Download Math 850 Algebra - San Francisco State University

Math 850
Algebra
David Meredith
Department of Mathematics
San Francisco State University
San Francisco, CA 94132
[email protected]
ii
Contents
Syllabus
0.1
0.2
0.3
0.4
Instructor .
Schedule . .
Projects . .
Assignments
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and Grading
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. vii
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Lectures
0.5 Sets and Maps . . . . . . . . . . . . . . . . . . . . .
0.6 Group Theory . . . . . . . . . . . . . . . . . . . . . .
0.6.1 Groups . . . . . . . . . . . . . . . . . . . . .
0.6.2 Subgroups . . . . . . . . . . . . . . . . . . . .
0.6.3 Normal Subgroups and Quotient Groups . . .
0.6.4 Homomorphism Theorems . . . . . . . . . . .
0.7 Ring Theory . . . . . . . . . . . . . . . . . . . . . .
0.7.1 Basic Definitions and Examples . . . . . . . .
0.7.2 Polynomials . . . . . . . . . . . . . . . . . . .
0.7.3 Greatest Common Divisors . . . . . . . . . .
0.7.4 Homomorphisms and Ideals . . . . . . . . . .
0.7.5 Euclidean Domains . . . . . . . . . . . . . . .
0.7.6 Linear Algebra . . . . . . . . . . . . . . . . .
0.7.7 Linear Transformations . . . . . . . . . . . .
0.7.8 Quotient Rings and Fields . . . . . . . . . . .
0.8 Commutative Rings . . . . . . . . . . . . . . . . . .
0.8.1 Prime and Maximal Ideals . . . . . . . . . . .
0.8.2 Unique Factorization Domains . . . . . . . .
0.8.3 Noetherian Rings . . . . . . . . . . . . . . . .
0.8.4 Algebraic Varieties . . . . . . . . . . . . . . .
0.9 Modules . . . . . . . . . . . . . . . . . . . . . . . . .
0.9.1 Introduction . . . . . . . . . . . . . . . . . .
0.9.2 Submodules . . . . . . . . . . . . . . . . . . .
0.9.3 Module Homomorphisms . . . . . . . . . . .
0.9.4 Quotient Modules . . . . . . . . . . . . . . .
0.9.5 Direct Sums and Products and Free Modules
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lxxii
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CONTENTS
0.9.6 Exact Sequences . . . . . . . . . . . . . . . . .
0.10 Categories . . . . . . . . . . . . . . . . . . . . . . . . .
0.10.1 Functors . . . . . . . . . . . . . . . . . . . . . .
0.10.2 Free, Projective and Injective modules . . . . .
0.11 Linear Algebra . . . . . . . . . . . . . . . . . . . . . .
0.11.1 Modules over PIDs . . . . . . . . . . . . . . . .
0.11.2 Applications of the Fundamental Theorem of
over PID’s to Linear Algebra . . . . . . . . . .
0.12 Some problems . . . . . . . . . . . . . . . . . . . . . .
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Modules
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Extra
0.13
0.14
0.15
0.16
0.17
0.12.1 Ring Homomorphisms . . . . . . . . . . . . .
0.12.2 Quotient Rings . . . . . . . . . . . . . . . . .
0.12.3 Homomorphism Theorems Again . . . . . . .
0.12.4 More on Ideals . . . . . . . . . . . . . . . . .
0.12.5 Rings of Fractions . . . . . . . . . . . . . . .
0.12.6 The Chinese Remainder Theorem . . . . . . .
Rings with Good Arithmetic . . . . . . . . . . . . .
0.13.1 Greatest Common Divisors . . . . . . . . . .
0.13.2 Euclidean Domains . . . . . . . . . . . . . . .
0.13.3 Principal Ideal Domains . . . . . . . . . . . .
0.13.4 Unique Factorization Domains . . . . . . . .
Polynomial Rings . . . . . . . . . . . . . . . . . . . .
0.14.1 Introduction . . . . . . . . . . . . . . . . . .
0.14.2 Factorization in Polynomial Rings . . . . . .
0.14.3 Polynomial Rings in One Variable . . . . . .
0.14.4 Monic Polynomials over general rings . . . .
0.14.5 Polynomials over Fields . . . . . . . . . . . .
Modules . . . . . . . . . . . . . . . . . . . . . . . . .
0.15.1 Basic Definitions and Examples . . . . . . . .
0.15.2 Submodules . . . . . . . . . . . . . . . . . . .
0.15.3 Module Homomorphisms . . . . . . . . . . .
0.15.4 Quotient Modules . . . . . . . . . . . . . . .
0.15.5 Direct Sums and Products and Free Modules
0.15.6 Tensor Products . . . . . . . . . . . . . . . .
0.15.7 Exact Sequences . . . . . . . . . . . . . . . .
Vector Spaces . . . . . . . . . . . . . . . . . . . . . .
0.16.1 Generalities . . . . . . . . . . . . . . . . . . .
0.16.2 Bases . . . . . . . . . . . . . . . . . . . . . .
0.16.3 Dual Spaces . . . . . . . . . . . . . . . . . . .
0.16.4 Tensor Algebras–New Rings from Modules .
Modules over Principal Ideal Domains . . . . . . . .
0.17.1 The Theory . . . . . . . . . . . . . . . . . . .
0.17.2 The Applications . . . . . . . . . . . . . . . .
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lxxii
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lxxxi
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cvii
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cxxix
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cxlviii
clii
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clxiii
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clxiv
clxiv
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clxxi
clxxi
clxxvi
CONTENTS
Appendix
0.18 Problem 2.3.21 . . . . . . . . . . . . . . . . . . . . . . . . . .
0.19 Problem 3.1.14 . . . . . . . . . . . . . . . . . . . . . . . . . .
0.20 Problem 3.4.? . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
clxxxi
. . clxxxi
. . clxxxii
. . clxxxiv
vi
CONTENTS
Syllabus
Algebra is one of the three main divisions of modern mathematics, the others being analysis and topology. Algebra studies algebraic structures, or sets
with operations. Algebra abstracts the operations of mathematics–addition,
multiplication, differentiation, integration–insofar as possible from their metric
context. In this course we begin with a brief review of groups, or sets with a single operation. The main focus will be subgroups and quotient groups. Following
our review of groups we will study rings, or sets with two operations. Rings are
most like the familiar objects of mathematics: integers, polynomials or matrices.
Included in our study will be special classes of rings including polynomial rings.
Polynomial rings are the main topic in computational algebra, which has been
the area of a several MA theses supervised by Dr. Hosten. The course will
conclude with an introduction to the theory of modules, or vector spaces over
rings. Our principal application will be a rapid development of the theory of
generalized eigenspaces over arbitrary fields.
0.1
Instructor
David Meredith
Department of Mathematics
San Francisco State University
1600 Holloway Ave.
San Francisco, CA 94132
E-mail: [email protected]
URL: http://online.sfsu.edu/~meredith
Ian Sammis
TH 941
Office Hours: Thursday, 2:00-4:00
vii
viii
0.2
SYLLABUS
Schedule
Date
February 3
February 10
February 17
February 24
March 3
March 10
March 24
March 31
April 7
April 14
April 21
April 28
May 5
0.3
Assignment
2.32, 2.46, 2.49, 2.71, 2.76
3.3, 3.6, 3.17, 3.21, 3.23, 3.24; extra credit 3.26, 3.27
3.28, 3.31, 3.32, 3.40, 3.43, 3.55; extra credit 3.47, 3.49
3.57, 3.61, 3.62, 3.72, 3.75, 3.77, 3.79; extra credit 3.80
3.81, 3.82, 3.83, 3.85, 3.88, 3.92, 3.93
6.2, 6.3, 6.4, 6.5, 6.13, 6.14; extra credit 6.6, 6.9
6.2, 6.4, 6.10, 6.11, 6.12, 6.17, 6.18, 6.19
6.36, 6.38 (if I is an ideal, does it have to be finitely generated?)
6.62, 6.67, 6.68; extra credit 6.33, 6.63
7.2, 7.3, 7.4, 7.7, 7.8, 7.12, 7.13
7.20, 7.23, 7.25, 7.26, 7.28, 7.34, 7.39
7.40, 7.43, 7.44, 7.47, 7.49, 7.50, 7.51
9.2, 9.3, 9.6, 9.8, 9.9
9.32, 9.33, 9.34, 9.35, 9.37, 9.38, 9.42
Projects
Every student will write a 5-10 page mathematical paper. Your work must be
written in correct mathematical English and typeset with proper formatting of
mathematical symbols, etc. You can work on one topic yourself, or groups of 2
(resp. 3) students can choose 2 (resp. 3) related topics and produce 10-20 (resp.
15-30) pages. Your paper should use a miniumum of three references. If you
reference a web site, document its reliability. You can choose one of the topics
below or propose your own topic. You may find yourself starting off with one
topic and ending with another.
1. Find examples of a principal ideal domain that is not a Euclidean domain.
What do the examples have in common? Is there a heirarchy: Euclidean
domain −→ PID −→ ?.
2. Connect the “definition” of tensor in physics with tensor algebras.
3. Connect the algebra of differential forms over a manifold to tensor algebras.
4. Define the group ring and group characters.
5. Discuss which theorems in Axler, Linear Algebra Done Right hold over
arbitrary fields, and which ones require a real or complex field.
The next two items would make a good group project
6. Define algebraic variety. How are varieties related to ideals?
7. State and prove the Hilbert Nullstellensatz. Give examples.
0.4. ASSIGNMENTS AND GRADING
ix
8. Discuss the concepts of prime and irreducible for ring elements.
9. Write about Noetherian rings and modules.
10. Find applications of Zorn’s lemma and the Axiom of Choice to group and
ring theory and linear algebra.
11. Is the existence of maximal ideals equivalent to Zorn’s lemma?
12. Define the functors Exti and/or T ori and present the basic properties.
13. What is a Gröbner basis for an ideal in a polynomial ring over a field.
14. Discuss localization of commutative rings and modules, starting with problem 10.4.8.
15. What is a local ring and what are its principal properties?
16. What is a valuation on a field?
17. What is a Gröbner basis for a polynomial ideal?
0.4
Assignments and Grading
Graded activities include written and oral homework and a term project.
1. There will be written homework assignments due weekly. Late papers will
not be accepted without prior arrangements. Each week you should do as
many problems as you can, then go on to the next assignment.
2. Every day homework is due, I will call on students more or less cyclically
for an oral presentations of the homework problems. I will try to give
everyone the same number of opportunities, and I’ll try to check that you
have done a problem before calling on you.
3. There will be a term project, as described above.
The relative weights of the different kinds of work are:
Written homework
Oral homework
Project
50%
15%
35%
x
SYLLABUS
Lectures
Don’t overdo on homework.
0.5
Sets and Maps
1. We assume naive set theory. You know what a set is and what subset and
membership mean.
(a) If A is a set, then 2A is the collection of subsets of A.
(b) 2A is another set.
2. Given two sets A and B, their cartesian product A×B = {(a, b) : a ∈ A, b ∈ B}
is the set of pairs .
3. Let A and B be sets. A function or map f with domain A and
codomain B is a subset grf ⊂ A × B such that if a ∈ A then there
exists a unique pair (a, b) ∈ grf . We write f (a) = b iff (a, b) ∈ grf .
(a) grf means graph of f .
f
(b) Use the notation f : A −→ B or A −→ B for “f is a function with
domain A and codomain B”
(c) Im f = {b : (a, b) ∈ grf } ⊂ B.
i. Analysts sometimes use the word range to mean image; algebraists use range to mean codomain. I will try to avoid using
range, but if I do use it, I mean codomain. This is an algebra
class.
4. For every set A there is a function idA : A −→ A given by idA (a) = a.
5. Suppose f : A −→ B.
(a) If C ⊂ A then f (C) = {f (c) : c ∈ C} ⊂ B. Sometimes we denote
this by f∗ (C).
xi
xii
LECTURES
(b) If D ⊂ B then f −1 (D) = {a ∈ A : f (a) ∈ D} ⊂ A. Sometimes we
denote this by f ∗ (D), reserving f −1 for the inverse of f when f is
bijective..
6. If f : A −→ B then f∗ : 2A −→ 2B and f ∗ : 2B −→ 2A
7. If we have two function f : A −→ B and g : B −→ C then we have the
composition gf = g ◦ f : A −→ C.
(g ◦ f ) (a) = g (f (a)) .
(a) We have a commutative diagram:
B
A
%f
g◦f
−→
g&
C
(b) (g ◦ f )∗ = g∗ ◦ f∗
(c) (g ◦ f )∗ = f ∗ ◦ g ∗
8. A function is 1 − 1 iff a 6= b =⇒ f (a) 6= f (b).
(a) Such maps are called injections or monomorphisms.
(b) f : A −→ B is 1 − 1 iff there exists a function g : B −→ A such that
gf = idA .
9. A function is onto if its image equals its codomain.
(a) Such maps are called surjections or epimorphisms.
(b) The Axiom of Choice assures us that f : A −→ B is onto iff there
exists a function g : B −→ A such that f g = idB .
10. If f : A −→ B and g : B −→ A are such that f g = idB , then we say that
f is a left inverse of g and g is a right inverse of f .
(a) Every right inverse is 1 − 1; every left inverse is onto.
(b) Every 1 − 1 function is a right inverse (has a left inverse), and every
onto function is a left inverse (has a right inverse).
11. A function that is 1 − 1 and onto is a bijection or isomorphism.
(a) For every set A the function idA : A −→ A is bijective.
(b) If f : A −→ B then the map A −→ grf given by a 7−→ (a, f (a)) is a
bijection.
(c) f is bijection if and only if there is a map g : B −→ A such that
f g = idB and gf = idA . This map is the unique left inverse of
f , and it is the unique right inverse of f . We call it an inverse or
two-sided inverse of f and denote it f −1 .The map f −1 is bijective.
0.6. GROUP THEORY
xiii
∼
=
(d) Bijections are sometimes denoted A −→ B
12. A cartesian product A × B has two projection functions πA : A × B −→ A
and π B : A × B −→ B defined by πA (a, b) = a and π B (a, b) = b.
(a) Both π A and π B are surjective.
13. If A ⊂ B there is an inclusion map denoted A → B.
(a) The inclusition map is injective.
14. This whole business can be into a diagram:
f
B
A −→
↓∼
↑
πB
=
F
→ A×B
15. If A ⊂ B and f : B −→ C then we have the function “f restricted to A”:
f |A : A −→ C.
∗
16. Let A be a set. A binary operation on A is a map A × A −→ A.
(a) We frequently denote ∗ (x, y) = x ∗ y
(b) An operation is associative iff x ∗ (y ∗ z) = (x ∗ y) ∗ z
(c) An operation is commutative iff x ∗ y = y ∗ x
0.6
0.6.1
Group Theory
Groups
1. A group (G, ∗) is a pair consisting of a set G and an associative operation
∗ on G such that:there exists an element e ∈ G with two properties:
(a) e ∗ a = e ∗ a = a all a ∈ G
(b) for each a ∈ G there exists a−1 ∈ G such that a ∗ a−1 = a−1 ∗ a = e
2. The element e is called the group identity; the element a−1 is called the
inverse of a.
3. The element e is unique.
(a) Proof: if e1 and e2 are both identities for a group G then e1 = e1 ∗e2 =
e2
4. In a group G, if a ∗ b = a ∗ c then b = c
xiv
LECTURES
(a) Proof:
b =
=
=
=
=
¡
¢
a−1 ∗ a ∗ b
a−1 ∗ (a ∗ b)
a−1 ∗ (a ∗ c)
¡ −1
¢
a ∗a ∗c
c
5. For each a ∈ G the element a−1 is unique because if b and c are both
inverses of a then a ∗ b = e = a ∗ c.
6. I’m tired of ∗. Let’s change to simpler notation a ∗ b = ab and e = 1
¡
¢−1
−1
7. for all a, b ∈ G, a−1
= a and (ab) = b−1 a−1 . If a2 = a then a = e.
−1
−1
If ab = 1then a = b and b = a .
(a) class prove
8. Examples: non-zero reals, non-zero rationals, non-zero complex numbers
with multiplication.
9. A group is abelian (after Abel) if its operation is commutative.
(a) Traditionally abelian groups are written additively, with + for the
operation and − for the inverse and 0 for the identity.
(b) If A is an abelian group, we would have a + 0 = 0 + a = a for all
a ∈ A, − (−a) = a, a + (−a) = a − a = 0.
(c) Examples: Z, Q, R, C with addition, non-zero reals, non-zero rationals, non-zero complex numbers with multiplication.
n
o
∼
=
10. Let A be a set, and let SA = p : A −→ A , a group under composition.
(a) If A is finite we can name the elements {1, . . . , n} and denote SA =
Sn , the symmetric group of permutations on n symbols.
(b) The symbol (m1 , . . . , mt ) denotes a permutation p where p (mi ) =
m(i+1) mod n .Such a permutation is called a cycle.
i. (m, n) = (n, m)
(c) (1, 2, 3) (2, 3, 4) =? You have to decide which permutation is applied
first. Let’s work like functions, apply the right-most first. Then
(1, 2, 3) (2, 3, 4) = (3, 4) (1, 2)
(d) Every permutation can be written as a product of disjoint cycles.
Disjoint cycles commute.
(e) Every permutation can be written as a product of 2-cycles. (m1 , . . . , mt ) =
(m2 , m1 ) · · · (mt , mt−1 )
0.6. GROUP THEORY
xv
i. the decomposition is not unique, but all possible decompositions
of a permutation into 2-cycles contain an odd number of factors
or all contain an even number of factors
ii. The parity of the permutation is either odd or even
(f) If n ≥ 3 then Sn is not abelian. (1, 2) (2, 3) 6= (2, 3) (1, 2). Class
check.
11. Groups from linear algebra:
(a) The n × m matrices are a group under addition.
(b) The n × 1 matrices are a group under addition.
(c) A vector space V is a group under addition
(d) The linear operators L (V, W ) from one vector space to another form
a group under addition.
(e) The set of invertible real or complex n × n matrices is a group under
matrix multiplication.
(f) The invertible linear operators on a vector space GL (V ) form a group
under composition.
12. A homomorphism ϕ : G −→ H from one group to another is a function
ϕ such that ϕ (ab) = ϕ (a) ϕ (b), all a, b ∈ G.
¡
¢
−1
(a) ϕ (1G ) = 1H and ϕ a−1 = ϕ (a) .
¡ ¢
i. Proof: ϕ (1G )2 = ϕ 12G = ϕ (1G ), therefore ϕ (1G ) = 1H .
¢
¡
¡ −1 ¢
ϕ (a) = ϕ a−1 a = ϕ (1G ) = 1H .
ii. Proof: ϕ a
(b) Example: det : invertible real n × n matrices −→ R×
(c) Example: Sn −→ {−1, 1}, p 7→ (−1)parity(p) .
r
(d) R× −→ {−1, 1}, r 7→ sgn (r) =
.
|r|
(e) R× −→ R× , x 7→ x2
(f) Z −→ Z, x 7→ nx for some fixed integer n.
13. A homomorphism ϕ : G −→ H is 1 − 1 iff ϕ (a) = 1 =⇒ a = 1.
¢
¡
(a) Proof: Suppose ϕ (a) = ϕ (b). Then ϕ ab−1 = ϕ (a) ϕ (b)−1 = 1 so
ab−1 = 1. Thus a = b.
14. An isomorphism ϕ : G −→ H is a homomorphism that is 1 − 1 and
onto. We say that ϕ makes G and H isomorphic, so when we say that two
groups are isomorphic we mean that there exists an isomorphism from one
to the other.
xvi
LECTURES
(a) It isn’t immediately obvious, but if ϕ : G −→ H is an isomorphism
then the inverse set map ϕ−1 : H −→ G is also a homomorphism and
therefore an isomorphism.
i. Proof: let a, b ∈ H. We must show ϕ−1 (a) ϕ−1 (b) = ϕ−1 (ab).
Since ϕ is a homomorphism:
¡
¢
¡
¢¡
¢
ϕ ϕ−1 (a) ϕ−1 (b) = ϕϕ−1 (a) ϕϕ−1 (b)
= ab
¡
¢
= ϕ ϕ−1 (ab)
Since ϕ is 1 − 1, ϕ−1 (a) ϕ−1 (b) = ϕ−1 (ab).
(b) If two groups are isomorphic, then they are identical in every grouptheoretic characteristic. They have the same number of elements, the
same number of solutions to a3 = 1, the same number of elements
equal to their own inverse, etc. From the point of view of group
theory, they are indistinguishable.
0.6.2
Subgroups
1. Let G be a group. A subset H ⊂ G is a subgroup iff H is not empty and
a, b ∈ H =⇒ ab−1 ∈ H.
(1)
(a) Equivalently H is not empty and
a, b ∈ H =⇒ ab ∈ H
a ∈ H =⇒ a−1 ∈ H.
(2)
(3)
i. If (2) and (3) hold and a, b ∈ H then b−1 ∈ H so ab−1 ∈ H.
conversely, if (1) holds and a, b ∈ H then bb−1 = e ∈ H so
¡ ¢−1
eb−1 = b−1 ∈ H and a b−1
= ab ∈ H.
(b) A subgroup with the inherited operation is a group.
2. Examples:
(a) trivial subgroups:{1} and the group itself..
(b) 2Z ⊂ Z
(c) nZ ⊂ Z
(d) R>0 ⊂ R×
(e) Z ⊂ Q ⊂ R ⊂ C
(f)
orthogonal n × n matrices
⊂ invertible real n × n matrices with determinant ± 1
0.6. GROUP THEORY
xvii
(g)
SL (n) = invertible real n × n matrices with determinant 1
⊂ invertible real n × n matrices with positive determinant
⊂ invertible real n × n matrices = GL (n) .
(h) even permutations (the alternating group) = An ⊂ Sn
(i) GL (V ) ⊂ SV
3. Let ϕ : G −→ H be a homomorphism of groups.
(a) ker ϕ = {a ∈ G : ϕ (a) = 1}
(b) The kernel of ϕ, is a subgroup of G.
(c) Im ϕ is a subgroup of H.
(d) Class prove.
4. Let S ⊂ G be a subset of a group. Then the subgroup generated by S is
hSi = {sn1 1 · · · snt t : si ∈ S, ni = ±1}. By convention hφi = {1}.
(a) a finitely generated subgroup is a subgroup that can be generated
by a finite set.
(b) every subgroup generates itself
(c) hSi is the smallest subgroup of G containing S.
i. That justifies hφi = {1}
5. A cyclic subgroup hbi ⊂ G is the subgroup generated by a single element
b ∈ G.
(a) For example h1i = Z ⊂ R.
(b) A group is cyclic if it is a cyclic subgroup of itself. E.g. Z.
(c) If G is cyclic and H is a subgroup of G then H is cyclic.
i. Proof: Let G = hai. Either H = {1} (a cyclic subgroup) or for
some n 6= 0, an ∈ H. Also (an )−1 = a−n ∈ H so we can assume
n > 0. Let m = min {n > 0 : an ∈ H}. I claim H = ham i.
Since am ∈ H, ham i ⊂ H. Conversely, Let an ∈ H and write
n = am + d where 0 ≤ d < m. (division algorithm–a and d
exist uniquely). Then an (am )−a = ad ∈ H. By the construction
of m, d = 0 and an = (am )a . so H ⊂ ham i.
6. The intersection of a collection of subgroups of a group is a subgroup.
It is the largest subgroup contained in all the components.
(a) Class prove.
xviii
LECTURES
7. The product of two subgroups A, B ⊂ G is AB = {ab : a ∈ A, b ∈ B}
(a) This is not a subgroup in general
(b) The product is associative: A (BC) = (AB) C for A, B, C subgroups
of G.
i. Also A2 = A for any subgroup A of G.
(c) If G is abelian the product becomes the sum A+B = {a + b : a ∈ A, b ∈ B}.
This is a subgroup.
(d) The product AB generates a subgroup, the same as generated by the
union A ∪ B, and is called the join of A and B. The join can be
denoted hABi.
©
ª
8. Centralizer subgroup: if A ⊂ G is a subset, then CG (A) = b ∈ G : bab−1 = a, all a ∈ A
(a)
CG (A) = ∩a∈A CG (a)
(b) CG (1) = G
(c) If A ⊂ B then CG (B) ⊂ CG (A).
(d) The center of G is Z (G) = CG (G).
(e) G is abelian iff Z (G) = G
(f) class prove centralizer is a subgroup..
©
ª
9. The normalizer of a subgroup H ⊂ G is NG (H) = a ∈ G : aHa−1 = H .
(a) Note this does not require aha−1 = h all h ∈ H.
(b) NG ({1}) = G, NG (G) = G
(c) Class prove this is a subgroup.
0.6.3
Normal Subgroups and Quotient Groups
1. We need to explore two ideas, normal subgroups and cosets, then pull
them together.
2. Class prove: if G is a group and K ⊂ G a subgroup and a ∈ G then
aKa−1 is a subgroup of G.
3. the set of such subgroups where a varies over the elements of G are called
the subgroups conjugate to K, including K itself.
(a)
i. It is possible for aKa−1 to be a proper subgroup of K.
4. If aKa−1 ⊂ K for all a ∈ G, then we say K is a normal subgroup.
(a) In that case we have aKa−1 = a−1 Ka = K for all a ∈ G.
0.6. GROUP THEORY
xix
(b) In other words, NG (K) = G.
(c) If K or K is a normal subgroup of G then hKKi = KK
5. If G is a group and K ⊂ G is a subgroup and a ∈ G then aK is a left
coset of K and Ka is a right coset of K.
(a) aK = bK are the same left coset iff a−1 b ∈ K
(b) Ka = Kb are the same right coset iff ab−1 ∈ K
(c) two left cosets are disjoint or identical, and similarly for right cosets.
i. The left coset containing a is aK
ii. The right coset containing a is Ka
(d) If K is finite then all left and right cosets of K have the same number
of elements as K.
(e) G is the (disjoint) union of all left cosets of K and the (disjoint) union
of all right cosets of K. That is, every element of G is in exactly one
left coset and exactly one right coset.of K.
(f) Lagrange’s Theorem: if G is finite then |K| divides |G|.
(g) Cosets partition a group into disjoint subsets
6. Now here’s the connection. A subgroup K of a group G is normal if and
only if the partition of G into left cosets is the same as the partition into
right cosets.
(a) Here’s a much better notation. If K is normal let ā = aK = Ka.
Read this as a mod K.
i. The symbol ā omits mention of K, and it only makes sense when
a normal subgroup K is specified.
¡
¢
(b) Proof : if K is normal then aK = a a−1 Ka = Ka. Conversely, if
aK is a right coset, then it must be the right coset containing a, so
aK = Ka or aKa−1 = K.
7. Assume K is a normal subgroup of a group G. Then:
(a) a = 1̄ iff a ∈ K . Class prove.
(b) (aK) (bK) = abK
i. Proof:
(aK) (bK) = abb−1 KbK
= abKK
= abK
This proof was done entirely without explicit reference to elements of the cosets.
xx
LECTURES
(c) In other words, āb̄ = ab
8. If G is a group and K ⊂ G is a normal subgroup then the cosets of K form
G
a group denoted
or G if you define ā ∗ b̄ = āb̄, which by the theorem
K
above is the coset ab.
(a) Proof: Since āb̄ is a coset, ∗ is a binary operation. Associativity,
identity and inverses are all inherited from G. The identity of G is
1̄, and the inverse of ā is (a−1 ).
(b) Ḡ is called the quotient group of G modulo K.
(c) The elements of G are {ā : a ∈ G}, BUT THESE SYMBOLS ARE
NOT ALL DIFFERENT.
(d) 1̄ is the identity in G. The inverse of ā, ā−1 = a−1
9. There is a canonical surjective homomorphism π : G −→ G given by
π (a) = ā. The kernel is K.
(a) Class prove.
10. Suppose, on the contrary, that we have a group H and a surjective homomorphism ρ : G −→ H with kernel K. Then there is a unique isomorphism
ϕ : G −→ H such that the diagram
G
G
.π
ϕ
−→
ρ&
H
commutes.
(a) Proof: Since π is surjective, the only possible definition of ϕ is:
ϕ (ā) = ρ (a). Thus ϕ, if it exists at all, is unique. To check ϕ is
well defined, suppose ā = b̄. Then a = bk for¡ some
k ∈ H. Thus
¢
ϕ (ā) = ρ (a) = ρ (bk) = ρ (b) ρ (k) = ρ (b) = ϕ b̄ , because ρ (k) = 1
since ker ρ = K.
Next we check that ϕ is a homomorphism.
¡ ¢
ϕ (ā) ϕ b̄ = ρ (a) ρ (b)
= ρ (ab)
¡ ¢
= ϕ ab
¡ ¢
= ϕ āb̄
To see that ϕ is 1 − 1, it suffices to assume ϕ (ā) = 1H and prove
ā = 1̄. But
ϕ (ā)
= 1H =⇒ ρ (a) = 1H
=⇒ a ∈ ker ρ = K
=⇒ ā = 1̄
0.6. GROUP THEORY
xxi
To see that ϕ is onto, let b ∈ H. Since ρ is onto, there exists a ∈ G
such that ρ (a) = b. But then ϕ (ā) = b.
(b) When working with the quotient group G all you need assume about
G is that is is the image of a homomorphism with domain G and
kernel K.
11. If G is an abelian group with operation +, then the cosets of a subgroup
K are denoted a + K.
G
, the operation in the
K
quotient group is +, and the operation satisfies ā + b̄ = a + b
(a) The quotient group is still denoted Ḡ =
0.6.4
Homomorphism Theorems
1. If ϕ : G −→ H is a homomorphism of groups then
(a) If K ⊂ G is a subgroup then ϕ (K) is a subgroup of H
i. If K is normal and ϕ is surjective then ϕ (K) is normal
(b) Proof: ϕ (K) is not empty since K is not empty. If c, d ∈ ϕ (K) we
must show cd−1 ∈ ϕ (K). But c = ϕ (a) and d¡ = ϕ
¢ (b) for
¡ some
¢
−1
a, b ∈ K. Thus cd−1 = ϕ (a) ϕ (b)
= ϕ (a) ϕ b−1 = ϕ ab−1 .
Since K is a subgroup, ab−1 ∈ K and cd−1 ∈ ϕ (K). Thus ϕ (K) is
a subgroup.
i. if c ∈ ϕ (K) and d ∈ H we must show dcd−1 ∈ ϕ (K). But c =
ϕ (a) for some a ∈ K, and–because
¡ ϕ is ¢surjective–d = ϕ (b)
for some b ∈ G. Thus dcd−1 = ϕ bab−1 . Since K is normal
in G, bab−1 ∈ K and therefore dcd−1 ∈ ϕ (K). Thus ϕ (K) is
normal.
(c) If K ⊂ H is a subgroup then ϕ−1 (K) is a subgroup of G
i. If K is normal then ϕ−1 (K) is normal.
(d) ϕ−1 (K)¡is not¢empty because 1G ∈ ϕ−1 (K). Suppose a, b ∈ ¡ϕ−1 (K).
¢
Then ϕ ab−1 = ϕ (a) ϕ (b)−1 . Since ϕ (a) , ϕ (b) ∈ K, ϕ ab−1 ∈
K. Thus ab−1 ∈ ϕ−1 (K) and K is a subgroup.
¡
¢
−1
