.10. Let M be a module over a P.I.D. A and let m be a torsion element
of M . Let (a) = Ann(m) and let a = upr11 . . . pkrk with u a unit and p1 , . . . , pk primes in
A. Then we say that the weight of m is r1 + · · · + rk .
If m1 , . . . , mk is a collection of torsion elements in M (the most interesting example
being a set of generators of M if M is a torsion module), we say that the weight of the
set m1 , . . . , mk is the sum of the weights of the elements in it.
Notice that the second form of the Chinese Remainder Theorem, while phrased in
terms of A-algebras, also expresses the interrelationship among certain cyclic modules:
If a1 , . . . , ak are pairwise relatively prime elements of the P.I.D. A, and if a = a1 . . . ak ,
then the cyclic module A/(a) breaks up as a direct sum:
A/(a) ∼
= A/(a1 ) ⊕ · · · ⊕ A/(ak ).
The next lemma shows that use of the Chinese Remainder Theorem will not affect
calculations of weight.
Lemma 8.9.11. Let a1 , . . . , ak ∈ A be pairwise relatively prime and let a = a1 . . . ak .
Let mi ∈ M = A/(a1 ) ⊕ · · · ⊕ A/(ak ) be the image in M of the standard generator of
A/(ai ) for i = 1, . . . , k. Then the generating set m1 , . . . , mk of M has the same weight
as the standard generator of the cyclic module A/(a).
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
336
We could now classify finitely generated torsion modules in exactly the same way that
we classified finite abelian groups. We shall use a different method, relying on assembling
smaller cyclic groups into larger ones.
Lemma 8.9.12. Let M be a module over the P.I.D. A and let m and n be torsion
elements of M . Let (a) and (b) be the annihilators of m and n, respectively, and suppose
that a and b are relatively prime. Then the submodule Am + An generated by m and n
is cyclic, with annihilator equal to (ab).
Proof By the Chinese Remainder Theorem, it suffices to show that Am + An is isomorphic to Am ⊕ An. We claim that Am + An is the internal direct sum of Am and
An, i.e., that the map μ : Am ⊕ An → Am + An given by μ(cm, dn) = cm + dn is an
isomorphism.
Clearly, μ is surjective, and (cm, dn) is in the kernel if and only if cm = −dn. But the
generator of Ann(cm) divides a and the generator of Ann(−dn) divides b by Lemma 8.9.8.
Since a and b are relatively prime, this forces Ann(cm) to be A, and hence cm = dn = 0,
so μ is injective.
Corollary 8.9.13. Let m1 , . . . , mk be a set of generators for a torsion module M over
the P.I.D. A. Then there is another generating set n1 , . . . , nl of M , whose weight is the
same as that of m1 , . . . , mk , such that the annihilator of nl is the same as that of M .
Proof First, apply the Chinese Remainder Theorem to each of the generators mi ,
replacing Ami with a direct sum of cyclic modules whose annihilators are generated by
elements of the form pj , where p is a prime in A. As shown in Lemma 8.9.11, replacing
mi with the set of generators of these cyclic modules does not change the weight of the
generating set. Thus, we may as well assume that we started out with a generating
set m1 , . . . , mk in which the annihilator of each mi is generated by a power of a prime
element in A.
Also, by Corollary 8.9.3, the annihilator of M is the least common multiple of the
annihilators of the mi . Thus, if Ann(M ) = (a), where a = pr11 . . . prkk , with p1 , . . . , pk
a pairwise inequivalent set of primes and with the exponents ri all positive, then there
r
must be a subcollection mi1 , . . . , mik of the generating set such that Ann(mij ) = (pj j )
for 1 ≤ j ≤ k.
Since the annihilators of the mij are pairwise relatively prime, an inductive application of Lemma 8.9.12 shows that the submodule generated by mi1 , . . . , min is a cyclic
module, generated by an element m, whose weight is the same as that of mi1 , . . . , min .
Moreover, the annihilator of m is precisely (a). Now take n1 , . . . , nl−1 to be the set of
elements of m1 , . . . , mk that are not in mi1 , . . . , min , and take nl = m.
We shall use the following analogue of Lemma 4.4.17.
Lemma 8.9.14. Let M be a finitely generated torsion module over the P.I.D. A. Suppose given m ∈ M with Ann(m) = Ann(M ). Let x ∈ M/Am. Then there exists
y ∈ M with y = x (i.e., x is the image of y ∈ M under the canonical map), such
that Ann(y) = Ann(x).
Proof Let (a) = Ann(M ). Since a annihilates every element of M , it is divisible by
the annihilator of every element of M .
If x ∈ M maps to x under the canonical map π, let Ann(x) = (b). Then as just noted,
b divides a. Also, since π(x) = x, b annihilates x, so that if Ann(x) = (c), then c divides
b.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
337
If (b) = (c), we just take y = x. Otherwise, it suffices to find an element y = x + a m,
for some a ∈ A, such that Ann(y) = (c). Note that for any element of the form
y = x + a m, Ann(y) ⊂ (c) because Ann(π(y)) = (c). Thus, it suffices to find an a such
that c annihilates y = x + a m.
Since π(cx) = cx = 0, cx ∈ ker π = Am, say cx = dm with d ∈ A. But then it suffices
to show that d is divisible by c, as, if d = cd , we can take a = −d to form y as above.
Write b = cb . Then Ann(cx) = (b ) by Lemma 8.9.8. But another application
of the same lemma shows that if a is the greatest common divisor of a and d, then
Ann(cx) = Ann(dm) = (a/a ). Since Ann(cx) = (b ), we may choose the greatest
common divisor a so that b a = a. But since b = b c divides a, c must divide a , which
divides d.
We are now ready to show how finitely generated torsion modules over P.I.D.s break
up as direct sums of cyclic modules.
Theorem 8.9.15. Let M be a finitely generated torsion module over the P.I.D. A. If
M is nonzero, then there is an isomorphism
M∼
= A/(a1 ) ⊕ · · · ⊕ A/(ak )
for a collection of ideals (ai ) which satisfies
(a1 ) ⊂ (a2 ) ⊂ · · · ⊂ (ak ),
where ak is not a unit in A. In particular, the annihilator of M is (a1 ).
Proof We argue by induction on the weight of a generating set for M . If M has a
generating set of weight 1, then M must be cyclic, and the result holds.
Suppose, then, that m1 , . . . , mn is a generating set for M and that the theorem is
true for all torsion modules that admit generating sets of lesser weight.
By Corollary 8.9.13, we may assume that Ann(m1 ) = Ann(M ). We write Ann(m1 ) =
Ann(M ) = (a1 ). If Am1 = M , then M is cyclic, and the theorem holds. Otherwise,
M/Am1 is generated by the images, m2 , . . . , mn , of m2 , . . . , mn under the canonical
map. Since the generator of Ann(mi ) divides the generator of Ann(mi ), the generating
set m2 , . . . , mn of M/Am1 has weight strictly less than the weight of m1 , . . . , mn , and
hence the induction hypothesis holds in M/Am1 .
Thus, by induction, there is an isomorphism M/Am1 ∼
= A/(a2 ) ⊕ · · · ⊕ A/(ak ) for
ideals (ai ) such that (a2 ) ⊂ (a3 ) ⊂ · · · ⊂ (ak ). Then clearly, Ann(M/Am1 ) = (a2 ). Note
that since M/Am1 is a quotient of M , (a1 ) = Ann(M ) must annihilate M/Am1 . But
that says (a1 ) ⊂ (a2 ). We claim that M is isomorphic to A/(a1 ) ⊕ · · · ⊕ A/(ak ).
To see this, let f : A/(a2 ) ⊕ · · · ⊕ A/(ak ) → M/Am1 be an isomorphism and let
mi ∈ M/Am1 be the image under f of the standard generator of A/(ai ) for i = 2, . . . , k.
But Lemma 8.9.14 says that we may choose an mi ∈ M whose image under the canonical
map is mi , such that the annihilator of mi is the same as that of mi (which is (ai )).
Define g : A/(a1 ) ⊕ · · · ⊕ A/(ak ) → M by
g(x1 , . . . , xk ) = x1 m1 + x2 m2 + · · · + xk mk .
Since the annihilator of mi is (ai ), g is a well defined homomorphism. Moreover, abbre-
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
338
viating with N = A/(a2 ) ⊕ · · · ⊕ A/(ak ), we have a commutative diagram
0
/ A/(a1 )
⊂ /
A/(a1 ) ⊕ N
∼
=
0
/ Am1
π
g
⊂
/ M
π
/ N
/ 0
f ∼
=
/ M/Am1
/ 0
where the maps π are the appropriate canonical maps. The result now follows from the
Five Lemma.
Of course, a free A-module is a direct sum of copies of A/(0), and we may now apply
Corollary 8.9.7.
Corollary 8.9.16. Let M be any finitely generated module (not necessarily a torsion
module) over the P.I.D. A, with M = 0. Then there is an isomorphism
M∼
= A/(a1 ) ⊕ · · · ⊕ A/(ak )
for a collection of ideals (ai ) which satisfies
(a1 ) ⊂ (a2 ) ⊂ · · · ⊂ (ak ),
where ak is not a unit in A.
We can now state and prove the Fundamental Theorem.
Theorem 8.9.17. (Fundamental Theorem of Finitely Generated Modules over a P.I.D.)
Let M be a finitely generated module over the P.I.D. A, with M = 0. Then M may be
written uniquely as a direct sum
M∼
= A/(a1 ) ⊕ · · · ⊕ A/(ak )
for a collection of ideals (ai ) which satisfies
(a1 ) ⊂ (a2 ) ⊂ · · · ⊂ (ak ),
where ak is not a unit in A.
Here, uniqueness means that if we’re given a collection of ideals
(b1 ) ⊂ (b2 ) ⊂ · · · ⊂ (bl )
such that bl is not a unit and
A/(a1 ) ⊕ · · · ⊕ A/(ak ) ∼
= A/(b1 ) ⊕ · · · ⊕ A/(bl ),
then k = l and (ai ) = (bi ) for i = 1, . . . , k.
∼ A/(a1 ) ⊕ · · · ⊕ A/(ak ) is shown in
Proof The existence of such an isomorphism M =
Corollary 8.9.16, so it suffices to show uniqueness.
Thus, let M = A/(a1 ) ⊕ · · · ⊕ A/(ak ) and let N = A/(b1 ) ⊕ · · · ⊕ A/(bl ), with the a’s
and b’s as above, and suppose given an isomorphism f : M → N .
Now A-module homomorphisms always carry torsion elements to torsion elements.
Applying this to both f and f −1 , we see that f induces an isomorphism f : Tors(M ) →
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
339
Tors(N ). Moreover, the induced homomorphism f : M/Tors(M ) −→ N/Tors(N ) will be
an isomorphism, with inverse given by the similarly induced map coming from f −1 .
Notice that Tors(M ) is the direct sum of those terms A/(ai ) for which ai = 0, and
hence M/Tors(M ) is isomorphic to the direct sum of those terms A/(ai ) for which ai = 0;
similarly for N . By Theorem 7.9.8, isomorphic free modules have the same rank. Since
f is an isomorphism, we may conclude that the same number of a’s as b’s are equal to
0. Thus, since f restricts to an isomorphism of torsion submodules, it suffices to assume
that both M and N are torsion modules, and hence that a1 and b1 are nonzero.
Let p be a prime of A which divides ak . Then p divides ai for all i ≤ k, and hence
[A/(ai )]/(p)[A/(ai )] is isomorphic to A/(p) by Corollary 8.9.9. In this case, M/(p)M is
isomorphic to (A/(p))k , and hence has dimension k as a vector space over A/(p).
On the other hand, if p does not divide ak , [A/(ak )]/(p)[A/(ak )] is isomorphic to
A/(1) = 0, since (p, ak ) = (1). In this case, M/(p)M has dimension strictly less than k
as a vector space over A/(p). Note that one way or the other, the dimension of M/(p)M
as a vector space over A/(p) is less than or equal to k. Of course, the same sort of
reasoning applies to calculating the dimension of N/(p)N as a vector space over A/(p).
Thus, suppose that p divides ak . Then M/(p)M is a k-dimensional vector space over
A/(p). But the dimension of N/(p)N (which is isomorphic to M/(p)M ), must be less
than or equal to l, and hence k ≤ l. But applying the same argument to a prime q that
divides bl shows l ≤ k, and hence l = k.
Note that this argument also shows that the primes dividing ak are precisely the
primes dividing bk . Let p be one such.
For an A-module M , write M [p] for the set of elements of M annihilated by p.
Thus, M [p] = {m ∈ M | pm = 0}. Note that since (p) is a maximal ideal of A, a
nonzero element m ∈ M is annihilated by p if and only if Ann(m) = (p).
Consider the cyclic module A/(d), where d = pc with c = 0. By Lemma 8.9.8, we
have Ann(x) = (p) if and only if (x, d) = (c), in which case Ax = Ac. In particular,
(A/(d))[p] = Ac, which is isomorphic to A/(p), as Ann(c) = (p). Moreover, we have
(A/(d))/(A/(d))[p] ∼
= A/(c).
Now p divides ak and bk , and hence it divides all the other a’s and b’s. Say ai =
pai and bi = pbi for i = 1, . . . , k. Clearly (M ⊕ M )[p] = M [p] ⊕ M [p] . Thus, the
isomorphism induced by f from M/M[p] to N/N[p] gives
A/(a1 ) ⊕ · · · ⊕ A/(ak ) ∼
= A/(b1 ) ⊕ · · · ⊕ A/(bk ).
Note that a1 has lower weight than a1 , so an argument by induction on the weight
of a1 is possible. Here, if a1 has weight 1, then M = (A/(p))k , and hence M and N are
both annihilated by p. Thus, N = N/(p)N = (A/(p))k and the result follows.
Thus, we may assume inductively that ai = bi for 1 ≤ i ≤ k. But then ai = bi as
desired.
This form of the Fundamental Theorem has an appealing formulation, but there’s
another form that is perhaps better from a practical standpoint. The point is that the
Chinese Remainder Theorem allows us to break up any cyclic module into a direct sum
of cyclic modules whose annihilators are prime powers.
Definitions 8.9.18. Let p be a prime element of the P.I.D. A and let M be an A-module.
We say that m ∈ M is a p-torsion element if m is annihilated by some power of p.
We write Mp ⊂ M for the collection of p-torsion elements in M .
We say that M is a p-torsion module if M = Mp . By a primary torsion module, we
mean a p-torsion module for some prime p of A.
CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS
340
Lemma 8.9.19. Let p be a prime element of the P.I.D. A and let M be an A-module.
Then a nonzero element m ∈ M is in Mp if and only if Ann(m) = (pr ) for some
r ≥ 1. Moreover, Mp is a submodule of M . Finally, if f : M → N is an A-module
homomorphism, then f (Mp ) ⊂ Np .
In particular, a cyclic module A/(a) is a p-torsion module if and only if (a) = (pr )
for some r.
Theorem 8.9.20. (Fundamental Theorem of Finitely Generated Modules over a P.I.D.,
second form) Let M be a finitely generated torsion module over the P.I.D. A. Then M
has a unique decomposition as a direct sum of primary torsion cyclic modules.
Here, uniqueness means that if M and N are two different direct sums of primary
torsion cyclic modules, then M and N are isomorphic if and only if for each prime p of
A and each r > 0, M and N have the same number of summands isomorphic to A/(pr ).
Proof The existence of such a decomposition may be obtained by applying the Chinese Remainder Theorem to break up the cyclic modules of the decomposition from
Theorem 8.9.15 into direct sums of primary torsion modules. Thus, it suffices to show
uniqueness.
Note that Lemma 8.9.19 shows that if f : M → N is an isomorphism of A-modules,
then it restricts to an isomorphism f : Mp → Np for each prime p of A. Thus, we
may assume that M = A/(pr1 ) ⊕ · · · ⊕ A/(prk ), where r1 ≥ · · · ≥ rk ≥ 1, and N =
A/(ps1 ) ⊕ · · · ⊕ A/(psl ), where s1 ≥ · · · ≥ sl ≥ 1. We are given an isomorphism from M
to N , and must deduce that k = l and ri = si for 1 ≤ i ≤ k.
But now (pr1 ) ⊂ · · · ⊂ (prk ) and (ps1 ) ⊂ · · · ⊂ (psl ), so we may apply the uniqueness
statement from the first form of the Fundamental Theorem.
Exercises 8.9.21.
1. Show that Q/Z is a torsion module over Z, but that no element of Z annihilates
all of Q/Z. (Of course, Q/Z is not finitely generated as a Z-module.)
2. Show that a finitely generated torsion module M over the P.I.D. A is cyclic if and
only if there are no primes p of A for which M contains a submodule isomorphic
to A/(p) ⊕ A/(p).
Chapter 9
Radicals, Tensor Products, and
Exactness
In this chapter, we develop some important tools for studying rings and modules. We
shall use a number of these tools in Chapter 12, where we shall study some particular
types of rings in detail. More generally, these tools are essential for anyone who works
in fields involving rings and modules.
In Section 9.1, we develop the Jacobson and nil radicals and the radical of an ideal in
a commutative ring. We also give Nakayama’s Lemma, an essential ingredient in certain
exactness arguments and in the study of local rings.
Section 9.2 develops the theory of primary decomposition, a weak analogue of the
unique factorization of ideals occurring in a P.I.D.
In Section 9.3, Hilbert’s Nullstellensatz is proven, using many of the tools so far
developed, and is used to define affine algebraic varieties.
Section 9.4 gives the basic theory of tensor products. Extension of rings is emphasized
as an application. Then, in Section 9.5, we use extension of rings and the theory of radicals to extend some of the basic results regarding homomorphisms of finite dimensional
vector spaces to the study of free modules over more general rings. In particular, if A
is a ring that admits a ring homomorphism to a division ring, we show that Am ∼
= An
m
n
if and only if m = n. We also show that if f : A → A , then the following conditions
hold.
1. If f is surjective, then m ≥ n.
2. If A is commutative, f is surjective, and m = n, then f is an isomorphism.
3. If A is commutative and f is injective, then m ≤ n.
We also discuss flat modules and characterize the flat modules over a P.I.D.
Section 9.6 gives the tensor product of algebras and verifies its universal property.
Then, in Section 9.7, we develop the module structures and exactness properties of the
Hom functors and describe their relationship to tensor products.
In Sections 9.8 and 9.9, we develop the theory of projective modules. These are
the modules P for which the functor HomA (P, −) preserves exact sequences, and are a
generalization of free modules. They play a crucial role in our analysis of semisimple
rings and Dedekind domains in Chapter 12, and hence are important in the study of
341
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
342
group representations and in number theory. They are also of fundamental importance
in homological algebra.
Section 9.9 introduces algebraic K-theory with the functor K0 , which classifies the
finitely generated projective A-modules modulo a sort of stabilization process. It plays
an important role in numerous mathematical problems.
Section 9.10 introduces the tensor algebra, TA (M ), on an A-module M . This is the
free A-algebra on M . We use it to construct the symmetric algebra and exterior algebra
on M , and to construct the Clifford algebra on a finite set of generators. The exterior
algebras are related to determinant theory, while the Clifford algebras have applications
in both topology and analysis. In the process of analyzing these algebras, we develop
the theory of skew commutative graded algebras, which has applications in homological
algebra and topology.
We shall assume a familiarity here with the notions of categories and functors, as
developed in the first two sections of Chapter 6. We shall also make use of some other
ideas from that chapter, but will not assume a familiarity with them.
9.1
Radicals
We return to the study of maximality considerations for ideals, but this time in the
context of noncommutative rings.
Definitions 9.1.1. A maximal left ideal is a maximal element of the partially ordered
set of proper left ideals. Maximal right ideals are defined analogously, and a maximal
two-sided ideal is a maximal element in the partially ordered set of proper two-sided
ideals.
The next lemma is proven by the argument that was given for Corollary 7.6.14.
Lemma 9.1.2. Let A be a nonzero ring. Then any proper left ideal of A is contained
in a maximal left ideal, any proper right ideal is contained in a maximal right ideal, and
any proper two-sided ideal is contained in a maximal two-sided ideal.
Maximal left ideals have important consequences to module theory. They relate to
the analogue for modules of simple groups.
Definition 9.1.3. A nonzero left A-module is called simple, or irreducible, if it has no
submodules other than 0 and itself.
The maximal left ideals are the key to understanding simple modules.
Lemma 9.1.4. Let M be a simple left A-module and let 0 = m ∈ M . Then Ann(m) is
a maximal left ideal m of A, and there is a module isomorphism between A/m and M .
Conversely, if m is a maximal left ideal of A, then A/m is a simple left A-module.
Proof Given 0 = m ∈ M , the submodule Am generated by m is nonzero, and hence
equal to M , since M is simple. But if m is the annihilator of m, we have an isomorphism
fm : A/m → Am = M , inducing a one-to-one correspondence between the submodules
of M and the left ideals of A containing m. Thus, the only left ideals containing m are
m and A, so that m is a maximal left ideal.
The converse comes precisely from the fact that there are only two left ideals containing a maximal left ideal m: m itself and A.
We shall also make use of the following fact about maximal left ideals.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
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Lemma 9.1.5. Let A be a ring and let x ∈ A. Then x lies outside all the maximal left
ideals if and only if x has a left inverse.
Proof Consider the principal left ideal Ax generated by x. Then x lies in no maximal
left ideal if and only if Ax lies in no maximal left ideal. But this is the case if and only
if 1 ∈ Ax, which happens if and only if 1 = yx for some y ∈ A.
Definition 9.1.6. Let A be a ring. The Jacobson radical, R(A), is the intersection of
the maximal left ideals of A.
Many of the rings we’re familiar with have R(A) = 0. Thus, it is valuable to keep in
mind that if A is a local ring, with maximal ideal m, then R(A) = m.
The next proposition gives a useful characterization of elements in R(A).
Proposition 9.1.7. Let A be a ring and let x ∈ A. Then the Jacobson radical of A is a
two-sided ideal, and the following statements are equivalent.
1. x is in the Jacobson radical, R(A).
2. For every simple left A-module M , x ∈ Ann(M ).
3. For any a, b ∈ A, 1 − axb is a unit in A.
Proof Since the annihilator of a module is a two-sided ideal, the fact that R(A) is
two-sided will follow from the equivalence of Conditions 1 and 2. We shall establish this
first.
We first show that the first condition implies the second. Let x ∈ R(A) and let M
be a simple left A-module. It suffices to show that for 0 = m ∈ M , xm = 0. But the
annihilator of m is a maximal left ideal of A because M is simple. Since R(A) is the
intersection of the maximal left ideals, x must annihilate m as required.
But now we can see that the second condition implies the first: if x annihilates A/m,
with m a maximal left ideal, then it must annihilate the element 1 ∈ A/m. Since the
annihilator of 1 is m, x ∈ m.
We now show that the first two conditions imply the third. Let x ∈ R(A) and let
a, b ∈ A. Since R(A) is a two-sided ideal, axb ∈ R(A) as well. But then 1 − axb cannot
lie in any maximal left ideal, as that would force 1 to be in the ideal as well.
By Lemma 9.1.5, this says that 1 − axb has a left inverse, y. But then y − yaxb = 1,
and hence −yaxb = 1 − y. But −yaxb is in R(A), and hence lies in every maximal left
ideal of A. But since 1−y is in every maximal left ideal, y can be in no maximal left ideal,
and hence y has a left inverse z. But a standard argument now shows that z = 1 − axb,
and hence y is a two-sided inverse for 1 − axb.
Finally, we show that the third condition implies the first. Suppose that 1 − axb is a
unit for each a, b ∈ A and let m be a maximal left ideal of A. Suppose that x is not in
m. Then we must have Ax + m = A. But then 1 = ax + m for some a ∈ A and m ∈ m,
so that 1 − ax ∈ m. But this contradicts the fact that 1 − ax is a unit, so x must have
been in m.
The following use of radicals is valuable in the study of local rings.
Proposition 9.1.8. (Nakayama’s Lemma) Let A be a ring and let a be a left ideal contained in R(A). Suppose that aM = M for some finitely generated left A-module M .
Then M = 0.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
344
Proof
Since a is contained in R(A), we may as well use R(A) in place of a. We
argue by contradiction. Assume that M = 0, and let m1 , . . . , mk be a set of A-module
generators for M , which is minimal in the sense that no proper subset will generate M .
Since R(A)M = M , we have m1 ∈ R(A)M , and hence m1 = a1 n1 + · · · + ar nr , with
ai ∈ R(A) and ni ∈ M for all i. But since since each ni is a linear combination of the
generators m1 , . . . , mk , and since R(A) is a right ideal, we may collect terms, writing
m1 = b1 m1 + · · · + bk mk , with bi ∈ R(A) for all i.
Rearranging terms, we get (1 − b1 )m1 = b2 m2 + · · · + bk mk . But since b1 ∈ R(A),
1 − b1 is a unit, so that m1 is in the submodule of M generated by m2 , . . . , mk . But then
m2 , . . . , mk generate M , contradicting the minimality of the generating set m1 , . . . , mk .
In the commutative case, there is another radical which is of interest.
Lemma 9.1.9. Let A be a commutative ring. Then the set of all nilpotent elements in
A forms an ideal, N(A), called the nil radical of A.
Proof Let x ∈ N(A) and let a ∈ A. Then xn = 0 for some n ≥ 1, and hence
(ax)n = an xn = 0, also. It suffices to show that N(A) is closed under addition. Thus,
let x, y ∈ N(A), with xn = y m = 0 for n, m ≥ 0. We claim that (x + y)n+m−1 = 0.
To see this, recall that the Binomial Theorem
(Theorem 4.5.16) holds in any commun+m−1 n+m−1 i n+m−i−1
xy
tative ring. Thus, (x + y)n+m−1 = i=0
. But if xi = 0, then
i
n+m−i−1
i < n, and hence n + m − i − 1 ≥ m, so that y
= 0.
The nil radical also has a characterization related to that of the Jacobson radical.
Proposition 9.1.10. Let A be a commutative ring. Then the nil radical of A is the
intersection of all the prime ideals of A.
Proof Let x ∈ N(A) and let p be a prime ideal of A. We claim that x ∈ p. To see this,
let n > 0 be the smallest exponent such that xn = 0. Then xn ∈ p. If n = 1, there’s
nothing to show. Otherwise, either x ∈ p or xn−1 ∈ p, so in either case, xn−1 ∈ p. But
then a downward induction shows that x ∈ p.
It suffices to show that if x ∈ A is not nilpotent, then there is a prime ideal that
doesn’t contain x. Let X = {xk | k ≥ 1} and let S be the set of all ideals a such that
a ∩ X = ∅. Since x is not nilpotent, 0 ∈ S, and hence S is nonempty. By an argument
similar to the one that shows the existence of maximal ideals, Zorn’s Lemma implies that
there are maximal elements in S. We claim that if p is a maximal element of S, then p
is prime.
It suffices to show that if a and b are in the complement of p, then so is ab. Since a
is not in p, p + (a) strictly contains p, and hence xm = p + ya for some p ∈ p, y ∈ A,
and m > 0. Similarly, there are p ∈ p, y ∈ A, and n > 0, with xn = p + y b. But
multiplying these two elements together, we get xn+m = p + yy ab for some p ∈ p. But
since p does not meet X, ab cannot lie in p.
We can generalize the nil radical to define a radical for any ideal of a commutative
ring.
Definition 9.1.11. Let a be an ideal in the commutative ring A. The radical rad(a) is
given by
rad(a) = {x ∈ A | xn ∈ a for some n > 0}.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
345
Notice that if π : A → A/a is the canonical map, then rad(a) = π −1 (N(A/a)). Thus,
rad(a) is an ideal of A. The following proposition is now immediate from the one-to-one
correspondence between the prime ideals of A/a and the primes of A that contain a.
Proposition 9.1.12. Let a be a proper ideal in the commutative ring A. Then rad(a) is
the intersection of the prime ideals of A containing a.
It is useful to be able to characterize the ideals a in a commutative ring with rad(a) =
a. As the reader may easily check, rad(rad(a)) = rad(a) for any ideal a. We obtain the
following lemma.
Lemma 9.1.13. An ideal a of a commutative ring A is its own radical if and only if
a = rad(b) for some ideal b of A.
Of course, if p is prime, then the intersection of the prime ideals containing it is
precisely p. Proposition 9.1.12 gives the following corollary.
Corollary 9.1.14. A prime ideal is its own radical.
The next lemma is easy and its proof is left as an exercise. It use is sufficiently
common to warrant a formal statement.
n
Lemma
9.1.15. Let a1 , . . . , an be ideals in a commutative ring A. Then rad( i=1 ai ) =
n
i=1 rad(ai ).
Corollary 9.1.14 now gives a corollary.
Corollary 9.1.16. A finite intersection of prime ideals is its own radical.
We shall show in Corllary 9.2.11 that a proper ideal in a Noetherian commutative
ring is its own radical if and only if it is a finite intersection of prime ideals.
Exercises 9.1.17.
1. What is R(A × B)? What is N(A × B)?
2. Let A be a P.I.D. Show that R(A) = 0 if and only if A has infinitely many prime
ideals.
3. Let A be the ring of Example 6 of Examples 7.8.2. What is R(A)?
4. Let A be a ring. Show that R(A) is equal to the intersection of the maximal right
ideals in A.
5. Let A be a ring. Show that R(A) is contained in the intersection of the maximal
two-sided ideals in A.
6. Give the proof of Lemma 9.1.15.
7. Let n > 1 in Z. What is rad((n))?
8. Let A be a commutative ring with only one prime ideal. Show that any injective
A-module homomorphism from A to itself is an isomorphism.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
9.2
346
Primary Decomposition
Primary decomposition of ideals is a weak analogue of the unique factorization of ideals
occurring in a P.I.D. The theory was first developed for polynomial rings by the chess
master Emanuel Lasker. Emmy Noether then showed that primary decompositions exist
for all proper ideals of a Noetherian commutative ring. We shall give her argument here.
The theory works as follows: There is a generalization of prime ideals called primary
ideals. In a Noetherian commutative ring, every proper ideal is a finite intersection of
primary ideals. But the decomposition is not unique, even if we insist that none of the
primary ideals contain the intersection of the others. However, under that assumption,
the set of ideals which are radicals of the primary ideals in the decomposition is unique.
As an application of this theory, we shall show that an ideal a in a Noetherian
commutative ring satisfies a = rad(a) if and only if a is a finite intersection of prime
ideals. We shall make use of this in Section 9.3 to give decompositions of affine varieties
over an algebraically closed field.
Definition 9.2.1. Let A be a commutative ring. A proper ideal q of A is primary if
xy ∈ q implies that either x ∈ q or y ∈ rad(q) (i.e., y n ∈ q for some n > 0).
If a is an ideal of A, then a primary decomposition of a is a presentation
a=
m
qi
i=1
of a as a finite intersection of primary ideals qi for i = 1, . . . , m.
Clearly, any prime ideal is primary. (Recall from Corollary 9.1.14 that a prime ideal
is its own radical.) We shall give further examples below.
Throughout this section, A denotes a commutative ring.
Since rad(q) is the inverse image under the canonical map of the nil radical of A/q,
the next lemma is immediate.
Lemma 9.2.2. A proper ideal q of A is primary if and only if every zero-divisor in A/q
is nilpotent.
The reader should verify the following lemma.
Lemma 9.2.3. Let q be a primary ideal of A. Then rad(q) is prime.
Definition 9.2.4. If q is primary with radical p, we say that q is p-primary, or that p is
the prime associated to q.
The converse to Lemma 9.2.3 is false: there are nonprimary ideals whose radical is
prime. Having a maximal ideal as radical is stronger.
Proposition 9.2.5. Let q be an ideal of A whose radical is a maximal ideal. Then q is
primary.
Proof Let m = rad(q). Since the radical is the intersection of the prime ideals containing
q (Proposition 9.1.12), any prime ideal containing q must contain m. Thus, m is the only
prime ideal of A containing q, and hence m/q is the only prime ideal in A/q.
Thus, any element of A/q outside m/q is a unit. So every zero-divisor of A/q lies in
m/q. But since m = rad(q), m/q is the nil radical of A/q, so all its elements are nilpotent.
Thus, q is primary by Lemma 9.2.2.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
347
Examples 9.2.6.
1. Let m be a maximal ideal of A and let mk = m . . . m be the ideal product of m
with itself k times. (See Definitions 7.2.21.) Let q be any proper ideal containing
mk . Then m ⊂ rad(q). Since q is proper, so is its radical. So rad(q) = m, and q is
primary by Proposition 9.2.5.
2. If K is a field and r1 , . . . , rn > 0, let q = (X1r1 , . . . , Xnrn ) ⊂ K[X1 , . . . , Xn ], the
ideal generated by X1r1 , . . . , Xnrn . Then rad(q) contains the maximal ideal m =
(X1 , . . . , Xn ). Since q is proper, it is m-primary.
3. Let A be a P.I.D. and let a = pr11 . . . prkk , where p1 , . . . , pk are pairwise inequivalent
primes of A. Then an element lies in rad((a)) if and only if it is divisible by
p1 , . . . , pk . Thus, rad((a)) = (p1 . . . pk ). Thus, a nonzero ideal (a) is primary if and
only if (a) = (pr ) = (p)r for some prime element p of A and r > 0.
The ideal 0 is also primary, being prime.
We shall now show that every proper ideal in a Noetherian commutative ring has a
primary decomposition. We shall make use of the following operation on ideals.
Definition 9.2.7. Let a and b be ideals of A, Then the ideal quotient (a : b) is the ideal
(a : b) = {x ∈ A | xb ⊂ a}.
Thus, x ∈ (a : b) if and only if xb ∈ a for all b ∈ b.
For a principal ideal (x) ⊂ A, we write (a : (x)) = (a : x). Note that y ∈ (a : x) if
and only if xy ∈ a.
We shall actually show that every proper ideal in a Noetherian commutative ring is
a finite intersection of irreducible ideals.
Definition 9.2.8. A proper ideal a of A is irreducible if a = b ∩ c implies that either
a = b or a = c.
Lemma 9.2.9. If A is Noetherian, then every irreducible ideal of A is primary.
Proof Let q be an irreducible ideal of A and let xy ∈ q with y ∈ q. Then we have an
ascending chain of ideals
(q : x) ⊂ (q : x2 ) ⊂ · · · ⊂ (q : xk ) ⊂ · · · .
Since A is Noetherian, the sequence is eventually constant, and hence (q : xn ) = (q :
xn+1 ) for some n > 0. We claim that (q + (y)) ∩ (q + (xn )) = q.
To see this, note that xy ∈ q implies that q + (y) ⊂ (q : x). Thus, if z + axn ∈ q + (y)
with z ∈ q, then zx + axn+1 = w ∈ q. So a ∈ (q : xn+1 ) = (q : xn ). But then z + axn ∈ q,
as desired.
Thus, (q+(y))∩(q+(xn )) = q. Since q is irreducible and y ∈ q, this forces q+(xn ) = q,
and hence x ∈ rad(q). Thus, q is primary.
We can now show the existence of primary decompositions in a Noetherian commutative ring. The technique we shall use is called Noetherian induction. We shall make
further use of it in Chapter 12.
Theorem 9.2.10. Every proper ideal in a Noetherian commutative ring A is a finite
intersection of irreducible ideals. In particular, Lemma 9.2.9 now shows that any proper
ideal of A has a primary decomposition.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
348
Proof Recall from Proposition 7.6.16 that if S is any nonempty set of ideals in a
Noetherian ring, ordered by inclusion, then S has maximal elements. Here, a is maximal
in S if a ∈ S and any ideal in S containing a must equal a.
Thus, we may argue by contradiction: Assuming the theorem false, let S be the set
of proper ideals which are not finite intersections of irreducible ideals and let a be a
maximal element in S.
Since a is in S, it cannot be irreducible, so we may write a = b ∩ c, where b and c
both properly contain a (and hence b and c are both proper ideals). Since a is a maximal
element in S, b and c cannot lie in S, and hence each of them is a finite intersection of
irreducible ideals. But a = b ∩ c, so this contradicts the assumption that a is not a finite
intersection of irreducible ideals.
Corollary 9.2.11. A proper ideal a in a Noetherian commutative ring A is its own
radical if and only if a is a finite intersection of prime ideals.
Proof Corollary 9.1.16 shows that a finite intersection of prime ideals is its own radical,
so it suffices to show the converse. As noted in Lemma 9.1.13, this amounts to showing
that the radical of any proper ideal of A is a finite intersection of prime ideals. n
Let a be a proper ideal of A. Theorem 9.2.10
n gives a decomposition a = i=1 qi
where qi is primary for all i. But then rad(a) = i=1 rad(qi ) (Lemma 9.1.15). The result
follows, since the radical of a primary ideal is prime (Lemma 9.2.3).
We shall now discuss the uniqueness properties of primary decompositions. As we
shall assume the existence of such a decomposition, A need not any longer be Noetherian.
Theorem 9.2.10 itself provides an example of nonuniqueness, in that not every primary
ideal is irreducible.
Example 9.2.12. Let K be a field and let m = (X, Y ), the ideal of K[X, Y ] generated
by X and Y . Then m2 = (X 2 , XY, Y 2 ) is primary by Proposition 9.2.5. As the reader
may check,
m2 = (m2 + (X)) ∩ (m2 + (Y )),
the intersection of two ideals properly containing it. Also, both m2 + (X) and m2 + (Y )
have radical m, and hence are primary.
We can eliminate this particular sort of nonuniqueness to our decompositions if we
insist that each of the primary ideals being intersected have a distinct radical. The
following lemma allows us to do this.
Lemma 9.2.13. Let q1 , . . . , qm be p-primary ideals
m in A for some prime ideal p (i.e.,
each qi is primary with rad(qi ) = p). Then q = i=1 qi is p-primary as well.
m
Proof By Lemma 9.1.15, rad(q) = i=1 rad(qi ) = p, so it suffices to show q is primary.
Suppose xy ∈ q and y ∈ rad(q). It suffices to show that x ∈ q. Since q is the intersection
of the qi , xy ∈ qi for all i. But y ∈ rad(qi ) = rad(q). Since each qi is primary, x must lie
in qi for all i, and hence in the intersection, q.
m
Definition 9.2.14. A primary decomposition a = i=1 qi of an ideal a is reduced if
1. The prime ideals rad(q1 ), . . . , rad(qm ) are all distinct.
2. No qi contains the intersection of the other q’s.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
349
Using Lemma 9.2.13 to obtain the first condition and then induction to obtain the
second, Theorem 9.2.10 gives the following corollary.
Corollary 9.2.15. Every proper ideal in a Noetherian commutative ring has a reduced
primary decomposition.
Even reduced primary decompositions are not unique.
Example 9.2.16. Let a = (X 2 , XY ) ⊂ K[X, Y ] with K a field. Let m = (X, Y ).
Then (X 2 , Y ) and m2 = (X 2 , XY, Y 2 ) both have radical m, and hence are primary by
Proposition 9.2.5. Thus, the following equalities, which the reader should verify, give two
distinct primary decompositions for a.
a
a
and
= (X) ∩ m2
2
= (X) ∩ (X , Y ).
Note that the radicals of the primary ideals in these two decompositions are given by
(X) and (X, Y ). Note that (X) ⊂ (X, Y ), even though there is no analogous inclusion
between the corresponding primary ideals in these decompositions.
We shall now develop some positive results regarding uniqueness of primary decompositions. First note that prime ideals are better behaved than primary ideals with respect
to irreducibility.
Lemma 9.2.17. Prime ideals are irreducible.
Proof Let p = a∩b with p prime. Since p contains a∩b, either a or b must be contained
in p by Corollary 7.6.10. But p is contained in both a and b, and hence must equal one
of them.
Lemma 9.2.18. Let p be a prime ideal of A and let
the ideal quotient (q : x) is given by
⎧
⎨ A
q
(q : x) =
⎩
a p-primary ideal
q be p-primary. Then for x ∈ A,
if x ∈ q
if x ∈ p
otherwise.
Proof The case x ∈ q is true for any ideal, while that of x ∈ p is immediate from the
definition of a primary ideal: If xy ∈ q and x ∈ p = rad(q), then y must lie in q.
Thus, it suffices to show that if x ∈ q then (q : x) is p-primary. Note, then that if
y n x ∈ q, then y n ∈ rad(q) = p, since x ∈ q and q is primary. But then y ∈ p, since p is
prime. Thus, rad((q : x)) ⊂ p. But q ⊂ (q : x), and hence p = rad(q) ⊂ rad((q : x)), so
p = rad((q : x)), and it suffices to show that (q : x) is primary.