i. If a ∈ ϕ−1 (K) and b ∈ G then ϕ bab−1 = ϕ (b) ϕ (a)
¡ ϕ (b)
¢ .
−1
Since ϕ (a) ∈ K and ϕ (b) ∈ H and K is normal in H, ϕ bab
∈
K. Thus bab−1 ∈ ϕ−1 (K) so ϕ−1 (K) is normal.
2. If ϕ : G −→ H is a homomorphism of groups with kernel K, then
(a) ϕ is injective iff ker ϕ = {1}. (Proven above)
(b) In all cases ϕ induces a 1 − 1 homomorphism ϕ̄ :
G
−→ H
K
xxii
LECTURES
i. First Homomorphism Theorem: image ϕ = image ϕ̄ so ϕ̄ :
G
−→ image ϕ is bijective
K
ii. We can diagram this result:
ϕ
G
−→
image ϕ → H
↓ ∼
=% ϕ̄
G
K
3. Second Homomorphism Theorem–simplified version for abelian groups:
If B and C are subgroups of an abelian group A then the canonical map:
B
B+C
−→
B∩C
C
is bijective. To construct the map, note first that we have the composiB+C
tion B → B + C −→
. The kernel of this map is B ∩ C., so by
C
the first homomorphism theorem, the induced map is injective. Showing
surjectivity is not hard.
4. Third Homomorphism Theorem: Let H ⊂ K ⊂ G be normal subgroups of a group. Then H is a normal subgroup of K and we can form
K
G
G G K
,
,
. Moreover,
⊂
is a normal subthree quotient groups
H K H
H
H
G/H
group, so we can form
. Then the composition
K/H
G −→
G
G/H
−→
H
K/H
sends K to the identiy and so induces a map:
G/H
G
−→
K
K/H
which is bijective. A diagram might help:
.
K
↓
K
H
−→
−→
H
↓
G
↓
G
H
−→
−→
G
K
↓∼
=
G/H
K/H
5. Let H be a normal subgroup of a group G. There is a 1 − 1 inclusionpreserving correspondence between subgroups of G that contain H and the
G
subgroups of G/H given by K ↔ π (K) where π : G −→
. Moreover
H
normal subgroups on one side correspond to normal subgroups on the
other.
0.7. RING THEORY
0.7
0.7.1
xxiii
Ring Theory
Basic Definitions and Examples
1. Examples of sets of things that have addition and multiplication
(a) Z, 2Z, Q, R, C
(b) Z/nZ, integers modulo n
(c) Mn (R) , the n × n matrices with coefficients in R, Also Mn (Z),
Mn (2Z), Mn (Q), Mn (C)
i. one set of matrices for each different n
(d) R [X], polynomials with coefficients in R. Also Z [X], Q [X], C [X] , R [X, Y ]
(e) If I = [0, 1] then real-valued functions on I that are continuous, ntimes differentiable, infinitely differentiable, “analytic” (representable
by power series)
2. Key differences:
(a) all but matrices have commutative multiplication
(b) all but 2Z have a multiplicative identity.
(c) In some, every non-zero thing has an multiplicative inverse
i. class: which ones
ii. In others some of the elements are invertible, others not.
A. In Z only ±1 are invertible
B. In Z/n, ā is invertible iff gcd (a, n) = 1. If n is prime then
all non-zero elements of Z/n are invertible.
C. some matrices are invertible, those with non-zero determinant
D. non-zero, constant polynomials are invertible
E. in the functions, f is invertible if f (x) 6= 0 for all x.
(d) some have zero-divisors: non-zero elements whose product (one way
at least) is zero
i. which examples do not have zero divisors? All but Z/n for composite n, matrices and functions
ii. If n = ab, a 6= 1, b 6= 1, then ā 6= 0̄, b̄ 6= 0̄ but āb̄ = 0̄ in Z/n.
·
¸
·
¸
·
¸
·
¸
0 1
0 0
0 0
0 0
iii. a =
,b=
, ab =
, ba =
0 0
1 0
0 0
0 1
iv. Continuous functions that are zero divisors:
½
0 if x ≤ 1/2
f (x) =
x − 1/2 otherwise
½
0 if x ≥ 1/2
g (x) =
1 − 2x otherwise
then f g = 0.
xxiv
LECTURES
A. It is possible to come up with infinitely differentiable zero
divisors, but not analytic ones. That’s the difference between
“infinitely differentiable” and “representable by power series”
3. A ring (R, +, ∗) is a set R with two associative operations. + and ∗
(a) (R, +) is an abelian group
(b) for all a, b, c ∈ R, a ∗ (b + c) = (a ∗ b) + (a ∗ c) and (a + b) ∗ c =
(a ∗ c) + (b ∗ c)
4. All the examples above are rings
5. Simple theorems that hold in all rings, Class prove:
(a) 0 ∗ a = 0
(b) (−a) ∗ b = a ∗ (−b) = − (a ∗ b)
(c) (−a) ∗ (−b) = a ∗ b
6. If ∗ is commutative, we say that the ring is a commutative ring. This
is all that our text deals with.
7. If there is an element 1 ∈ R such that 1 ∗ a = a ∗ 1 = a for all a ∈ R then
we say that R is a ring with identity
(a) Class: prove identity unique
(b) −a = (−1) ∗ a
(c) In our text, all rings have an identity.
8. From here on out, "ring" = "commutative ring with identity".
9. Sometimes the condition 1 6= 0 is added to the definition of a ring with
identity. The only ring with 1 = 0 is the zero ring, consisting of a single
element usually denoted 0.Our text does not do this.
10. Usually ∗ is dropped in favor of adjacency. Write ab not a ∗ b.
(a) To be even more conventional, “multiplication” usually precedes “addition”
i. a + bc means a + (b ∗ c) instead of (a + b) ∗ c,
ii. Compare to a − b − c, which does mean (a − b) − c instead of
a − (b − c). Operations are usually performed left-to-right unless
some other convention intervenes.
iii. exponentiation preceding multiplication preceding addition is one
such convention.
0.7. RING THEORY
xxv
iv. a lesser known convention is that exponentiation is performed
right-to-left. Compare
a − b − c = (a − b) − c
c
c
ab = a(b )
11. An element u in a ring with identity is a unit iff there exists another
element v such that uv = 1.
(a) Of course, then v is a unit also. Units come in pairs.
(b) If uv = wu = 1 then v = w, so inverses of units are unique. We can
denote v = u−1
i. v = 1v = (wu) v = w (uv) = w1 = w
(c) In any ring with identity R, the units with multiplication form an
abelian group R× or R∗
(d) Examples: R× = {a 6= 0}, Z× = {−1, 1}
12. If all the non-zero elements in a ring with identity are units, we say that
R is a field.
(a) i.e. the ring has a multiplicative identity 1 and for every a 6= 0 there
is a b such ab = ba = 1.
(b) Examples include R, C and Q
(c) Z/p where p is prime.
£√ ¤
√
(d) Q 2 , Q [ n], lots of other "number fields" very important for
algebraic number theory
13. In a field 1 6= 0 because if 1 = 0 then we have the zero ring and the
non-zero elements form an empty set not a group!
14. A non-zero element a in a ring is called a zero divisor iff there exists a
non-zero element b such that ab = 0 .
(a) b is a zero divisor too.
(b) An element cannot simultaneously be a zero divisor and a unit if
1 6= 0
i. if a is a unit and ab = 0 then 0 = a−1 0 = a−1 ab = 1b = b so a
cannot be a left zero divisor. A similar argument shows that a
unit cannot be a right zero divisor either, so a unit cannot be a
zero divisor.
15. A commutative ring with 1 6= 0 and no zero divisors is called an integral
domain.
£√ ¤
(a) Example: any field, Z, Z 2 .
xxvi
LECTURES
16. If R is an integral domain, then there is a field F such that R ⊂ F and
f ∈ F =⇒ ∃a, b ∈ R such that f = ab−1 in F .
(a) F is unique in the sense that there is exactly one map F1 −→ F2 that
is the identity on the subring R, and it is bijective.
(b) F can be constructed as equivalence classes in the set of pairs {(a, b) ∈ R × R, b 6= 0}.
The equivalence relation is (a, b) ∼ (a0 , b0 ) iff ab0 = a0 b. The addition
rule is (a, b) + (c, d) = (ad + bc, bd), and the multiplication rule is
(a, b)(c, d) = (ac, bd). The zero is (0, 1) and the identity is (1, 1).
The inclusion of R in F is given by r −→ (r, 1).
i. Lots of checking is required.
a
ii. We usually denote (a, b) by or ab−1 .
b
£√ ¤
£√ ¤
(c) Examples: Q 2 is the quotient field of Z 2 .
17. Important example for applications: Let E be a set and R a ring. Then
the set of functions F (E, R) from E to R is a ring.
(a) addition and multiplication are defined pointwise.
i. That is, if f, g ∈ F (E, R) then f +g is the function (f + g) (e) =
f (e) + g (e) and f g is the function (f g) (e) = f (e) g (e).
(b) Check that (F (E, R) , +) is an abelian group. For f, g, h ∈ F (E, R)
and e ∈ E we have
i.
((f + g) + h) (e) =
=
=
=
=
(f + g) (e) + h (e)
(f (e) + g (e)) + h (e)
f (e) + (g (e) + h (e))
f (e) + (g + h) (e)
(f + (g + h)) (e)
Since (f + g) + h and f + (g + h) are functions that take the
same value at every point e of their common domain, they are
the same function: (f + g) + h = f + (g + h).
ii. The function z (e) = 0 for all e ∈ E is the zero in the group of
function.
(z + f ) (e) = z (e) + f (e)
= 0 + f (e)
= f (e)
Therefore z + f = f for all functions f . Similarly f + z = f .
iii. The function o (e) = 1 for all e ∈ E is the identity.
0.7. RING THEORY
xxvii
iv. Given a function f the function −f defined by (−f ) (e) = − (f (e))
is the additive inverse of f in the group of functions.
v. And multiplication of functions is also associative and commutative and distributes over addition, so F (E, R) is a ring.
0.7.2
Polynomials
1. Let R be a ring. Consider the set of sequences (r0 , r1 , r2 , . . .) with componentwise addition, a zero element (0, 0, 0, . . .), and a strange multiplication
rule:
(r0 , r1 , r2 , . . .) (s0 , s1 , s2 , . . .) = (t0 , t1 , t2 , . . .)
ti
=
i
X
rj si−j
j=0
The sequences form a ring.
(a) the only non-obvious ring properties to check are multiplicative associativity and distributivity. Let r = (r0 , r1 , r2 , . . .), s = (s0 , s1 , s2 , . . .)
and t = (t0 , t1 , t2 , . . .). To show (rs) t = r (st), we must show
((rs) t)i = (r (st))i for all i ≥ 0.
((rs) t)i
=
i
X
(rs)j ti−j
j=0
=
j
i X
X
rk sj−k ti−j
j=0 k=0
X
=
rm sn tp
m,n,p≥0
m+n+p=i
= (r (st))i
Similarly, to show r (s + t) = rs + rt,we must show (r (s + t))i =
(rs + rt)i for all i ≥ 0:
(r (s + t))i
=
i
X
rj (s + t)i−j
j=0
=
i
X
rj (si−j + ti−j )
j=0
=
i
X
rj si−j +
j=0
= (rs)i + (rt)i
= (rs + rt)i
i
X
j=0
rj ti−j
xxviii
LECTURES
(b) (1, 0, 0, . . .) is the identity of the ring of sequences.
(c) If you write the sequence (r0 , r1 , rt , . . .) as r0 + r1 X + r2 X 2 + · · · for
some variable X, you can call the ring of sequences the ring R [[X]]
of power series over R. The multiplication rule is the usual one for
power series.
2. The collection of power series with a finite number of non-zero coefficients
is a subring. The only thing to check is that the set is closed under addition
and multiplication. We call this subring R [X], the polynomials over R.
(a) A polynomial that is just a power of X, X n , is a monomial. Every
polynomial is a unique linear combination of monomials with coefficients from R.
(b) The degree of a polynomial, deg p, is the largest index of a non-zero
coefficient, or equivalently the highest power of X that appears with
a non-zero coefficient. By convention, deg 0 = −∞.
(c) We say a polynomial p is constant if deg p < 1.
3. Suppose R is an integral domain.
(a) Class prove: deg pq = deg p + deg q.
i. Degree is like a logarithm.
ii. The product of non-zero polynomials cannot be 0, so R [X] is an
integral domain.
(b) A unit in R [X] is a constant polynomial c that is a unit in R.
Z
[X], (1 + 2X) (1 − 2X) = 1. so non-constant polynomials can be
(4)
units when the ground ring is not an integral domain.
4. In
5. For any field k, the function field of the integral domain k [X] is called the
field of rational functions (in one variable) over k.
(a) These are all the expressions of the form
teach in Math 70.
p (X)
with the algebra you
q (X)
0.7. RING THEORY
xxix
6. Evaluation: f ∈ R [X] and a ∈ R then f (a) is just what you think:
f = f0 + · · · + fn X n , f (a) = f0 + · · · = fn an .
(a) f (a) + g (a) = (f + g) (a), f (a) g (a) = (f g) (a).
(b) That’s the reason polynomial multiplication is defined as it is.
0.7.3
Greatest Common Divisors
1. If f ∈ k [X], f 6= 0, let f = fn X n + · · · + f0 where fn 6= 0. Then
LT (f ) = fn X n .
LT (g)
(a) If f, g ∈ k [X] and f 6= 0 and deg (g) ≥ deg (f ). Then
∈ k [X]
LT (f )
¶
µ
LT (g)
f < deg (g).
and deg g −
LT (f )
2. Division algorithm: If k is a field and f, g ∈ k [X] with f =
6 0, then
exists unique q, r ∈ k [X] such that g = qf + r and either r = 0 or
deg (r) < deg (f ).
(a) this proof is more algorithmic than the proof in the text. Let r0 = g,
LT (ri )
So long as ri 6= 0 and deg (ri ) ≥ deg (f ) define ri+1 = ri −
f.
LT (f )
Since deg (ri+1 ) < deg (ri ), this process must
µ eventually stop with
¶
LT (r0 ) LT (r1 )
LT (rt−1 )
rt = 0 or deg (rt ) < deg (f ). We have g = f
+
+ ··· +
+
LT (f )
LT (f )
LT (f )
rt .
To show that q, r are unique, suppose f q +r = f q 0 +r0 where r = 0 or
deg (r) < deg (f ) and r0 = 0 or deg (r0 ) < deg (f ). Then f (q − q 0 ) =
r0 − r0 . If q 6= q 0 then deg (f (q − q 0 )) ≥ deg f > deg (r − r0 ). Thus
q = q 0 . But then r − r0 = 0 so r = r0 .
(b) The process that constructs q and r is just long division of polynomials.
(c) In particular, for any a ∈ k we have g = (X − a) q + r where r is a
constant polynomial.
i. And r = g (a).
3. If R is a ring and f, g ∈ k [X] with f a non-zero monic polynomial, then
exists q, r ∈ k [X] such that g = qf + r and either r = 0 or deg (r) <
deg (f ). If R is a domain then the polynomials q and r are unique.
(a) The construction above can be performed in R [X] provided that the
leading coefficient of f is a unit in R. The uniqueness part of the
proof goes through if R is a domain
4. If k is a field and f ∈ k [x], then a ∈ k is a root of f iff f (a) = 0.
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LECTURES
(a) Since f = (X − a) q + f (a) for any a ∈ k, a is a root of f iff X − a
divides f.
(b) Let a1 , . . . , at be distinct roots of f ∈ k [X]. Then (X − a1 ) · · · (X − at )
divides f .
i. This does NOT follow immediately from the fact that all the
factors divide f individually
ii. Note that k [X] is an integral domain, and for i 6= j, (ai − aj ) 6=
0. We proceed by induction. If t = 1 then by the above X − a1
divides f . Suppose (X − a1 ) · · · (X − at−1 ) divides f , so f =
(X − a1 ) · · · (X − at−1 ) q for some q ∈ k [X]. Substituting at for
X we get 0 = (at − a1 ) · · · (at − at−1 ) q (at ). Thus q (at ) = 0 so
q = (X − at ) r for some r ∈ k [X], and f = (X − a1 ) · · · (X − at−1 ) (X − at ) r.
(c) A polynomial of degree n over a field has at most n roots.
i. Otherwise a polynomial (X − a1 ) · · · (X − at ) of degree larger
than degree f divides f .
µ
¶
2πk
th
2πik/n
(d) Corollary: the n roots of 1 in C are e
= cos
+
n
µ
¶
2πk
i sin
, k = 0..n − 1. since these are all roots of X n − 1 and
n
there cannot be any others.
(e) Corollary: if k is infinite and f, g ∈ k [X] and f (a) = g (a) for all
a ∈ k then f = g.
i. Because f − g cannot be other than 0 since it has infinite roots.
ii. This is certainly false if k is finite.
5. If k is a field and G ⊂ k∗ is a finite subgroup (under multiplication), then
G is cyclic.
(a) Very important for number theory
(b) See text for proof—requires result from group theory
0.7. RING THEORY
xxxi
6. For the next while, we are working over a field k. All polynomials have
coefficients in k.
7. Notation: we have defined LT (f ) for a polynomial f . Define LC(f ) to be
f
the leading coefficient of f . so g =
is a monic polynomial with
LC (f )
the same divisibility properties as f . That is h divides f iff h divides g,
and f divides h iff g divides h.
8. Remark: if f and g are both monic polynomials, and if f divides g and
g divides f then f = g. In fact if f and g are monic polynomials of the
same degree and f divides g then f = g.
9. The common divisor of two polynomials f, g ∈ k [X] is a polynomial
h ∈ k [X] dividing both f and g.
(a) A greatest common divisor is a monic common divisor of maximal
degree
(b) Given two non-zero polynomials, at least one greatest common divisor must exist.
(c) This common divisor is unique and is denoted gcd (f, g). Every common divisor of f and g divides gcd (f, g). Moreover there exist polynomials p, q ∈ k [X] such that
gcd (f, g) = pf + qg
i. Proof (constructive): Let r0 = f and r1 = g. Define ri+1 as the
remainder when ri−1 is divided by ri . That is, ri=1 = ri hi +
ri+1 and if ri+1 6= 0 then deg (ri+1 ) < deg (ri ). Thus for some i,
ri+1 = 0. Moreover, for all i, ri = pi f + qi g for some pi , qi . This
follows by induction:
r0
ri+1
=
=
=
=
1f + 0g
ri−1 − hi ri
(pi−1 f + qi−1 g) − hi (pi f + qi g)
(pi−1 − hi pi ) f + (qi−1 − hi qi ) g
ri
=
LC(ri )
gcd (f, g). I claim that ri divides rj for 0 ≤ j ≤ i. The result holds for j = i, so we proceed by backwards induction:
If ri 6= 0 and ri+1 = 0 then we will show that
rj = hj+1 rj+1 + rj+2
If ri divides rj+1 and divides rj+2 then ri divides rj . Thus ri
divides r1 = g and r0 = f ., so ri is a common divisor of f and g.
Since ri = pf +qg, any common divisor of f and g also divides ri .
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LECTURES
ri
is a greatest comon divisor of f and g. Moreover
LC (ri )
(as we have just shown) every common divisor of f and g divides
ri
ri
. It remains to show that
is the only greatest
LC (ri )
LC (ri )
common divisor of f and g.
If h is a greatest common divisor of f and g, then h is monic
ri
. But we have showed above that
of the same degree as
LC (ri )
ri
every common divisor of f and g divides ri , so h divides
.
LC (ri )
ri
Thus h =
and the greatest common divisor of f and g
LC (ri )
is unique.
Thus
10. Factorization of polynomials
(a) A polynomial f ∈ k [X] is irreducible if f 6= 0, deg (f ) > 0, and
f cannot be factored into two factors of positive degree. I.e. f is
irreducible iff f cannot be factored into a product of polynomials
both having degree smaller than f .
(b) More generally, in an integral domain, an element p is irreducible if
p 6= 0, p is not a unit, and whenever p = uv, either u or v is a unit.
(c) Euclid’s Lemma: if p is irreducible and p divides f g then p divides f
or p divides g.
i. Proof: Let r = gcd (p, f ). Since r divides p either r = p and p
divides f or r = 1 and ap + bf = 1 for some polynomials a and
b. Thus g = apg + bf g, which is divisible by p.
(d) A polynomial f ∈ k [X] is prime if whenever f divides gh then f
divides g or f divides h.
i. The result above shows that an irreducible polynomial is prime.
The converse is also true. Suppose f is prime and f = gh. Since
f divides gh, either f divides g or f divides h. Thus g and h
cannot both have smaller degree than f .
(e) Every monic polynomial factors into a product of prime monic polynomials. Moreover the factorization is unique up to order of factors.
i. Thus every polynomial factors into a product of prime polynomials which are unique up to constant factors.
ii. Proof (this is not contructive—I don’t know of a constructive
proof): The existence of a prime factorization is easy. Either
f is prime or it has monic factors f = gh of smaller but positive
degree. Since you cannot keep factoring the factors forever, you
eventually get a prime factorization.
To show that the factorization is unique, suppose you had two
factorizations: p1 · · · pr = q1 · · · qt . Since p1 divides q1 · · · qt , p1
0.7. RING THEORY
xxxiii
must divide one of the factors qi . But the factors are irreducible,
so p1 = qi for some i. Remove the equal factors from both sides
and repeat. The last factor has to be removed from both sides
simultaneously, showing that both sides have the same factors.
(f) gcd (f, g) is the product of the common prime factors of f and g.
i. Given two polynomials f and g, f = p gcd (f, g) and g = q gcd (f, g),
where gcd (p, q) = 1.
f
and be written with numerator and deg
nominator having no common factors: gcd (f, g) = 1. This is called
"lowest terms".
(g) Every rational function
xxxiv
0.7.4
LECTURES
Homomorphisms and Ideals
Delete from homework 3.57, 3.61 and 3.62. We will skip section on Euclidean
rings
1. If R and S are rings and ϕ : R −→ S is a function, then ϕ is a homomorphism iff ϕ (a + b) = ϕ (a)+ϕ (b) and ϕ (ab) = ϕ (a) ϕ (b) and ϕ (1R ) = 1S .
(a) Examples:
i. ϕ : Z −→ Q
Z
ii. ϕ : Z −→
(n)
¡ ¢
iii. ϕ : R [X] −→ R [X], ϕ (f (X)) = f X 2
iv. Given any ring R and an element r ∈ R there is a homomorphism
R [X] −→ R given by f −→ f (r)
v. Given a homomorphism R −→ S there is induced a homomorphism R [X] −→ S [X]
¢
¡
(b) If u is a unit of R then ϕ (u) is a unit of S; moreover ϕ u−1 =
ϕ (u)−1 .
i. because if uv = 1 then ϕ (uv) = ϕ (u) ϕ (v) = 1
2. The kernal of a homomorphism is the set ker (ϕ) = {r ∈ R : ϕ (r) = 0}
(a) The image of a homomorphism is the set image (ϕ) = ϕ (R).
(b) The image is a subring of S. It contains 0 and 1 and is closed under
addition and multiplication, and addition and multiplication in S
inherit all the necessary properties like associativity from R.
3. Definition: an ideal of a ring R is a subset I ⊂ R such that
I =
6
φ
a, b ∈ I =⇒ a + b ∈ I
a ∈ I, b ∈ R =⇒ ab ∈ I
(a) That is, I is a subgroup satisfying the third property.
(b) Ideals are not in general subrings, because we do not require 1 ∈ I.
i. In fact, if 1 ∈ I then I = R
(c) Examples
i.
ii.
iii.
iv.
2Z ∈ Z
nZ ∈ Z
polynomials without constant term in R [X]
Fix a ∈ k. The set {f ∈ R [X] : f (a) = 0} is an ideal of R [X].
0.7. RING THEORY
xxxv
4. Let R be a ring and E ⊂ R. Then (E) = {r1 e1 + · · · + rt et : ri ∈ R, ei ∈ E}
is an ideal of R
(a) By convention, (φ) = {0}. We have E ⊂ (E) so (E) is never empty.
If a, b ∈ (E) let a = r1 e1 + · · · + rs es and b = rs+1 es+1 + · · · + rt et .
Then a + b = r1 e1 + · · · + rt et ∈ (E). If c ∈ R then ca = (cr1 ) e1 +
· · · + (crs ) es ∈ (E).
5. If I is an ideal of R and if there exists a finite set E such that I = (E),
then we say that I is a finitely generated ideal.
(a) If there exists a single element r ∈ R and I = (r) we say that I is a
principal ideal.
(b) e.g. 2Z ⊂ Z
6. A Principal Ideal Domain (PID) is a domain where every ideal is
principal
7. Z is a PID
8. Let k be a field. Then every ideal in k [X] is principal.
(a) Proof: let I be an ideal of k [X]. If I = {0} then I = (0) is principal.
Otherwise choose a non-zero element p of minimal degree in I.
First we show I ⊂ (p). If f ∈ I let f = pq + r where r = 0 or
deg (r) < deg (p). But r = f − pq ∈ I, so r cannot be non-zero. Thus
r = 0 and f ∈ (p).
Conversely, we show (p) ⊂ I. But that’s obvious. For any element
f ∈ I, (f ) ⊂ I.
(b) To prove that Z is a PID, use the proof above, but replace deg (p) by
|p|.
9. Let ϕ : R −→ S be a homomorphism of rings. Then ker ϕ is an ideal of R.
(a) Proof: 0 ∈ ker ϕ. If a, b ∈ ker ϕ then ϕ (a + b) = ϕ (a) + ϕ (b) =
0 + 0 = 0, so a + b ∈ ker ϕ. Also if a ∈ ker ϕ and b ∈ R then
ϕ (ab) = ϕ (a) ϕ (b) = 0ϕ (b) = 0 so ab ∈ ker ϕ. Thus ker ϕ is an
ideal.
(b) image (ϕ) is a subring.
(c) Later we will prove that I ⊂ R is an ideal iff there is a homomorphism
with domain R and kernal I.
10. Let R be a ring. If a, b ∈ R then a common divisor of a and b is
any element c ∈ R such that c | a and c | b. (Equivalently a, b ∈ (c),
or (a, b) ⊂ (c).) A greatest common divisor of a and b is a common
divisor that is divisible by all other common divisors. In terms of ideals, a
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LECTURES
greatest common divisor is (a generator of) a principal ideal containing a
and b and contained in all other principal ideals containing a and b. That
is g is a greatest common divisor of a and b iff (a, b) ∈ (g) and (a, b) ⊂ (c)
implies (g) ⊂ (c).
(a) Two elements always have a common divisor, since 1 works, but there
need not be a greatest common divisor (I think).
(b) In a PID, two elements always have a greatest common divisor,
namely any generator for the ideal (a, b).
0.7.5
Euclidean Domains
1. We won’t study this section, but I want to review the main ideas. Rotman
says "Euclidean Ring", but he always means a domain.
2. A Euclidean domain is an integral domain R together with a function
∂ : R∗ −→ Z (called the degree function) such that
(a) ∂ (f g) ≥ ∂ (f ), all f, g ∈ R∗
(b) for all f ∈ R∗ , g ∈ R there exist q, r ∈ R such g = qf + r and either
∂ (r) < ∂ (f ) or r = 0.
i. q, r are not necessarily unique—at least Rotman doesn’t prove
that they are, and I cannot.
3. The main examples are Z with ∂ =absolute value and k [X] for k a field
and ∂ =degree.
4. Every Euclidean domain is a PID (but not conversely)
(a) Proof: Every non-zero ideal I is generated by a non-zero element of
least degree. The proof is exactly the same as for polynomials
0.7. RING THEORY
0.7.6
xxxvii
Linear Algebra
1. A vector space over a field k (in case you’ve forgotten) is an abelian
group V together with a “product” k × V −→ V satisfying
(a) a (u + v) = au + av
(b) (a + b) v =av + bv
(c) (ab) v =a (bv)
(d) 1v = v
2. Traditionally the elements of V are called vectors and the elements of k
are called scalars.
3. Example: k n
(a) (ai ) + (bi ) = (ai + bi )
(b) a (bi ) = (abi )
4. Example k [X]
(a) or any ideal in k [X]
5. C is a vector space over R, and over Q
6. A subspace of a vector space V is a subgroup W ⊂ V such that kW ⊂ W
(a) Examples are lines and planes through the origin in R2 and R3
(b) Another example is an ideal in k [X] over k
7. Subspaces are vector spaces
8. A linear combination of vectors from V is an expression of the form
a1 v1 + · · · + at vt where ai ∈ k and vi ∈ V . The empty linear combination
is said to have the value 0 ∈V .
9. Subspaces usually arise in one of four ways. Let V be a vector space over
a field k
(a) Let E ⊂ V be a subset of a vector space. The subspace hEi =
{a1 e1 + · · · + at et : ai ∈ k, ei ∈ E} =all linear combinations of vectors from E.
i. We say that E generates hEi
ii. If E ⊂ F ⊂ V then hEi ⊂ hF i ⊂ hV i
iii. A set of vectors E is said to span V if hEi = V .
A. If E ⊂ F ⊂ V and E spans V then F spans V .
iv. The set of all linear combinations of v1 , . . . , vt is hv1 , . . . , vt i.
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LECTURES
(b) The kernal of a linear transformation. See next section.
(c) If U and W are subspaces of V then so is U ∩ W .
i. The intersection of any collection of subspaces is also a subspace.
(d) If U and W are subspaces of V , then so is U +W = {u + w : u ∈ U, w ∈ W } =
hU ∪ W i.
i. The sum of any collection of subspaces is defined to be the subspace generated by their union.
10. A subset E of a vector space V is linearly independent if no non-empty
linear combination of distinct vectors with non-zero coefficients is 0.
(a) That is, if e1 , . . . , et are distinct elements of E, and if a1 e1 + · · · +
at et = 0, then a1 = · · · = at = 0.
(b) Example: two non-colinear vectors in R2 , three non-colinear vectors
in R3 .
(c) The empty set is linearly independent.
(d) A set of vectors that is not linearly independent is said to be linearly
dependent.
(e) If F ⊂ E ⊂ V and E is linearly independent, then F is linearly
independent.
11. Let u1 , . . . , un ∈ V and v1 , . . . , vm ∈ hu1 , . . . , un i.
v1 , . . . , vm is a linearly dependent list of vectors.
If m > n then
(a) The book gives two proofs of this most important theorem; here’s a
third using matrix algebra. The notation is useful sometimes.
(b) Let vi = ai1 u1 + · · · + ain un . That is
 