Thus, suppose that ab ∈ (q : x) with a ∈ rad((q : x)) = p. We wish to show b ∈ (q : x).
We have abx ∈ q, so bx ∈ (q : a) = q by the second case above. But then b ∈ (q : x).
We can use the ideal quotients to show that the primes associated to the primary
ideals in a primary decomposition are unique.
m
Proposition 9.2.19. Let a = i=1 qi be a reduced primary decomposition of the ideal a
and let pi = rad(qi ) for i = 1, . . . , m. Then {p1 , . . . , pm } is the set of prime ideals which
occur among the ideals {rad((a : x)) | x ∈ A}.
k
Thus, if a = i=1 qi is another reduced primary decomposition of a, then k = m, and
we may reindex the qi so that rad(qi ) = pi for i = 1, . . . , m.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
350
m
m
Proof As the reader may easily check, ( i=1 qi : x) = i=1 (qi : x). Since radicals also
commute with intersections (Lemma 9.1.15), Lemma 9.2.18 shows that
m
if x ∈ a = i=1 qi
{i | x
∈qi } pi
rad((a : x)) =
A
if x ∈ a.
If rad((a : x)) is prime, then x ∈ a, and rad((a : x)) is the finite intersection x
∈qi pi .
But prime ideals are irreducible (Lemma 9.2.17), so rad((a : x)) must equal one of the
pi .
Conversely, since
the decomposition is reduced, for each i we can find an element
xi which lies in j
=i qj but not in qi . But then the displayed equation shows that
rad((a : xi )) = pi .
Definitions 9.2.20. Let a be an ideal admitting a primary decomposition. Then the
primes associated to a (or, more simply, the primes of a) are the radicals of the primary
ideals in any reduced primary decomposition of a.
Let p1 , . . . , pm be the primes of a. If pi contains pj for some j = i, we say that pi is
an embedded
m prime of a. Otherwise, we say that pi is an isolated prime of a.
If a = i=1 qi is a reduced primary decomposition of a, with pi the associated prime
of qi , we call the qi the primary components of the decomposition, and say that qi is
isolated (resp. embedded) if pi is isolated (resp. embedded).
Thus, Example 9.2.16 shows that (X) is an isolated prime of (X 2 , XY ) ⊂ K[X, Y ]
and (X, Y ) is embedded.
The word “embedded” expresses the fact that if K is an algebraically closed field and
if p ⊂ p is a proper inclusion of prime ideals of K[X1 , . . . , Xn ], then the affine variety
V(p ) induced by p embeds as a subvariety of V(p). We shall investigate this situation
in Section 9.3.
We can obtain a uniqueness result for the isolated primary components of a decomposition.
m
Proposition 9.2.21. Let a = i=1 qi be a reduced primary decomposition for a with
associated primes p1 , . . . , pm . Suppose that pi is an isolated prime of a. Then
qi = {x ∈ A | (a : x) ⊂ pi }.
Since the right hand side is independent of the choice of reduced primary decomposition,
so is qi .
Proof Write S = {x ∈ A | (a : x) ⊂ pi }. We first show that S ⊂ qi . Since a ⊂ qi ,
(a : x) ⊂ (qi : x) for all x. If x ∈ qi , then (qi : x) ⊂ pi by Lemma 9.2.18. But then
(a : x) ⊂ pi , and hence x ∈ S. So S mustbe contained in qi .
Since pi is isolated, it cannot contain j
=i pj : If it did, then it would have to contain
some pj for j = i by Corollary 7.6.10. Let y be an element of j
=i pj not in pi . Since
pj = rad(qj ) for all j, we can choose an k > 0 such that y k ∈ j
=i qj . Since pi is a prime
not containing y, y k ∈ p.
Let x ∈ qi . Then xy k lies in all of the q’s, and hence in a. Thus, y k ∈ (a : x). Since
k
y ∈ pi , (a : x) ⊂ pi , and hence x ∈ S. Thus, qi ⊂ S, and hence qi = S.
The situation for embedded primary components is much more complicated. Indeed,
if A is Noetherian, then one may make infinitely many different choices for each embedded
component.1
1 See
p. 231 of Zariski and Samuel, Commutative Algebra, Vol. I , Van Nostrand, 1958.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
351
But reduced primary decompositions are unique if all primes are isolated. The most
obvious case is for a p-primary ideal, q. Here, there is only one associated prime, p, and
hence q = q is the only reduced primary decomposition of q.
A more interesting case is that of a finite intersection of prime ideals. (Recall from
Corollary 9.2.11 that if A isNoetherian, then an ideal has this form if and only if it is its
m
own radical.) Here, if a = i=1 pi , then we can obtain a reduced primary decomposition
for a by inductively eliminating primes containing the intersection of all the other primes.
The result is a primary decomposition in which every primary ideal is prime and every
prime is isolated.
Corollary 9.2.22. Let a be a finite intersection of prime ideals. Then a may be written
uniquely as a finite intersection of prime ideals none of which contain the intersection of
the others. Here, the uniqueness is up to permutation of the primes.
9.3
The Nullstellensatz and the Prime Spectrum
Recall that an A-algebra of finite type is one which is finitely generated as an A-algebra.
Lemma 9.3.1. Suppose given inclusions A ⊂ B ⊂ C of commutative rings, where A
Noetherian and C is an A-algebra of finite type. Suppose that C is finitely generated as
a B-module. Then B is finitely generated as an A-algebra.
Proof Suppose that C = A[c1 , . . . , cn ] and that d1 , . . . , dm generate C as a B-module.
Then we can find elements bij and bijk in B such that
cj
dj dk
=
=
m
i=1
m
bij di
bijk di
for j = 1, . . . , n
and
for 1 ≤ j, k ≤ m.
i=1
Write A = A[bij , birs | 1 ≤ i, r, s ≤ m, 1 ≤ j ≤ n] ⊂ B. Since A is finitely generated
as an A-algebra, it is Noetherian by Corollary 7.8.15.
Using the displayed equations and induction, it is easy to see that any monomial in
the cj with coefficients in A lies in the A-submodule of C generated by 1, d1 , . . . , dm .
Thus, C is finitely generated as an A-module.
But then C is a Noetherian A-module (Proposition 7.8.9), so its submodule B is also
finitely generated as an A-module (Proposition 7.8.3). Since A is finitely generated as
an A-algebra, B must be, also.
We obtain a recognition principle for finite extensions.
Theorem 9.3.2. (Hilbert’s Nullstellensatz) Suppose the extension field L of K is finitely
generated as a K-algebra. Then L is a finite extension of K.
Proof Let L = K[α1 , . . . , αn ]. If each αi is algebraic over K, then we are done by
Lemma 8.2.9.
Otherwise, we wish to derive a contradiction. By Proposition 8.3.6, we may reorder
the α’s so that α1 , . . . , αr is a transcendence basis for L over K for some r ≤ n. Thus,
L is algebraic over K1 = K(α1 , . . . , αr ). Since L is obtained by adjoining finitely many
algebraic elements to K1 , Lemma 8.2.9 shows that L is a finite extension of K1 .
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By Lemma 9.3.1, K1 is finitely generated as a K-algebra. We write K1 = K[β1 , . . . , βs ].
Since α1 , . . . , αr are algebraically independent, K[α1 , . . . , αr ] is a unique factorization
domain, by Proposition 8.5.9. Since K1 is its field of fractions, we may write βi =
fi (α1 , . . . , αr )/gi (α1 , . . . , αr ), where fi and gi are relatively prime for i = 1, . . . , s.
But then any polynomial in β1 , . . . , βs has a denominator whose prime factors all
divide the product of the gi . Thus, it suffices to show that K[α1 , . . . , αr ] has infinitely
many primes, as, if h has a prime factor that does not divide any of the gi , then the
element 1/h of K(α1 , . . . , αr ) cannot lie in K[β1 , . . . , βs ].
But if p1 , . . . , pk are primes in K[α1 , . . . , αr ], then p1 . . . pk + 1 is relatively prime to
each pi , and hence has a prime factor distinct from the pi .
The Nullstellensatz has a number of important consequences for commutative algebras
of finite type over a field.
Corollary 9.3.3. Let K be a field and let f : A → B be a K-algebra homomorphism
between commutative K-algebras of finite type. Then for each maximal ideal m of B, the
inverse image f −1 (m) is maximal in A.
Proof Let m be a maximal ideal of B and let p = f −1 (m). Then there is an inclusion
of K-algebras A/p ⊂ B/m. But the Nullstellensatz shows B/m to be a finite extension
of K. Since A/p is a subalgebra of B/m, every element α ∈ A/p is algebraic over K
(Proposition 8.2.14). Thus, (Proposition 8.2.5) K[α] ⊂ A/p is a field, and hence α is
invertible in A/p. Since α was arbitrary, A/p is a field, and hence p = f −1 (m) is maximal
in A.
This allows us to deduce an important fact about radicals in commutative algebras
of finite type over a field.
Proposition 9.3.4. Let K be a field and let a be a proper ideal in K[X1 , . . . , Xn ]. Then
the radical, rad(a), of a is the intersection of the maximal ideals of K[X1 , . . . , Xn ] containing a.
In consequence, if A is a commutative algebra of finite type over a field K, then the
Jacobson radical and nil radical of A coincide:
N(A) = R(A).
Proof The second assertion is immediate from the first: if A is a commutative algebra of
finite type over K and if f : K[X1 , . . . , Xn ] → A is a surjective K-algebra homomorphism,
then f −1 (N(A)) = rad(ker f ), while f −1 (R(A)) is the intersection of the maximal ideals
of K[X1 , . . . , Xn ] containing ker f .
Thus, let a be a proper ideal of K[X1 , . . . , Xn ]. By Proposition 9.1.12, rad(a) is the
intersection of the prime ideals of K[X1 , . . . , Xn ] containing a. It suffices to show that if
f (X1 , . . . , Xn ) is not in rad(a), then there is a maximal ideal containing a which doesn’t
contain f .
Since f is not in rad(a), there is a prime ideal p containing a but not f . Thus, f
represents a nonzero element f of the domain A = K[X1 , . . . , Xn ]/p. Let B = A[1/f ],
the subring of the field of fractions of A obtained by adjoining 1/f to A. Write g :
K[X1 , . . . , Xn ] → B for the composite
π
→ K[X1 , . . . , Xn ]/p = A ⊂ B,
K[X1 , . . . , Xn ] −
where π is the canonical map. Then g is a K-algebra homomorphism between commutative K-algebras of finite type. Thus, if m is a maximal ideal of B, then g −1 (m) is a
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
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maximal ideal of K[X1 , . . . , Xn ] by Corollary 9.3.3. Since g(f ) is a unit in B, f ∈ g −1 (m).
Since a ⊂ ker g ⊂ g −1 (m), the result follows.
In a domain A, there are no nonzero nilpotent elements, and hence N(A) = 0.
Corollary 9.3.5. Let A be an integral domain finitely generated as an algebra over a
field K. Then R(A) = 0.
We can use the Nullstellensatz to give a complete determination of the maximal ideals
of a polynomial ring over an algebraically closed field. This consitutes the starting point
for algebraic geometry.
Corollary 9.3.6. (Nullstellensatz, weak version) Let K be an algebraically closed field
and let m be maximal ideal of K[X1 , . . . , Xn ]. Then m is the kernel of the evaluation
map
εa1 ,...,an : K[X1 , . . . , Xn ] → K
for some a1 , . . . , an ∈ K.
Thus, by Proposition 7.3.22, m = (X1 − a1 , . . . , Xn − an ), the ideal generated by
X1 − a1 , . . . , Xn − an .
Proof Let L = K[X1 , . . . , Xn ]/m. Since the polynomial ring maps onto L, L is a
K-algebra of finite type, and hence is a finite extension of K, by the Nullstellensatz.
Since K is algebraically closed, it has no irreducible polynomials of degree > 1. Thus,
any element algebraic over K has a minimal polynomial of degree 1, and hence lies in
K. Thus, the inclusion of K in L is an isomorphism.
∼
=
We obtain a K-algebra isomorphism f : K[X1 , . . . , Xn ]/m −→ K. Note that m is the
kernel of the composite
π
f
→ K[X1 , . . . , Xn ]/m −
K[X1 , . . . , Xn ] −
→ K,
where π is the canonical map. But if f ◦ π(Xi ) = ai for i = 1, . . . , n, then f ◦ π must
be the evaluation map εa1 ,...,an : K[X1 , . . . , Xn ] → K by the universal property of a
polynomial algebra (Corollary 7.3.20).
Definitions 9.3.7. Let A be a commutative ring. We write Spec(A) for the set of prime
ideals of A and write Max(A) for the set of maximal ideals of A. We call them the prime
spectrum and the maximal ideal spectrum of A, respectively.
If K is a field and (a1 , . . . , an ) ∈ K n , we write ma1 ,...,an ∈ Max(A) for the kernel of
the evaluation map εa1 ,...,an : K[X1 , . . . , Xn ] → K:
ma1 ,...,an = {f ∈ K[X1 , . . . , Xn ] | f (a1 , . . . , an ) = 0}.
By Proposition 7.3.22, ma1 ,...,an = (X1 − a1 , . . . , Xn − an ), the ideal generated by X1 −
a1 , . . . , Xn − an .
We write ν : K n → Max(K[X1 , . . . , Xn ]) for the map given by ν(a1 , . . . , an ) =
ma1 ,...,an .
Proposition 9.3.8. The maps ν : K n → Max(K[X1 , . . . , Xn ]) are injective for any field
K and are surjective if K is algebraically closed.
Proof If ma1 ,...,an = mb1 ,...,bn , then Xi − ai and Xi − bi lie in this ideal for each i. But
(Xi − ai ) − (Xi − bi ) is a unit unless ai = bi , so ν is injective.
If K is algebraically closed, then ν is surjective by the weak Nullstellensatz.
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354
If K is algebraically closed, we can think of ν as a way of providing a geometry for the
maximal ideals of K[X1 , . . . , Xn ]. In particular, when K = C,2 this provides a standard
sort of geometry to associate to algebraically significant subsets of the set of maximal
ideals. Some subsets of K n which have algebraic significance are the affine varieties.
Definitions 9.3.9. Let K be an algebraically closed field. We shall refer to K n as affine
n-space over K. If S ⊂ K[X1 , . . . , Xn ] is any subset, then the affine variety, V(S),
determined by S is the subset of affine n-space consisting of the common zeroes of the
elements of S:
V(S) = {(a1 , . . . , an ) ∈ K n | f (a1 , . . . , an ) = 0
for all f ∈ S}.
As we shall verify in Lemma 9.3.12, the affine variety V(S) is mapped by ν : K n ∼
=
Max(K[X1 , . . . , Xn ]) to the set of maximal ideals of K[X1 , . . . , Xn ] containing S.
There is a geometry to affine varieties intrinsic to the algebra. It does not coincide
with ordinary complex geometry when K = C. It is called the Zariski topology, and may
be defined not only on affine varieties, but also on Spec(A) for any commutative ring A.
Definitions 9.3.10. Let A be a commutative ring and let S ⊂ A. The Zariski open set
U (S) of Spec(A) is given by
U (S) = {p ∈ Spec(A) | S ⊂ p}.
The Zariski topology on Spec(A) is the one in which the open sets are precisely the
Zariski open sets. The Zariski topology on Max(A) is the subspace topology of that on
Spec(A).
Let K be an algebraically closed field. Then Proposition 9.3.8 shows that
ν : K n → Max(K[X1 , . . . , Xn ])
is a bijection. We define the Zariski topology on K n by declaring ν to be a homeomorphism.
That this gives a topology, of course, requires justification.
Proposition 9.3.11. Let A be a commutative ring. Then the Zariski open sets form a
topology on Spec(A). Moreover, they have the following properties.
1. U ({1}) = Spec(A).
2. U ({0}) = ∅.
3. If S ⊂ T ⊂ A, then U (S) ⊂ U (T ).
4. If a is the ideal generated by S, then U (S) = U (a).
5. If Si ⊂ A for i ∈ I, then i∈I U (Si ) = U ( i∈I Si ).
6. U ({s}) ∩ U ({t}) = U ({st}) for all s, t ∈ A.
7. For S, T ⊂ A, U (S) ∩ U (T ) = U ({st | s ∈ S and t ∈ T }).
2 We
shall show in Theorem 11.6.7 that C is algebraically closed.
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355
Proof The first condition holds because the elements of Spec(A) are proper ideals. The
next four conditions are clear, and the sixth is a restatement of the definition of a prime
ideal: st ∈ p if and only if either s ∈ p or t ∈ p. The last condition follows from the
others.
If K is algebraically closed, then the Zariski topology on affine n-space over K is
determined by the affine varieties.
Lemma 9.3.12. Let K be an algebraically closed field. Then the affine varieties are precisely the closed subspaces in the Zariski topology on the affine n-space K n . Specifically,
if S ⊂ K[X1 , . . . , Xn ], then the homeomorphism ν : K n → Max(K[X1 , . . . , Xn ]) carries
the affine variety V(S) onto the complement of the Zariski open set U (S).
Proof An n-tuple (a1 , . . . , an ) lies in V(S) if and only if f (a1 , . . . , an ) = 0 for all
f ∈ S. But f (a1 , . . . , an ) = 0 if and only if f lies in the maximal ideal ν(a1 , . . . , an ) =
ker εa1 ,...,an , so ν(V(S)) is the set of maximal ideals of K[X1 , . . . , Xn ] containing S. But
this is precisely the complement of U (S).
By analogy, for an arbitrary commutative ring A and a subset S ⊂ A, we shall write
V (S) for the complement of U (S) in either Spec(A) or Max(A). In this more general
context, the spaces are called schemes, rather than varieties.
From a topologist’s point of view, the topologies on Spec(A) and Max(A) are bizarre.
Note that a point p in Spec(A) is closed if and only if there is a set S such that p
contains S, but no other prime ideal contains S. Clearly, such an S exists if and only if
p is maximal.
Thus, Max(A) is a T1 space, but Spec(A) is not. So why bother with Spec(A)?
For one thing, prime ideals have some interest in their own right. For another, as we
shall see presently, Spec is a functor of commutative rings, but Max is not. However,
Corollary 9.3.3 shows that Max is a functor on algebras of finite type over a field K:
Definition 9.3.13. Let f : A → B be a homomorphism between commutative rings.
Then the induced map f ∗ : Spec(B) → Spec(A) is given by f ∗ (p) = f −1 (p).
If K is a field and f : A → B is a K-algebra homomorphism between commutative
K-algebras of finite type, then the induced map f ∗ : Max(B) → Max(A) is given by
f ∗ (m) = f −1 (m).
The next lemma is left as an exercise.
Lemma 9.3.14. Let f : A → B be a homomorphism of commutative rings. Then f ∗ :
Spec(B) → Spec(A) is continuous. Moreover, if g : B → C is another homomorphism of
commutative rings, then f ∗ ◦ g ∗ = (g ◦ f )∗ . Thus, Spec is a contravariant functor from
the category of commutative rings to the category of topological spaces.
The same conclusions hold for the maps on Max induced by K-algebra homomorphisms between commutative algebras of finite type over a field K.
Let a be an ideal of the commutative ring A and let π : A → A/a be the canonical
map. Then π ∗ : Spec(A/a) → Spec(A) is a closed map, inducing a homeomorphism of
Spec(A/a) onto the closed subset V (a) of Spec(A). Moreover, since π −1 (p) is maximal
in A if and only if p is maximal in A/a, π ∗ restricts to a homeomorphism of Max(A/a)
onto the closed subset V (a) of Max(A).
The next lemma is obvious, but important.
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Lemma 9.3.15. Let a be the ideal generated by S in A, then V (S) = V (a) in Spec(A).
Thus, the closed subspaces of Spec(A) are the homeomorphic images of the prime spectra
of the quotient rings of A.
Of course, the ideals in a Noetherian ring are all finitely generated.
Corollary 9.3.16. Let A be a Noetherian ring. Then every closed subset of Spec(A) or
Max(A) has the form V ({a1 , . . . , ak }) for some finite subset {a1 , . . . , ak } of A.
Similarly, if K is an algebraically closed field, then every affine variety has the form
V({f1 , . . . , fk }) for some finite subset {f1 , . . . , fk } of K[X1 , . . . , Xn ].
Examples 9.3.17.