a11 · · · a1n
u1
v1
 ..   ..
..   .. 
..
 . = .
.
.  . 
vm
am1
···
amn
un
or v =Au. There exists an invertible m × m matrix U such that U A
is row reduced. Since m > n, the bottom row of U A is all zeros. Let


u11 · · · u1m

.. 
..
U =  ...
.
. 
um1
···
umm
Since U v =U Au, the last entry in this product is


v1
£
¤

um1 · · · umm  ...  = um1 v1 + · · · + umm vm
vm
= 0
0.7. RING THEORY
xxxix
But the last row of U is not all zeros, since U is invertible. Thus
v1 , . . . , vm is linearly dependent.
12. A basis of V is a linearly independent spanning set for V .
©
ª
(a) Example: 1, X, X 2 , . . . ⊂ k [X] over k.
(b) Example: {(1, 0, . . . , 0) , (0, 1, 0, . . . , 0) , . . . , (0, . . . , 0, 1)} is a basis of
kn over k.
(c) This is not a basis of k [[X]]. A basis of k [[X]] probably cannot be
written down, just like a basis for R over Q (a Hamel basis) cannot
be written down.
(d) The empty set is the basis for the zero vector space.
13. The Axiom of Choice implies that every vector space has a basis.
(a) The result is obvious if you add the assumption (as our author does)
that the vector space has a finite spanning set.
14. If a vector space V has a finite basis B, then all bases have the same size.
(a) We will prove that if V has a finite basis then all bases are finite of
the same size. Let B be a finite basis of V , and let C be any other
basis. Since B spans V and C is linearly independent, any finite
subset of C is linearly independent. By the lemma above, no finite
subset of C can be larger than B, so |C| ≤ |B|. Conversely, C spans
V and B is linearly independent, so |B| ≤ |C|.
15. This size of B is the dimension of the vector space
(a) The dimension of kn over k is n.
16. If V has a countable basis then all bases are countable and we say that V
has countable dimension.
(a) k [X] has countable dimension over k
17. If V does not have a finite or countable basis, then the dimension of V is
uncountable.
(a) R has uncountable dimension over Q
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LECTURES
18. One more theorem about dimension. Suppose B is a subset of a finite
dimensional vector space V . If any two of the following statements is
true, so is the third:
(a) B is linearly independent in V
(b) B spans V
(c) |B| = dim V
These statements are true if and only if B is a basis of V .
0.7.7
Linear Transformations
1. Let V and W be vector spaces over a field k. A linear transformation
T : V −→ W is function satisfying
(a) T (x + y) = T (x) + T (y)
(b) T (ax) = aT (x)
2. Examples:
(a) I : V −→ V
(b) 0 : V −→ W
(c) left multiplication by a m × n matrix with coefficients from k is a
linear transformation k n −→ k m
i. Later we prove that this is the only kind of linear map kn −→ k m .
(d) differentiation is a linear transformation k [X] −→ k [X]
i. Any differential operator a0 + a1 D + · · · + at Dt : k [X] −→ k [X],
ai ∈ k, is a linear transformation.
3. The set of linear transformations from V to W : L (V, W ), is a vector space
over k
(a) addition is defined (S + T ) (v) = S (v) + T (v)
(b) scalar multiplication is defined (aT ) (v) = a (T (v))
4. Composition of linear transformations is a linear transformation.
(a) The set of linear transformations L (V, V ) = L (V ) is a k-vector space
and a ring, or a k-algebra.
5. If T : V −→ W is bijective, then the inverse map T −1 : W −→ V is also
linear.
0.7. RING THEORY
xli
(a) The group of units (invertible transformations) in L (V ) is denoted
GL (V )
6. Let V and W be vector spaces, and suppose B is a basis of V and C
is a subset of W . Moreover, suppose there is a bijective map of sets
ϕ : B −→ C. Then ϕ extends to a unique linear map ϕ̄ : V −→ W .
(a) Proof: the only possible extension is ϕ̄ (a1 b1 + · · · + at bt ) = a1 ϕ (b1 )+
· · · + at ϕ (bt ). This map is well-defined because B is a basis, and it
is clearly linear.
(b) If C is also a basis then ϕ̄ is bijective.
(c) If two maps agree on a basis, they are the same.
7. A most important linear transformation is a basis transformation (my
terminology).
(a) Let B = {b1 , . . . , bn } be a basis of a finite-dimensional vector space
V over k. The map B : kn −→ V given by:


a1


B  ...  = a1 b1 + · · · + an bn
an
is a bijective linear transformation. B −1 (v) gives the coordinates
of v with respect to the basis B.
(b) There is a correspondence between the bases of V and isomorphisms
kn −→ V .
8. If T : kn −→ k m is linear, then there is a unique m × n matrix M such
that T (x) = M x.


0
 .. 
 . 




(a) Let coli M = T 
 1i .
 . 
 .. 
0
9. Given a vector v and a basis B in a finite-dimensional vector space V , we
know what the coordinates of v with respect to B are. We can also find
the matrix of a linear transformation T : V −→ W with respect to bases
B of V and C of W :
(a) Using the basis transformations we get a commutative diagram:
V
↓B
kn
T
−→
M
−→
W
↓C
km
xlii
LECTURES
The given maps B, T, C give a map CT B −1 : kn −→ km , which is
multiplication by a unique matrix M . We say that M is the matrix
of the map T with respect to the bases B and C.
(b) Sometimes we say M =C [T ]B .
(c) The map
C
[]B : L (V, W ) −→ (m × n matrices) is linear
(d) If we have vector spaces U, V, W with bases B, C, D and linear maps
S
T
U −→ V −→ W , then
D
[T ]C
C
[S]B =D [T S]B
i. Corollary: matrix multiplication is associative.
ii. So if B is a basis of a vector space V then the map B []B :
L (V ) −→ (n × n matrices) is a bijective homomorphism of rings.
10. The kernal or null-space of a linear transformation is a subspace. So is
the image.
11. If T : V −→ W , then linearly dependent subsets of V map to linearly
dependent subsets of W .
(a) If the image of a set is linearly independent, then the set is linearly
independent.
(b) If T : V −→ W is 1 − 1, then linearly independent sets in V map to
linearly independent sets in W .
12. Let V be a finite dimensional vector space, and suppose T : V −→ V is
linear. Then T is 1 − 1 iff T is onto.
(a) In this case T is bijective.
(b) Proof: as we saw above, a linear map is invertible iff it maps a basis to
a basis. If T is 1 − 1, T maps a basis to a linearly independent set of
size dim V . The image of the basis is therefore a basis. Conversely, if
T is onto, Let v1 , . . . , vn be a basis of V , and choose ui ∈ V such that
T (ui ) = vi . Since {T (ui )} is linearly independent, {ui } is linearly
independent and therefore a basis. Since T maps a basis to a basis,
it is bijective and therefore 1 − 1.
0.7. RING THEORY
0.7.8
xliii
Quotient Rings and Fields
1. Let R be a ring and I and ideal. Then R is an abelian group and I is a
R
subgroup, so we can form the group , which is presented as the image
I
of a homomorphism with domain R and kernal I:
π : R −→
R
I
R
R
as āb̄ = ab we have that
is a ring
(a) If we define multiplication in
I
I
and π is a homomorphism of rings.
i. We must check that if π (a) = π (a0 ) and π (b) = π (b0 ) then
π (ab) = π (a0 b0 ), But a − a0 ∈ I and b − b0 ∈ I, so
ab − a0 b0
= a(b − b0 ) + (a − a0 ) b0
∈ I
so π (ab) = π (a0 b0 ).
2. We call
R
the quotient ring of R modulo I
I
3. Examples:
(a)
Z
(n)
ª
k [X] ∼ ©
= a + bx : a, b ∈ k, x2 = 0
(X 2 )
¤
k [X, Y ] ∼ £
(c)
= k X, X −1 .
(XY )
(b)
4. Now we see that the following are equivalent for a subset I ⊂ R
(a) I is an ideal
(b) I is the kernal of a ring homomorphism with domain R
(c) I is the kernal of a surjective ring homomorphism with domain R
i. Proof: (a) =⇒ (c) =⇒ (b) =⇒ (a)
5. In other words, if ϕ : R −→ S is a homomorphism of rings then image (ϕ) ∼
=
R
ker (ϕ)
6. Now we turn to fields, which will be contructed as quotient rings.
xliv
LECTURES
7. The intersection of all the subfields of a field is called the prime field. It
Z
is always isomorphic to either Q or
= Fp
(p)
(a) Note that Q and Fp have no proper subfields.
(b) Let k be a field, and consider the homomorphism ϕ : Z −→k. There
is only one homomorphism possible, the one that takes 1Z to 1k , and
that map is a homomorphism. Either ϕ is 1 − 1 or it is not.
If ϕ is 1 − 1 then all the non-zero elements of Z map to non-zero
elements of k, and the map ϕ may be extended to a map ϕ : Q −→k.
The prime field of k is contained in ϕ (Q), but Q cannot have a proper
subfield. Therefore ϕ : Q −→prime field of k is an isomorphism.
Z
If ϕ is not 1 − 1 then let ker ϕ = (n). Thus
is (isomorphic to) a
(n)
subring of k. But then n must be prime or k would have zero divisors,
Z
for some prime p.
and k contains a subfield isomorphic to
(p)
8. If Q ⊂k we say that k has characteristic 0; otherwise k has characteristic p. In the first case, 1 + · · · + 1 is never zero; in the second case
p times
z }| {
1 + ··· + 1 = 0
(a) Q, R, C and C (X) are fields of characteristic 0.
(b) Fp , Fp (X) are fields of characteristic p
9. If k is a finite field, then k has characteristic p for some prime p and
|k| = pn .
(a) Since k is finite, Q is not a subfield. Thus Fp ⊂ k and k is a vector
space over Fp , which has p elements. Since k is finite, it ¯has¯ a finite
basis and is therefore isomorphic to Fnp for some n. But ¯Fnp ¯ = pn .
10. Suppose k is a field, for example Q or Fp . Next we see how to construct
a lot of different fields containing k. These are called extensions of k.
11. First a triviality: if p is a non-constant polynomial then the natural map
k [X]
is injective
k −→
(p)
(a) If any non-zero element of k mapped to 0, then every element of
k would map to 0. Thus 1 ∈ (p), which is not possible for a nonconstant polynomial p.
(b) Now we can think of k as a subfield of
k [X]
.
(p)
0.7. RING THEORY
xlv
k [X] ∼
(c) If p = p0 + · · · + pn−1 X n−1 + X n is monic of degree n, Then
=
(p)
ª
©
a0 + · · · + an−1 xn−1 : ai ∈ k . Addition is componentwise, and multiplication is given by the rule xn = −p0 − · · · − pn−1 xn−1 .
i. So
k [X]
is a vector space of dimension n over k.
(p)
12. If p ∈ k [X] then
k [X]
is a field iff p is irreducible (prime).
(p)
k [X]
is a ring; to prove
(p)
it is a field we must only show that every non-zero element has an
k [X] ¯
, f 6= 0̄, then f ∈
/ (p). Since p is irreducible,
inverse. Iff¯ ∈
(p)
gcd (p, f ) = 1. Thus there exist a, b ∈ k [X] such that af + bp = 1.
k [X]
Therefore āf¯ = 1̄ in
.
(p)
k [X]
Conversely, suppose
is a field. If f, g ∈ k [X] and f g ∈ (p) then
(p)
f¯ḡ = 0̄. Thus either f¯ = 0̄ or ḡ = 0̄, that is either f ∈ (p) or g ∈ (p).
Therefor p is prime.
(a) Suppose p is irreducible. We know that
13. WARNING–potential confusion ahead. Suppose p ∈ k [X] is irrek [X] ∼
ducible. If p = x − a then
= k, but if deg (p) > 1 then p has no
(p)
roots in k. (If p (a) = 0 then x − a divides p so p cannot be irreducible.)
k [X]
, we have created a field larger than k that contains
In constructing
(p)
a root of p, as we explain.
k [X]
. Since L is a field, we can construct the polynomial
(p)
ring L [X]. There are two maps from k [X] into L [X].
(a) Let L =
i. The projection k [X] −→ L extends to k [X] −→ L [X]. For
every polynomial f ∈ k [X], the image under this map is f¯ ∈
L ⊂ L [X]. That is, for all polynomials f , f¯ is a constant in
L [X]. We use the bar to denote this map.
ii. Since k ⊂ L, we have k [X] ⊂ L [X]. This map is denoted by the
absence of any symbol, since it is really just subset inclusion.
(b) So what happens to the critical polynomials X and p under these
maps?
i. p̄ = 0̄ and X̄ is an element of L
ii. p ∈ L [X] is the same polynomial as it was in k [X], and X ∈
L [X] is the variable.
xlvi
LECTURES
¡ ¢
(c) What have we gained? p X̄ = 0. The element X̄ ∈ L is a root of p
over L.
Q
∼
(d) Example: p = X 2 + 1 is irreducible over Q, and L =
=
(X 2 + 1)
Q (i). The polynomial p can be considered as a polynomial over
L. Moreover, under this isomorphism,
X̄ = ±i (you can set up the
¡ ¢
isomorphism either way), so p X̄ = p (i) = 0.
14. We have just shown that every irreducible polynomial (and therefore every
polynomial) in k [X] has a root in L, where L is some overfield of k. There
is an important uniqueness statement to be made.
(a) If p, q ∈ k [X] are monic, irreducible polynomials, and if k ⊂ L, and
if there exists a ∈ L such that p (a) = q (a) = 0 then p = q.
i. We begin with three important Lemmas:
A. Lemma 1: Let k ⊂ L be a tower of fields, and let a ∈ L. Suppose there exists a non-zero polynomial f ∈ k [X] ⊂ L [X]
such that f (a) = 0. Construct the map ϕ : k [X] −→ L
by ϕ|k = id, ϕ (X) = a. Then image (ϕ) is a ring that is a
finite-dimensional k-algebra. (The next lemma proves that
image (ϕ) is a field.) We denote image (ϕ) = k [a].
k [X]
B. Proof: The ring
is a finite-dimensional k-algebra, and
(f )
we have a commutative diagram:
k [X]
↓
k [X]
(f )
ϕ
−→ L
% ϕ̄
Thus im (ϕ) = im (ϕ̄) is finite dimensional over k.
C. Lemma 2: Let k ⊂ D with k a field and D a domain.. If
dimk D < ∞ then D is a field.
D. Proof: we need to show that for any d ∈ D, d 6= 0, The
set dD = D. Then there will be e ∈ D such that de = 1.
But the map D −→ D given by e −→ de is linear over k.
Moreover the map is 1 − 1 because D is a domain. But D is
finite dimensional over k, so every injective k-linear map is
also onto..
E. Lemma 3: Let k ⊂ L be fields, and let a ∈ L. Construct
a homomorphism ϕ : k [X] −→ L by ϕ|k = id and ϕ (X) =
a. If p ∈ k [X] is irreducible, then p (a) = 0 if and only if
ker (ϕ) = (p).
F. Proof: the map ϕ is given by ϕ (f ) = f (a), so ker (ϕ) = (p)
certainly implies p (a) = 0. Conversely, suppose p (a) = 0.
0.7. RING THEORY
xlvii
Then p ∈ ker (ϕ). We know that ker (ϕ) = (f ) for some
polynomial f which cannot be zero. On the other hand, ϕ is
not the zero map since ϕ (1) = 1, so f is not a unit. Since f
divides p, which is irreducible, we must have (f ) = (p).
ii. now for the proof: the map ϕ from lemma 3 has ker (ϕ) = (p) =
(q). Since p and q are both monic irreducibles, they must be
equal.
15. If k ⊂ K are fields, then we say that K is an extension of k and write
K/k. In this case K is also a vector space over k. We define [K : k] =
dimk K , and we say that K is a finite extension of k if [K : k] < ∞.
The number [K : k] is called the degree of the extension
·
¸
k [X]
(a) For an irreducible polynomial p ∈ k [X], deg (p) =
:k .
(p)
16. Suppose K/k is a field extension. An element a ∈ K is algebraic over k
iff there exists a non-zero polynomial f ∈ k [X] such that f (a) = 0.
(a) K is algebraic over k if every element of K is algebraic.
17. If K/k and [K : k] < ∞ then K is algebraic over k.
(a) Proof: suppose a ∈ K. The map k [X] −→ K given by X −→ a
cannot be 1 − 1, because K is finite-dimensional over k and k [X] is
infinite dimensional over k. Therefore the map has a non-zero kernal
ideal whose generator is the desired non-zero polynomial with root
a.
18. If k ⊂ E ⊂ K with k, K fields, then there is a unique smallest subfield L
of K containing E. We denote this field k (E).
(a) To construct L, first construct the integral domain D ⊂ K consisting
of all polynomial combinations of finite subsets of E with coefficients
in k:
nX
o
D=
ai1 ...it ei11 · · · eitt : ai1 ...it ∈ k, e1 , . . . , et ∈ E
Then k ⊂ D ⊂ K, the the quotient field of D is L.
19. Let K/k and a ∈ K be algebraic. Then there exists a unique irreducible
monic polynomial p ∈ k [X] such that p (a) = 0. Moreover, there is an
k [X]
isomorphism
−→ k (a) given by X̄ −→ a. We say that p is the
(p)
minimial polynomial of a over k.
xlviii
LECTURES
(a) By definition of algebraic, there exists a polynomial f ∈ k [X] such
that f (a) = 0. One of the monic irreducible factors p of f over k
must have a for a root. By the lemma above, p is unique and the map
k [X]
k [X] −→ k (a) mapping X −→ a has kernal (p), so
−→ k (a)
(p)
is bijective.
20. It follows that if p is irreducible over k and K/k and a, b ∈ K are such
that p (a) = p (b) = 0, then k (a) ∼
= k (b) via a map that is constant on k
and sends a to b.
21. If k ⊂ K ⊂ L are fields then [L : k] = [L : K] [K : k].
(a) Proof: assume all dimensions are finite. Let K1 , . . . , Ks be a basis for
K/k and let L1 , . . . , Lt be a basis for L/K. Then it is easy to prove
that Ki Lj is a basis for L/k.
22. So far the only examples we have of finite algebraic extensions are of the
¡ ¢ k [X]
form k X̄ =
. One might think that all finite algebraic extensions
(p)
were generated by one element, but that is not true. More advanced theorems show that a surprising number of algebraic extensions are generated
by one element, but not all of them.
23. Kronecker proved: let k be a field and f ∈ k [X], deg (f ) > 0. There exists
a finite field extension K/k such that f factors into linear factors over k
(a) Proof: proceed by induction on deg (f ), since the result is clearly
true with K = k if deg (f ) = 1. We may assume f is monic. Let
k [X]
. L is a finite
p be a monic irreducible factor f , and L =
(p)
extension of k and L contains a root X̄ of p, which
is ¢a root of f .
¡
Therefore, over L, f can be factored into f = X − X̄ g for some
g ∈ L [X]. Since deg g < deg f, by the induction hypothesis there
exists a finite extension K/L such that g factors into linear factors
g = (X − g1 ) ·¡· · (X −¢gt ) in K [X]. Therefore K is a finite extension
of k and f = X − X̄ (X − g1 ) · · · (X − gt ) in K [X].\
(b) By unique factorization of polynomials, X̄, g1 , . . . , gt are the only
roots of f in K. A polynomial cannot have more roots than it has
degree, no matter how far you extend the field to look for roots.
24. If f ∈ k [X] factors into linear factors over an extension field K/k, then
we say that f splits over K. We have just proved that every polynomial
splits over some finite field extension K. I claim there is a smallest subfield
E such that k ⊂ E ⊂ K and f splits over E. We call E a splitting field
for f . You can prove its existence by taking the subfield of K generated
over k by the roots of f in K.
0.7. RING THEORY
xlix
25. Suppose we have two isomorphic fields k −→ k 0 . The isomorphism extends to an isomorhism k [X] −→ k 0 [X]. Given a polynomial f ∈ k [X]
corresponding to f 0 ∈ k 0 [X] and a splitting field E for f over k and a splitting field E 0 for f 0 over k 0 , then there exists a non-unique isomorphism
E −→ E 0 extending k −→ k0 .
(a) So long as there is one isomorphism, then the isomorphisms E −→
F can be matched up with the isomorphisms E −→ E which are
the identity on k. These form a group: the Galois group of the
polynomial f over k.
(b) Proof: proceed by induction on deg (f ). Let p be an irreducible
factor of f over k, let p0 be the corresponding factor of f 0 over k 0 ,
and let α ∈ E, β ∈ F be roots of p and p0 respectively. Then
k [X] ∼ k0 [X] ∼
k (α) ∼
=
=
= k (β), and we have the diagram:
(p)
(p0 )
k −→ k (α) −→ E
↓∼
↓∼
=
=
k 0 −→ k (β) −→ E 0
E is the splitting field of f / (X − α) over k (α) and F is the splitting
field of f 0 / (X − β) over k0 (β). Since the isomorphism k (α) −→ k (b)
maps f / (X − α) into f 0 / (X − β), we have the desired isomorphism
E −→ F by induction.
26. It is important that splitting are not unique but are all isomorphic. Since
they are all the same degree (dimension) over k, you cannot have a nonisomorphism E −→ F over k. All maps of fields are 1 − 1, and any 1 − 1
map of finite extensions of the same degree over k is also onto.
27. One key application: for every prime number p and natural number n there
exists a field, unique up to non-unique isomorphism, with pn elements.
n
(a) It is the splitting field E for f = X p − X over Fp . Such a field
exists, but we have to show that it has the right number of elements.
It will turn out that all of them are roots of our polynomial, so that
the field consists precisely of the roots of f .
n
By an earlier problem, since f 0 = pn X p −1 − 1 = −1 (since p = 0 as
a multiplier in Fp ), f 0 has no roots and in particular f and f 0 have no
common roots. That means that any complete factorization of f will
involve pn distinct roots. The splitting field E contains pn distinct
roots; it remains only to show that the set of roots R is a subfield.
n
n
If a, b ∈ R then ap = a and bp = b. By a well-known property of
n
n
n
binomial coefficiens, (a + b)p = ap + bp = a + b, so a + b ∈ R.
n
n
n
p
Products are even easier: (ab) = ap bp = ab so ab ∈ R. Finally if
¡ −1 ¢pn ¡ pn ¢−1
= a
= a−1 so a−1 ∈ R.
a 6= 0 then a
l
LECTURES
0.8
Commutative Rings
This section applies primarily to the ring k [X1 , . . . , Xn ] and rings derived from
it. It is a matter of aesthetics and efficiency to make all definitions as general
as possible.
0.8.1
Prime and Maximal Ideals
1. The ideals in a ring form a lattice under inclusion.
(a) An interesting sublattice is the set of ideals containing a fixed ideal.
(b) An isomorphism of rings induces an isomorphism of their lattices of
ideals.
(c) A homomorphism of rings ϕ : R −→ S induces a map ϕ−1 from the
lattice of ideals of S to the lattice of ideals of R.containing ker (ϕ).
i. To prove this, you have to show that whenever I is an ideal of
S then ϕ−1 (I) is an ideal of R containing ker (ϕ), and that if
I ⊂ J are ideals of S than ϕ−1 (I) ⊂ ϕ−1 (J).
A. First, 0 ∈ ϕ−1 (I), so ϕ−1 (I) 6= φ. If a, b ∈ ϕ−1 (I) then
ϕ (a + b) = ϕ (a)+ϕ (b) ∈ I, so a+b ∈ ϕ−1 (I). If a ∈ ϕ−1 (I)
and b ∈ R then ϕ (ab) = ϕ (a) ϕ (b) ∈ I, so ab ∈ ϕ−1 (I).
Thus ϕ−1 (I) is an ideal of R.
B. It is a property of set maps that I ⊂ J implies ϕ−1 (I) ⊂
ϕ−1 (J). In particular ϕ−1 (I) ⊃ ϕ−1 (0) = ker (ϕ) for every
ideal I ⊂ S.
ii. The map ϕ−1 is not necessarily 1 − 1 or onto on the lattices of
ideals.
iii. If ϕ is surjective then ϕ−1 is 1 − 1 on ideals.
iv. For every ideal I ⊂ S, ϕ induces a commutative diagram:
R
↓
R
ϕ−1 (I)
ϕ
−→
ϕ̄
−→
S
↓
S
I
where the bottom arrow is injective.
A. Proof: it suffices to note that the kernal if the composite map
S
R −→ S −→
is ϕ−1 (I).
I
(d) Caution: if I ⊂ R is an ideal then ϕ (I) ⊂ S is not necessary an
ideal.
i. Example: k [X] −→ k [X, Y ], the image of (X) is not an ideal.
ii. But if ϕ is onto, then ϕ (I) is an ideal.
0.8. COMMUTATIVE RINGS
li
iii. Proof: since I is not empty, ϕ (I) is not empty. If a, b ∈ ϕ (I)
then a = ϕ (α) and b = ϕ (β) for some α, β ∈ I. Thus
a + b = ϕ (α) + ϕ (β)
= ϕ (α + β)
∈ ϕ (I) .
If a ∈ ϕ (I) and b ∈ S then choose α ∈ I and β ∈ R such that
ϕ (α) = a and ϕ (β) = b. We can find β because ϕ is onto. Thus
ab = ϕ (α) ϕ (β)
= ϕ (αβ)
∈ ϕ (I)
R
(e) In particular the projection ϕ : R −→
induces a bijection from the
I
R
lattice of ideals of
to the sublattice of ideals of R containing I.
I
i. Let LR (I) be the lattice of ideals of R containing I. Thus LR (0)
is the entire lattice of ideals. By what we have said above, π−1 :
LR/I (0) −→ LR (I). We must show that π−1 : LR/I (0) −→
LR (I) is 1 − 1 and onto.
Since π is surjective, π −1 is 1 − 1. To show that π−1 is onto, it
suffices to show that if J is an ideal of R and I ⊂ J then π (J)
R
is an ideal of
and π−1 π (J) = J.
I
We have already proven that surjective maps send ideals to ideals,
R
so π (J) is an ideal of . It is always true that π −1 π (J) ⊃ J.
I
If a ∈ π−1 π (J) then there exists b ∈ J such that π (a) = π (b).
Thus π (a − b) = 0 so a − b ∈ I ⊂ J. Therefore a ∈ J.
Z
ii. The ideals of
are the projections of the ideals (m) where
(n)
(n) ⊂ (m) or the set of m dividing n.
2. An proper ideal I in a ring R is prime if ab ∈ I implies a ∈ I or b ∈ I.
(a) In Z an ideal (n) is prime if and only iff n is a prime number or 0.
i. Proof: suppose (n) is a prime ideal. If n 6= 0 we will show that
n is a prime number. Assume n = ab; we will prove that a = ±1
or b = ±1. Since ab ∈ (n) either a ∈ (n) or b ∈ (n). Suppose
a ∈ (n), or a = nc. Then n = ncb. Since n 6= 0, cb = 1 or b = ±1.
Conversely, (0) is prime (if ab = 0 then a = 0 or b = 0), and if
n is a prime number then suppose ab ∈ (n). Thus n divides ab;
since n is prime either n divides a or n divides b; that is either
a ∈ (n) or b ∈ (n). Thus (n) is a prime ideal.
lii
LECTURES
(b) In k [X] an ideal (f ) is prime if and only if f = 0 or f is an irreducible
(prime) polynomial.
i. Proof: suppose (f ) is a prime ideal. If f 6= 0 we will show that
f is irreducible. Assume f = ab; we will prove that deg (a) = 0
or deg (b) = 0. Since ab ∈ (f ) either a ∈ (f ) or b ∈ (f ). Suppose
a ∈ (f ), or a = f c. Then f = f cb. Since f 6= 0, cb = 1 or
deg (b) = 0. Conversely, (0) is prime (if ab = 0 then a = 0
or b = 0), and if f is an irreducible polynoomial then suppose
ab ∈ (f ). Thus f divides ab; since f is prime either f divides a
or f divides b; that is either a ∈ (f ) or b ∈ (f ). Thus (f ) is a
prime ideal.
3. An ideal I in a ring R is prime if and only if
R
is an integral domain.
I
R
(a) Suppose I is prime; we must show that
is an integral domain.
I
R
R
Clearly
is a commutative ring. Let ā, b̄ ∈
such that āb̄ = 0̄.
I
I
Then ab ∈ I so a ∈ I or b ∈ I. Thus either ā = 0̄ or b̄ = 0̄. Conversely
R
is an integral domain. We must show I is a prime ideal.
suppose
I
R
Let a, b ∈ R, ab ∈ I. We must show a ∈ I or b ∈ I. But in ,
I
ab = āb̄ = 0̄, so ā = 0̄ or b̄ = 0̄. That is, either a ∈ I or b ∈ I.
4. If ϕ : R −→ S is a ring homomorphis, then ϕ−1 takes prime ideals to
prime ideals.
(a) Let I ⊂ S be prime, and suppose a, b ∈ R, ab ∈ ϕ−1 (I). We must
show a ∈ ϕ−1 (I) or b ∈ ϕ−1 (I). But ϕ (ab) = ϕ (a) ϕ (b) ∈ I, which
is prime, so either ϕ (a) ∈ I or ϕ (b) ∈ I. That is, either a ∈ ϕ−1 (I)
or b ∈ ϕ−1 (I).
5. An ideal I in a ring R is a maximal ideal if I 6= R and I ⊂ J ⊂ R
implies I = J or J = R.
(a) Every proper ideal in R is contained in a maximal ideal, but that
requires Zorn’s Lemma or the Axiom of Choice to prove.
(b) The text proves this theorem in some special cases.
(c) I would like a non-Axiom of Choice proof for k [X1 , . . . , Xn ].
(d) Here’s a proof for Z: (n) ⊂ (m) if and only if m | n. Therefore an
ideal is maximal iff it is generated by a prime number. In particular,
for any non-unit m, (m) ⊂ (p) where p is a prime divisor of m.
(e) Here’s a proof for k [X], in fact the same proof. Just change “number”
to “polynomial”.
0.8. COMMUTATIVE RINGS
liii
(f) Over an algebraically closed field k, the maximal ideals of k [X] are
(X − a) for a ∈ k.
(g) The same is true but harder to prove is several variables. The
maximal ideals of k [X1 , . . . , Xn ] are (X1 − a1 , . . . , Xn − an ), where
ai ∈ k.
(h) Over an algebraically closed field, the maximal ideals of k [X1 , . . . , Xn ]
correspond to the points of k n
liv
LECTURES
R
6. Let I be an ideal in a ring R. Then I is maximal if and only if
is a
I
field.
R
(a) Suppose I is maximal. Then (0) is a maximal ideal in . Therefore,
I
R
R
R
if ā ∈ , ā 6= 0̄, then the ideal (ā) = . Thus for some b̄ ∈ ,
I
I
I
R
R
āb̄ = 1̄. Thus every non-zero element of
is a unit and
is a field.
I
I
R
Conversely, if
is a field then there are no ideals between I and R
I
so I is maximal.
7. A maximal ideal is prime
(a) If I is maximal then R/I is a field and therefore an integral domain,
so I is prime.
0.8.2
Unique Factorization Domains
1. Let R be a ring.
(a) Suppose r is a non-zero, non-unit element of R. If r = ab with a, b
non-units then r is reducible. If a factorization into non-units is not
possible, we say r is irreducible.
i. In Z, 6 is reducible and 5 is irreducible.
(b) A non-zero element p ∈ R is prime iff (p) is a prime ideal.
i. A prime cannot be a unit because R is not a prime ideal of R.
ii. Another way to say this is that p is prime iff p|ab =⇒ p|a or p|b.
(c) Two elements a, b ∈ R are associates in R iff a = bu where u is a
unit.
i. Class prove: the property of being associates is an equivalence
relation.
2. In the integers or polynomials over a field, primes and irreducible are the
same. But this is not true in general.
(a) In an integral domain R, a prime element p is irreducible.
i. If p is prime and p = ab then ab ∈ (p) so either a ∈ (p) or b ∈ (p).