1. Let A be a P.I.D. and let 0 = a ∈ A. Then V ({a}) is the set of equivalence classes
of prime elements that divide a, and hence is finite. In fact, we can see that the
proper closed subsets of Spec(A) are precisely the finite subsets of Max(A).
p ), then
2. If A is a P.I.D. with only one nonzero prime ideal (e.g., A = Z(p) or A = Z
Spec(A) is a space consisting of two points, one of which is open and the other not.
This space is sometimes called Sierpinski space.
3. Let K be algebraically closed. Then in affine n-space K n , we have V({Xn }) =
{(a1 , . . . , an ) ∈ K n | an = 0}, the image of the standard inclusion of K n−1 in K n .
In particular, not all Zariski closed sets are finite.
Remarks 9.3.18. There is an important interplay between the study of affine varieties
over the complex numbers and differential topology. Given a finite collection f1 , . . . , fk
of C ∞ functions from Rm to R, let f : Rm → Rk be given by f (x) = (f1 (x), . . . , fk (x))
for x ∈ Rn . Then the set of common zeros of f1 , . . . , fk is precisely f −1 (0). Using the
Implicit Function Theorem of advanced calculus, one can show that f −1 (0) is a smooth
(i.e., C ∞ ) submanifold of Rm , provided that the Jacobian matrix of f has rank k at
every point of f −1 (0).
This construction is generic for smooth manifolds: Every smooth manifold is diffeomorphic (i.e., equivalent as a smooth manifold) to a manifold obtained by the above
procedure.
Now, a complex affine variety is the set of common zeros of a finite collection of
complex polynomials f1 , . . . , fk ∈ C[X1 , . . . , Xn ], and hence is the zero set of the function
f : Cn → Ck whose i-th component function is fi for i = 1, . . . , n.
Regarding f as a function from R2n to R2k , we can compute the Jacobian matrix
of f . If it has rank 2k at each point of V(f1 , . . . , fk ), then V(f1 , . . . , fk ) is a smooth
manifold. (Here, we use the subspace topology from R2n , of course, and not the Zariski
topology.)
In this manner, complex varieties have been an important source of examples of
manifolds. Such examples have been useful in a number of contexts, especially in real
dimension 4, where some of the standard techniques of smooth handle theory fail.
Complex varieties that are not manifolds are also interesting topologically as a sort
of generalized manifolds.
From the other side, topological techniques provide invariants for complex varieties.
These invariants are generally weaker than their structure as algebraic varieties, but can
be useful nevertheless.
Returning to the study of algebraic varieties, let K be an algebraically closed field
and let a be an ideal in K[X1 , . . . , Xn ]. By Lemma 9.3.14, we may identify the affine
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
357
variety V(a) with Max(K[X1 , . . . , Xn ]/a), the maximal ideal spectrum of a commutative
K-algebra of finite type. Since Max is a functor on such algebras, we see that the
structure of the affine variety V(a) depends only on the K-algebra K[X1 , . . . , Xn ]/a.
Indeed, if A is a commutative K-algebra of finite type, we can think of Max(A) as an
abstract affine variety: Any choice of K-algebra generators, a1 , . . . , an , of A induces a
surjective K-algebra homomorphism εa1 ,...,an : K[X1 , . . . , Xn ] → A, and hence induces
a homeomorphism ε∗a1 ,...,an : Max(A) → V (ker εa1 ,...,an ) ∼
= V(ker εa1 ,...,an ).
Abstract varieties may be compared by means of algebraic maps.
Definitions 9.3.19. Let K be an algebraically closed field and let A and B be commutative K-algebras of finite type. An algebraic map Max(B) → Max(A) is a function of
the form f ∗ for some K-algebra homomorphism f : A → B. Here, the map depends only
on its effect as a function Max(B) → Max(A), and not on the homomorphism inducing
it.
We need to take a slightly different approach in order to define mappings between
affine varieties. The point is that while there’s a standard identification of V(a) ⊂ K n
with Max(K[X1 , . . . , Xn ]/a), the variety V(a) does not determine the ideal a: There
exist proper inclusions a ⊂ b of ideals such that V(b) = V(a). To resolve this, we shall
develop the notions of polynomial mappings and coordinate rings.
Definition 9.3.20. Let K be an algebraically closed field and let a be an ideal of
K[X1 , . . . , Xn ]. Write X ⊂ K n for the affine variety determined by a:
X = V(a) = {(a1 , . . . , an ) | f (a1 , . . . , an ) = 0
for all f ∈ a}.
We define the ideal of X, written IX , to be the set of all polynomials which vanish on
every element of X:
IX = {f ∈ K[X1 , . . . , Xn ] | f (a1 , . . . , an ) = 0
for all (a1 , . . . , an ) ∈ X}.
Then the coordinate ring, O(X), of X is defined by
O(X) = K[X1 , . . . , Xn ]/IX .
The coordinate ring catalogues the polynomial functions from X to K. Recall from
Lemma 7.3.25 that the passage from polynomials f (X1 , . . . , Xn ) ∈ K[X1 , . . . , Xn ] to
the induced functions f : K n → K gives a ring homomorphism φ : K[X1 , . . . , Xn ] →
Map(K n , K), where Map(K n , K) is the ring of all functions from K n to K. Indeed,
restricting the effect of these polynomials to the affine variety X ⊂ K n gives a ring
homomorphism
φX : K[X1 , . . . , Xn ] → Map(X, K).
The ideal, IX , of X is easily seen to be the kernel of φX . Thus, we may identify O(X)
with the image of φX , the ring of functions from X to K induced by polynomials. Thus,
two polynomials, f and g, are equal in O(X) if and only if they agree as functions when
restricted to X.
By definition of X = V(a), we see that a ⊂ IX .
Proposition 9.3.21. Let X = V(a) be an affine variety over the algebraically closed
field K and let IX be the ideal of X. Then IX is the intersection of all the maximal
ideals of K[X1 , . . . , Xn ] containing a. Thus, by Proposition 9.3.4, IX = rad(a).
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
358
Moreover, V(IX ) = V(a) = X, and the canonical map
π : K[X1 , . . . , Xn ]/a → K[X1 , . . . , Xn ]/IX = O(X)
induces a homeomorphism from Max(O(X)) to Max(K[X1 , . . . , Xn ]/a).
Proof Recall that the element (a1 , . . . , an ) ∈ K n corresponds to the maximal ideal
ma1 ,...,an ⊂ K[X1 , . . . , Xn ], which is the kernel of the evaluation map εa1 ,...,an : K[X1 , . . . , Xn ] →
K. An element (a1 , . . . , an ) lies in X if and only if every element of a vanishes on
(a1 , . . . , an ), meaning that a ⊂ ma1 ,...,an . In particular, since K is algebraically closed,
the points of X correspond to the maximal ideals of K[X1 , . . . , Xn ] containing a.
But an element of K[X1 , . . . , Xn ] lies in IX if and only if it vanishes on each point of
X, and hence is contained in every maximal ideal of K[X1 , . . . , Xn ] containing a.
Since a ⊂ IX , V(IX ) ⊂ V(a). But since every element of IX kills every element of
X = V(a), the two varieties coincide. The result now follows from Lemma 9.3.14.
In particular, an ideal b has the form IX for an affine variety X if and only if b = rad(a)
for some ideal a. Since rad(rad(a)) = rad(a), this is equivalent to saying that b is its own
radical (Lemma 9.1.13).
Theorem 9.3.22. Let K be an algebraically closed field. There is a one-to-one correspondence between the affine varieties in K n and the ideals in K[X1 , . . . , Xn ] which are
their own radicals. The correspondence takes a variety X to IX and takes an ideal a that
is its own radical to V(a).
Since each direction of this correspondence is order reversing, we see that X ⊂ Y
if and only if IY ⊂ IX , and if either inclusion is proper, so is the other. Similarly, if
each of a and b is its own radical, then a ⊂ b if and only if V(b) ⊂ V(a), and if either
inclusion is proper, so is the other.
Proof If a is its own radical, then IV(a) = rad(a) = a. Conversely, if X is a variety in
K n , then X = V(IX ) by Proposition 9.3.21.
It is easy to see (Corollary 9.1.16) that a finite intersection of prime ideals is its own
radical.
Corollary 9.3.23. Let K be an algebraically closed field and let a be a finite intersection
of prime ideals of K[X1 , . . . , Xn ]. Then O(V(a)) = K[X1 , . . . , Xn ]/a.
In particular, affine n-space K n is V(0). Since 0 is prime, we obtain a calculation of
O(K n ).
Corollary 9.3.24. O(K n ) = K[X1 , . . . , Xn ] for any algebraically closed field K. Thus,
polynomials in n variables over K are determined by their effect as functions on K n .
We could have given simpler argument using Proposition 7.3.26 and Lemma 8.4.3.
Since K[X1 , . . . , Xn ] is Noetherian, the theory of primary decomposition shows (Corollary 9.2.11) that an ideal a of K[X1 , . . . , Xn ] is its own radical if and only if it is a finite
intersection of prime ideals. Using this, we can show that every affine variety over an
algebraically closed field has a unique decomposition as a union of irreducible varieties.
Definition 9.3.25. An affine variety X over an algebraically closed field is irreducible
if whenever we have affine varieties Y and Z with X = Y ∪ Z, either X = Y or X = Z.
Recall from Corollary 7.6.10 that a prime ideal contains an intersection a ∩ b of ideals
if and only if it contains at least one of a and b. We obtain the following lemma.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
359
Lemma 9.3.26. Let a and b be ideals in the commutative ring A. Then V (a ∩ b) =
V (a) ∪ V (b) in either Spec(A) or Max(A).
If a and b are ideals in K[X1 , . . . , Xn ] for K algebraically closed, then the affine
varieties V(a ∩ b) and V(a) ∪ V(b) are equal.
Proposition 9.3.27. An affine variety X over an algebraically closed field K is irreducible if and only if IX is prime.
Proof Suppose that IX is prime and that X = Y ∪ Z. Clearly, IY∪Z = IY ∩ IZ . But
prime ideals are irreducible ideals (Lemma 9.2.17), so either IX = IY or IX = IZ . By
Theorem 9.3.22, either X = Y or X = Z, so X is irreducible.
Conversely, supposethat X is irreducible. By Corollary 9.2.11 and Theorem 9.3.22,
m
X = V(a),where a = i=1 pi , where pi is prime for i = 1, . . . , m. Now Lemma 9.3.26
m
gives X = i=1 V(pi ). Since X is irreducible, X = V(pi ) for some i, and hence IX = pi
by Theorem 9.3.22.
Corollary 9.2.22 shows that
mevery self-radical ideal of K[X1 , . . . , Xn ] may bewritten
uniquely as an intersection i=1 pi of prime ideals such that no pi contains j
=i pj .
Theorem 9.3.22 and Lemma 9.3.26 give the following proposition.
Proposition 9.3.28. Every affine variety X over the algebraically closed field has a
unique decomposition
X=
m
Xi
i=1
where Xi is irreducible for i = 1, . . . , m, and no Xi is contained in
uniqueness is up to a reordering of the Xi .
j
=i
Xj . Here, the
We now define the mappings appropriate for comparing two affine varieties.
Definition 9.3.29. Let K be an algebraically closed field and let X ⊂ K n be an affine
variety. A polynomial function f : X → K m is a function of the form
f (x) = (f1 (x), . . . , fm (x))
for all x ∈ X, where f1 , . . . , fm ∈ O(X). If the image of f is contained in an affine variety
Y ⊂ K m , we may consider f to be a polynomial function from X to Y.
The next lemma is quite elementary, but occurs frequently in the subsequent material.
Lemma 9.3.30. Let A be a commutative ring and let g(X1 , . . . , Xm ) ∈ A[X1 , . . . , Xm ].
Suppose given polynomials f1 , . . . , fm ∈ A[X1 , . . . , Xn ]. Then evaluating each Xi at fi
gives a polynomial g(f1 , . . . , fm ) ∈ A[X1 , . . . , Xn ]. The evaluation of this polynomial at
an element x ∈ An is given by
g(f1 , . . . , fm )(x) = g(f1 (x), . . . , fm (x)),
where the right hand side is the evaluation of g at the element (f1 (x), . . . , fm (x)) ∈ Am .
In other words, if εf1 ,...,fm : A[X1 , . . . , Xm ] → A[X1 , . . . , Xn ] is the A-algebra homomorphism obtained by evaluating Xi at fi for i = 1, . . . , m, then for g ∈ A[X1 , . . . , Xm ],
the function from An to A induced by εf1 ,...,fm (g) is given by
(εf1 ,...,fm (g))(x) = (g ◦ f )(x),
where f : An → Am is the function given by f (x) = (f1 (x), . . . , fm (x)) for all x ∈ An .
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
360
Proof If f : B → C is a homomorphism of commutative A-algebras and b1 , . . . , bm ∈ B,
then the composite
εb
,...,bm
f
1
A[X1 , . . . , Xm ] −−−
−−−→ B −
→C
is just the evaluation map εf (b1 ),...,f (bm ) . Taking f to be the map from A[X1 , . . . , Xn ]
obtained from evaluation at x ∈ An , the result follows.
A first application is to composites of polynomial functions.
Corollary 9.3.31. Let K be an algebraically closed field and let X ⊂ K n and Y ⊂ K m
be affine varieties. Then for any polynomial mapping f : X → Y, there is a polynomial
mapping f : K n → K m making the following diagram commute.
X
∩
Kn
f
/ Y
∩ f / m
K
If Z ⊂ K k is another affine variety and if g : Y → Z is a polynomial mapping, then
the composite g ◦ f is a polynomial mapping as well.
Finally, let ξi be the image of Xi under the canonical map π : K[X1 , . . . , Xn ] → O(X)
for i = 1, . . . , n. Then ξi : X → K is the restriction to X of the projection of K n onto
its i-th factor. Thus, the polynomial mapping ι : X → K n given by
ι(x) = (ξ1 (x), . . . , ξn (x))
is just the inclusion map of X in K n . The image of ι is, of course X, so the identity
map of X is a polynomial mapping.
Proof Since f is a polynomial mapping, there are elements f1 , . . . , fm ∈ O(X) such
that f (x) = (f1 (x), . . . , fm (x)) for all x ∈ X. But O(X) = K[X1 , . . . , Xn ]/IX , so
there are polynomials fi ∈ K[X1 , . . . , Xn ] representing fi for i = 1, . . . , n. Setting
f(x) = (f1 (x), . . . , fm (x)) for all x ∈ K n , we obtain a polynomial mapping f : K n → K m
making the diagram commute.
Similarly, there are polynomials gi ∈ K[X1 , . . . , Xm ] for i = 1, . . . , k such that the
polynomial mapping g : K m → K k given by g(y) = (
g1 (y), . . . , gk (y)) for y ∈ K m
restricts on Y to g : Y → Z.
The component functions of g ◦ f are the maps gi ◦ f, which satisfy
(
gi ◦ f)(x) = gi (f1 (x), . . . , fm (x))
for all x ∈ K n . By Lemma 9.3.30, this is the map induced by the polynomial gi (f1 , . . . , fm ) ∈
K[X1 , . . . , Xn ]. Thus, the composite g ◦ f : K n → K k is a polynomial mapping which
clearly restricts on X to the composite function (g ◦ f ) : X → Z. Since the restriction to
X of any element of K[X1 , . . . , Xn ] determines an element of O(X), the the polynomial
mappings are closed under composites.
Let ξi be the image in O(X) of the polynomial Xi . Then Xi : K n → K restricts to
ξi on X. But Xi (a1 , . . . , an ) = ai , and the result follows.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
361
In particular, Corollary 9.3.31 shows that the affine varieties and polynomial mappings
form a category.
Definition 9.3.32. Let K be an algebraically closed field. We write VarK for the
category whose objects and are the affine varieties over K and whose morphisms are
given as follows. If X and Y are affine varieties over K, then the set of morphisms,
MorVarK (X, Y), from X to Y in VarK is the set of polynomial mappings from X to Y.
We now show that polynomial mappings induce homomorphisms of coordinate rings.
Proposition 9.3.33. Let X ⊂ K n and Y ⊂ K m be affine varieties over the algebraically
closed field K and let f : X → Y be a polynomial mapping, given by
f (x) = (f1 (x), . . . , fm (x))
for x ∈ X, where fi ∈ O(X) for i = 1, . . . , m. Then there is a unique K-algebra
homomorphism O(f ) : O(Y) → O(X) with the property that O(f )(ξi ) = fi for i =
1, . . . , m, where ξi is the image of Xi under the canonical map π : K[X1 , . . . , Xm ] →
O(Y).
If g : Y → Z is another polynomial mapping, then O(f ) ◦ O(g) = O(g ◦ f ). Also,
O(1X ) = 1O(X) . Thus, O is a contravariant functor from the category, VarK , of affine
varieties and polynomial mappings over K to the category of commutative K-algebras of
finite type and K-algebra homomorphisms.
Proof Since ξ1 , . . . , ξm generate O(Y) as a K-algebra, there is at most one K-algebra
homomorphism carrying ξi to fi for all i. We must show that such a homomorphism
exists.
For each i, let fi be a lift of fi under the canonical map π : K[X1 , . . . , Xn ] → O(X).
Then there is a K-algebra homomorphism εf1 ,...,fm : K[X1 , . . . , Xm ] → K[X1 , . . . , Xn ]
obtained by evaluating Xi at fi for i = 1, . . . , m. By the universal property of quotient
rings, it suffices to show that εf1 ,...,fm (IY ) ⊂ IX : We may then define O(f ) to be the
unique homomorphism making the following diagram commute.
εf1 ,...,fm
/ K[X1 , . . . , Xn ]
K[X1 , . . . , Xm ]
π
O(Y)
π
O(f )
/ O(X)
Thus, let h(X1 , . . . , Xm ) ∈ IY . We wish to show that εf1 ,...,fm (h) ∈ IX , meaning
that it vanishes on each element of X. By Lemma 9.3.30, ε (h) = h ◦ f as a
f1 ,...,fm
function on K n , where f(x) = (f1 (x), . . . , fm (x)) for x ∈ K n . Since fi is a lift of fi to
K[X1 , . . . , Xn ], fi (x) = fi (x) for all x ∈ X. Thus, as a function on X, f = f . Since
f (X) ⊂ Y and since h vanishes on Y, εf1 ,...,fm (h) ∈ IX , as desired.
If g : Y → Z ⊂ K k is a polynomial map, write g(y) = (g1 (y), . . . , gk (y)) for y ∈ Y,
where gi ∈ O(Y) for all i. Let gi be a lift of gi to K[X1 , . . . , Xm ] for i = 1, . . . , k. Then
the following diagram commutes.
ε
εg1 ,...,gk
/ K[X1 , . . . , Xm ] f1 ,...,fm/ K[X1 , . . . , Xn ]
K[X1 , . . . , Xk ]
π
O(Z)
π
O(g)
/ O(Y)
π
O(f )
/ O(X)
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
362
Thus, O(f ) ◦ O(g)(ξi ) is the image in O(X) of εf1 ,...,fm (
gi ) = gi (f1 , . . . , fm ). Now, an
element of O(X) is determined by its value as a function on X. But for x ∈ X, we have
gi (f1 (x), . . . , fm (x))
= gi (f (x))
= gi (f (x)),
since f (x) ∈ Y. Since they have the same effect on elements of X, O(f ) ◦ O(g)(ξi ) must
be equal to the i-th component function of g ◦ f . Thus, the uniqueness statement used
to define O(g ◦ f ) gives O(f ) ◦ O(g) = O(g ◦ f ).
Finally, Corollary 9.3.31 shows that the identity map of X is given by 1X (x) =
(ξ1 (x), . . . , ξn (x)) for all x ∈ X, so O(1X ) is characterized by O(1X )(ξi ) = ξi for i =
1, . . . , n. This forces O(1X ) = 1O(X) .
Proposition 9.3.34. Let X ⊂ K n and Y ⊂ K m be affine varieties over the algebraically
closed field K. Then the passage from f : X → Y to O(f ) : O(Y) → O(X) gives a oneto-one correspondence between the polynomial mappings from X to Y and the K-algebra
homomorphisms from O(Y) to O(X).
Proof By Proposition 9.3.33, it suffices to show that every K-algebra homomorphism
γ : O(Y) → O(X) is equal to O(f ) for some polynomial mapping f : X → Y.
Thus, let γ : O(Y) → O(X) be a K-algebra homomorphism, and let fi = γ(ξi ) ∈
O(X) for i = 1, . . . , m. (As above, ξi is the image of Xi under the canonical map
π : K[X1 , . . . , Xm ] → O(Y).) Let f : X → K m be the polynomial mapping given by
f (x) = (f1 (x), . . . , fm (x))
for x ∈ X. By the characterization of the maps O(f ) given in Proposition 9.3.33, it
suffices to show that f (X) ⊂ Y.