Suppose a ∈ (p). Then a = pc so p = ab = pcb and, since p 6= 0
and we are in a domain, bc = 1. Thus b is a unit and p cannot
be factored into a product of non-units.
ii. Examples of irreducible elements that are not prime are complicated.
0.8. COMMUTATIVE RINGS
lv
(b) In a PID, an irreducible element is prime.
i. Suppose r is irreducible and ab ∈ (r). We must show a ∈ (r) or
b ∈ (r). Let I = {c : ac ∈ (r)}. Thus b ∈ I. Let I = (d). Since
r ∈ I, r = de. But r is irreducible, so either d is a unit or e is a
unit. If d is a unit then I = R, 1 ∈ I, 1a = a ∈ (r). Otherwise e
is a unit and (d) = (r). Since b ∈ I = (d), b ∈ (r).
(c) PID’s where primes are the same as irreducibles include Z and k [X].
3. Domains in which prime is the same as irreducible are not only the PID’s.
4. A unique factorization domain (UFD) is an integral domain in which
every non-zero, non-unit element r can be factored into a product of irreducibles uniquely up to associates.
(a) That is, R is a unique factorization domain iff a ∈ R implies:
i. a = 0 or a is a unit or a = r1 · · · rt where ri are irreducible (but
not necessarily distinct)
ii. if ri , sj are irreducible and r1 · · · rm = s1 · · · sn then m = n and,
after reordering, ri and si are associates.
5. We shall demonstrate eventually that all PID’s are UFD’s, and that all
polynomial rings in several variables over a field of coefficients are UFD’s.
This is a rich and important class of rings.
6. In a UFD R, every irreducible element is prime, so primes and irreducibles
are the same.
(a) Suppose p is irreducible in R and p | ab for a, b ∈ R. We must show
p | a or p | b. Let ab = pc for some c ∈ R. Let the irreducible decompostions of a, b, c and p be
a
b
c
p
=
=
=
=
a1 · · · am
b1 · · · bn
c1 · · · ct
p
Thus we have equal products of irreducibles
a1 · · · am b1 · · · bn = pc1 · · · ct
Therefore one of the ai or one of the bi is an associate of p, and either
p | a or p | b.
7. Henceforth we will spreak of the prime decomposition of an element in
a UFD, rather than the decomposition into irreducibles.
lvi
LECTURES
(a) If an element factors into powers of distinct primes:
n1
ms
ns
1
a = pm
1 · · · ps = q1 · · · qs
then, after reordering, mi = n1 and (pi ) = (qi ).
2
i. Example: 60 = 22 × 3 × 5 = (−2) × (−3) × (−5)
(b) For some purposes it is convenient to consider the set of all principal
prime ideals (not all prime elements) in a UFD R, and choose P be
a set of prime elements, one generating each principal prime ideal in
R.
Q
i. If a ∈ R then we can factor a uniquely as a = u p ∈P pnp
where u is a unit and almost all np = 0. Sometimes will will
say np Q
= vp (a), the number of times the prime p divides a, and
a = u p ∈P pvp (a) . For example in Z, v2 (40) = 3. For any a
and almost all p ∈ P, vp (a) = 0.
ii. If a, b ∈ R then vp (gcd (a, b)) = min (vp (a) , vp (b)) and vp (lcm (a, b)) =
max (vp (a) , vp (b)). Thus
a = u
Y
pvp (a)
p ∈P
b = v
Y
pvp (b)
p ∈P
gcd (a, b) =
Y
pmin(vp (a),vp (b))
p ∈P
lcm (a, b) =
Y
pmax(vp (a),vp (b))
p ∈P
Thus lcm (a, b) gcd (a, b) is an associate of ab.
A. Proof of gcd property. Think what the prime factorization
of gcd (a, b) must be if the gcd divides a and b and is divided
by anything dividing a and b.
8. The definition of UFD is complex and powerful; for that reason it is hard
to demonstrate that a particular class of rings are UFD’s.
9. If R is a PID then R is a UFD.
(a) That’s what they taught you about integers in grade school and polynomials in high school. Integers and polynomials can be factored into
powers of prime factors.
i. This theorem proves the assumptions of grade-school arithmetic
and high-school algebra because the integers and the polynomials
in one variable over a field are PID’s.
0.8. COMMUTATIVE RINGS
lvii
(b) This is our second long proof. It is also deeper than most seen so far.
The methods are not obvious.
(c) It will go better if we first prove two lemmas:
i. Let I0 ⊂ Ii ⊂ I2 ⊂ · · · be a chain of ideals in a ring R. Then
J = ∪i Ii is a ideal.
A. J is a union of non-empty sets, so J is not empty. If a, b ∈ J
then for some i, a, b ∈ Ii . Thus a + b ∈ Ii ⊂ J. If a ∈ J and
b ∈ R then for some i, a ∈ Ii . Thus ab ∈ Ii ⊂ J.
ii. If a = bc and b and c are not units and b 6= 0 then (a) Ã (b)
A. Certainly (a) ⊂ (b) because ax = b (cx) for all x ∈ R. If
(b) ⊂ (a) then b = ax for some x ∈ R so b = bcx and since
b 6= 0, cx = 1. But c is not a unit.
(d) First we show that every non-zero, non-unit element is a product of
irreducibles. Suppose, on the contrary, that a0 ∈ R is not zero, not
a unit, and not a product of irreducibles. In particular, a0 is not
irreducible, so a0 = a1 b1 , where a1 and b1 are not units. One of the
factors, say a1 , must not be a product of irreducibles, because if both
a1 and b1 were products of irreducibles then a0 would be a product
of irreducibles. Moreover, since a1 and b1 are both non-zero and not
units, (a0 ) Ã (a1 ). Proceeding in this way, we can construct two lists
of non-zero, non-unit elements a0 , a1 , · · · and b1 , b2 , · · · such that
i. ai = ai+1 bi+1
ii. ai is not a product of irreducibles.
iii. (ai ) Ã (ai+1 )
Let I = ∪i (ai ). I claim 1 ∈
/ I, because otherwise 1 ∈ (ai ) for some i,
which means ai is a unit. But I is an ideal, so it is principal: I = (q).
But then q ∈ (ai ) for some i. Therefore, (q) ⊂ (ai ) ⊂ (ai+1 ) ⊂ (q),
so (ai ) = (ai+1 ).. This contradicts (ai ) Ã (ai+1 ). Therefore all nonzero, non-unit elements of R are products of irreducibles..
(e) Next we show that factorizations are unique up to associates. The
proof is by induction.on the number of factors. Suppose an element
a ∈ R is a product of one irreducible: a = p. Then if p can be
factored some other way into a product of irreducibles, if p = q1 · · · qt
then since p is irreducible t = 1 and p = q1 . The factorization is
unique. Suppose any element that can be factored into fewer than n
factors has a unique factorization (up to associates). Suppose further
that a = p1 · · · pn = q1 · · · qm , where pi , qj are irreducible and m ≥
n. Since pn divides a, (remember, we are working in a PID) , pn
divides one of the factors qi . We can assume pn divides qm . Since
pn and qm are irreducibles, they are associates. Thus p1 · · · pn−1 and
q1 · · · qm−1 are associates. By the induction hypothesis, m = n and
after rearranging pi and qi are associates.
lviii
LECTURES
10. But k [X, Y ] is not a PID (homework) but, as we will see, it is still a UFD
11. Remark: if R is an integral domain, then so is R [X]
12. Theorem (Gauss): if R is a UFD then R [X] is a UFD
(a) Therefore, if k is a field, k is a UFD and so is k [X1 , . . . , Xn ].
(b) Gauss’ Lemma: Let R be a UFD with quotient field k. Suppose
A, B ∈ k [X] are such that AB ∈ R [X]. Then there exists c ∈ k such
that cA, c−1 B ∈ R [X]. That is, if a polynomial in R [X] factors in
k [X]. then it factors in R [X].
i. To understand what this says, let’s look at a corollary before
doing the proof.
X
ii. Corollary: if p ∈ R [X], p =
pi X i , is such that gcd (p0 , . . . , pn ) =
1, then p is irreducible in R [X] iff p is irreducible in F [X]. In
particular, a monic polynomial in R [X] is irreducible iff it cannot
be factored in F [X].
iii. This corollary applies to Z [X] and Q [X]. The hypothesis avoids
the factorization 4 + 2X = 2 (2 + X), which is a factorization
into non-units in Z [X] but not in Q [X]. The polynomial 4 + 2X
is reducible in Z [X] but not in Q [X].
iv. Proof of Corollary: we show p is reducible in R [X] iff p is reducible in F [X]. By Gauss’ Lemma, if p is reducible in F [X]
then it is reducible in R [X]. Conversely, if p is reducible in R [X]
then p = qr for q, r ∈ R [X]. If q or r was a non-unit constant,
then it would be a common factor of the pi , which by hypothesis
have no common factors. Thus both q and r are non-constant
polynomials in R [X] and are therefore non-unit factors of p in
F [X].
v. Now for the proof of Gauss’ Lemma. Choose d, d ∈ R, d 6= 0,
such that dA, dB ∈ R [X]. If d is a unit of R then we are done,
since then A = d−1 dA ∈ R [X] and B = d−1 dB ∈ R [X], Just
take c = 1.
If d is not a unit, let A0 = dA and B 0 = dB. Then A0 , B 0 ∈ R [X]
and d2 AB = A0 B 0 .We will show that constants a and b can be
factored out of A0 and B 0 in R [X] such that ab = d2 . That is,
A0 = aA00 and B 0 = bB 00 where ab = d2 . Thus¡ AB ¢= A00 B 00 and
A00 = da−1 A and B 00 = db−1 B. Since (da−1 ) db−1 = 1, we can
take c = da−1 .
To construct a and b, we proceed by induction on the number
of prime divisors of d2 . We have already shown that the result
holds if the number of prime divisors is 0, or if d is a unit. Let
p be a prime divisor of d2 . I claim either A0 = pA000 or B 0 =
pB 000 .Suppose this is true. Without loss of generality, we can
assume (d2 /p)AB = A000 B 0 . By induction there exist a0 , b such
0.8. COMMUTATIVE RINGS
lix
that a0 b = d2 /p and A000 = a0 A00 and B 0 = ¡bB 00 . ¢ Therefore
A0 = pA000 = pa0 A00 = aA00 where ab = pa0 b = p d2 /p = d2 .
It remains to prove the claim. Working modulo pR [X],
2
0 = d AB
0
= A0 B
Since pR is a prime ideal,
R
is an integral domain and so is
pR
R
R [X]
[X] =
. Therefore either A0 = 0 or B 0 = 0. Thus
pR
pR [X]
either A0 = pA000 or B 0 = pB 000 in R [X].
(c) Gauss’ Lemma implies that R [X] is a UFD if R is a UFD
i. Let F be the quotient field of R. If f ∈ R [X] ⊂ F [X] then
in F [X] we can factor f uniquely into a product of primes f =
Y
pi . By Gauss’ lemma and induction on the number of factors,
we can assume pi ∈ R [X]. Factoring out the greatest common
divisor of the coefficients of pi results in pi = gi qi where gi ∈ R
and qi ∈ R [X] is irreducible (since qi is already irreducible in
F [X]). After factoring gi into primes in R we have a complete
factorization of f in R [X].
Were there two such factorizations of f , divide them into their
constant and non-constant parts. The non-constant part is a
factorization of f in F [X] so both non-constant parts must be
the same up to units. The factorization of the constant parts
must be the same because R is a UFD.
lx
LECTURES
0.8.3
Noetherian Rings
1. Named for Emmy Noether 1882-1935
2. A ring satisfies the ascending chain condition (ACC) if every ascending chain of ideals I1 ⊂ I2 ⊂ · · · ⊂ It ⊂ · · · stops.
3. An ideal I in a ring R is finitely generated if there is a finite subset
{x1 , . . . , xt } ⊂ I such that I = (x1 , . . . , xt ).
4. Theorem: for a ring R the following conditions are equivalent
(a) R has the ACC
(b) Every non-empty subset of ideals of R has maximal elements
(c) Every ideal of R is finitely generated
5. Proof: (a) =⇒ (b): Given a non-empty collection S of ideals, pick I1 ∈ S.
If I1 isn’t maximal in S, pick I1 ∈ S, I2 % I1 . By the ACC you can’t do
this forever, so there must be maximal ideals in I. (b) =⇒ (c). Suppose
I is an ideal of R, and let S be the collection of finitely generated ideals
contained in I. Let J ∈ S be maximal. If J 6= I there must be an element
x ∈ I, x ∈
/ J. But then K = J + (x) is finitely generated and K ⊂ I
so K ∈ S. But K % J, contradicting the maximality of J. (c) =⇒ (a).
Suppose I1 ⊂ I2 ⊂ · · · . Let I = ∪Ii . I is an ideal, and therefore a finitely
generated ideal: I = (x1 , . . . , xt ). Each xi ∈ Ij for some j, so for some
j, x1 , . . . , xt ∈ Ij . Therefore Ij = I and the sequence of ideals does not
increase forever.
6. Rings satisfying these conditions are called Noetherian rings
7. A field is Noetherian. The only ideals are (0) and (1), and both are finitely
generated.
8. Z is Noetherian
(a) Let I1 $ I2 $ · · · , and let Ii = (ni ). Then ni+1 divides ni and ni
does not divide ni+1 , so the sequence ni , n2 , . . . cannot go on forever.
9. Every PID is Noetherian, by the same proof.
10. Example of non-Noetherian ring: polynomials in a countable number of
variables over a field: k [X1 , X2 , . . .]. An infinitely ascending chain of ideals
is (X1 ) ⊂ (X1 , X2 ) ⊂ · · ·
(a) UFD’s are not necessarily Noetherian, since the polynomial ring in
several variables is a UFD
11. Hilbert Basis Theorem: if R is Noetherian, then the polynomial ring
R [X] is also Noetherian.
0.8. COMMUTATIVE RINGS
lxi
(a) Proof: a mess, but some of the ideas in the proof are related to
the ideas developed in Groebner bases, the modern computational
theory of polynomial ideals. Suppose I ⊂ R [X] is an ideal that is
not finitely generated. We will derive a contradiction. Of course
I 6= (0). Choose f1 ∈ I such that f1 is non-zero of minimal degree.
Construct a sequence of elements f1 , f2 , f3 , . . .such that fi+1 ∈ I −
(f1 , . . . , fi ) has minimal degree. Thus deg f1 ≤ deg f2 ≤ · · · . If
I is not finitely generated, then I 6= (f1 , . . . , fi ), I − (f1 , . . . , fi ) is
not empty, and the sequence f1 , f2 , . . . can be continued forever. To
derive a contradiction, we will show, for i >> 1, that there exists g ∈
I − (f1 , . . . , fi ) and deg (g) < deg (fi+1 ). Thus the desired sequence
cannot be constructed.
Let aj ∈ R be the leading coefficient of fj (the coefficient of the
highest degree term in fj ). Since R is Noetherian, the ascending
sequence of ideals (a1 ) ⊂ (a1 , a2 ) ⊂ · · · eventually stops. That is,
there exists i such that j > i implies aj ∈ (a1 , . . . , ai ). In particular
a
ai for some¢ rj ∈ R. Let deg (fj ) = dj . Then
1 a1¢ + · · · + ri¡
¡ i+1 d= r−d
di+1 −di
r1 x i+1 1 f1 + · · · + ri x
coefficient
¡ d −dfi ¢has the same
¡ leading
¢
di+1 −di
i+1
1
as fi+1 , so if g = fi+1 − r1 x
f1 + · · · + ri x
fi then
g ∈ I − (f1 , . . . , fi ) and deg (g) < deg (fi+1 ).
12. If R is a Noetherian ring and I ⊂ R is an ideal, then
R
is Noetherian.
I
R
(a) Proof: Let J¯ be an ideal of
that pulls back to J ⊂ R. Since J is
I
¯
finitely generated, so is J. The images of the generators of J generate
¯
J.
13. So k [X1 , . . . , Xn ] and Z [X1 , . . . , Xn ] are Noetherian rings.
14. So any quotient of a polynomial ring over a field is Noetherian, and that’s
an important class of rings. They are the rings of algebraic geometry.
0.8.4
Algebraic Varieties
1. Descartes invented graphing y = f (x) about 1650
(a) By the 18th century the idea of graphing with coordinates was routine. The notion of the graph of f (x, y, z) = 0 was well-understood.
(b) The study of graphs of polynomial functions grew deeper and deeper
through the 19th and 20th centuries, developing into the massive
body of work called algebraic geometry. Incredibly deep and elaborate tools have been developed, all of them seemingly required to
answer the simplest questions like Fermat’s Last Theorem.
lxii
LECTURES
2. Note that the graph of f (x, y, z) = 0 in R3 is usually a surface. To
get a curve we need to take the simultaneous solutions to two equations
f (x, y, z) = 0 and g (x, y, z) = 0.
3. Let k be a field. Then k n is a vector space, and a function f ∈ k [X1 , . . . , Xn ]
has its zeros in k n .
(a) From each function we get a graph
(b) A set of functions F ⊂ k [X1 , . . . , Xn ] gives a graph, the intersection
of the individual graphs.
i. We call this set an algebraic variety (some authors would say
affine algebraic set, affine variety) or just a variety, denoted
Var (F ) = {x | f (x) = 0, all f ∈ F } ⊂ k n
¡
¢
ii. Example: in R3 , Var x2 + y 2 − 4, x + y − 1 is a small circle,
the intersection of a plane and a sphere.
iii. More examples: Var (1) = φ and V (0) = k n .
4. If F ⊂ G ⊂ k [X1 , . . . , Xn ] then Var (G) ⊂ Var (F ) ⊂ k n .
(a) more equations means fewer solutions (more precisely: no fewer equations means no more solutions)
5. Usually we don’t try to solve infinite systems of equations, but there is
one case where this is possible. Let F ⊂ k [X1 , . . . , Xn ] and let I = (F ),
the ideal generated by F . Then Var (F ) = Var (I).
(a) Proof: since F ⊂ I, Var (I) ⊂ Var (F ). Suppose x ∈ Var (F ). If
g ∈ I then there exists f1 , . . . , ft ∈ F and r1 , . . . , rt ∈ k [X1 , . . . , Xn ]
such that g = r1 f1 + . . . + rt ft . Thus g (x) = r1 (x) f1 (x) + · · · +
rt (x) ft (x) = r1 (x) 0 + · · · rt (x) 0 = 0. Therefore x ∈ Var (I).
(b) In applications F is usually a finite set of polynomials. Conversely
, since every ideal I ⊂ k [X1 , . . . , Xn ] is generated by a finite set
F ⊂ k [X1 , . . . , Xn ], the variety Var (I) is the variety of a finite set
Var (F ).
(c) In the future we will usually discuss the variety of an ideal rather
than the solution set to a finite collection of polynomials. They are
the same anyway.
6. Let I, J ⊂ k [X1 , . . . , Xn ] be ideals. Then
Var (IJ) = Var (I ∩ J) = Var (I) ∪ Var (J)
(a) Think in two variables: the solutions to f (x, y) g (x, y) = 0 is the
union of the solutions to f (x, y) = 0 and the solutions to g (x, y) = 0.
In other words, Var (f g) = Var (f ) ∪ Var (g).
0.8. COMMUTATIVE RINGS
lxiii
(b) Result extends to finite collections of ideals but not infinite collections
of ideals.
(c) Proof: since IJ ⊂ I ∩ J ⊂ I, Var (I) ⊂ Var (I ∩ J) ⊂ Var (IJ).
By symmetry, Var (I) ∪ Var (J) ⊂ Var (I ∩ J) ⊂ Var (IJ). We must
show Var (IJ) ⊂ Var (I) ∪ Var (J). Suppose x ∈ Var (IJ). Either
f (x) = 0 for all f ∈ I or there exists f0 ∈ I such that f0 (x) 6= 0. In
the first case x ∈ Var (I). In the second case, for all g ∈ J, f0 g ∈ IJ
so (f0 g) (x) = f0 (x) g (x) = 0. Since f0 (x) , g (x) ∈ k, a field, and
f (x) 6= 0, we have g (x) = 0 for all g ∈ J. Thus in the second case
x ∈ Var (J). Thus if x ∈ Var (IJ) then x ∈ Var (I) ∪ Var (J).
7. P
Let A be a collection of ideals of k [X1 , . . . , Xn ]. We can form the ideal
I ∈A I. Then
!
Ã
\
X
I =
Var (I)
Var
I ∈A
I ∈A
(a) This is like the result for a finite set of polynomials. The solution to
a set of equations is the intersection of the solutions to the individual
equations.
¡P
¢
P
(b) Proof: since for all I ∈ A we have I ⊂ I ∈A I, Var
I ⊂
I
∈A
\
¡P
¢
Var (I). Therefore Var
Var (I). To prove the conI ∈A I ⊂
verse, suppose x ∈
\
I ∈A
I ∈A
Var (I) and suppose f ∈
P
I ∈A
I. We must
show f (x) = 0. There exists a finite number of ideals I1 , . . . , It ∈ A
and elements fj ∈ Ij such that f = f1 + · · · + ft . Moreover,
x ∈ Var (Ij ) so fj (x) = 0 for all j. Thus f (x) = 0.
8. Let T = {Var (I) | I is an ideal of k [X1 , . . . , Xn ]}, the collection of all varieties of all ideals in k [X1 , . . . , Xn ]. (We are not suggesting that I 6= J
implies Var (I) 6= Var (J).) Then T is the collection of closed sets for a
topology on k n , the Zariski topology.
(a) We must show
i. φ, kn ∈ T
ii. if C, D ∈ T then C ∪ D ∈ T
\
C∈T
iii. if C is a subset of T then
C∈C
and we have just established all these properties.
9. Suppose I ⊂ k [X1 , . . . , Xn ] is an ideal. Then there is a natural injection
k [X1 , . . . , Xn ] −→ k-valued functions on k n . The text proves that this
map is 1 − 1 if k is infinite.
lxiv
LECTURES
(a) The map restricts to a map k [X1 , . . . , Xn ] −→ k-valued functions on
Var (I).
i. The image is called the polynomial functions on Var (I)
(b) Moreover I is contained in the kernal of this map, so we have a map
k [X1 , . . . , Xn ]
−→ k-valued functions on Var (I).
I
(c) Looking ahead: suppose k is algebraically closed. Then the√Hilbert
Nullstellensatz
says that this map is 1 − 1 if and only I = I. We
√
define I below, but we will not prove this theorem. It’s in the text.
¡ ¢
(d) Here’s an example where the map is not 1−1. Let I = X 2 ⊂ C [X].
Then Var (I) = {0} ⊂ C. The functions on a one point set like {0} are
C [X]
isomorphic to C. Moreover we have a non-injective map
−→ C
(X 2 )
¡ ¢ C [X]
given by aX + b −→ b. The kernal is the ideal X̄ ⊂
.
(X 2 )
√
10. Suppose I ⊂ k [X1 , . . . , Xn ] is an ideal. Then I = {f ∈ k [X1 , . . . , Xn ] | f n ∈ I for some n ≥ 1}.
√
(a) You have a homework problem proving that I is an ideal.
√
(b) I ⊂ I.
√
i. If I = I then we say that I is a radical ideal.
³√ ´
(c) Var (I) = Var
I
³√ ´
i. Proof: clearly Var
I ⊂ Var (I). Suppose x ∈ Var (I). If
√
n
n
f ∈ I then f ∈ I for some n ≥ 0. Thus f n (x) = f³ (x)´ = 0.
√
Since f (x) ∈ k, a field, f (x) = 0. Therefore x ∈ Var
I .
11. We can do the variety think backwards. Given a subset A ⊂ k n , define
Id (A) = {f ∈ k [X1 , . . . , Xn ] | f (x) = 0, all x ∈A}.
(a) Id (A) is a radical ideal
(b) If A ⊂ B ⊂ kn then Id (B) ⊂ Id (A).
i. more points mean fewer functions vanish everywhere.
(c) Id (φ) = k [X1 , . . . , Xn ] and Id (kn ) = (0) if k is infinite.
(d) The map k [X1 , . . . , Xn ] −→ k-valued functions on k n restricts to a
map k [X1 , . . . , Xn ] −→ k-valued functions on A. The kernal of this
map is Id (A), so we have an injection:
k [X1 , . . . , Xn ]
−→ k-valued functions on A
Id (A)
i. The image is called the polynomial functions on A.
0.8. COMMUTATIVE RINGS
lxv
ii. We use quotient rings to construct the ring of polynomial functions on any subset of k n
12. What is the relationship between the operators Var () and Id () that go
between subset of k [X1 , . . . , Xn ] and subsets of kn .
(a) The value of Var () is always a Zariski-closed set, the full set of solutions to some equations.
(b) The value of Id () is always an ideal.
(c) The natural question to ask is: what is Var (Id (A)) or Id (Var (F )),
where A ⊂ k n and F ⊂ k [X1 , . . . , Xn ]. If we assume k is algebraically
closed then
i. Var (Id (A)) is the Zariski closure of A, the smallest variety containing A.
p
ii. Id (Var (F )) = (F )
iii. Thus Id (Var (Id (A))) = Id (A) and Var (Id (Var (F ))) = Var (F )
13. There is a lot more in 6.5. I recommend reading it. If you get interested,
I have other references
lxvi
0.9
0.9.1
LECTURES
Modules
Introduction
1. A (unitary) module over R is an abelian group M together with a
“product” R × M −→ M such that
(a) (rs) m = r (sm)
(b) r (m1 + m2 ) = rm1 + rm2 .
(c) (r + s) m = rm + sm
(d) 1m = m (this is the axiom that makes the module unitary)
2. In old-fashioned language, an R-module M is an R-action on an abelian
group M
(a) sometimes we say “R-module” instead of “module over R”.
(b) Two nothing results to get out of the way
i. 0R m = 0M
A. 0R m = (0R + 0R ) m = 0R m + 0R m. Since M is an abelian
group, 0R m = 0M .
ii. r0M = 0M
A. r0M = r (0M + 0M ) = r0M + r0M . Since M is an abelian
group, r0M = 0M .
3. Examples:
(a) Q over Z or R over Q or R over Z etc.
(b) Vector spaces over their defining fields
i. kn over k
(c) Abelian groups over Z
(d) Any commutative ring with 1 over itself
i. Rn over R, free modules
(e) Any ideal in a ring
4. If R ⊂ S is a subring, then every S-module is an R-module
5. If ϕ : R− > S is a ring homomorphism, then
(a) S is an R-module: rs = ϕ (r) s
R
is an R-module: rs̄ = r̄s̄ = π (r) s̄
I
(b) every S-module M is an R-module: rm = ϕ (r) m
R
i. Every -module is an R-module
I
i. So for every ideal I we have
0.9. MODULES
0.9.2
lxvii
Submodules
1. A submodule of an R-module M is a subgroup N ⊂ M such that RN ⊂ N .
(a) Submodules are R-modules
2. Examples:
(a) Q is a Z-submodule of R, and R is a Q-submodule of C.
(b) A subspace of an k-vector space V is an k-submodule of V
(c) A subgroup of an abelian group is a Z-submodule
(d) An ideal is a submodule of R considered as an R-module. In fact,
“ideal” could be defined this way.
(e) If I ⊂ R is an ideal, then IM = {i1 m1 + · · · it mt : ij ∈ I, mj ∈ M }
is a submodule of M .
3. Let M be an R-module. The intersection of any collections of R-submodules
of M is an R-submodule of M .
(a) Class prove
(b) The intersection of subspaces is a subspace, the intersection of subgroups is a subgroup
4. Let M be an R-module and E ⊂ M . Then the submodule generated by
E is
(
)
X
RE =
ri ei : ri ∈ R, ei ∈ E
i
(a) This set is obviously closed under subtraction and multiplication by
R.
(b) An R-module M is finitely generated if M = RE for some finite
subset E ⊂ M .
i. An R-module M is cyclic if it is generated by a single element:
M = Rm.
(c) The finitely generated modules over a field are the finite dimensional
vector spaces over the field
(d) R is finitely generated as a module over itself with the generating set
being {1}.
(e) Finitely generated ideals in R are finitely generated submodules of
R.
5. If N is a collection of submodules of the R-module M , then
[ the sum of
P
the submodules N ∈N N is the submodule generated by
N.
N ∈N
lxviii
LECTURES
(a) N1 + · · · + Nt = {n1 + · · · + nt : ni ∈ Ni }
(b) The sum of subspaces is a subspace, the sum of subgroups is a subgroup
0.9.3
Module Homomorphisms
1. If M and N are R-modules, then an R-module homomorphism from M
to N is a group homomorphism ϕ : M −→ N such that ϕ (rm) = rϕ (m)
for all r ∈ R, m ∈ M .
2. Examples
(a) Linear maps between F -vector spaces are F -module homomorphisms
(b) Abelian group homomorphisms are Z-module homomorphisms
(c) If N is a R-submodule of M then the inclusion N −→ M is an
R-module homomorphism
R
is an R-module homomorphism
(d) The structure map R −→
I
3. The kernel of an R-homomorphism is an R-submodule of the domain
(a) Proof: Let ϕ : M −→ N be an R-module homomorphism. Then
ker ϕ is a subgroup of M . If k ∈ ker ϕ and r ∈ R then ϕ (rk) =
rϕ (k) = r0 = 0 so rk ∈ ker ϕ and ker ϕ is a submodule of M .
4. The image of an R-homomorphism is an R-submodule of the codomain.
(a) Proof: Let ϕ : M −→ N be an R-module homomorphism. Then
image ϕ is a subgroup of M . If n ∈ image ϕ then n = ϕ (m) for some
m ∈ M . Then rn = rϕ (m) = ϕ (rm) ∈ image ϕ and image ϕ is a
submodule of N ..
5. Kernels and images are symmetric in some categories but not others.
(a) The kernel and image of a homomorphism of abelian groups are normal subgroups, and the kernel and image of a module homomorphism
are submodules.
(b) The kernel of a ring homomorphism is an ideal, but the image is only a
subring in general. The kernel of a group homomorphism is a normal
subgroup, but the image is a non-necessarily normal subgroup.
6. The composition of R-module homomorphisms is an R-module homomorphism.
7. Here’s an abstact, very general construction that is sometimes used. We
need a special case when we study eigenvalues and canonical forms.
0.9. MODULES
lxix
(a) Let M be an R-module, and let S = HomR (M, M ). (Sometimes
elements of S are called endomorphisms.)
i. S is a ring (non-commutative with 1). Multiplication is composition of operators.
ii. M is an S-module.
(b) Let ϕ : M −→ M be an R-homomorphism, so ϕ ∈ S.
i. There exists a ring homomorphism R [X] −→ S such that X 7−→
ϕ.
ii. M can be made into an R [X]-module via this homomorphism
(which depends on the choice of ϕ).
A. If p = a0 + · · · + at X t ∈ R [X] and m ∈ M , then
pm = a0 m + a1 ϕ (m) + · · · + at ϕt (m)
= p (ϕ) m
iii. The R [X]-module M is denoted M ϕ .
8. The special case we need is this: let k be a field and A an n × n matrix
over k. Then kn is a k [X]-module via px =p (A) x.
(a) Every module over k [X] is a vector space over k, and multiplication
by X is an k-linear operator. Therefore the theory of vectors spaces
and linear operators over k is the same as the theory of k [X]-modules.
We will see that this observation leads to a whole new approach to
eigenvalues and generalized eigenspaces.
9. Theorem: Let ϕ, ψ : M −→ M be R-homomorphisms. Then α : M ϕ −→
M ψ is an isomorphism of R [X]-modules iff α : M −→ M is an isomorphism of R-modules such that αϕ = ψα.
(a) Proof: Suppose α : M ϕ −→ M ψ is an isomorphism of R [X]-modules.
Since R ⊂ R [X], α:M −→ M is an isomorphism of R-modules.
Moreover, for all m ∈ M :
α (Xm)
α (ϕ (m))
(αϕ) (m)
αϕ
=
=
=
=
X (α (m))
ψ(α (m))
(ψα) (m)
ψα
Conversely, suppose α : M −→ M is an isomorphism of R-modules
such that αϕ = ψα. We must show that, for any polynomial f ∈
R [X], α (f m) = f (α (m)), where the first multiplication is defined by
lxx
LECTURES
f m = f (ϕ) (m), and the second is defined by f (α (m)) = f (ψ) (α (m)).
Since αϕ = ψα and α is bijective,
αϕα−1
αϕt α−1
αf (ϕ) α−1
αf (ϕ)
α (f (ϕ) m)
=
=
=
=
=
ψ
ψt
f (ψ)
f (ψ) α
f (ψ) (α (m))
B
10. Corollary: if A and B are similar n × n matrices, then (k n )A ∼
= (k n ) .
0.9.4
Quotient Modules
M
is defined as an
1. If M is an R-module and N an R-submodule, then
N
abelian group. This abelian group has in a natural way the structure of
an R-module if we define rm̄ = rm. It suffices to check that m̄ = n̄
implies rm = rn. But m − n ∈ N implies r (m − n) = rm − rn ∈ N .
M
The associative and distributive properties for
follow from the same
N
properties for M .
M
as the codomain of a surjective homomorN
M
phism (called the structure homomorphism) M −→
with
N
kernel N .
(a) You should think of
(b) All the homomorphism theorems hold because they hold for abelian
groups and all the objects and maps consructed in those theorems
are submodules and module homorphisms. We copy them here:
2. If ϕ : M −→ N is a homomorphism of R-modules then
(a) If K ⊂ M is a submodule then ϕ (K) is a submodule of N
(b) Proof: ϕ restricts to a module homomorphism ϕ : K −→ N , and the
image of a module homomorphism is a submodule of the codomain.
(c) If K ⊂ N is a submodule then ϕ−1 (K) is a submodule of M
(d) ϕ−1 (K) is not empty because 0M ∈ ϕ−1 (K). Suppose a, b ∈ ϕ−1 (K).
Then ϕ (a − b) = ϕ (a) − ϕ (b). Since ϕ (a) , ϕ (b) ∈ K, ϕ (a − b) ∈ K.
Thus a − b ∈ ϕ−1 (K). Morover if a ∈ ϕ−1 (K) and r ∈ R , then
ϕ (ra) = rϕ (a) ∈ rK ⊂ K, so ra ∈ ϕ−1 (K). Thus K is a submodule.
3. If ϕ : M −→ N is a homomorphism of R-modules with kernel K, then
(a) ϕ is injective iff ker ϕ = {0}.
0.9. MODULES
lxxi
M
−→ N
K
i. First Homomorphism Theorem: image ϕ = image ϕ̄ so ϕ̄ :
M
−→ image ϕ is bijective
K
ii. We can diagram this result:
(b) In all cases ϕ induces a 1 − 1 homomorphism ϕ̄ :
ϕ
−→
image ϕ
∼
=% ϕ̄
M
↓
M
K
→ N
4. Second Homomorphism Theorem–If N and N 0 are submodules of
an R-module M then the canonical map:
N + N0
N
−→
N ∩ N0
N0
is bijective. Since the result holds for groups, all one has to show is that
the canonical map is a module homomorphism.
5. Third Homomorphism Theorem: Let H ⊂ K ⊂ M be submodules
M M K
of an R-module. Then we can form three quotient modules
,
,
.
H K H
K
M
M/H
Moreover,
⊂
is a submodule, so we can form
. Then the
H
H
K/H
composition
M/H
M
−→
M −→
H
K/H
sends K to the identiy and so induces a map:
M/H
M
−→
K
K/H
which is bijective. A diagram might help:
.
K
↓
K
H
−→
−→
H
↓
M
↓
M
H
−→
−→
M
K
↓∼
=
M/H
K/H
6. Let H be a submodule of an R-module M . There is a 1 − 1 inclusionpreserving correspondence between submodules of M that contain H and
M
the submodules of M/H given by K ↔ π (K) where π : M −→
.
H
lxxii
LECTURES
0.9.5
Direct Sums and Products and Free Modules
1. Let M1 , · · · , Mt be R-modules. Then the direct sum or direct product is:
M1 ⊕ · · · ⊕ Mt
= M1 × · · · × Mt
= {(m1 , . . . , mt ) : mi ∈ Mi }
We give this set of tuples the structure of an R-module with componentwise addition and scalar multiplication.
2. Important examples: the direct sum of copies of R: R ⊕ R, R ⊕ R ⊕ R,
etc.
(a) Use notation Rn
(b) Text gives condition on when a submodule is a direct summand
i. If N ⊂ M , when is M ∼
=N ⊕T
0.9.6
Exact Sequences
An important computational tool for proving theorems about submodules, quotient modules, etc.
ϕ
ϕ
ϕt−t,t
ϕ
01
12
23
1. Let M0 −→
M1 −→
M2 −→
· · · −→ Mt be a sequence of modules
and homomorphisms
over
a
ring
R. We say that the squence is exact if
¡
¢
¡
¢
image ϕi−1,i = ker ϕi,i+1 for 0 < i < t .
(a) Any subsequence of an exact sequence is exact.
(b) 0 −→ M −→ N is exact iff M −→ N is 1 − 1
(c) M −→ N −→ 0 is exact iff M −→ N is onto
ϕ
ψ
(d) M −→ N −→ P is exact iff image (ϕ) = ker (ψ)
2. A short exact sequence is an exact sequence of the form:
ϕ
ψ
0 −→ M −→ N −→ P −→ 0
(a) ϕ is 1 − 1, ψ is onto and image(ϕ) = ker (ψ).
(b) If M is a submodule of M then we have an exact sequence:
0 −→ M −→ N −→
M
−→ 0
N
(c) Is M ⊂ N ⊂ T is a tower of submodules, then the third homomorphism theorem says that we have an exact sequence:
0 −→
T
T
N
−→
−→
−→ 0
M
M
N
0.9. MODULES
lxxiii
3. Here’s a sample theorem using short exact sequences. Suppose the following diagram is commutative with exact rows:
0 −→
0 −→
ϕ
M −→ N
↓α
↓β
M0
ϕ0
−→
N0
ψ
−→
ψ0
−→
P
↓γ
P0
−→ 0
−→ 0
If α and γ are bijective, then so is β.
(a) Proof by diagram chasing, which is sometimes not written down. Let
n ∈ N be such that β (n) = 0. To show β is 1 − 1, we must show
n = 0. First, γψ (n) = ψ 0 β (n) = 0. Since γ is bijective, ψ (n) = 0.
Therefore there exists m ∈ M such that n = ϕ (m) .Moreover 0 =
β (n) = βϕ (m) = ϕ0 α (m). Since ϕ0 is 1 − 1, α (m) = 0. But α is
bijective, so m = 0 and therefore n = ϕ (m) = 0.
Next let n0 ∈ N 0 . To show β is onto we must show n0 ∈ image (β).
Choose p ∈ P such that γ (p) = ψ 0 (n0 ). Choose n” ∈ N such that
ψ (n”) = p. Then ψ 0 (n0 ) = γ (p) = γψ (n”) = ψ 0 β (n”). Therefore
ψ 0 (n0 − β (n”)) = 0. Choose m0 ∈ M 0 such that n0 −β (n”) = ϕ0 (m0 ),
and choose m ∈ M such that α (m) = m0 . Thus n0 − β (n”) =
ϕ0 α (m) = βϕ (m). Therefore n = β (n” + ϕ (m)).
lxxiv
LECTURES
0.10
Categories
1. The idea is to formalize the notion that what matters is not specific sets
that form groups or rings or modules but rather isomorphism classes of
such objects.
(a) We want to generalize the idea of algebraic object and homomorphism, capturing concepts like 1-1 and onto,.isomorphism and quotient, direct sum, etc. in the most general way possible.
(b) The trick is to drop the assumption that basic objects are sets!
(c) We are going to have to use classes, which are bigger than sets
and don’t satisfy all the properties of sets. You can assume that
something is in a class, but you may not be able to tell if two things
are the same or not.
(d) History of category theory would be a great project
2. A category C has three parts:
(a) Obj (C), a class (of objects)
(b) For every A, B ∈ Obj (C), there is a set Hom (A, B) called the
morphisms\A dioa from A to B
f
i. Sometime we write f : A −→ B or A −→ B for f ∈ Hom (A, B).
(c) For every A, B, C ∈ Obj (C) there is a "composition map" Hom(A, B)×
Hom (B, C) −→ Hom (A, C) denoted (f, g) −→ gf
i. Hom sets are disjoint. Every homomorphism has a unique domain and codomain.
ii. For every A ∈ C there is a unique element 1A ∈ Hom (A, A) such
that f 1A = f for all morphisms f with domain A and 1A g = g.for
all morphisms g with codomain A.
h
g
f
iii. Composition is associative: if D −→ C −→ B −→ A then
(f g) h = f (gh)
3. Examples:
(a) Sets
(b) Groups
(c) abelian groups
(d) CommRings
(e)
R Mod
(f)
Z Mod
(g) Top
= Ab
0.10. CATEGORIES
lxxv
4. f : A −→ B is an equivalence or isomorphism in a category if there
exists g : B −→ A such that f g = 1B and gf = 1A
(a) g is the inverse of f
(b) g is unique because if g and h were both inverses than g = g1B =
g (f h) = (gf ) h = 1A h = h
5. A category is pre-additive if
(a) Hom (A, B) is an abelian group
(b) h (f + g) = hf + hg and (f + g) h = f h + gh
(c)
R Mod
is pre-additive
6. We can draw diagrams in categories
(a) formally, they are directed multigraphs with vertices consisting of
objects of the category and arrows being morphisms
(b) A diagram is commutative if any two paths between the same vertices define the same morphism (by composition).
7. In a category C the coproduct of two objects A and B, denoted A t B, is
an object Q together with morphisms α : A −→ Q and β : B −→ Q (called
the injection morphisms) such that, for any object X and morphisms
f : A −→ X and g : B −→ X there exists a unique morphism h : Q −→ X
such that the following diagram commutes:
A
Q
α.
&f
h
99K
β-
X
%g
B
(a) We really should say that Q, α, β is a coproduct of A and B.
(b) If Q, α, β is a coproduct of A and B, then the only map h that fills
in the diagram is 1Q
A
Q
α.
β-
h
99K
B
&α
%β
Q
lxxvi
LECTURES
(c) If Q, α, β and Q0 , α0 , β 0 are coproducts for A and B, then the map h
that fills in the diagram is an equivalence of Q and Q0
A
Q
α.
& α0
h
99K
β-
%β
B
0
Q0
i. The inverse map is the unique map h0 : Q0 −→ Q that fills in the
opposite diagram.
8. In Sets, the coproduct is the disjoin union
(a) The injection map is the usual A −→ A ∪ B
9. In R Mod the coproduct is the direct sum
(a) The injection map for A is A −→ A ⊕ B, a −→ (a, 0).
10. This kind of definition is called a universal mapping definition
11. In a category C a product of objects A and B, denoted A u B, is an
object P together with morphisms α : P −→ A and β : P −→ B (called
the projection morphisms) such that, for any object X and morphisms
f : X −→ A and g : X −→ B there exists a unique morphism h : X −→ P
such that the following diagram commutes:
A
X
f%
g&
h
99K
B
-α
P
.β
(a) We really should say that P, α, β is a product of A and B.
(b) Products, when they exists, are unique up to unique isomorphism
commuting with the projection maps.
12. In Sets the product is the Cartesian product (so the product and coproduct are different)
(a) the projection A × B −→ A is (a, b) −→ a.
13. In R Mod the product is the direct product (so the product and coproduct
are the same).
(a) The projection map A ⊕ B −→ A is (a, b) −→ a.
0.10. CATEGORIES
lxxvii
14. Products and coproducts can be extended to any set of objects (with no
guarantee that they exists, but if they do they are unique up to unique isomorphism commuting with the structure (injection and projection) maps.
G
(a) If {Ai : i ∈ I} is the set of objects, then the coproduct is
Ai
i∈I
and the product is ui∈I Ai .
15. In Sets, the coproduct and the product are the disjoint union and the
cartesian product respectively (provided you believe they exist).
lxxviii
LECTURES
16. In R M od,
(a)
G
i∈I
Ai =
(b) ui∈I Ai =
X
Ai =
i∈I
Y
i∈I
M
i∈I
©
ª
Ai = (ai )i∈I : almost all ai = 0
©
ª
Ai = (ai )i∈I
(c) So, in a natural way,
X
i∈I
Ai ⊂
Y
Ai
i∈I
17. Theorem: for R-modules Ai and B
homR
Ã
homR
B,
Y
i∈I
³X
Ai
!
Ai , B
(a) Proof of the first: let πj :
Y
i∈I
Y
´
∼
=
∼
=
Y
homR (B, Ai )
i∈I
Y
homR (Ai , B)
i∈I
Ai −→ Aj . Define T : homR
Ã
B,
Y
i∈I
Ai
!
−→
homR (B, Ai ) by T (ϕ)i = π i ϕ. Conversely define . Define S :
Ã
!
Y
Y
Y
homR (B, Ai ) −→ homR B,
Ai as follows: if ψ ∈
homR (B, Ai )
i∈I
i∈I
i∈I
i∈I
then S (ψ) is defined by S (ψ) (b) = (ψ i (b))i∈I . It is easy
à to see that
!
Y
ST = Id and T S = Id . Let’s check ST . If ψ ∈ homR B,
Ai ,
i∈I
We must show ST (ϕ) = ϕ,or, for each b ∈ B, ST (ϕ) (b) = ϕ (b), or
for each b ∈ B and i ∈ I, (ST (ϕ) (b))i = ϕ (b)i .
(ST (ϕ) (b))i
=
=
=
=
(S (T (ϕ)) (b))i
T (ϕ)i (b)
(π i ϕ) (b)
ϕ (b)i
´
Y
homR (Ai , B)
Ai , B −→
i∈I
´
³X
Y
by T (ϕ)i = ϕ|Ai . Conversely define S :
homR (Ai , B) −→ homR
Ai , B
(b) Proof of the second: Define T : homR
by S (ψ) (ai ) =
X
i
³X
i∈I
ψ i (ai ). Since only a finite number of ai are non-
0.10. CATEGORIES
lxxix
zero, this last sum is defined. Let’s show that ST = Id.
X
ST (ϕ) (a) =
T (ϕ)i (ai )
i
X
=
ϕ|Ai (ai )
i
= ϕ (a)
18. Given
B
C
↓g
f
−→
A
a pullback of this diagram is a triple D, α, β such that
(a) the following diagram commutes:
α
D −→ C
β↓
↓g
B
f
−→
A
(b) Given any commutative diagram
α0
X −→ C
β0 ↓
↓g
B
f
−→
A
θ
there exist unique morphism X −→ D such that the following diagram commutes:
α0
X
β0 ↓
B
&θ
β.
−→
D
C
α%
f
−→
↓g
A
19. In Sets the pullback is {(b, c) ∈ B × C : f (b) = g (c)}. The maps α and
β are restrictions of the projection maps.
20. In R Mod, the pullback is {(b, c) : f (b) = g (c)} ⊂ B ⊕ C. This is a submodule.
(a) The pullback of
B
f
−→
0
↓g
A
lxxx
LECTURES
is ker f .
21. Given
B
C
↑g
f
←−
A
the pushout is the same as the pullback with all the arrow reversed.
22. In Sets, the pushout is defined as follows: in B ∪C identify (glue together)
b and c if there exists a such that b = f (a) and c = g (a).
23. In R Mod the pushout is defined as follows: let S = {(f (a) , −g (a)) : a ∈ A} ⊂
B⊕C
B ⊕ C. This is a submodule. The pushout is
.
S
(a) The pushout of
B
is coker (f ) =
f
←−
B
.
image (f )
24. I don’t understand Example 7.39 (i).
0
↑g
A
0.10. CATEGORIES
0.10.1
lxxxi
Functors
1. Let C and D be categories. A covariant functor T : C Ã D is
(a) If A ∈ Ob (C) then T (A) ∈ Ob (D)
(b) T : homC (A, B) −→ homD (T (A) , T (B))
(c) Given
A
f
−→
B
gf & ↓ g
C
then
T (A)
T (f )
−→
T (B)
T (gf ) & ↓ T (g)
T (C)
is commutative. I.e. T (g) T (f ) = T (gf )
(d) For all A ∈ Ob (C), T (1A ) = 1T (A)
2. Covariant functors map commutative diagrams to commutative diagrams
3. Covariant functors map equivalences to equivalences
f
T (f )
g
(a) If A −→ B −→ A is such that f g = 1B and gf = 1A , then T (A) −→
T (g)
T (B) −→ T (A) and T (f ) T (g) = T (f g) = T (1B ) = 1T (B) and
T (g) T (f ) = T (gf ) = T (1A ) = 1T (A) .
4. Examples of covariant functors
(a) identity functor
(b) forgetful functor T : Group à Sets
i. T (G) = G as a set (forget the operation)
(c) For some category C, choose A ∈ Ob (C). The Hom functor TA :
C Ã Sets is given by:
i. For any B ∈ Ob (C), TA (B) = homC (A, B)
ii. For any B, B 0 ∈ Ob (C), we must define TA : homC (B, B 0 ) −→
homSets (TA (B) , TA (B 0 )) = homSets (homC (A, B) , homC (A, B 0 )).
A. for f : B −→ B 0 , TA (f ) : homC (A, B) −→ homC (A, B 0 ) is
"composition with f on the left"
A
α
−→
B
& TA (α) = f ◦ α ↓ f
B0
lxxxii
LECTURES
(d) The example above, applied to R Mod gives a function TA :R Mod ÃR
Mod.
5. Let C and D be categories. A contravariant functor T : C Ã D is
(a) If A ∈ Ob (C) then T (A) ∈ Ob (D)
(b) T : homC (A, B) −→ homD (T (B) , T (A))
(c) Given
f
A
−→
B
gf & ↓ g
C
then
T (C)
T (f )
−→
T (B)
T (gf ) & ↓ T (g)
T (A)
is commutative. I.e. T (f ) T (g) = T (gf )
(d) For all A ∈ Ob (C), T (1A ) = 1T (A)
6. Contravariant functors map commutative diagrams to backwards commutative diagrams
7. Contravariant functors map equivalences to equivalences.
8. Example of a contravariant functor:
(a) B ∈ Ob (C). Then T B : C ÃSets
i. For any A ∈ Ob (C), T B (A) = homC (A, B)
0
ii. For any ¡A, A0 ∈ Ob (C), we
C (A, A ) −→
¢ must define TB : hom
B
0
B
0
homSets T (A ) , T (A) = homSets (homC (A , B) , homC (A, B)).
A. for f : A −→ A0 , T B (f ) : homC (A0 , B) −→ homC (A, B) is
"composition with f on the right"
f
A
−→
& T (α) = α ◦ f
A0
↓α
B
9. How about exact sequences?
f
g
(a) In R Mod, if 0 −→ A −→ B −→ C is exact then for any R-module
X we have the exact sequence:
f∗
g∗
0 −→ homR (X, A) −→ homR (X, B) −→ homR (X, C)
0.10. CATEGORIES
lxxxiii
(b) Similarly, if A −→ B −→ C −→ 0 is exact, then we have the exact
sequence:
g∗
f∗
0 −→ homR (C, X) −→ homR (B, X) −→ homR (A, X)
(c) We will check (a). We must show first that if f is 1−1, then f∗ is 1−1.
Since f∗ is a homomorphism of R-modules, it suffices to show that
f∗ (α) = 0 (as a map X −→ B) implies α = 0 (as a map X −→ A).
Suppose for all x ∈ X, f∗ (α) (x) = (f ◦ α) (x) = f (α (x)) = 0. Since
f is 1 − 1, α (x) = 0 for all x ∈ X. Thus α = 0 as a map X −→ A.
Next we must check that image (f∗ ) = ker (g∗ ). First, to show
image (f∗ ) ⊂ ker (g∗ ), we will show g∗ ◦ f∗ = 0. But this is obvious since for any α ∈ homR (X, A), (g∗ ◦ f∗ ) (α) = g ◦ f ◦ α = 0
because g ◦ f = 0. Conversely, we must show ker (g∗ ) ⊂ image (f∗ ).
Suppose β ∈ ker (g∗ ). That is, β ∈ homR (X, B) and g ◦ β = 0.
We must find α ∈ homR (X, A) such that f∗ (α) = f ◦ α = β. But
image (β) ⊂ ker (g) = image (f ), Since f is 1 − 1 we have a map f −1 :
image (f ) −→ A. Let α = f −1 ◦ β. Obviously f ◦ α = f ◦ f −1 ◦ β = β.
lxxxiv
LECTURES
10. A covariant functor T :R Mod à Ab is called left-exact if 0 → A →
B → C is exact implies 0 → T (A) → T (B) → T (C) is exact
(a) example: T (B) = homR (X, B) for X fixed
(b) The functor T is exact if 0 → A → B → C → 0 is exact implies
0 → T (A) → T (B) → T (C) → 0 is exact
(c) If X = R then the functor homR (X, .) is exact
11. A contravariant functor T :R Mod à Ab is called left-exact if A →
B → C → 0 is exact implies 0 → T (C) → T (B) → T (A) is exact
(a) example: T (B) = homR (B, X) for X fixed
(b) The functor T is exact if 0 → A → B → C → 0 is exact implies
0 → T (C) → T (B) → T (A) → 0 is exact
(c) In the next section, we will construct modules X making the functor
homR (., X) exact. These are the injective modules
12. In R mod, if A → B → C and for all M the induced sequence 0 →
homR (M, A) → homR (M, B) → homR (M, C) is exact, then 0 → A →
B → C is exact.
(a) Proof: homR (R.A) ∼
= A, so let M = R.
13. In
R mod,
f
g
if A → B → C and for all M the induced sequence 0 →
g
∗
f∗
f
g
homR (C, M ) → homR (B, M ) → homR (A, M ) is exact, then A → B →
C → 0 is exact.
(a) Proof: it is all a matter of choosing the right values for M . To show
C
g is surjective, let M =
, and let π ∈ homR (C, M ) be the
image (g)
projection. Then g ∗ (π) = π ◦ g = 0. Since g ∗ is injective , π = 0, or
image (g) = C. Thus g is surjective.
To show that image (f ) ⊂ ker (g), let M = C and φ ∈ homR (M, C),
∗
φ = 1C . Then 0 = (g ∗ f ∗ ) (φ) = (f g) (φ) = φgf = gf . Thus
image (f ) ⊂ ker (g).
B
To show that ker (g) ⊂ image (f ), let M =
and let π ∈
image (f )
homR (B, M ) be the projection. Then f ∗ (π) = πf = 0. Therefore
there exists a map ρ ∈ homR (C, M ) such that g ∗ (ρ) = π. If b ∈
ker (g), b ∈
/ image (f ), then gb = 0 and π (b) 6= 0. But this is a
contradiction, since π (b) = g ∗ (ρ) (b) = ρg (b) = 0.
0.10. CATEGORIES
0.10.2
lxxxv
Free, Projective and Injective modules
Throughout this section, R is a ring (commutative with identity, as always).
1. A free module over R is (isomorphic to) a direct sum of (possibly infinite)
copies of R. (The text says a direct sum of modules isomorphic to R.)
∼ P Ri , and if we define xi = (δ ij )j∈I , then the
(a) If we let F =
i∈I
X = {xi } form a basis for the free module F over R.
P
(b) Every f ∈ F can be written f = i∈I ri xi , where ri ∈ R are unique
and only a finite number are non-zero.
(c) If M is an R-module, then every (set) function X → M extends
uniquely to a homomorphism F → M .
(d) Given any set X whatsoever, we can ©
form
ª
P a free module F with X as
its basis. Simply form all expressions
x∈X rx x : rx ∈ R, almost all rx = 0 .
(e) The 0 modules is free with the empty set for a basis.
F
R
2. If I ⊂ R is an ideal then
is a free module over , and X = {Xi } is a
IF
I
F
R
basis. Thus F and
have bases of the same cardinality over R and
IF
I
respectively.
3. For a free module F , the cardinality of the basis X is the rank of F . This
number is well-defined:
(a) Proof: If R is a field, this is a theorem of linear algebra. In general,
(watch out–here comes the Axiom of Choice), choose a maximal
ideal M ⊂ R. Since a basis for F has the same cardinality as a
F
basis for
, the cardinality of any basis for F is the cardinality of
MF
F
R
any basis for the vector space
over the field
, and all these
MF
M
cardinalities are the same.
4. Two free modules over R have the same rank if and only if they are
isomorphic.
(a) The only if part is obvious. Rank is an algebraic property and so