Now, Y = V(IY ) is the set of elements of K m which vanish under every element of
the ideal of Y, so it suffices to show that if g ∈ IY and x ∈ X, then g vanishes on f (x).
Let fi be a lift of fi to K[X1 , . . . , Xn ] for i = 1, . . . , m and let f : K n → K m be
the polynomial mapping whose i-th component function is fi for i = 1, . . . , m. Then
f(x) = f (x) for all x ∈ X, so it suffices to show that the polynomial inducing g ◦ f lies
in IX .
Lemma 9.3.30 shows that g ◦ f is induced by εf1 ,...,fm (g). Note that the fi were chosen
to make the following diagram commute.
εf1 ,...,fm
/ K[X1 , . . . , Xn ]
K[X1 , . . . , Xm ]
π
O(Y)
π
γ
/ O(X)
The kernels of the vertical maps π are precisely IY and IX . Since g ∈ IY , its image
under εf1 ,...,fm must lie in IX .
Thus, the relationships between affine varieties given by polynomial mappings and
the relationships between their corrdinate rings given by K-algebra homomorphisms are
precise mirrors of one another.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
363
Corollary 9.3.35. Let X and Y be affine varieties over the algebraically closed field K.
Then there is a polynomial isomorphism between X and Y (i.e., a bijective polynomial
mapping f : X → Y whose inverse function is polynomial) if and only if the coordinate
rings O(X) and O(Y) are isomorphic as K-algebras.
Proof If f : X → Y is a polynomial isomorphism, then O(f ) is an isomorphism by the
functoriality of the coordinate ring: If f −1 : Y → X is the inverse function of f , then
O(f ) ◦ O(f −1 ) = O(1X ) = 1O(X) and O(f −1 ) ◦ O(f ) = O(1Y ) = 1O(Y) .
−1
If O(f ) is an isomorphism, then Proposition 9.3.34 shows that O(f ) = O(g) for
some polynomial mapping g : Y → X. But then O(f ◦ g) = 1O(Y) , so Proposition 9.3.34
shows that f ◦ g = 1Y . Similarly, g ◦ f = 1X .
We next show that the polynomial functions between affine varieties may be identified
with the algebraic maps between the maximal ideal spectra of their coordinate rings.
Proposition 9.3.36. Let X ⊂ K n and Y ⊂ K m be affine varieties over the algebraically
closed field K. Let f : X → Y be a polynomial mapping. Then there is a commutative
diagram
∗
Max(O(X))
O(f )
∼
=
/ Max(O(Y))
∼
=
X
/ Y
f
where the vertical maps are the standard isomorphisms. Thus, the algebraic maps from
Max(O(X)) to Max(O(Y)) correspond precisely under the standard isomorphisms to the
polynomial mappings from X to Y.
Proof The standard isomorphism from Max(O(X)) to X is the composite
π∗
ν −1
∼
=
∼
=
Max(O(X)) −→ V (IX ) −−→ X,
where π : K[X1 , . . . , Xn ] → O(X) is the canonical map. The analogous result holds for
Y.
Let f : K n → K m be a polynomial mapping which restricts to f on X. Then the top
and bottom squares of the following diagram commute.
∗
Max(O(X))
O(f )
π∗
/ Max(O(Y))
π∗
O(f)∗
/ Max(K[X1 , . . . , Xm ])
Max(K[X1 , . . . , Xn ])
O
O
ν ∼
=
KO n
ν ∼
=
f
∪
X
/ Km
O
∪
f
/ Y
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
364
It suffices to show that the middle square commutes.
For x = (a1 , . . . , an ) ∈ K n , ν(x) is the kernel of the evaluation map εa1 ,...,an , and
hence O(f)∗ (ν(x)) is the kernel of εa1 ,...,an ◦ O(f). Let fi be the i-th component function of f. Then O(f) is precisely the evaluation map εf1 ,...,fm : K[X1 , . . . , Xm ] →
K[X1 , . . . , Xn ]. Thus, εa1 ,...,an ◦ O(f) is the evaluation map from K[X1 , . . . , Xm ] → K
which evaluates Xi at fi (a1 , . . . , an ). The kernel of this evaluation map is precisely
ν(f(x)), so the middle square commutes.
Corollary 9.3.37. Polynomial mappings are continuous in the Zariski topology.
The polynomial mappings give a much more rigid notion of equivalence of varieties
than is provided by the Zariski topology alone. Indeed, it is quite possible for varieties to
be homeomorphic as spaces but inequivalent as varieties. Consider the following example.
Example 9.3.38. Let K be an algebraically closed field. Let κ : K[X, Y ] → K[X] be
the K-algebra homomorphism taking X to X 2 and Y to X 3 , and let p = ker κ. Since
im κ is a domain, p is prime, so O(V(p)) = K[X, Y ]/p by Corollary 9.3.23. (We leave it
as an exercise to show that p = (Y 2 − X 3 ).)
Write κ : O(V(p)) = K[X, Y ]/p → K[X] = O(K) for the homomorphism induced
by κ. Then Proposition 9.3.34 shows that κ = O(f ), where f : K → V(p) is given by
f (x) = (x2 , x3 ) for all x ∈ K. By inspection, f : K → V(p) is injective. We claim it is
a homeomorphism. But the coordinate ring of V(p) is isomorphic to im κ = K[X 2 , X 3 ],
which cannot be generated as a K-algebra by a single element. Thus, Corollary 9.3.35
shows that V(p) is not isomorphic to K as a variety (i.e., through polynomial mappings).
The coordinate ring O(K) = K[X] is a P.I.D. Thus, as shown in Example 1 of
Examples 9.3.17, the proper closed subsets of K are precisely the finite subsets of K.
Since affine varieties are T1 spaces, f (S) is closed in V(p) for every proper closed subspace
S of K. Thus, to show that f is a homeomorphism, it suffices to show it is onto. Thus,
by Proposition 9.3.36, it suffices to show that i∗ : Max(K[X]) → Max(K[X 2 , X 3 ]) is
onto, where i : K[X 2 , X 3 ] ⊂ K[X] is the inclusion.
For this, we shall make use of the theory of integral dependence developed in Section 12.7. Note first that since K[X] is generated by 1, X as a K[X 2 , X 3 ]-module,
it is integral over K[X 2 , X 3 ] by Proposition 12.7.2. Now Proposition 12.7.22 shows
that for each maximal ideal m of K[X 2 , X 3 ] there is a prime ideal p of K[X] with
m = p ∩ K[X 2 , X 3 ]. But every nonzero prime ideal of K[X] is maximal, so m lies in the
image of i∗ : Max(K[X]) → Max(K[X 2 , X 3 ]).
Exercises 9.3.39.
1. Give the proof of Lemma 9.3.14.
2. Let A be a commutative ring. Show that A[X 2 , X 3 ] is free as an A[X 2 ]-module,
with basis 1, X 3 . Deduce that the kernel of the evaluation map εX 2 ,X 3 : A[X, Y ] →
A[X 2 , X 3 ] is the principal ideal generated by Y 2 − X 3 .
3. Let K be a field and fix n > 0. Suppose the map ν : K n → Max(K[X1 , . . . , Xn ])
is onto. Show that K is algebraically closed.
4. Let K be any field. Show that a maximal ideal m lies in the image of ν : K n →
Max(K[X1 , . . . , Xn ]) if and only if there is an element (a1 , . . . , an ) ∈ K n such that
f (a1 , . . . , an ) = 0 for all f ∈ m.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
365
5. Show that we may identify Max(R[X]) with the upper half plane in C. What is
the map Max(C[X]) → Max(R[X]) induced by the natural inclusion?
6. What is Spec(A × B)?
7. Let A be a commutative ring. Show that the topology on Spec(A) is T0 , i.e., that
for any pair of points p, q, at least one of the two points is open in the subspace
topology on {p, q}.
8. Let A be a commutative ring. Show that the sets U ({a}) for a ∈ A form a basis
for the Zariski topology on Spec(A). It is customary to write X = Spec(A) and
Xa = U ({a}).
9. Let A be a commutative ring. Show that Spec(A) is compact in the sense that any
open cover of Spec(A) has a finite subcover. (Many writers call a non-Hausdorff
space with this property quasi-compact.)
9.4
Tensor Products
The theory of tensor products is one of the most important tools in module theory. We
have already seen it used implicitly in more than one argument. One especially important
application, which we shall emphasize here, is extension of rings, whereby, if M is an
A-module and f : A → B is a ring homomorphism, we create an associated B-module
from the tensor product B ⊗A M . This construction is functorial, and has important
applications to exactness arguments.
Tensor products are a good example of an important concept that really cannot be
adequately understood except via universal mapping properties.
We shall use some somewhat nonstandard terminology in the next definition, as there
are two definitions of bilinearity common in the literature, and we wish to distinguish
between them.
Definitions 9.4.1. Let A be a ring and suppose given a right A-module M and a left
A-module N . Let G be an abelian group. A function f : M × N → G is said to be
weakly A-bilinear if
f (m + m , n)
= f (m, n) + f (m , n)
f (m, n + n ) = f (m, n) + f (m, n )
f (ma, n) = f (m, an)
and
for all choices of m, m ∈ M , n, n ∈ N , and a ∈ A.
If A is commutative, then we need not distinguish between left and right modules.
Let M1 , M2 , and N be A-modules. We say that f : M1 × M2 → N is A-bilinear if for
each m1 ∈ M1 and each m2 ∈ M2 , the maps
f (m1 , −) : M2 → N
and
f (−, m2 ) : M1 → N
are A-module homomorphisms.
If A is commutative, then an A-bilinear map is easily seen to be weakly A-bilinear.
Note also that for A-arbitrary, any weakly A-bilinear map is Z-bilinear.
We shall give a formal construction of the tensor product, but its purpose is that
it is the universal group for bilinear functions. Thus, we shall have an abelian group
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
366
M ⊗A N and a weakly A-bilinear map ι : M × N → M ⊗A N with the following universal
property: For each abelian group G and each weakly A-bilinear map f : M × N → G,
there is a unique group homomorphism f : M ⊗A N → G such that the following diagram
commutes.
f
/ G
M ×N
:
JJ
tt
JJ
t
t
J
t
tt
ι JJJ$
tt f
M ⊗A N
Surprisingly, it turns out that if A is commutative, then M ⊗A N is an A-module that is
universal in exactly the same manner for A-bilinear mappings. Thus, the two different
notions of bilinearity are “solved” by exactly the same construction.
Indeed, almost everything we know about tensor products will be derived from this
universal property, rather than the actual construction of the tensor product. Probably
the most important fact coming out of the construction itself is that M ⊗A N is generated
as an abelian group by the image of ι.
We now construct the tensor product. First, let F (M, N ) be the free abelian group
(i.e., free Z-module) on the set M × N . Thus, using the notation of Section 7.7,
,
F (M, N ) =
Z.
(m,n)∈M ×N
Continuing this notation, we write e(m,n) for the canonical basis element corresponding
to the (m, n)-th summand. We then have a function i : M × N → F (M, N ) with
i(m, n) = e(m,n) .
Now let H ⊂ F (M, N ) be the subgroup generated by
{e(m+m ,n) − e(m,n) − e(m ,n) | m, m ∈ M, n ∈ N }
∪{e(m,n+n ) − e(m,n) − e(m,n ) | m ∈ M, n, n ∈ N }
∪{e(ma,n) − e(m,an) | m ∈ M, n ∈ N, a ∈ A}
Definition 9.4.2. With the conventions above, we set the tensor product M ⊗A N
i
equal to F (M, N )/H, and set ι : M × N → M ⊗A N equal to the composite M × N −
→
π
→ M ⊗A N , where π is the canonical map onto F (M, N )/H.
F (M, N ) −
We write m ⊗ n for the element ι(m, n) ∈ M ⊗A N .
The map ι : M × N → M ⊗A N is easily seen to be weakly A-bilinear. That is
important in showing that the following proposition gives universality for weakly bilinear
maps.
Proposition 9.4.3. Suppose given a right A-module M , a left A-module N , and a weakly
A-bilinear map f : M ×N → G. Then there is a unique group homomorphism f : M ⊗A N
such that the following diagram commutes.
f
/ G
M ×N
:
JJ
tt
JJ
t
t
J
t
tt
ι JJJ$
tt f
M ⊗A N
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
367
Proof The function f : M × N → G is defined on the generators of the free abelian
group F (M, N ). By the universal property of free abelian groups (Lemma 7.7.32), there
is a unique group homomorphism f : F (M, N ) → G such that f(e(m,n) ) = f (m, n) for
all (m, n) ∈ M × N .
Since f is weakly A-bilinear, f is easily seen to vanish on the generators of H ⊂
F (M, N ). So the Noether Isomorphism Theorem, which gives the universal property of
factor groups, shows that there is a unique group homomorphism f : M ⊗A N → G with
f ◦ π = f. Here, π : F (M, N ) → M ⊗A N is the canonical map. The composite of the
uniqueness of f with respect to f and the uniqueness of f with respect to f gives the
desired result.
Warning: Not every element of M ⊗A N has the form m ⊗ n. The generic element has
the form m1 ⊗ n1 + · · · + mk ⊗ nk for k ≥ 0 (the empty sum being 0), with mi ∈ M and
ni ∈ N for 1 ≤ i ≤ k.
Remarks 9.4.4. In practice, in defining homomorphisms out of tensor products, one
often neglects to explicitly write down a weakly A-bilinear function from M × N to G,
but does so implicitly, by a statement such as “setting f (m ⊗ n) = expression defines
a homomorphism f : M ⊗A N → G.” Of course, f ◦ ι is the desired weakly A-bilinear
function, and is specified in a statement such as the above, since m ⊗ n = ι(m, n). It is
left to the reader to verify that the expression that is supplied in such a statement does
in fact define a weakly A-bilinear function.
We shall follow this convention of implicitly defined bilinear functions from now on,
other than in explicit statements about universal properties of the tensor product. We
reiterate, as pointed out in the above warning, that specifying f on terms of the form
m⊗n only gives the value of f on a generating set for M ⊗A N , and not on every element.
The following fundamental observation is immediate from the fact that 0 + 0 = 0 in
either M or N . We shall use it without explicit mention.
Lemma 9.4.5. Let M and N be right and left A-modules, respectively. Then for m ∈ M
and n ∈ N , the elements m ⊗ 0 and 0 ⊗ n are 0 in M ⊗A N .
The next lemma says that the tensor product is a functor of two variables.
Lemma 9.4.6. Let f : M → M and g : N → N be left and right A-module homomorphisms, respectively. Then there is a homomorphism, which we shall denote by
f ⊗ g : M ⊗A N → M ⊗A N whose effect on generators is given by
(f ⊗ g)(m ⊗ n) = f (m) ⊗ g(n).
If f : M → M and g : N → N are left and right A-module homomorphisms,
respectively, then
(f ⊗ g ) ◦ (f ⊗ g) = (f ◦ f ) ⊗ (g ◦ g).
Finally, 1M ⊗ 1N is the identity map of M ⊗ N .
Proof That setting (f ⊗ g)(m ⊗ n) = f (m) ⊗ g(n) gives rise to a well defined homomorphism follows precisely as in Remarks 9.4.4. The rest follows since any two homomorphisms that agree on the generators of a group must be equal.
The functoriality of tensor products is all that we need to establish their universality
for A-bilinear mappings when A is commutative.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
368
Proposition 9.4.7. Let A be a commutative ring and let M and N be A-modules. Then
there is a natural A-module structure on M ⊗A N which is defined on generators by
a · (m ⊗ n) = (am) ⊗ n.
Under this convention, ι : M × N is A-bilinear, and is universal with respect to that
property: For any A-bilinear map f : M × N → N , there is a unique A-module homomorphism f : M ⊗A N → N such that the following diagram commutes.
f
/ N
M ×N
:
JJ
tt
JJ
t
t
JJ
t
tt
ι JJ$
tt f
M ⊗A N
Proof For a ∈ A, write μa : M → M for the map induced by multiplication by a.
Because A is commutative, μa is an A-module homomorphism for each a ∈ A. Thus,
μa ⊗ 1N : M ⊗A N → M ⊗A N is a group homomorphism. We use this to define
multiplication by a on M ⊗A N . That this gives an A-module structure on M ⊗A N may
be deduced from the basic properties of tensor products and of A-module structures,
together with the fact that
(μa ⊗ 1N ) ◦ (μb ⊗ 1N ) = (μa ◦ μb ) ⊗ 1N = μab ⊗ 1N .
We leave the rest as an exercise for the reader.
There is a very useful generalization of Proposition 9.4.7 in which M has module
structures coming from two different (and possibly non-commutative) rings. It will give
us a way to turn A-modules into B-modules, induced by any ring homomorphism from
A to B.
Definition 9.4.8. Let A and B be rings. We say that M is B-A-bimodule if the following
hold.
1. M is a left B-module.
2. M is a right A-module.
3. These two module structures interact as follows. For m ∈ M , b ∈ B, and a ∈ A,
we have b(ma) = (bm)a.
Examples 9.4.9.
1. A itself is an A-A-bimodule via ordinary multiplication. Its A-A-submodules are
its two-sided ideals.
2. If f : A → B is a ring homomorphism, then any B-B-bimodule may be pulled back
to an A-B, a B-A, or an A-A-bimodule, via f .
3. If A is commutative, then any A-module M is an A-A-bimodule, where both actions
are just the usual action on M .
4. For any ring A, identify An with the space of 1 × n row matrices. Then An is a
right module over Mn (A) by setting x · M equal to the matrix product xM for
x ∈ An and M ∈ Mn (A). This gives An the structure of an A-Mn (A)-bimodule.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
369
If M is a B-A-bimodule and N is a left A-module, we shall put a left B-module
structure on M ⊗A N which will be universal for weakly A-bilinear maps with the following
additional property.
Definition 9.4.10. Let M be a left B-module and let N be any set. We say that a
function f : M × N → N is B-linear in M if N is a left B-module and for each n ∈ N
the function
f (−, n) : M → N is a B-module homomorphism.
Let M be a B-A-bimodule and let b ∈ B. Then multiplication by b, μb : M → M
is easily seen to be a homomorphism of right A-modules. Thus, the proof of the next
proposition proceeds along the lines of that of Proposition 9.4.7.
Proposition 9.4.11. Let M be a B-A-bimodule and let N be a left A-module. Then
there is a natural B-module structure on M ⊗A N which is defined on generators by
b · (m ⊗ n) = (bm) ⊗ n.
Under this convention, ι : M × N is B-bilinear in M and satisfies the following universal
property: if N is a left B-module and if f : M × N → N is weakly A-bilinear as well as
being B-linear in M , then there is a unique B-module homomorphism f : M ⊗A N → N such that the following diagram commutes.
f
/ N
M ×N
:
JJ
tt
JJ
t
t
JJ
t
tt
ι JJ$
tt f
M ⊗A N
Similarly, if M is a right A-module and N is an A-B-bimodule, then the tensor
product M ⊗A N inherits a right B-module structure and satisfies a similar universal
property.
An important application of the role of bimodule structures in tensor products comes
from the B-A-bimodule structure on B that arises from a ring homomorphism f : A → B.
Given a left A-module M , we obtain a left B-module B ⊗A M . The passage from M
to B ⊗A M is functorial, and is known as “extension of rings” or “base change,” among
other names. It satisfies an important universal property:
Proposition 9.4.12. Let f : A → B be a ring homomorphism. Let M be a left Amodule and let N be a left B-module. Regarding N as an A-module via f , let g : M → N
be an A-module homomorphism. Then there is a unique B-module homomorphism g :
B ⊗A M → N such that the following diagram commutes.
g
/ N
M J
t:
JJ
t
J
tt
αM JJ$
tt g
B ⊗A M
Here, αM : M → B ⊗A M is the A-module homomorphism defined by αM (m) = 1 ⊗ m
for m ∈ M . Explicitly, g is defined on generators by setting
g(b ⊗ m) = bg(m).
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
370
Proof Setting g(b⊗m) = bg(m) is easily seen to induce a left B-module homomorphism
from B ⊗A M to N . Uniqueness follows from the fact that 1 ⊗ m = αM (m), and hence
the commutativity of the diagram requires that g(1 ⊗ m) = g(m).
Recall that if M and N are A-modules, then HomA (M, N ) is the group (under addition of homomorphisms) of A-module homomorphisms. The following may be expressed
in category theoretic language by saying that passage from M to B ⊗A M is a left adjoint
to the forgetful functor from B-modules to A-modules induced by f : A → B.
Corollary 9.4.13. Let f : A → B be a ring homomorphism. Let M be a left A-module
and N a left B-module. Let αM : M → B ⊗A M be the A-module homomorphism defined
by αM (m) = 1 ⊗ m for all m ∈ M . Then there is an isomorphism of abelian groups
α∗
→ HomA (M, N )
HomB (B ⊗A M , N ) −−M
∼
=
∗
defined by αM
(h) = h ◦ αM .
Additionally, the isomorphism is natural in the sense that if g : M → M is an
A-module homomorphism and h : N → N is a B-module homomorphism, then the
following diagram commutes.
HomB (B ⊗A M , N )
∗
αM
/ HomA (M, N )
γ
∗
αM / HomA (M , N )
HomB (B ⊗A M , N )
β
Here, β and γ are defined by
β(k) = h ◦ k ◦ (1B ⊗ g)
and
γ(l) = h ◦ l ◦ g
for k ∈ HomB (B ⊗A M , N ) and l ∈ HomA (M, N ).
∗
is well defined, and is clearly
Proof Since αM is an A-module homomorphism, αM
a homomorphism between the Hom groups. Let h ∈ HomB (B ⊗A M , N ). Then the
following diagram commutes.
∗
αM
(h)
/ N
M J
t:
JJ
t
J
tt
αM JJ$
tt h
B ⊗A M
∗
So αM
is an isomorphism by Proposition 9.4.12. The reader may verify the naturality
statement.
The simplest type of base change is the trivial one: B = A and f is the identity map.
Here, Corollary 9.4.13 gives us an isomorphism
α∗
HomA (A ⊗A M , N ) −−M
→ HomA (M, N )
∼
=
∼
=
for any pair of A-modules M and N . For formal reasons, this implies that αM : M −→
A ⊗A M for all A-modules M . We can also prove it directly, with a direct construction
of the inverse homomorphism:
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
371
Lemma 9.4.14. Let M be a left A-module and let αM : M → A ⊗A M be the A-module
homomorphism given by αM (m) = 1 ⊗ m for all m ∈ M . Then αM is an isomorphism
whose inverse, εM : A ⊗A M → M , is defined on generators by εM (a ⊗ m) = am.
Additionally, the isomorphisms εM and their inverses are natural in the sense that if
f : M → M is an A-module homomorphism, then the following diagram commutes.
εM /
A ⊗A M
M
1A ⊗ f f εM / A ⊗A M M
Proof It is easy to see that εM is well defined and that εM ◦ αM = 1M . So εM and
αM are inverse isomorphisms, providing that αM is onto. But each generator a ⊗ m is
equal to αM (am), so the image of αM contains a generating set for A ⊗A M . The rest is
left to the reader.
We’ve already seen a couple of examples of extension of rings without knowing it.
Corollary 9.4.15. Let S be a multiplicative subset of the commutative ring A. Then
for any A-module M there is an isomorphism of S −1 A-modules
ϕM : S −1 A ⊗A M → S −1 M
whose effect on generators is given by ϕM ((a/s) ⊗ m) = am/s for a ∈ A, s ∈ S, and
m ∈ M.
This isomorphism is natural in the sense that if f : M → N is an A-module homomorphism, then the following diagram commutes,
S −1 A ⊗A M
1⊗f
S −1 A ⊗A N
ϕM /
S −1 M
S −1 (f )
ϕN /
−1
S N
where 1 is the identity of S −1 A.
Proof The point is that Lemma 7.11.14 shows that the canonical map η : M → S −1 M
satisfies exactly the same universal property in terms of A-module maps into S −1 Amodules that Proposition 9.4.12 shows for the mapping αM : M → S −1 A ⊗A M . In
particular, the universal property for η shows that there exists a unique S −1 A-module
homomorphism ψ : S −1 M → S −1 A ⊗A M such that the following diagram commutes.
αM
/ S −1 A ⊗ M
M OO
A
OOO
7
OOO
ooo
o
o
O'
η
oo ψ
S −1 M
And ϕM : S −1 A ⊗A M → S −1 M , if placed in the diagram in the opposite direction from
that of ψ, would make it commute as well.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
372
Thus, the composite S −1 A-module homomorphisms ϕM ◦ ψ and ψ ◦ ϕM restrict to
the identity maps on the images of αM and η, respectively. Since these images generate
S −1 A ⊗A M and S −1 M , respectively, as modules over S −1 A, ϕM and ψ must be inverse
isomorphisms.
The naturality statement holds by an easy diagram chase.
Similarly, Lemma 7.7.19 shows that if a is a two-sided ideal in the ring A, then the
passage from a left A-module M to the A/a-module M/aM satisfies the same universal
property as the extension of rings. We leave the proof of the following to the reader.
Corollary 9.4.16. Let a be a two-sided ideal in the ring A. Then for any left A-module
M , there is an isomorphism of A/a-modules
βM : A/a ⊗A M → M/aM
whose effect on generators is given by βM (a ⊗ m) = am for a ∈ A and m ∈ M .
This isomorphism is natural in the sense that if f : M → N is an A-module homomorphism, then the following diagram commutes,
A/a ⊗A M
1⊗f
A/a ⊗A N
βM /
M/aM
f
βN /
N/aN
where 1 is the identity map of A/a, and f is the map induced by f .
We may now reinterpret Nakayama’s Lemma in terms of extension of rings.
Corollary 9.4.17. (Nakayama’s Lemma, second form) Let a be a two-sided ideal which
is contained in the Jacobson radical of the ring A and let M be a finitely generated
A-module. Then A/a ⊗A M = 0 if and only if M = 0.
Proof M/aM = 0 if and only if M = aM . Now apply Nakayama’s Lemma.
We’d like to know that extension of rings takes free modules to free modules. For
this purpose, it suffices to understand what tensoring does to a direct sum.
*k
Proposition 9.4.18. Let N1 , . . . , Nk be left A-modules and let ιi : Ni →
i=1 Ni be
the canonical inclusion for i = 1, . . . , k. Then for any right A-module M , there is an
isomorphism
" k
#
k
,
,
ω:
(M ⊗A Ni ) −→ M ⊗A
Ni
i=1
i=1
whose restriction to M ⊗A Ni is 1M ⊗ ιi for i = 1, . . . , k. If M is a B-A-bimodule, then
ω is an isomorphism of left B-modules; if each Ni is an A-B-bimodule, then ω is an
isomorphism of right B-modules.
Similarly, given
A-modules M*
1 , . . . , Mk and a left A-module N , there is an
*right
k
k
isomorphism ω : i=1 (Mi ⊗A N ) → ( i=1 Mi ) ⊗A N whose restriction to Mi ⊗A N
is ιi ⊗ 1N for i = 1, . . . , k. Once again, appropriate bimodule structures give module
isomorphisms.
These isomorphisms are natural in all variables.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
373
*k
*k
Proof Let β : M ⊗A ( i=1 Ni ) → i=1 (M ⊗A Ni ) be the homomorphism whose effect
on generators is given by
β(m ⊗ (n1 , . . . , nk )) = (m ⊗ n1 , . . . , m ⊗ nk ).
Then the composites ω *
◦ β and β ◦ ω are
*k easily seen to give the identity maps on genk
erating sets for M ⊗A ( i=1 Ni ) and i=1 (M ⊗A Ni ), and hence ω and β are inverse
isomorphisms.
We leave the rest to the reader.
Corollary 9.4.19. Let f : A → B be a ring homomorphism. Then B ⊗A An is a free
B-module of rank n with basis 1 ⊗ e1 , . . . , 1 ⊗ en , where e1 , . . . , en is the canonical basis
of An .
Similarly, if we regard An as a right A-module, then An ⊗A B is a free right B-module
with basis e1 ⊗ 1, . . . , en ⊗ 1.
Proof We have B ⊗A An ∼
=B
= (B ⊗A A)n . Now notice that the isomorphism B ⊗A A ∼
from Lemma 9.4.14 is one of left B-modules, and takes 1 ⊗ 1 to 1.
The proof for right modules is identical.
Recall that Mm,n (A) is the set of m×n matrices over A. Given a ring homomorphism
f : A → B, we write f∗ : Mm,n (A) → Mm,n (B) for the function that takes a matrix (aij )
to the matrix whose ij-th coordinate is f (aij ).
Recall from Section 7.10 that if g : An → Am is a homomorphism of right A-modules
(or left ones, of course, if A is commutative), then g is induced by left-multiplying a
column vector by the m × n matrix whose i-th column is g(ei ).
Similarly, if g : An → Am is a left A-module map, then g is induced by rightmultiplying a row vector by the n × m matrix whose i-th row is g(ei ). In particular,
there are two different conventions possible for the matrix of g if A is commutative.
It is now an easy verification to show the following.
Corollary 9.4.20. Let f : A → B be a ring homomorphism. Let g : An → Am be a
homomorphism of right A-modules, represented by the matrix M . Then the matrix of
g ⊗ 1B with respect to the bases coming from Corollary 9.4.19 is precisely f∗ (M ).
Similarly, if the left A-module homomorphism g : An → Am is represented (under
whichever convention one chooses to use in the case where both A and B are commutative)
by the matrix M , then the matrix of 1B ⊗ g with respect to the bases of Corollary 9.4.19
is f∗ (M ).
f
g
Suppose given ring homomorphisms A −
→ B −
→ C. We’d like to know that the
extension of rings from A to B and then from B to C coincides with the extension in
one step from A to C. The verification of this depends on an understanding of the
associativity properties of iterated tensor product.
In the study of cartesian products, the existence of a single n-fold product greatly
simplifies the discussion of iterated products and their associativity properties. We shall
make use of the same device in the study of iterated tensor products.
Definitions 9.4.21. Let A1 , . . . , An−1 be rings. Suppose given a right A1 -module M1 ,
a left An−1 -module Mn , together with Ai−1 -Ai -bimodules, Mi for 2 ≤ i ≤ n − 1. Let G
be an abelian group. Then a function
f : M1 × · · · × Mn → G
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
374
is weakly A1 , . . . , An−1 -multilinear if the maps
f (m1 , . . . , mi−1 , −, −, mi+2 , . . . , mn ) : Mi × Mi+1 → G
are weakly Ai -bilinear for each i = 1, . . . , n − 1 and for all possible choices of elements
mj ∈ Mj for j = i, i + 1.
If A is commutative and if M1 , . . . , Mn are A-modules, then a function
f : M1 × · · · × M n → N
is said to be A-multilinear if N is an A-module and the maps
f (m1 , . . . , mi−1 , −, mi+1 , . . . , mn ) : Mi → N
are A-module homomorphisms for each i = 1, . . . , n and all possible choices of elements
mj ∈ Mj for j = i.
A-multilinear maps are easily seen to be weakly A, . . . , A-multilinear. Note that the
word “weakly” may be safely omitted in the A1 , . . . , An -multilinear case if the rings
A1 , . . . , An are not all equal, as there’s no stronger concept with which to confuse it.
Note that if A is commutative, then a weakly A, . . . , A-multilinear map is A-multilinear
if and only if it is A-linear on M1 in the sense of Definition 9.4.10.
The construction of the n-fold tensor product has no surprises. Suppose given a right
A1 -module M1 , a left An−1 -module Mn , and Ai−1 -Ai -bimodules Mi for i = 2, . . . , n −
1. Let F (M1 , . . . , Mn ) be the free abelian group on M1 × · · · × Mn , where e(m1 ,...,mn )
is the basis element corresponding to (m1 , . . . , mn ) ∈ M1 × · · · × Mn . Let H be the
subgroup of F (M1 , . . . , Mn ) generated by all elements of the form e(m1 ,...,mi +mi ,...,mn ) −
e(m1 ,...,mi ,...,mn ) −e(m1 ,...,mi ,...,mn ) , for i = 1, . . . , n and all possible choices of mj ∈ Mj for
j = 1, . . . , n and mi ∈ Mi , together with all elements of the form e(m1 ,...,mi a,mi+1 ,...,mn ) −
e(m1 ,...,mi ,ami+1 ,...,mn ) , for i = 1, . . . , n−1, and all possible choices of a ∈ Ai , and mj ∈ Mj
for j = 1, . . . , n.
Definition 9.4.22. Under the conventions above, set
M1 ⊗A1 · · · ⊗An−1 Mn = F (M1 , . . . , Mn )/H
and write ι : M1 × · · · × Mn → M1 ⊗A1 · · · ⊗An−1 Mn for the map that takes (m1 , . . . , mn )
to the image in M1 ⊗A1 · · · ⊗An−1 Mn of the basis element e(m1 ,...,mn ) .
The proof of the next proposition is analogous to the proofs of Propositions 9.4.3,
9.4.7, and 9.4.11.
Proposition 9.4.23. Suppose given a right A1 -module M1 , a left An−1 -module Mn ,
together with Ai−1 -Ai -bimodules, Mi for i = 2, . . . , n − 1. Let f : M1 × · · · × Mn → G be
a weakly A1 , . . . , An−1 -multilinear function. Then there is a unique group homomorphism
f : M1 ⊗A1 · · · ⊗An−1 Mn → G such that the following diagram commutes.
f
M1 × · · · × M n
j/4 G
TTTT
jjjj
j
j
TTTT
j
jjj
ι TTT*
jjjj
f
M1 ⊗A1 · · · ⊗An−1 Mn
Suppose in addition M1 is an A0 -A1 -bimodule. Then M1 ⊗A1 · · · ⊗An−1 Mn is a left
A0 -module by an action in which
a · m1 ⊗ · · · ⊗ mn = (am1 ) ⊗ · · · ⊗ mn .
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
375
Moreover, if G is a left A0 -module and f is A0 -linear in M1 , then the induced map
f : M1 ⊗A1 · · · ⊗An−1 Mn → G is a homomorphism of left A0 -modules.
The analogous statement holds for the case where Mn is an An−1 -An -bimodule and
f is An -linear in Mn .
In particular, if A is commutative and M1 , . . . , Mn are A-modules, then there is an
A-module structure on M1 ⊗A · · · ⊗A Mn induced by
a · m1 ⊗ · · · ⊗ mn = (am1 ) ⊗ · · · ⊗ mn .
If G is an A-module and f : M1 × · · · × Mn is an A-multilinear map, then the induced
map f : M1 ⊗A · · · ⊗A Mn → G is an A-module homomorphism.
As in the bilinear case, we shall generally define our multilinear maps implicitly,
rather than explicitly.
Corollary 9.4.24. Suppose given a right A-module M1 , a left B-module M3 , and an
A-B-bimodule M2 . Then there are isomorphisms
γ1
γ2
∼
=
∼
=
(M1 ⊗A M2 ) ⊗B M3 ←−− M1 ⊗A M2 ⊗B M3 −→ M1 ⊗A (M2 ⊗B M3 )
whose effects on generators are given by
γ1 (m1 ⊗ m2 ⊗ m3 )
γ2 (m1 ⊗ m2 ⊗ m3 )
= (m1 ⊗ m2 ) ⊗ m3
= m1 ⊗ (m2 ⊗ m3 ).
and
Proof The formulæ above induce well defined homomorphisms γ1 and γ2 as stated, so
it suffices to construct their inverse homomorphisms. We shall confine ourselves to the
case of γ1 .
Here, we note that for m3 ∈ M3 , there is a homomorphism from M1 ⊗A M2 to
M1 ⊗A M2 ⊗B M3 that carries each generator m1 ⊗ m2 to m1 ⊗ m2 ⊗ m3 . This is enough
to construct a homomorphism δ : (M1 ⊗A M2 ) ⊗B M3 → M1 ⊗A M2 ⊗B M3 whose effect
k
k
on generators is given by δ(( i=1 mi ⊗ mi ) ⊗ m ) = i=1 (mi ⊗ mi ⊗ m ). And δ is
easily seen to give an inverse to γ1 .
Note that if M1 is a C-A-bimodule, then γ1 and γ2 are isomorphisms of left C-modules.
Similarly, we get right module isomorphisms if M3 is a B-C-bimodule.
f
g
→B−
→ C. Then there is a
Corollary 9.4.25. Suppose given ring homomorphisms A −
natural isomorphism of left C-modules
C ⊗B (B ⊗A M ) ∼
= C ⊗A M
for A-modules M .
Proof The isomorphisms γi are clearly natural in all three variables. It suffices to note
that
(C ⊗B B) ⊗A M ∼
= C ⊗A M
by Lemma 9.4.14.
The theory of localization, together with Nakayama’s Lemma, now gives a valuable
application of extension of rings.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
376
Corollary 9.4.26. Let M be a finitely generated module over the commutative ring A.
Then M = 0 if and only if A/m ⊗A M = 0 for every maximal ideal m of A.
Proof Let m be a maximal ideal of A. Recall from Proposition 7.11.24 that A/m is
isomorphic to the quotient ring Am /mm of the localization Am of A. Thus,
A/m ⊗A M ∼
= Am /mm ⊗A M ∼
= Am /mm ⊗A (Am ⊗A M ) = Am /mm ⊗A Mm .
m
m
Since M is finitely generated over A, there is a surjection A → M for some k ≥ 0. Since
localization is an exact functor, we see that Mm is finitely generated over Am . Thus, if
A/m ⊗A M = 0, then Mm = 0 by the second form of Nakayama’s Lemma.
Proposition 7.11.25 shows that M = 0 if and only if Mm = 0 for every maximal ideal
of M , so the result follows.
k
Exercises 9.4.27.
1. Let a and b be two-sided ideals of A. Show that
A/a ⊗A A/b ∼
= A/(a + b).
2. Let f : A → B be a ring homomorphism and let M and N be a right and a left Bmodule, respectively. Then we may consider them to be A-modules as well. Show
that there is a natural surjection from M ⊗A N onto M ⊗B N . In particular, every
tensor product is a quotient group of the appropriate tensor product over Z.
† 3. Let a be a two-sided ideal of A, and let M and N be a right and a left A/a module,
respectively. Show that the natural map M ⊗A N → M ⊗A/a N of the preceding
problem is an isomorphism.
† 4. Show that if A and B are commutative rings and f : A → B is a ring homomorphism, then there is a natural B-module isomorphism from (B ⊗A M ) ⊗B (B ⊗A N )
to B ⊗A (M ⊗A N ) for A-modules M, N .
5. Let A be a commutative ring and let m, n > 0. Show that Am ⊗A An is a free
A-module of rank mn, with basis ei ⊗ ej for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Here,
e1 , . . . , em and e1 , . . . , en are the canonical bases of Am and An , respectively.
† 6. Give the generalization of Proposition 9.4.18 to infinite direct sums, and deduce
that extension of rings carries infinitely generated free modules to free modules.
7. Let f : A → B be a ring homomorphism. Show that passage from a left A-module
M to the extended B-module B⊗A M provides a left adjoint to the forgetful functor
from left B-modules to left A-modules (i.e., the functor that considers a B-module
to be an A-module via f ).
8. Let A be a ring. Show that if M is a left A-module, then the action of A on M
induces a homomorphism α : A ⊗Z M → M . Conversely, if M is an abelian group,
then any homomorphism α : A⊗Z M → M defines the function (α◦ι) : A×M → M ,
which we may then investigate to see if it gives a module action.
Show that α : A ⊗Z M → M induces an A-module structure on M if and only if
the following diagrams commute:
ν ⊗ 1M/
1A ⊗ α/
A ⊗Z M
Z ⊗Z M
A ⊗Z M
A ⊗Z A ⊗Z M
JJ
JJ
JJ
J
μ ⊗ 1M
α
α
εM JJJJ
J
$ α
/ M
A ⊗Z M
M
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
377
Here, μ is induced by the multiplication of A, ν is the unique ring homomorphism
from Z to A, and εM is the isomorphism induced by the action of Z on M as in
Lemma 9.4.14.
9.5
Tensor Products and Exactness
Here, we shall explore the relationships between tensor products and exact sequences,
with particular attention paid to extension of rings. As an application, we shall study
the A-module homomorphisms f : An → Am between finitely generated free modules,
and extend, as much as we can, the sort of results that are known to hold for finitely
generated vector spaces over a division ring.