is preserved by isomorphism. (The text gives a proof that I think
is unnecessary.) To prove that equal rank implies isomorphic, the F
and F 0 be the free modules with bases X and X 0 of the same cardinality. Let φ : X → X 0 be bijective. Then
uniquely to an
¡P φ extends
¢
P
isomorphism φ : F → F 0 given by φ
r
x
=
x
x∈X
x∈X rx φ (x).
(Remember, the set {φ (x) : x ∈ X} is the basis X 0 for F 0 .)
5. Every R-module M is the quotient of a free R-module F . If M is finitely
generated, then F can be chosen to be finitely generated.
lxxxvi
LECTURES
(a) Choose a basis X for M , choosing X to be finite if M is finitely
generated. Construct a free module F on X considered as a set. The
inclusion map X → M induces a homomorphism F → M that is
surjective.
6. If F is a free module and we have a diagram with an exact row:
f
A →
F
↓h
A00
→ 0
then we can find a map g : F → A making the diagram commute:
A
F
g. ↓h
f
→
A00
→ 0
(a) Choose a basis X of F and–using the surjectivity of f –choose elements ax ∈ A, f (ax ) = h (x). Map X → A by sending x to ax and
extend to a homomorphism g : F → A. Since f g (x) = h (x) for all
x ∈X, a basis of F , f g = h and the diagram commutes
7. An R-module P is projective if in every diagram with an exact row:
f
A →
P
↓h
A00
→ 0
the map h : P → A00 can be lifted to a map g : P → A so that the
following diagram commutes:
A
P
g. ↓h
f
→
A00
→ 0
8. Free modules are projective.
(a) Are there any non-free projective modules?
(b) Examples come later. Serious examples are difficult.
9. The following statements are equivalent for an R-module P :
(a) P is projective
(b) the functor homR (P, .) is exact.
0.10. CATEGORIES
lxxxvii
g
f
(c) Given a short exact sequence 0 → A → B → P → 0 then B ∼
= A⊕P
in the following sense: there is a diagram with exact rows and the
top row has the usual maps:
0 →
0 →
A
→ A⊕P
1A ↓
↓∼
=
g
A
→
→
f
B
→
P
↓ 1P
P
→ 0
→ 0
i. To prove C, it is only necessary to construct a homomorphism
B → A ⊕ P making the diagram commutative. Then the middle
map must be bijective.
(d) P is a direct summand of a free module.
10. And here’s the proof.
(a) (a) =⇒ (b) Since homR (P, .) is left exact, it suffices to show that
if A → A00 → 0 is exact then homR (P, A) → homR (P, A00 ) → 0
is exact; i.e. if f : A → A00 is surjective then f∗ : homR (P, A) →
homR (P, A00 ) is surjective. Let h : P → A00 be a homomorphism.
P
↓h
f
A00
A →
→ 0
i. Since P is projective, we get a commutative diagram
P
g. ↓h
f
A
A00
→
→ 0
where f∗ (g) = f g = h.
(b) (b) → (c) We have the diagram
f
B
→
P
↓ 1P
P
→ 0
so we get a map h : P → B lifting 1P
B
P
h . ↓ 1P
f
→
P
→ 0
Thus we get a map (g, h) : A ⊕ P → B. We need only show that this
map yields a commutative diagram:
0 →
0 →
A
→ A⊕P →
P
1A ↓
↓ (g, h)
↓ 1P
A
g
→
B
f
→
P
→ 0
→ 0
lxxxviii
LECTURES
That is, we must show that f (g, h) (a, p) = p and g (a) = (g, h) (a, 0).
But
f (g, h) (a, p) =
=
=
=
f (g (a) + h (p))
f g (a) + f h (p)
0 + 1P (p)
p
(g, h) (a, 0) = g (a) + h (0)
= g (a)
i. Corollary: Let P be projective and f : A → P be a surjective
homomorphism. Then P is a direct summand of A.
A. Proof: 0 → ker (f ) → A → P → 0 is exact so A ∼
= ker (f ) ⊕
P.
(c) (c) =⇒ (d) There is a free module F and a surjective homomorphism
F → P.
(d) (d) → (a) Let F = P ⊕ Q be a free module. Given a diagram:
P
↓h
f
A00
A →
→ 0
add P ⊕ Q and the usual maps between P ⊕ Q and P :
π
P ⊕Q À
P
i
A
f
→
↓h
A00
→ 0
Since P ⊕ Q is free, the map hπ : P ⊕ Q → A00 can be lifted to
g : P ⊕ Q → A:
π
P ⊕Q À
P
i
g↓
A
f
→
↓h
A00
→ 0
That is f g = hπ. Thus f gi = hπi = h, so gi lifts h and P is
projective.
0.10. CATEGORIES
lxxxix
11. An R-module E is injective if the left-exact contravariant function homR (., E)
is exact.
(a) If 0 → A0 → A → A00 → 0 then 0 → homR (A00 , E) → homR (A, E) →
homR (A0 , E) → 0
(b) if
then
0 → A0
↓
E
→ A
0 → A0
↓
E
→ A
.
(c) Every map A0 → E extends to a map A → E
(d) if i : A0 → A is injective then i∗ : homR (A, E) → homR (A0 , E) is
surjective
(e) Slogan: any theorem that is true about projective modules is true
about injective modules with the arrows reversed.
12. A module is injective if and only if every short exact sequence 0 → E →
B → C → 0 splits, i.e.if there is a commutative diagram
0 →
E
→
B
∼
1E ↓
=↓
0 →
E
→ E⊕C
→
C
↓ 1C
→
C
→ 0
→ 0
(a) Every injective submodule is a direct summand. E ⊂ M and E injective implies M = E ⊕ E 0 for some E 0 (which need not be injective).
13. The direct sum of projective modules is projective
(a) Given
A →
X
Pi
↓
A00
→ 0
we get for each i
Pi
. ↓
A → A00
and therefore:
.
A →
X
→ 0
Pi
↓
A00
→ 0
xc
LECTURES
(b) Why doesn’t this work for the direct product of projective modules?
14. The direct product of injective modules is injective.
(a) Given
0 →
we get for each i
A0
Y↓
Ei
A0
↓
Ei
0 →
and therefore:
0 →
A0
Y↓
Ei
→ A
→ A
.
→ A
.
(b) Why doesn’t this work for the direct sum of injective modules?
15. Baer’s Criterion Let’s take the following result for granted. Proof is in
the book and takes a whole page. An R-module E is injective iff for each
ideal I ⊂ R, every R-homomorphism I → E can be extended to R → E.
0 →
I
↓
E
→ R
.
(a) Clearly this is true for an injective module. The hard part is to prove
that a module is injective if it is injective for ideals.
(b) Proof requires Axiom of Choice in the form of Zorn’s Lemma.
16. Assume for the rest of this section that D is an integral domain.
17. F = f rac (D) is an injective D-module
(a) It suffices to show that if I is an ideal of R and f : I → D then f
extends to a map g : R → D. First we prove a remarkable fact: if
f (a)
f (b)
a, b ∈ I. a, b 6= 0, then in D,
=
. Since both f (a) and
a
b
f (b) are defined, and ab ∈ I so f (ab) is also define. Then
af (b) = f (ab) = bf (a)
f (a)
for all a ∈ I, a 6= 0. Then define g : R → D by
a
g (r) = rc. It remains to show that g (a) = f (a) for all a ∈ I, a 6= 0.
Let c =
0.10. CATEGORIES
xci
(It is obvious that g (0) = g (0) = 0.)
g (a) = ac
f (a)
= a
a
= f (a)
18. An D-module M is divisible if for every m ∈ M and d ∈ D, d 6= 0, there
exists an element m0 ∈ M such that dm0 = m. That is, we can divide by
d in M .
(a) f rac (D) is divisible
(b) An abelian group A is divisible iff a ∈ A and n ∈ Z, n 6= 0 implies
a = na0 for some a0 ∈ A.
(c) Every direct sum and quotient of divisible modules is divisible.
19. Every injective D-module E is divisible
(a) Choose e ∈ E and d ∈ D, d 6= 0. We must find e0 ∈ E with de0 = e.
Define f : (d) → E by f (da) = ae. Since D is a domain, this is well
defined. Since E is injective, we can extend f to g : D → E. Let
e0 = g (1). Then
de0
= dg (1)
= g (d)
= e
20. If D is a P ID and E is a divisible module, then D is injective.
(a) So divisible abelian groups are injective Z-modules
Q
(b) Q, R, C,
are all injective over Z
Z
(c) Proof: suppose E is divisible and I ⊂ D is an ideal and f : I → E
. We must find g : D → E extending f . Let I = (d0 ). Choose
e0 ∈ E such that d0 e0 = f (d0 ). Define g : D → E by g (d) = de., a
homomorphism. To check that g extends f , suppose d0 r ∈ I. Then
f (rd0 ) = rf (d0 )
= rd0 e0
= g (rd0 )
xcii
LECTURES
0.11
Linear Algebra
0.11.1
Modules over PIDs
1. Why would a section on vector spaces begin with a section on modules
over PID’s
(a) Remember if T : V −→ V was a linear map of k-vector spaces, then
V became a k [X]-module by Xv = T (v).
(b) and k [X] is a PID
(c) So maybe studying modules over PIDs will tell us something about
vector spaces and linear maps
2. For the rest of this section, let R be a PID and R∗ = R˜ {0}. Since R is a
domain, R∗ is closed under products. Let M be a finitely generated (fg)
R-module. The theorem we aim for is:
(a) Let MT = {m ∈ M : ∃t ∈ R∗ , tm = 0j}. MT is the torsion submodule of M . and is finitely generated.
(b) There exists a finitely generated free submodule F ⊂ M such that
M = MT ⊕ F . The submodule F is not unique.
©
ª
(c) For each prime element p ∈ R let Mp = m ∈ M : pi m = 0 ⊂ MT .
i. Mp = 0 for almost all primes p
M
Mp . We call Mp the p-primary component of MT .
ii. MT =
primes p
It is uniquely determined by MT .
iii. For each p there exist unique integers e1 , . . . , etp such that Mp ∼
=
R
R
⊕ · · · ⊕ e . The isomorpohism is not unique.
(pe1 )
(p )
(d) Therefore (to be justified later) there exist r1 , . . . , rt ∈ R∗ such that
R
R
ri | ri+1 and MT ∼
⊕···⊕
. The isomorphism is not unique
=
(r1 )
(rt )
but the ideals (ri ) are uniquely determined by MT .
3. Let’s start by proving that MT ⊂ M is a submodule. The element 0 ∈ MT
so MT 6= φ. If a, b ∈ MT and r, s ∈ R∗ are such that ra = sb = 0 then
rs ∈ R∗ and rs(a + b) = 0. Therefore a + b ∈ MT . Moreover for any t ∈ R,
r (ta) = tra = 0 so ta ∈ MT . Thus MT is a submodule of M .
4. We say M is a torsion module if M = MT , if for every m ∈ M we can
find r ∈ R∗ such that rm = 0. We say M is torsion-free if MT = {0}.
(a)
M
M
is torsion-free. Proof: suppose m̄ ∈
and there exists r ∈ R∗
MT
MT
such that rm̄ = 0̄. We must show m̄ = 0̄. Since rm̄ = rm = 0̄,
rm ∈ MT so there exists s ∈ R∗ such that s (rm) = 0. But sr ∈ R∗
and (sr) m = 0 so m ∈ MT and m̄ = 0̄.
0.11. LINEAR ALGEBRA
(b)
xciii
M
is also finitely generated, because it is the quotient of a finitely
MT
generated module.
5. For any finitely generated module M we have an exact sequence:
0 −→ MT −→ M −→
M
−→ 0
MT
M
were free, imagine what we could
(a) If the torsion-free module
MT
prove.
M
i. The sequence would split, so M ∼
= MT ⊕
MT
ii. MT is a quotient of M so MT is finitely generated (we knew that
already–all submodules of a fg modules over a Noetherian ring
are finitely generated).
6. So I guess we better prove that a finitely generated torsion-free module
over a PID is free. Let N be the module.
(a) Begin with a lemma: let Q be the quotient field of R and suppose S
is a fg submodule of Q. Then S ∼
= R.
a1
at
i. Let s be generated by
, . . . , , where ai ∈ R, bi ∈ R∗ . Let
b1
bt
b = b1 · · · bt , and map S −→ R by s −→ bs. Since S is torsionfree because Q is torsion-free, the map is injective. The image is
a submodule of R, and since R is a PID all submodules are free
of rank 1. Therefore S is free of rank 1.
(b) The proof that N is free is by induction on the number of generators
R
of N. If N has one generator then N ∼
for some r ∈ R. Since N
=
(r)
∼
is torsion-free, r = 0 and N = R and is therefore free.
Let N have t generators v1 , . . . , vt , with vt 6= 0. (It wouldn’t hurt to
assume all the vi 6= 0, but we only need vt 6= 0.) Assume by induction
that every torsion-free module with fewer than t generators is free.
N
We have a projection π : N −→
= P . P is generated by
Rvt
P
v1 , . . . , vt−1 . Thus
also has t − 1 generators and is torsion-free,
PT
P
N ∼ P
so
is free. Let S = π−1 (PT ). Then
is free. Therefore
=
PT
S
PT
N
is projective and the sequence
S
0 −→ S −→ N −→
N
−→ 0
S
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LECTURES
N
splits: N ∼
=S ⊕ . If we can prove that S is free, we are done.
S
The trick is to map S −→ Q injectively, where Q =qf(R). By the
lemma above then, S is free of rank 1. If s ∈ S then π (s) ∈ PT and
for some a ∈ R∗ we have aπ (s) = 0. Thus π (as) = 0 or as ∈ Rvt .
Thus there exists a0 ∈ R such that as = a0 vt . Define φ : S → Q by
a0
φ (s) = . It isn’t too hard to show that φ is a homomorphism once
a
you have shown that φ is well-defined. We’ll do that and quit.
If as = a0 vt and bs = b0 vt where a, b ∈ R∗ and a0 , b0 ∈ R, then
abs = ba0 vt = ab0 vt . These expressions all represent elements in N .
Since (ba0 − ab0 ) vt = 0 and vt 6= 0 and N is torsion-free, ba0 − ab0 = 0
ba0 − ab0
a0
b0
a0
b0
in R. Therefore
=
−
= 0 in Q, or
=
and φ is
ab
a
b
a
b
well-defined.
7. Every submodule G of a fg free module F over R is free, and rank (G) ≤
rank (F ).
(a) The proof is by induction n. If n = 1 then F ∼
= R and all submodules
are free of rank no more than 1. If F has a basis f1 , . . . , fn let F 0
be the free submodule generated by f1 , . . . , fn−1 . Consider the exact
sequence:
G
0 → F0 ∩ G → G →
→0
G ∩ F0
By induction, F 0 ∩ G is free on no more than n − 1 generators. On
F
G
G + F0
G
∼
⊂ 0 ∼
the other hand
is
=
= R. Therefore
0
0
G∩F
F
F
G∩F
G
free of rank no more than 1.Since G ∼
, G is free
= (F 0 ∩ G) ⊕
G∩F
of rank no more than n.
0.11. LINEAR ALGEBRA
xcv
8. Every fg torsion R-module is the direct
M sum of its primary components.
That is, if M = MT then M =
Mp .
prime ideals (p)
(a) Proof: for each m ∈ M we must show that there exist
P a finite number
of primes pi ∈ R and elements mi ∈ M such that
mi = m and for
some exponents ri , pri i m = 0. Let ann (m) = {r ∈ R : rm = 0}. This
is an ideal of R, non-zero because M is torsion. Suppose ann (m) =
(d) and d = upr11 · · · prt t , where
Y r pi are primes and u is a unit. We
pj j , so pri i qi = d. The qi are relatively
can assume u = 1. Let qi
j6=i
P
ai qi = 1. Let mi =Pai qi m.
prime, so there exists ai ∈ R such that
P
Then pri i mi = 0, so mi ∈ Mpi , and
mi = m. Thus M = p Mp .
It remains to show that the sum is direct.
X
If q ∈ R is prime, we must show that Mq ∩
Mp = 0. If m ∈
Mq ∩
X
(p)6=(q)
Mp then q r m = 0 for some r, and m =
P
f inite
mi where
(p)6=(q)
Y
pri i mi = 0 for primes pi 6= q. Let c =
pri i , so cm = 0. But q r and c
have no common prime factors, so they are relatively prime. Choose
a, b such that aq r + bc = 1. Then m = (aq r + bc) m = 0.
(b) We still have not shown that only a finite number of Mp are non-zero.
Let’s do that now. Since M is finitely generated, M has generators
{m1 , . . . , mt }. Since M is torsion, we
Ycan choose non-zero elements
ri ∈ R such that ri mi = 0. Let r =
ri . Then r 6= 0 and rm = 0.
If Mp 6= 0 then I claim p divides r. Thus only a finite number of Mp
are non-zero. To prove the claim, suppose p does not divide r. I will
show that Mp = 0.Choose m ∈ Mp . Then for some e, pe m = 0. Also
rm = 0. Since p is prime, (pe , r) = (1), so there exists a, b ∈ R such
that ape + br = 1. Therefore m = (mpe + br) m = 0.
9. So far we have shown that any fg R-module M = F ⊕MT = F ⊕(
P R
It only remains to prove that Mp ∼
= i ei .
(p )
P
⊕Mp ).
10. Let p be a prime of R. A p-primary R-module is a module M such that,
for all m ∈ M , there exists e such that pe m = 0. If M is also fg, then for
some e, pe M = 0.
11. We must show that a fg p-primary R-module is a direct sum of cyclic
R
modules of the form f for different values of f . Let’s assume that
(p )
M 6= 0, or the theorem is vacuously true. There exists a positive integer
e such that pe M = 0, pe−1 M 6= 0. The proof has two parts:
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(a) Every fg p-primary module has a non-zero cyclic direct summand of
maximal order:
R
M∼
= e ⊕ M0
(p )
(b) If (a) is true, then every fg p-primary modules is a direct sum of
R
cyclic submodules of the form f for different values of f .
(p )
First we assume (a) and prove (b). Note that pe M 0 = 0 and M 0 is a
quotient of M so M 0 is a fg p-primary module. Thus M 0 has a direct
R
summand of the form f . Continuing the process assumed possible, we
(p )
get
R
R
R
M∼
= e ⊕ f1 ⊕ · · · ⊕ ft ⊕ M 00
(p ) (p )
(p )
Consider
t+1 factors
z
}|
{
M ∼ R
R
M 00
⊕ ··· ⊕
⊕
=
pM
(p)
(p) pM 00
M
R
R
R
is a field, and
is a fg
-module, or a finite dimensional
(p)
pM
(p)
(p)
M
vector space. Thus t ≤ dim
and the process of extracting cyclic
pM
factors cannot continue forever. Eventually M 00 = 0. In fact the number
M
of cyclic factors of M will always be dim
.
pM
Now we prove (a), and here’s where all the work is. Choose an element s ∈
M such that pe−1 s 6= 0. Let S = Rs ⊂ M . Using the Noetherian property
of M , let M 0 be the a maximal submodule subject to the condition that
M
S ∩ M 0 = 0. I claim S + M 0 = M . Replacing M by
, we still have
M0
e
p M = 0, S ⊂ M and 0 is the maximal submodule not intersecting S.
That is, every non-zero submodule of M intersects S non-trivially. We
must show S = M . If S 6= M we can derive a contradiction. Choose
m ∈ M, m ∈
/ S. Choose the smallest value of f such that pf m ∈ S, and
replace m by pf −1 m. Then m ∈
/ S but pm ∈ S. Let pm = as for some
a ∈ R. Then 0 = pe m = pe−1 as. I claim a ∈ (p); otherwise we can find
b, c ∈ R such that 1 = ab + pc and then pe−1 s = pe−1 (ab + pc) s = 0,
contradicting the construction of s.
Let a = pd. Then pm = pds. Note that m − bs ∈
/ S but p (m − bs) = 0.
Replacing m with m − bs we have m ∈
/ S but pm = 0. I claim S ∩ Rm = 0
which is our contradiction since Rm 6= 0. If n ∈ S ∩ Rm, n 6= 0, then
n = as = bm. If b ∈ (p) then n = 0, so b ∈
/ (p) and there are c, d ∈ R
such that bc + dp = 1. Then m = (bc + dp) m = cn ∈ S, which is also a
contradiction.
0.11. LINEAR ALGEBRA
xcvii
12. Suppose M is a p-primary R-module. We know
z
te factors
}|
{
te−1 factors
z
}|
{
t1 factors
z
}|
{
R
R
R
R
R
R
M∼
⊕ ··· ⊕
= e ⊕ · · · ⊕ e ⊕ e−1 ⊕ · · · ⊕ e−1 ⊕ · · · ⊕
(p )
(p ) (p )
(p )
(p)
(p)
It is possible that ti = 0 for some values of i. We just need a notation
that allows for multiple components of the same size. There is a formula
that allows us to compute the numbers ti . which are uniquely determined
by M even though the isomorphism is not unique.
(a) t1 + · · · + te = dimR/(p)
M
pM
(b) If Mi = pi M then ti + · · · + te = dimR/(p)
i. in particular, te = dimR/(p)
Mi−1
Mi
¢
¡
pe−1 M
= dimR/(p) pe−1 M
e
p M
13. Every fg torsion module M determines a unique set of prime elements
p1 , . . . , pt ∈ R and exponents ei,j , i = 1, . . . , t and j = 1, . . . , s such that
eij ≥ ei,j+1 and M is the direct sum of components:
R
(pe111 )
R
(pe221 )
..
.
R
(pet t1 )
R
(pe112 )
R
(pe222 )
..
.
R
(pet t2 )
···
···
···
R
(pe11s )
R
(pe22s )
..
.
R
(pet ts )
Note: some of the exponents eij may be 0.
(a) The direct sum of each row is the component Mpi .
(b) If you take the direct sum down column i, you get:
R ∼
R
R
⊕ · · · ⊕ eti =
(pe11i )
(pt )
(pe11i · · · pet ti )
The isomorphism follows from the Chinese Remainder Theorem. Define ri = pe11i · · · pet ti . Then ri+1 | ri and
R
R
M∼
⊕ ··· ⊕
=
(r1 )
(rs )
(c) Conversely, if M has the form immediately above, then factoring the
ri in to powers of primes yields a table as above the the primary
decomposition of M .
xcviii
LECTURES
14. The fundamental theorem of abelian groups. Suppose A is a fg
abelian group. Then
(a) A ∼
=
(b) A ∼
=
M
Z
¡ ekj ¢ for a finite collection of primes pi and exponents eij
pi
M Z
where ri+1 | ri
(ri )
0.11. LINEAR ALGEBRA
xcix
Let R be a PID, F a finitely generated, free R-module and G ⊂ F a pure
submodule. That means rF ∩G = rG for all r ∈ R. Then G is a direct summand
of F .
F
F
Proof: it suffices to show that
is torsion-free, because then
is free and
G
G
the exact sequence
F
0→G→F →
→0
G
splits.
F
F
is torsion free, let f¯ ∈ . Suppose r ∈ R, r 6= 0, and rf¯ = 0.
To show
G
G
We will show f¯ = 0. Since rf¯ = rf = 0, rf ∈ G. Thus rf ∈ G ∩ rF = rG, so
rf = rg for some g ∈ G. Thus r (f − g) = 0 in F . Since F is free and r 6= 0,
f − g = 0 or f = g. But g ∈ G so f¯ = 0.
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LECTURES
0.11.2
Applications of the Fundamental Theorem of Modules over PID’s to Linear Algebra
Throughout this section k is a field and V , W , etc. are finite dimensional
k-vector spaces. Occasionally we may assume that it is algebraically closed.
1. A basis for a finite dimensional k-space V is a set B ⊂V that is linearly
independent and spans V .
(a) If B has n vectors v1 , . . . , vn , then we can construct a linear isomorphism k n → V mapping the standard basis vector ei to vi .
(b) If we have a linear isomorphism k n → V then the images of the
standard basis vectors form a basis of V .
(c) Therefore a basis of V is equivalent to a linear isomorphism k n → V .
i. We will abuse our notation and use the same symbol for the
collection of vectors that makes a basis and the corresponding
linear isomorphism.
ii. Actually we will let a symbol like B denote the map and Bi =
B (ei ) be the ith basis vector.
2. This notation allows us to simplify the always troublesome topic: “matrix
of a map”
(a) Every linear map S : k n → k m is (left) multiplication by a matrix
(where the objects in k n and k m are thought of as column vectors).
The matrix is [S (e1 ) · · · S (en )].
(b) If V , W are vector spaces with basis B : k n → V and C : k m → W ,
and if T : V → W is linear, then the matrix of T with respect to the
bases B and C is:
−1
TB
C [T ]B = C
Thus the following diagram is commutative:
V
B↑
kn
T
→
→
C [T ]B
W
↑C
kn
(c) This proves that if we have three spaces U, V, W and three bases
B, C, D and two maps and their composition:
V
U
%S
→
TS
T &
W
Then
D
because:
[T ]C
C
[S]B =D [T S]B
D−1 T C C −1 SB = D−1 T SB
0.11. LINEAR ALGEBRA
ci
3. Suppose V is a k-space and T : V → V is linear. If B is a basis of V , we
will NOT use any shorthand for the matrix B [T ]B . We will not call this
[T ]B for short.
4. Suppose V has two bases B and C. What is the relation between B [T ]B
and C [T ]C . A commutative diagram will show us:
kn
B↓
V
C↑
kn
B [T ]B
→
T
→
→
C [T ]C
kn
↓B
V
↑C
kn
(a) Thus
C
[T ]C
= C −1 BB [T ]B B −1 C
= M −1 B [T ]B M
where M = B−1 C.
(b) Conversely, if B is a basis of V and T : V → V and M is an invertible
n × n matrix then M −1 B [T ]B M =C [T ]C where C = BM : k n → V
is a basis.
(c) We say that two n × n matrices X and Y are similar if there is an
invertible matrix M such that Y = M −1 XM
i. similarity is an equivalence relationship
ii. If X and Y are similar, there exists an n-dimensional vector
space V , a map T : V → V and bases B and C for V such that
X =B [T ]B and Y =C [T ]C .
A. In fact, we can take V = kn , B the standard basis, C the
columns of M and T multiplication by X.
5. Suppose V = U ⊕ W is a direct sum of spaces, and suppose S : U → U
and T : W → W are linear. Then S ⊕ T : V → V .
(a) If B is a basis for U and C is a basis for W then B ∪ C is a basis of
V . Moreover:
·
¸
0
B [S]B
B∪C [S ⊕ T ]B∪C =
0
C [T ]C
6. All this is complex trivia. The real problem is to show that, given an
operator T : V → V , we can find a basis so that the matrix of T with
respect to the basis is simple.
(a) For example, if V is a real or complex inner product space and T is
a self-adjoint operator then T can be represented by a real diagonal
matrix (the spectral theorem).
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(b) We will show that any operator on a space over an algebraically
closed field (e.g. C) can be represented by a special matrix called the
Jordan Canonical Form for the operator. This matix will well us all
the eigenvalues and generalized eigenvalues of the operator
7. Recall how, if T : V → V is a linear operator on a k-space, we can view
V as a module over k [X] and recover the action of T .
(a) The action of k on V is already defined. Extemd to all polynomials
by defining Xv =T (v).
(b) If V is finite dimensional, then it is a fg torsion k [X]-module.
8. Example: the companion matrix
(a) Let T : V ©→ V and supposeªV is n-dimensional and for some vector
v the set v,T v, . . . ,T n−1 v is linearly
independent. Then
©
ª this set
is a basis B of V . Moreover, since v,T v, . . . ,T n−1 v,T n v is linearly
dependent, there exist constants ai such that:
T n + an−1 T n−1 + · · · + a0 I = 0.
(b) The matrix of T is