We’ve seen one example, tensoring with S −1 A, where S is a multiplicative subset of
the commutative ring A, where extension of rings is an exact functor. However, extension
of rings is often not an exact functor. For instance, as we shall see in Exercises 9.5.21,
extension of rings from Z to Z2 fails to be exact. Thus, the next definition has significance.
Definition 9.5.1. A right A-module M is flat if the passage from left A-modules N to
M ⊗A N and from A-module homomorphisms f to 1M ⊗ f gives an exact functor.
Similarly, a left A-module N is flat if − ⊗A N gives an exact functor.
As we noted above, Corollary 9.4.15, together with the proof that passage to modules
of fractions is exact, gives the following corollary.
Corollary 9.5.2. Let S be a multiplicative subset of the commutative ring A. Then
S −1 A is a flat A-module.
Similarly, the next corollary follows from Lemma 9.4.14.
Corollary 9.5.3. A itself is flat as either a left or a right A-module.
While extension of rings is not in general an exact functor, it does have some exactness
properties.
Definitions 9.5.4. Let F be a covariant functor between two categories of modules.
1. We say that F is right exact if for each short exact sequence
g
f
M −
→M −
→ M → 0
in the domain category of F , the resulting sequence
F (f )
F (g)
F (M ) −−−→ F (M ) −−−→ F (M ) → 0
is exact.
2. We say that F is left exact if for each short exact sequence
f
g
0 → M −
→M −
→ M in the domain category of F , the resulting sequence
F (f )
F (g)
0 → F (M ) −−−→ F (M ) −−−→ F (M )
is exact.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
378
One can also study partial exactness properties of contravariant functors. We shall
do so (without giving formal definitions of what constitutes “left” and “right”) in our
study of the Hom functors in Section 9.7.
Proposition 9.5.5. Let M be a right A-module. Then the functor which takes N to
M ⊗A N and takes f to 1M ⊗ f is a right exact functor from the category of left Amodules to abelian groups.
Similarly, if N is a left A-module, then − ⊗A N is a right exact functor on right
A-modules.
Proof We treat the case of M ⊗A −. The other is similar. Suppose given a short exact
sequence
f
g
N −
→N −
→ N → 0.
We wish to show that
1M ⊗f
1M ⊗g
M ⊗A N −−−−→ M ⊗A N −−−−→ M ⊗A N → 0
is exact.
Consider first the exactness at M ⊗A N . This amounts to showing that 1M ⊗ g is
onto. But if n = g(n), then m ⊗ n = (1M ⊗ g)(m ⊗ n) for all m ∈ M , and hence the
image of 1M ⊗ g contains a set of generators for M ⊗A N , so 1M ⊗ g is onto.
Clearly, (1M ⊗ g) ◦ (1M ⊗ f ) = 0. Thus, if B ⊂ M ⊗A N is the image of 1M ⊗ f , then
B is contained in the kernel of 1M ⊗ g. We obtain an induced map
h : (M ⊗A N )/B → M ⊗A N .
It suffices to show that h is an isomorphism.
To show this, we construct a map the other way. As usual, we use the universal
mapping property of the tensor product. This time, we shall explicitly define a weakly
A-bilinear function k : M ×N → (M ⊗A N )/B, as follows. For a pair (m, n ) ∈ M ×N ,
choose n ∈ N with g(n) = n , and set k(m, n ) = m ⊗ n, the element of (M ⊗A N )/B
represented by m ⊗ n. To see that k(m, n ) does not depend on which n ∈ g −1 (n )
we choose, suppose that g(n) = g(n1 ) = n . By the exactness of the original sequence,
n − n1 = f (n ) for some n ∈ N . But then
(m ⊗ n) − (m ⊗ n1 ) = m ⊗ (n − n1 ) = (1M ⊗ f )(m ⊗ n )
lies in B. Thus, k is a well defined function, and is easily seen to be weakly A-bilinear.
Thus, there is a homomorphism k : M ⊗A N → (M ⊗A N )/B extending k. Now if
g(n) = n and m ∈ M , then
(h ◦ k)(m ⊗ n ) = h(m ⊗ n) = (1M ⊗ g)(m ⊗ n) = m ⊗ n .
Since h ◦ k is the identity on a set of generators for M ⊗A N , it must be the identity
map everywhere.
For m ⊗ n ∈ (M ⊗A N )/B, we have (k ◦ h)(m ⊗ n) = k(m ⊗ g(n)) = m ⊗ n, since any
lift to N of g(n) may be used in defining k. So k ◦ h is the identity on a set of generators,
and hence k and h are inverse to one another.
The right exactness of extension of rings has some immediate applications, but let us
first develop some tools for recognizing flat modules.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
379
Proposition 9.5.6. Let F be a covariant functor from A-modules to abelian groups.
Then F is exact if and only if it takes short exact sequences to short exact sequences.
Proof To see that an exact functor, F , preserves short exact sequences, it suffices to
0
0
show that F (0) = 0. Note that 0 −
→0−
→ 0 is an exact sequence in which the morphisms
are both isomorphisms. Since any functor preserves isomorphisms, the same is true of
F (0)
F (0)
F (0) −−−→ F (0) −−−→ F (0). Thus, F (0) = 0, as claimed.
f
g
For the converse, suppose given an exact sequence M −
→M −
→ M of A-modules.
Then we have a commutative diagram
0 JJ
t: 0
JJ
tt
JJ
t
t
J$
tt
M/ ker g
JJ g
:
JJ
πtttt
JJ
t
t
$
t
f
g
/
/ M M
M J
t:
JJ
JJ
ttt
t
J
t
t f
π $
M / ker f
JJ
:
JJ
tt
JJ
tt
t
J$
tt
0
0
in which all the straight lines are exact.
Suppose then that F takes short exact sequences to short exact sequences. Then it
is easy to see that it must take injections to injections and take surjections to surjections. Thus, if we apply F to the above diagram, the diagonal straight lines will remain
exact. Thus, im F (f ) = im F (f ) and ker F (g) = ker F (π), and hence the exactness
F (f )
F (g)
of F (M ) −−−→ F (M ) −−−→ F (M ) follows from the exactness of five-term diagonal
sequence.
Corollary 9.5.7. Let F be a right exact functor from A-modules to abelian groups. Then
F is exact if and only if for any injective A-module homomorphism f : M → M , the
induced map F (f ) is also injective.
The following consequence of Corollary 9.5.7 has an easy direct proof, as well.
Proposition 9.5.8. Let N1 and N2 be left A-modules. Then N1 ⊕ N2 is a flat module
if and only if both N1 and N2 are flat. The analogous statement holds for right modules
as well.
Proof Suppose given an injective right A-module homomorphism f : M → M . The
naturality of the isomorphism of Proposition 9.4.18 shows that we may identify f ⊗A
1(N1 ⊕N2 ) with (f ⊗A 1N1 ) ⊕ (f ⊗A 1N2 ). But clearly, a direct sum of homomorphisms is
injective if and only if each one of them is injective, and hence the result follows.
The next corollary now follows from Lemma 9.4.14 and induction on n.
Corollary 9.5.9. For n ≥ 1, the free module An is flat as either a right or a left Amodule.
The next lemma can be useful in recognizing flatness.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
380
Lemma
9.5.10. Let M and N be left and right A-modules, respectively, and suppose that
k
m
⊗n
i
i = 0 in M ⊗A N . Then there are finitely generated submodules M0 ⊂ M and
i=1
k
N0 ⊂ N such that mi ∈ M0 and ni ∈ N0 for all i, and i=1 mi ⊗ ni = 0 in M0 ⊗A N0 .
Proof As in Section 9.4, we write F (M, N ) for the free abelian group on M × N and
write e(m,n) for the basis element corresponding to the ordered pair (m, n) ∈ M × N .
k
Then our construction of tensor products shows that i=1 e(mi ,ni ) must be equal to a
sum of integral multiples of terms of the form e(xi +xi ,yi ) − e(xi ,yi ) − e(xi ,yi ) , e(xi ,yi +yi ) −
e(xi ,yi ) −e(xi ,yi ) , and e(xi ai ,yi ) −e(xi ,ai yi ) . So take M0 to be the submodule of M generated
by m1 , . . . , mk together with the x’s and x ’s and take N0 to be the submodule of N
generated by n1 , . . . , nk together with the y’s and y ’s.
Corollary 9.5.11. Let N be a left A-module with the property that every finitely generated submodule of N is flat. Then N is flat.
Proof Let f : M → M be an injective homomorphism of right A-modules and let
k
9.5.10 provides a
i=1 mi ⊗ ni be an element of the kernel of f ⊗ 1N . Then Lemma
k
finitely generated submodule N0 containing n1 , . . . , nk such that i=1 f (mi ) ⊗ ni = 0 in
M ⊗A N0 .
k
By our hypothesis, N0 is flat, and hence i=1 mi ⊗ ni = 0 in M ⊗A N0 , and hence
also in M ⊗A N .
We can now characterize the flat modules over a P.I.D.
Corollary 9.5.12. Let A be a P.I.D. Then an A-module is flat if and only if it is torsionfree.
Proof Let M be a torsion-free A-module. Then every finitely generated submodule
of M is also torsion-free, and hence free by Proposition 8.9.6. By Corollary 9.5.9, the
finitely generated submodules of M are flat, so M is flat by Corollary 9.5.11.
1⊗η
Conversely, suppose that M is a flat A-module. Then M ⊗A A −−→ M ⊗A K is
injective, where η : A → K is the canonical inclusion of A in its field of fractions. By
Corollary 9.4.15, we may identify 1 ⊗ η with the canonical map η : M → M(0) from M
to its localization at (0). But the kernel of η is easily seen to be the torsion submodule
of M , and hence M is torsion-free.
There are times when tensoring with even a non-flat module will preserve short exactness. Recall that a short exact sequence
g
f
0 → N −
→N −
→ N → 0
splits if f admits a retraction, meaning an A-module homomorphism r : N → N such
that r ◦ f = 1N .
Corollary 9.5.13. Suppose given a split short exact sequence of left A-modules:
f
g
→N −
→ N → 0.
0 → N −
Then for any right A-module M , the sequence
1⊗f
1⊗g
0 → M ⊗A N −−→ M ⊗A N −−→ M ⊗A N → 0
is exact and splits.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
381
Proof Note that any map that admits a retraction is injective. Thus, since tensoring
with M is right exact, it suffices to show that 1 ⊗ f admits a retraction. But if r is a
retraction for f , then 1 ⊗ r is a retraction for 1 ⊗ f .
We now consider the applications of this material to homomorphisms of free modules.
We’ve already seen, in Theorem 7.9.8, that the rank of a free module is well defined over
a commutative ring. The argument there implicitly used extension of rings from A to
the field A/m, where m is a maximal ideal of A. Now, we’re given a general extension of
rings functor, so we can give a more general result:
Proposition 9.5.14. Let A be any ring that admits a ring homomorphism f : A → D,
with D a division ring. Then the free left A-modules Am and An are isomorphic if and
only if m = n.
Proof Let g : Am → An be an A-module isomorphism and let g −1 be the inverse
isomorphism. Since extension of rings is a functor, 1D ⊗g is a D-module isomorphism from
Dm to Dn , with inverse 1D ⊗ g −1 . Since dimension is well defined for finite dimensional
vector spaces over a division ring (Corollary 7.9.5), m = n.
This does indeed cover the case of commutative rings by precisely the same argument
as that given earlier, as the commutative ring A maps to the field A/m if m is a maximal
ideal of A.
Here is another result about vector spaces that extends to a more general context.
Proposition 9.5.15. Let A be a commutative ring, or more generally, any ring that
admits a ring homomorphism f : A → D, with D a division ring. Let g : Am → An be a
surjective homomorphism. Then m ≥ n.
Proof By the right exactness of the tensor product, 1D ⊗ g : Dm → Dn is surjective.
Now apply Proposition 7.9.2.
Of course, if D is a division ring and if f : Dm → Dn is surjective, then this determines
the isomorphism type of ker f . In particular, any surjection from Dn to itself must be
an isomorphism. We shall extend this latter result to commutative rings:
Proposition 9.5.16. Let A be a commutative ring. Then any surjective homomorphism
f : An → An is an isomorphism.
Proof Let K = ker f . Then the exact sequence
i
f
0→K−
→ An −
→ An → 0
splits by Proposition 7.7.51. Thus, for each maximal ideal m of A, the sequence
1⊗i
1⊗f
0 → A/m ⊗A K −−→ A/m ⊗A An −−→ A/m ⊗A An → 0
is exact by Corollary 9.5.13. In particular, the dimension of the kernel of 1 ⊗ f must be
0, and hence A/m ⊗A K = 0 for each maximal ideal m of A.
Note that if r : An → K is a retraction for i, then r is surjective, and hence K is
finitely generated. Thus, K = 0 by Corollary 9.4.26.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
382
We shall now show that if A is a commutative ring, and if f : Am → An is injective,
then m ≤ n.
If A is an integral domain, then this is almost immediate, because tensoring with
its field of fractions gives an exact functor to a category of vector spaces. If A is not
a domain, we shall need some sort of substitute for the field of fractions. The thing to
notice is that in an integral domain, 0 is minimal in the partially ordered set of all prime
ideals under inclusion.
So the technique we shall use in general is to localize at a minimal prime ideal.
Before we can do that we need to know that minimal primes exist. If A has finite Krull
dimension, this is obvious. For the general case, we need further argument.
Lemma 9.5.17. A commutative ring A has minimal prime ideals.
Proof A minimal prime is a maximal element in the partially ordered set of prime ideals
under reverse inclusion. We apply Zorn’s Lemma. Given a totally ordered collection of
prime ideals, an “upper bound” in this ordering would be given by their intersection,
provided it is prime.
Normally, intersections of prime ideals are not prime.(Look at Z, for instance.) But
if we’re intersecting a totally ordered collection, say p = i∈I pi , suppose that ab is in p.
Then if a is not in p, it must not be in pi for some i. But then b ∈ pj whenever pj ⊂ pi .
But since the collection {pj | j ∈ I} is totally ordered, for j ∈ I we either have pj ⊂ pi
or pi ⊂ pj , so in either case b ∈ pj . So b must be in p, the intersection of the pj . So p is
prime, and the hypotheses of Zorn’s Lemma are satisfied.
If f : Am → An is injective, so is its localization at a minimal prime ideal. Thus,
we may assume that A has exactly one prime ideal, p, which therefore is its nil radical
(Proposition 9.1.10). In other words, every element of p is nilpotent. The following will
be useful.
Lemma 9.5.18. Let A be a commutative ring with only one prime ideal, p. Let p1 , . . . , pk
be nonzero elements of p. Then there is a nonzero element p ∈ p which annihilates each
of the pi .
Proof We claim that p may be taken to have the form pr11 . . . prkk , with ri ≥ 0 for
, where s is the
1 ≤ i ≤ k. We show this by induction on k. If k = 1, just take p = ps−1
1
smallest positive integer for which ps1 = 0. For k > 1, assume by induction that we’re
rk−1
given an element p = pr11 . . . pk−1
, such that p = 0, and p annihilates p1 , . . . , pk−1 .
Let t be the smallest integer such that ptk = 0. If p pt−1
= 0, then that’s our value
k
of p. Otherwise, let rk be the largest non-negative integer such that p prkk = 0, and set
p = p prkk .
Proposition 9.5.19. Let A be a commutative ring with only one prime ideal. Let f :
Am → M be an injective A-module homomorphism for some A-module M . Then 1 ⊗ f :
A/p ⊗A Am → A/p ⊗ M is also injective.
Proof We may identify 1 ⊗ f with the unique A/p-module homomorphism that makes
the following diagram commutative.
f
/ M
Am
πm
(A/p)m
π
f
/ M/pM,
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
383
Where π : A → A/p and π : M → M/pM are the canonical maps.
Suppose that (a1 , . . . , am ) ∈ ker f , where ai = π(ai ) for 1 ≤ i ≤ m. Then f (a1 , . . . , am ) ∈
pM . Thus, we can write f (a1 , . . . , am ) = p1 m1 +· · ·+pk mk , where p1 , . . . , pk are nonzero
elements of p, and m1 , . . . , mk ∈ M . By Lemma 9.5.18, there is a nonzero element p ∈ p
which annihilates p1 , . . . , pk . But then pf (a1 , . . . , am ) = 0.
Since f is injective, this says p annihilates (a1 , . . . , am ). But since A is local, with
maximal ideal p, every zero divisor of A lies in p. In particular, we must have ai ∈ p for
all i.
But then π(ai ) = 0 for all i, and hence (a1 , . . . , am ) = π n (a1 , . . . , am ) = 0. Thus, the
kernel of f is trivial.
We can now put it all together.
Theorem 9.5.20. Let A be a commutative ring and let f : Am → An be an injective
A-module homomorphism. Then m ≤ n.
Proof Since localization is exact, we may localize at a minimal prime ideal. Thus,
we may assume that A is a local ring with a unique prime ideal p. But now, Proposition 9.5.19 shows that the homomorphism obtained by extension of rings to A/p is also
injective. Since A/p is a field, the result follows.
Exercises 9.5.21.
1. Deduce that Z2 is not a flat Z-module by tensoring it with the exact sequence
f2
π
0 → Z −→ Z −
→ Z2 → 0
from Problem 3 of Exercises 7.7.52. Is Zn a flat Z-module for n > 2?
2. Show that Z2 is not a flat Z2r -module for r > 1. Note that Z2r is a local ring with
only one prime ideal.
3. Let A be a domain. For which ideals a is A/a flat as an A-module?
† 4. Show that an infinite direct sum of flat modules is flat. Deduce that all free modules
are flat.
f0
f1
5. Suppose given a direct system M0 −→ M1 −→ · · · of right A-modules, together
with a left A-module M .
(a) Show that −
lim
→(Mi ⊗A N ) is naturally isomorphic to (lim
−→ Mi ) ⊗A N .
(b) Deduce that if each Mi is flat, so is lim
−→ Mi .
9.6
Tensor Products of Algebras
Let A be a commutative ring and let B and C be A-algebras, via homomorphisms f : A →
B and g : A → C. We shall show that B ⊗A C is then an A-algebra, and in fact satisfies
an important universal mapping property with respect to A-algebra homomorphisms out
of B and C.
A ring called A, in this section, will be assumed to be commutative.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
384
Proposition 9.6.1. Let B and C be A-algebras, via f : A → B and g : A → C. Then
there is an A-algebra structure on B ⊗A C whose product is given on generators by
(b ⊗ c) · (b ⊗ c ) = bb ⊗ cc .
In particular, B ⊗A C is commutative if and only if both B and C are commutative.
In addition, if we define ι1 : B → B ⊗A C and ι2 : C → B ⊗A C by ι1 (b) = b ⊗ 1
and ι2 (c) = 1 ⊗ c, then the ιi are A-algebra homomorphisms, and satisfy the following
universal property.
1. Every element of the image of ι1 commutes with every element of the image of ι2 .
2. Suppose given an A-algebra D, together with A-algebra homomorphisms h1 : B →
D and h2 : C → D such that every element of the image of h1 commutes with
every element of the image of h2 . Then there is a unique A-algebra homomorphism
h : B ⊗A C → D such that the following diagram commutes.
ι1 /
ι2
B ⊗A C o
B ?
C
??


??

??

?? h 
h1 ??  h2
? 
D
If B, C, and D are commutative, this says simply that B ⊗A C, together with the maps
ιi , is the coproduct of B and C in the category of commutative A-algebras.
Proof First note that the A-algebra structures on B and C imply that there is an
A-module homomorphism
μ : B ⊗A C ⊗A B ⊗A C → B ⊗A C
whose effect on generators is given by μ(b ⊗ c ⊗ b ⊗ c ) = bb ⊗ cc . By the associativity of
the tensor product (e.g., Corollary 9.4.24), the iterated tensor product B ⊗A C ⊗A B ⊗A C
is naturally isomorphic to (B ⊗A C) ⊗A (B ⊗A C), so that μ induces a product operation
on B ⊗A C via x · y = μ(x ⊗ y) for x, y ∈ B ⊗A C. By the laws of the tensor product
(B ⊗A C) ⊗A (B ⊗A C), this product satisfies both distributive laws.
We claim that this product makes B ⊗A C a ring, with multiplicative identity 1 ⊗ 1.
On generators, we have (1 ⊗ 1) · (b ⊗ c) = b ⊗ c = (b ⊗ c) · (1 ⊗ 1). Since 1 ⊗ 1 acts as the
identity on a set of generators, it must act as the identity everywhere. Similarly, it suffices
to show associativity of products of generators, where it follows from the associativity of
the products in B and C.