0
 1

 0

[T
]
=
 0
B
B

 ..
 .
then:
0
0
1
0
..
.
0
0
0
1
..
.
···
···
···
···
..
.
0 0 0 ···
g
0
0
0
0
..
.
−a0
−a1
−a2
−a3
..
.
1 −an−1









= C (g) the companion matrix of the monic polynomial
= X n + an−1 X n−1 + · · · + a0
(c) V is a cyclic k [X]-module with generator v.
i. Proof: suppose w ∈V . Then w =c0 v+c1 T (v)+· · ·+cn−1 T n−1 (v)
for some scalars ci . If f = c0 +c1 X +· · ·+cn−1 X n−1 then w =f v.
k [X]
ii. In fact as a k [X]-modules, V ∼
where g = X n +an−1 X n−1 +
=
(g)
· · · + a0 . The isomorphism maps v to 1̄.
(d) Conversely, if T : V → V makes V into a cyclic k [X]-module isomork [X]
, where g = X n + an−1 X n−1 + · · · + a0 , then we can
phic to
(g)
make a basis of V corresponding to 1̄, X̄, . . . , X n−1 , and the matrix
of T with respect to this basis if the companion matrix above.
0.11. LINEAR ALGEBRA
ciii
9. Let T : V → V . A subspace W ⊂ V is T -cyclic iff there exists a vector
w ∈ W such that W is spanned by w,T w, T 2 w, . . .. The first segment is
a basis for W .
(a) We will show that V is a direct sum of T -cyclic subspaces.
10. Rational Canonical Form: suppose T : V → V . Then there exists
monic polynomials g1 |g2 | · · · |gt such that, as k [X]-modules
k [X]
k [X]
V ∼
⊕ ··· ⊕
=
(g1 )
(gt )
and T corresponds to multiplication by X.
(a) There is a basis B for V such that the matrix of T is

C (g1 )
0
···
0

0
C
(g
)
·
·
·
0
2

..
..
..
B [T ]B = 
..

.
.
.
.
0
0
· · · C (gt )





i. the corresponding basis in the direct sum isna basis for each sum-o
mand, and the basis for the ith summand is 1̄, X̄, . . . , X deg(gi )−1
(b) Every operator T has a unique rational canonical form. In particular,
every matrix is similar to exactly one rational canonical form..(That
is, the rational canonical form of the operator "multiplication by M "
is the unique rational canonical matrix similar to M .)
i. The rational canonical form of T is unique because if there were
two then the k [X]-module V would have two different rational
decompositions.
(c) Two matrices A and B are similar iff they have the same rational
canonical form C
i. Proof: similarity is transitive. If A and B are both similar to C
then A is similar to B. Conversely, if A is similar to B and the
rational canonical form of A is C then A is similar to C so B is
similar to C or C is the rational canonical form of B.
11. The characteristic polynomial of T is g1 · · · gn . The minimal polynomial of T is gt .
(a) We could take these as definitions, but let’s show
det (XI − T ) = g1 · · · gt
p (T ) = 0 ⇐⇒ gt | p
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i. A corollary is the Cayley-Hamilton Theorem: if c is the characteristic polynomial of T then c (T ) = 0.
ii. The minimal and the characteristic polynomials have the same
roots, but the characteristic polynomial may have higher multiplicities.
iii. Generically (over the complex or real numbers) a random matrix
has minimal polynomial equal to its characteristic polynomial.
A. That is, the rational canonical form has only one component.
iv. Also generically, the minimum polynomial of a random matrix
has distinct roots.
A. A matrix for which the minimal polynmial has repeated roots
is called defective.
·
¸
¡ ¢
0 0
B. Example:
= C X2 .
1 0
(b) To prove det (XI − T ) = g1 · · · gt .it suffices to prove det (XI − C (g)) =
g, but that follows from the row expansion of the determinant of
XI − C (g).
(c) Now we’ll prove that p (T ) = 0 ⇐⇒ gt | p. Let’s assume V =
k [X]
k [X]
⊕ ··· ⊕
and T is multiplication by X. Then p (T ) is the
(g1 )
(gt )
operator: multiply by p = p (X). But multiplication by p is zero on
all factors iff gi | p for all i iff gt | p.
12. Recall: c ∈ k is an eigenvalue of T : V → V if T − cI is not invertible. An
operator on an n-dimensional space has at most n distinct eigenvalues.
(a) the eigenspaces of T are the kernals of T − cI for all the eigenvalues
c.
(b) the non-zero elements of the eigenspaces are the eigenvectors of T .
(c) the generalized eigenspaces of T are the kernals of (T − cI)dim V
for the eigenvalues c.
(d) the generalized eigenvectors of T are the non-zero vectors in the
generalized eigenspaces.
(e) every generalized eigenspace contains an eigenspace.
i. If T − cI is 1 − 1 then (T − cI)n is 1 − 1.
(f) If you can find a matrix representing an operator, then the eigenvalues
of the matrix are the eigenvalues of the operator; the dimensions of
the corresponding eigenspaces of the matrix and operator are the
same, as are the dimensions of the generalized eigenspaces.
0.11. LINEAR ALGEBRA
cv
13. What are the eigenvalues, eigenspaces and generalized eigenspaces of the
k [X]
operator “multiplication by X” on the vector space
n
(X − a)
(X − a)n−1
n , and the general(X − a)
k [X]
ized eigenspace is the whole thing
(X − a)n
(a) The eigenvalue is a, the eigenspace is
14. What is the matrix of “multiplication by X” on the vector space
k [X]
n
(X − a)
(a) It depends on what basis you use
h
i
(b) So let’s use 1, X − a, . . . (X − a)n−1 .
(c) Then the matrix of multiplication by

0 0 ···
 1 0 ···

 0 1 ···

 .. .. . .
 . .
.
0 0 ···
X − a is

0 0
0 0 

0 0 

.. .. 
. . 
1 0
and the matrix of multiplication by X is multipliplication by
plus multiplication by a:
 
 

a 0 0 ··· 0
0 0 ··· 0 0
 1 0 ··· 0 0   0 a 0 ··· 0  
 
 

 0 1 ··· 0 0   0 0 a ··· 0  
J (n, a) = 
=
+
 .. .. . .
..  
.. ..   .. .. .. . .
 . .
. .  
. . .   . . .

0 0 0 ··· a
0 0 ··· 1 0
X −a
a 0 ···
1 a ···
..
.
0 1
.. .. . .
.
. .
0 0 ···
15. Over an algebraically closed field k, the only monic prime polynomials are
X − a for a ∈ k.
16. Let V be a finitely generated torsion k [X]-module over an algebraically
closed field k. Then by the fundamental theorem of modules over PID’s:
V ∼
=
M
i,j
k [X]
(X − ai )nij
h
i
n −1
(a) If we choose a basis for V over k by choosing 1, . . . , (X − ai ) ij
in each direct summand, then the matrix for multiplication by X is
a block matrix with blocks J (nij , ai ) down the diagonal.
0
0
0
..
.
1

0
0 


0 

.. 
. 
a
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(b) So if T : V −→ V is a linear operator on a finite dimensional vector
space, then V has a basis so that the matrix of T is a block matrix
with blocks J (nij , ai ) down the diagonal.
i. The big question is: how do we find the ai and the nij
ii. A secondary question is: how do we find the basis.
17. The ai are the eigenvalues of T
(a) Vai =
ai
M
j
k [X]
is the generalized eigenspace for the eigenvalue
(X − ai )nij
(b) But how do we break up the generalized eigenspace into the direct
summands. How do we find the nij for each i?
(c) The solution uses the same ideas as the problem 9.9 from the previous homework:.
Observe first for a fg torsion R-module M that
M
since M =
Mp and qMp = Mp if p 6= q are primes of R. Thus
p
pn M
p n Mp
= Up (n, Mp ),
= dimR/(p) n+1
n+1
p
M
p
Mp
n = 0, 1, 2, . . . .In general, if M is a fg R-module whose free part
pn M ∼ pn MT
pn R r
⊕ n+1 r ∼
has rank r, then M ∼
= MT ⊕ Rr and n+1
=
= n+1
MT pµ R¶
µ
µ pµ M
¶r
¶r ¶ p
r
n
R
R
R
p Mp
.⊕
Moreover Up n,
=
= dimR/(p)
n+1
p
Mp
(p)
(p)
(p)
r. Thus
Up (n, M ) = dimR/(p)
Up (n, M ) = Up (n, MT ) + r
= Up (n, Mp ) + r
Now for the problem.
i. We prove this part assuming only that M and N are fg, not fg and
pn (M ⊕ N ) ∼ pn M ⊕ pn N
pn M
∼
torsion. Since n+1
⊕
= n+1
=
p
(M ⊕ N )
p
M ⊕ pn+1 N
pn+1 M
n
p N
, we have Up (n, M ⊕ N ) = Up (n, M ) + Up (n, N ).
n+1
p
N
Before proceeding to the next parts we need a lemma. The isomorphism class of a fg R-module M is determined by the integers
Up (n, M ) for all n. Thus two modules have
the same Up values
M
r
∼
iff they are isomorphic.Proof: M = R ⊕
Mp , where Mp ∼
=
p
rp
M
j=1
R
n . For all p and n sufficiently large, Up (n, M ) = r, so
(p) pj
r is determined by Up . Let µM (p, n) = # {npj > n}. The numbers µM (p, n) for all primes p and natural numbers n determine
0.12. SOME PROBLEMS
cvii
all npj . But the parameters µM (p, n) are determined by the
the numbers Up (n, M ), because Up (0, Mp ) = µM (p, 0) and for
n > 0 we have Up (n, Mp ) = µM (p, n) − µM (p, n + 1).
ii. Suppose A ⊕ B ∼
= A ⊕ C. Then for all p and n, Up (n, A) +
Up (n, B) = Up (n, A ⊕ B) = Up (n, A ⊕ C) = Up (n, A)+Up (n, C).
Thus Up (n, B) = Up (n, C) for all p and n, and therefore B ∼
= C.
∼
iii. Suppose A ⊕ A = B ⊕ B. Then for all p and n, 2Up (n, A) =
Up (n, A ⊕ A) = Up (n, B ⊕ B) = 2Up (n, B). Thus Up (n, A) =
Up (n, B) for all p and n, so A ∼
= B.
(d) In the vector space case, there is no free part, so r = 0. The primes
p correspond to polynomials X − ai , and to find the numbers nij it
(X − ai )n V
suffices to find UX−ai (n, Vai ) = dimk
.
n+1
(X − ai )
V
0.12
Some problems
\
Suppose R is a PID and S ⊂ M is a module and submodule. If pn M
S = pn S
\
for all primes p ∈ R and exponents n, then rM
S = rS for all r ∈ R.
Proof: it suffices to show that if x ∈ M is such that rx ∈ S for some r ∈ R,
then rx ∈ rS.
t
Y
r
Let r =
pni i , and define qi = ni . Since (q1 , . . . , qt ) = (1), there exists
p
i
i=1
X
α1 qi = 1. Since rx ∈ S we have
αi ∈ R such that
\
S = pni i S
rx = pni i qi x ∈ pni i M
Let qi x = xi . Then pni i xi = pni i si for some si ∈ S. Then
X
rx = r
αi qi x
X
= r
αi xi
X
=
αi rxi
X
=
αi qi pni i xi
X
=
αi qi pni i si
X
=
αi rsi
X
= r
αi si ∈ rS
We want to find an example of a pure submodule that is not a direct summand. Since every pure submodule of a fg module over a PID is a direct sumM
mand, the example must not be fg. First a little lemma: If N ⊂ M and
is
N
cviii
LECTURES
torsion-free, then N is pure. Proof: if m ∈ M and rm ∈ N then rm = 0. Since
M
is torsion-free, either m̄ = 0 or r = 0. If r = 0 then rm = 0 ∈ rN = (0). If
N
m̄ = 0 then m ∈ N so rm ∈ rN . Either way, rM ∩ N = rN .
∞
M
Z. Define
Now for the example: over the integers Z consider M =
i=0
na
o
: a ∈ Z, i ∈ N ⊂ Q. (You’ve never seen me use the notation from
Z2 =
2i
Z
elementary books Z2 =
because, in more advanced texts and papers, Z2
(2)
means the object just defined.) Construct a homomorphism φ : M → Z2 by
φ (ei ) = 2−i . Let S = ker φ. I claim S 6= 0, S is pure and S is not a direct
summand of M .
First of all, (1, −2, 0, 0, . . .) ∈ ker φ.
M ∼
Second,
= Z2 , which is torsion-free, so S is pure.
S
Finally, if S is a direct summand of M , then Z2 is isomorphic to a complementary direct summand. But Z2 cannot be isomorphic to a submodule of
M because Z2 contains a sequence of non-zero elements x0 = 1, x1 = 2−1 ,
x2 = 2−2 , . . . such that 2xi+1 = xi , i = 0, 1, 2, . . .. No such sequence can exist
in M because, if it did, then the for some i the ith component of x0 would
be a non-zero integer z0 . But then the ith component of xi , zi , would satisfy
2zi+1 == zi , and it is impossible to have such a non-zero sequence of integers.
0.12. SOME PROBLEMS
cix
Suppose M is a fg.module over a PID R, and S ⊂ M is a pure submodule.
We want to show that S is a direct summand of M .
M
Proof: Choose a spanning sets for S and
as follows. For S choose
S
a1 , . . . , ak a basis for the free part of S, and choose b1 , . . . , bm with each bi
generating one of the primary direct summands of S. That is, for each i, there
i
is X
a prime pi and exponent mi such that pimi −1 bi 6= 0 and pm
i bi = 0. Moreover,
P
i
if
ri ai + si bi = 0 then ri = 0 and si is divisible by pm
i . Choose a similar
M
, with q ni d¯i = 0. For each i, qini di = qini si for some si ∈ S.
basis c̄i , d¯i for
S
Replace di by di − si so qini di = 0. We still have qini −1 di 6= 0 because this is true
modulo S. I claim that the M is a direct sum of the cyclic submodules generated
by ai , bi , ci , di . Since S is the direct sum of the cyclic modules generated by ai
and bi , S is a direct summand of M .
It suffices to show that ai , bi , ci , di span M
as
P and are
Pas linearly
P independent
P
possible. That is, we must also show that if ri ai + si bi + ti ci + ui di = 0
then ri ai = 0, si bi = 0, ti ci = 0, and ui di = 0. That is we must show that
i
ri = 0, ti = 0, pm
| si and qini | ui . Working modulo S P
(so the first
i
P two sums
vanish), the required
conditions
hold
on
t
and
u
.
Thus
t
c
+
ui di = 0 in
i
i
i
i
P
P
M , and therefore
ri ai + si bi = 0. But this is just an expression in a basis
for S