The homomorphisms ιi defined above are easily seen to be ring homomorphisms. Note
that if ν and ν give the A-algebra structures of B and C, respectively, then ιi ◦ ν(a) =
a ⊗ 1 = 1 ⊗ a = ι2 ◦ ν (a) for all a ∈ A, and that this element lies in the center of
B ⊗A C. Since the ιi , ν, and ν are ring homomorphisms, this composite is also, and
gives B ⊗A C an A-algebra structure such that the ιi are A-algebra homomorphisms. We
have (b ⊗ 1) · (1 ⊗ c) = b ⊗ c = (1 ⊗ c) · (b ⊗ 1), and hence the elements of the image of ι1
commute with the elements of the image of ι2 .
Let h1 : B → D and h2 : C → D be A-algebra homomorphisms with the property
that every element of the image of h1 commutes with every element of the image of h2 .
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
385
Then, as the reader may check, there is an A-module homomorphism h : B ⊗A C → D
whose effect on generators is given by h(b ⊗ c) = h1 (b)h2 (c).
The fact that each h1 (b) commutes with each h2 (c) shows that h(x · y) = h(x)h(y)
as x and y range over a set of generators of B ⊗A C. Since h is a homomorphism of
abelian groups, this is enough to show that h(xy) = h(x)h(y) for all x, y ∈ B ⊗A C.
Since h(1 ⊗ 1) = 1, h is a ring homomorphism, which is easily seen to be an A-algebra
homomorphism as well. It is easy to see that h is the unique A-algebra homomorphism
that makes the diagram commute.
We’ve already seen some examples of tensor products of algebras.
Proposition 9.6.2. Let A be a commutative ring and let B be an A-algebra via f : A →
B. Then there is an isomorphism of A-algebras
η : B ⊗A Mn (A) → Mn (B)
which makes the following diagram commute.
ι1 /
ι2
B ⊗A Mn (A) o
Mn (A)
B J
JJ
t
JJ
t
t
JJ
t
J
tt
tt
ι JJJJ η
t
JJ
tt f∗
$ ztt
Mn (B)
Here ι : B → Mn (B) is given by ι(b) = bIn , while f∗ takes (aij ) to the matrix whose ij-th
entry is f (aij ). The maps ιi are the structure maps of the tensor product of algebras.
Proof Because the square
A
ι
Mn (A)
f
/ B
ι
f∗ /
Mn (B)
commutes, Mn (B) has an A-algebra structure such that both f∗ and ι : B → Mn (B) are
A-algebra homomorphisms.
It is easy to see that every element in ι(B) commutes with every element in the image
of f∗ . So the universal property of the tensor product of algebras provides an A-algebra
homomorphism η : B ⊗A Mn (A) → Mn (B) that makes the above diagram commute. It
suffices to show that η is an isomorphism.
Write eij for the matrix whose ij-th coordinate is 1 and whose other coordinates are
all 0. Recall from Lemma 7.10.5 that if C is any ring, then {eij | 1 ≤ i, j ≤ n}, in any
order, gives a basis for Mn (C) as either a left or a right C-module.
By extension of rings, B ⊗A Mn (A) is free as a left B-module, with basis {1 ⊗ eij | 1 ≤
i, j ≤ n}. But η(1 ⊗ eij ) = eij , so, viewing η as a left B-module homomorphism, we see
that it carries a basis of B ⊗A Mn (A) onto a basis of Mn (B). By Lemma 7.7.36, this is
sufficient to conclude that η is an isomorphism.
There is an analogous result for monoid rings, which we shall treat as an exercise.
We shall see other examples later.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
386
Exercises 9.6.3.
‡ 1. Let B be an A-algebra, via f : A → B, and let M be a monoid. Show that there
is a natural A-algebra isomorphism from B ⊗A A[M ] to B[M ].
2. Combining Proposition 9.6.2 with the preceding problem, we see that if A is commutative and M is a monoid, then there is an isomorphism between Mn (A[M ]) and
Mn (A)[M ]. Write down the isomorphism explicitly and give a proof of isomorphism
that doesn’t use tensor products.
3. Let M and M be monoids. Show that there is a natural A-algebra isomorphism
from A[M ] ⊗A A[M ] to A[M × M ].
4. Let K be a subfield of the fields L and L . Is L ⊗K L a field?
5. Let B be an A-algebra with structure map ν : A → B. Show that there is a
homomorphism of A-modules (but not necessarily of A-algebras), μ : B ⊗A B → B
induced by μ(b ⊗ b ) = bb . Show also that the following diagrams commute.
B ⊗A B ⊗A B
1⊗μ
μ ⊗ 1/
B ⊗A B
μ
μ
/ B
B ⊗A B
ν ⊗ 1/
1⊗ν
A ⊗A B
B ⊗A B o
B ⊗A A
JJ
t
JJ
t
JJ
tt
t
J
μ
tt
εB JJJJ
tt εB
J$ zttt
B
Here, the maps εB are the natural isomorphisms of Lemma 9.4.14.
Conversely, suppose given an A-module B, together with A-module homomorphisms ν : A → B and μ : B ⊗A B → B such that the above diagrams commute.
Show that B is an A-algebra with structure map ν and with ring product given by
bb = μ(b ⊗ b ).
6. Let B be an A-algebra and let μ : B ⊗A B → B be induced by the ring product of B.
Show that B is a commutative ring if and only if μ is an A-algebra homomorphism.
9.7
The Hom Functors
Let M and N be A-modules. Recall that HomA (M, N ) is the abelian group of A-module
homomorphisms from M to N . Here, the group structure on HomA (M, N ) is given by
(f +g)(m) = f (m)+g(m). Without extra hypotheses (e.g., A being commutative), there
is no module structure on HomA (M, N ).
We shall use the same notation HomA (M, N ) whether we are discussing the homomorphisms of left modules or of right modules, and the theory proceeds identically. Thus,
in general, we shall not make reference to the sidedness of the module structures. In
applications, it must either be made clear by the context or be explicitly stated.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
387
We should immediately discuss the functoriality of Hom, as it is good to keep it
in mind as we go. Here, for an A-module M , HomA (M, −) gives a covariant functor
from A-modules to abelian groups. For f : N → N , the induced map HomA (M, f ) :
HomA (M, N ) → HomA (M, N ) is generally written as f∗ , and is given by f∗ (g) = f ◦ g.
Similarly, for an A-module N , HomA (−, N ) is a contravariant functor. Here, if f :
M → M , the induced map HomA (f, N ) : HomA (M, N ) → HomA (M , N ) is generally
written as f ∗ , and is given by f ∗ (g) = g ◦ f .
We shall first establish the exactness properties of the Hom functors.
Lemma 9.7.1. For any A-module M , the functor HomA (M, −) is left exact; i.e., if
we’re given an exact sequence
g
f
0 → N −
→N −
→ N ,
then the induced sequence
f∗
g∗
0 → HomA (M, N ) −→ HomA (M, N ) −→ HomA (M, N )
is exact.
Proof Clearly, f∗ is injective because f is, and g∗ ◦ f∗ = (g ◦ f )∗ = 0. Let h : M → N
be in the kernel of g∗ . Then im h ⊂ ker g. But the kernel of g is equal to the image of f .
Thus, since f is injective, the induced map f : N → ker g is an isomorphism. Thus, h is
f −1
h
→ ker g −−→ N .
the image under f∗ of the composite M −
The contravariant direction of Hom also has exactness properties.
Lemma 9.7.2. Let N be an A-module and let
f
g
M −
→M −
→ M → 0
be an exact sequence of A-modules. Then the following sequence is exact.
g∗
f∗
0 → HomA (M , N ) −→ HomA (M, N ) −→ HomA (M , N )
Proof The injectivity of g ∗ follows directly from the surjectivity of g. It is also immediate that f ∗ ◦ g ∗ = 0. Let h : M → N be in the kernel of f ∗ . Thus, h ◦ f = 0. Thus,
h vanishes on the image of f . By the exactness of the original sequence, the image of f
is the kernel of g. But then h factors through the canonical map from M to M/(ker g).
But the Noether Theorem says we may identify this canonical map with g itself. In other
words, there is a homomorphism h : M → N such that h ◦ g = h. But this is just
another way of saying that h = g ∗ h for some h ∈ HomA (M , N ).
We have so far avoided a discussion of module structures on the Hom groups. The
proof of the next lemma is easy, and is left to the reader.
Lemma 9.7.3. Bimodule structures may be used to put module structures on Hom groups
in any of the following ways.
1. If M is an A-B-bimodule and N is a left A-module, then there is a left B-module
structure on HomA (M, N ) via (b · f )(m) = f (mb).
2. If M is a B-A-bimodule and N is a right A-module, then there is a right B-module
structure on HomA (M, N ) via (f · b)(m) = f (bm).
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
388
3. If N is an A-B-bimodule and M is a left A-module, then there is a right B-module
structure on HomA (M, N ) via (f · b)(m) = f (m)b.
4. If N is a B-A-bimodule and M is a right A-module, then there is a left B-module
structure on HomA (M, N ) via (b · f )(m) = bf (m).
If A is commutative, then any A-module is an A-A-bimodule. So each of the procedures above puts an A-module structure on HomA (M, N ). In fact, each of the above
procedures puts the same A-module structure on HomA (M, N ).
In particular, for any ring A, if M is a left A-module, then so is HomA (A, M ), by the
first item above. Similarly, if M is a right A-module, then the second item puts a right
A-module structure on HomA (A, M ). Recall from Corollary 7.7.9 that the evaluation
map ε : HomA (A, M ) → M is an isomorphism of abelian groups, where ε(f ) = f (1). In
fact, it is more:
Proposition 9.7.4. The evaluation map ε : HomA (A, M ) → M gives a natural isomorphism of A-modules.
The various module structures on Hom allow us to demonstrate the relationship
between Hom and the tensor products.
Proposition 9.7.5. Suppose given a right B-module M1 , a B-A-bimodule M2 , and a
right A-module M3 . Then there is an isomorphism of abelian groups, natural in all three
variables:
ϕ : HomB (M1 , HomA (M2 , M3 )) → HomA (M1 ⊗B M2 , M3 ).
Explicitly, if f ∈ HomB (M1 , HomA (M2 , M3 )), then the effect of ϕ(f ) on generators is
given by
ϕ(f )(m1 ⊗ m2 ) = f (m1 )(m2 ).
If M1 is a C-B-bimodule, then ϕ is an isomorphism of right C-modules.
Proof For f ∈ HomB (M1 , HomA (M2 , M3 )), it is easy to see that the displayed formula
for ϕ(f ) gives rise to a well defined homomorphism. And ϕ itself is then easily seen to
be a homomorphism.
To see that ϕ is an isomorphism, we construct its inverse. Define
ψ : HomA (M1 ⊗B M2 , M3 ) → HomA (M1 , HomA (M2 , M3 ))
as follows. For g ∈ HomA (M1 ⊗B M2 , M3 ), we set
((ψ(g))(m1 ))(m2 ) = g(m1 ⊗ m2 ).
We leave it to the reader to check that ϕ and ψ are inverse isomorphisms, and to
verify the claims regarding naturality and C-module structures.
Corollary 9.7.6. Let A be a commutative ring and let Mi be an A-module for i = 1, 2, 3.
Then there is a natural isomorphism
ϕ : HomA (M1 , HomA (M2 , M3 )) → HomA (M1 ⊗A M2 , M3 )
of A-modules.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
389
As in the case of tensor products, it is useful to understand the behavior of Hom
groups with respect to direct sums.
Proposition 9.7.7. Let M, M1 , . . . , Mk and N, N1 , . . . , Nk be left A-modules. Then
there are natural isomorphisms
k
k
,
,
α : HomA (
Mi , N ) →
HomA (Mi , N ),
i=1
i=1
whose i-th coordinate map is ι∗i , where ιi is the natural inclusion of Mi in
β : HomA (M,
k
,
i=1
Ni ) →
k
,
*k
i=1
Mi , and
HomA (M, Ni ),
i=1
*k
whose i-th coordinate map is (πi )∗ , where πi is the natural projection map from i=1 Ni
onto Ni .
These isomorphisms are natural in all variables. If each Mi is an A-B-bimodule or
if N is an A-B-bimodule, then α is an isomorphism with respect to the appropriate Bmodule structure. Similarly, if each Ni is an A-B-bimodule or if M is an A-B-bimodule,
then β is an isomorphism with respect to the appropriate B-module structure.
Proof In fact, these follow from the universal properties of the coproduct and product
in the category of A-modules. Thus, α is an isomorphism because a homomorphism out
of a direct sum is determined uniquely by its restrictions to the individual summands.
Similarly, β is an isomorphism because finite direct sums coincide with finite direct
products. Thus, a homomorphism into a finite direct sum is determined uniquely by its
component functions. We leave the statements about naturality and module structures
to the reader.
Definitions 9.7.8. Let M be an A-module. Define the dual module, M ∗ , by M ∗ =
HomA (M, A). Since A is an A-A-bimodule, M ∗ is a right A-module when M is a left
A-module, and M ∗ is a left A-module when M is a right A-module. In either case, the
dual of M ∗ , M ∗∗ = HomA (HomA (M, A), A) is defined, and is an A-module from the
same side that M is.
The next proposition is fundamental in our understanding of the duality of free modules.
Proposition 9.7.9. Let M be a finitely generated free left A-module, with basis m1 , . . . , mn .
Then the dual module M ∗ is a free right A-module, with basis m∗1 , . . . , m∗n , where m∗i (a1 m1 +
· · · + an mn ) = ai . We call m∗1 , . . . , m∗n the dual basis of m1 , . . . , mn .
Proof Let g : An → M be the isomorphism defined by the basis m1 , . . . , mn : g(a1 , . . . , an ) =
a1 m1 + · · · + an mn . Then g ∗ : M ∗ → (An )∗ is an isomorphism (because (g −1 )∗
is its inverse). But Proposition 9.7.7 provides an isomorphism α : (An )∗ → (A∗ )n .
And A∗ = HomA (A, A) is isomorphic to A (Proposition 9.7.4) by the correspondence,
ε : HomA (A, A) → A, which takes a left A-module homomorphism f : A → A to f (1).
In Proposition 9.7.4 it was shown that ε is a left A-module map with respect to the
left module structure on A∗ induced by the A-A-bimodule structure on the A serving as
the domain. Here, we are using the right module structure on A∗ which comes from the
bimodule structure of the A that receives the maps f ∈ HomA (A, A). But we see that
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
390
ε is a module map with respect to this structure also: ε(f a) = (f a)(1) = f (1)a by the
definition of f a. But this last is (ε(f ))a.
Thus, we have a composite isomorphism of right A-modules:
g∗
εn
→ (A∗ )n −→ An .
M ∗ −→ (An )∗ −
α
But this composite is easily seen to carry the dual basis m∗1 , . . . , m∗n to the canonical
basis of An .
Note that the homomorphism m∗i depends on the entire basis m1 , . . . , mn , and not
just on the element mi . A different basis whose i-th term is also mi could give rise to a
different homomorphism m∗i . Explicitly, m∗i is the composite
g −1
π
i
M −−→ An −→
A
∼
=
where g is the isomorphism induced by the basis m1 , . . . , mn , and πi is the projection
onto the i-th factor.
We see from Proposition 9.7.9 that if A is commutative and if M is a finitely generated
free A-module, then M and M ∗ are isomorphic. However, even in the case where A is a
field, there is no natural isomorphism between M and M ∗ : if we’re given two different
bases for M and map each one of them to its dual basis, we will get two different
isomorphisms from M to M ∗ . But if we dualize twice, some naturality returns. The
reader may verify the following lemma.
Lemma 9.7.10. Let M be an A-module for any ring A. Then there is an A-module
homomorphism κM : M → M ∗∗ defined as follows. For m ∈ M , κM (m) is the homomorphism from M ∗ to A defined by evaluating at m: κM (m)(f ) = f (m).
In addition, κM is natural in M in the sense that if g : M → N is an A-module
homomorphism, then the following diagram commutes.
M
κM M ∗∗
g
g ∗∗
/ N
κN / N ∗∗
Here, for h ∈ M ∗∗ and f ∈ N ∗ , g ∗∗ (h)(f ) = h(f ◦ g).
We shall borrow a definition from functional analysis.
Definition 9.7.11. An A-module M is reflexive if κM : M → M ∗∗ is an isomorphism.
Proposition 9.7.12. Finitely generated free modules are reflexive.
Proof Let M be a finitely generated free module with basis m1 , . . . , mn . Then the dual
module M ∗ is free, with basis m∗1 , . . . , m∗n , by Proposition 9.7.9.
∗∗
∗
∗
∗∗
But then M ∗∗ is free on the dual basis, m∗∗
1 , . . . , mn , to m1 , . . . , mn . Here, mi :
∗
∗∗
∗
∗
M → A is given by mi (m1 a1 + · · · + mn an ) = ai .
Thus, M and M ∗∗ are both free modules of the same rank, and hence are abstractly
isomorphic. But as the reader may check, κM (mi ) = m∗∗
i . Thus, κM carries a basis of
M to a basis of M ∗∗ , and hence is an isomorphism.
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
391
The fact that M is finitely generated here is essential. Infinitely generated free modules are not reflexive, because their duals are infinite direct products of copies of A,
rather than infinite direct sums.
The proof of Proposition 9.7.12 brings up a useful notation that originated in exactly
this context, called the Kronecker delta.
Definition 9.7.13. The Kronecker delta, δij , is defined to be 0 or 1, depending on the
values of i and j. Explicitly,
1 if i = j
δij =
0 otherwise.
Thus, if m1 , . . . , mn is the basis of a free module and if m∗1 , . . . , m∗n is its dual basis,
then m∗j (mi ) = δij . Indeed, this fact uniquely determines the dual basis m∗1 , . . . , m∗n .
Exercises 9.7.14.
1. Let a be a two-sided ideal of A and let M be a left A-module. Show that there is
a natural isomorphism of left A-modules
ε : HomA (A/a, M ) → {m ∈ M | a ⊂ Ann(m)}
given by ε(f ) = f (1).
Note the use of the modules HomA (A/(a), M ) in the proof of the Fundamental
Theorem of Finitely Generated Modules Over a P.I.D.
2. Let M, M and N be A-modules, with M a submodule of M . Show that HomA (M/M , N )
is isomorphic to the subgroup of HomA (M, N ) consisting of those homomorphisms
that factor through the canonical map π : M → M/M .
3. Give an example of a ring A and an A-module M for which HomA (M, −) is not an
exact functor.
4. Give an example of a ring A and an A-module N for which HomA (−, N ) is not an
exact functor.
5. Let A be an integral domain and let M be an A-module. Let π : M → M/Tors(M )
be the canonical map. Show that the induced map of dual modules, π ∗ : (M/Tors(M ))∗ →
M ∗ is an isomorphism.
6. Let M be a finitely generated module over the principal ideal domain A. Show
that the double dual module M ∗∗ is naturally isomorphic to M/Tors(M ).
7. Suppose given A-modules M , {Mi | i ∈ I}, N and {Ni | i ∈ I}.
(a) Show that there is a natural isomorphism
,
.
α
HomA (
Mi , N ) −
→
HomA (Mi , N ).
i∈I
i∈I
Here,
* the i-th factor of α(f ) is f ◦ ιi , where ιi is the canonical inclusion of Mi
in i∈I Mi .
CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS
392
(b) Show that there is a natural isomorphism
.
HomA (M,
β
Ni ) −
→
i∈I
.
HomA (M, Ni ).
i∈I
Here,
the i-th factor of β(f ) is πi ◦ f , where πi is the canonical projection of
!
N
i onto its i-th factor.
i∈I
(c) Suppose that M is finitely generated. Show that the preceding map β restricts
to an isomorphism
,
β ,
Ni ) −
→
HomA (M, Ni ).
HomA (M,
i∈I
i∈I
8. Interpret the relationship between Hom and tensor in terms of adjoint functors.
Which of the results on these functors would then have automatic proofs?
9.8
Projective Modules
Unless it’s explicitly stated to the contrary, all modules in this section are left A-modules,
and all homomorphisms are A-module homomorphisms.
Recall that a short exact sequence
g
f
→M −
→ M → 0
0 → M −
splits if g admits a section, meaning an A-module homomorphism s : M → M such
that g ◦ s = 1M . Recall from Proposition 7.7.49