Download MATH20212: Algebraic Structures 2

MATH20212: Algebraic Structures 2
This follows on from Algebraic Structures 1 MATH20201, so some basic
knowledge and understanding of groups and rings is assumed. This course
concentrates rather more on rings than on groups. The central aims of the
course are:
• that you become more acquainted with the basic examples and the general
ideas, to the point where you have internalised the main concepts and can use
them in new situations;
• to introduce the general notion of a homomorphism (a structure-preserving
map) and its kernel;
• to introduce the factor, or quotient, construction.
It is not expected that the average student will master these two rather sophisticated new concepts (especially the factor/quotient construction) by the end of
the course: this is just an introduction; you can renew your acquaintance with
them, and have the opportunity to understand them better, in later algebra
courses.
Like Algebraic Structures 1, there is a great deal of emphasis on examples:
subrings of Z, Q and C; polynomial rings; matrix rings; groups of permutations
and symmetries; groups of units of rings. Polynomials are the central examples.
Algebraic Structures 1 and 2 together give a foundation for further courses
in algebra, as well as those (e.g. some in geometry and topology) which use
algebraic structures.
These notes cover the material of the whole course but not everything is
here. Lectures will be used for explaining the concepts, giving the proofs (only
a few proofs are given in these notes) and presenting details of examples. There
are a lot(!) of exercises in these notes. Some will be done in class; you should
try some of the rest, giving priority to those marked **, then those marked *.
It is really important to keep up to date with your work on examples. Work
on the examples relevant to what has been covered recently in lectures: don’t
get behind (you can always come back and use those you skipped over for
revision). The Friday 11.00 slot will be used for going over examples and should
be regarded as being as important as the lectures. Prepare for this (try the
examples beforehand) and be ready to participate (asking questions, suggesting
solutions).
Further reading is not necessary but might be interesting and/or useful.
There is the course text, Fraleigh. I will maintain a list other good sources on
the website for the course.
Finally, please let me know of any errors/typos you find in the notes, examples, solutions.
1
Part I
Rings
1
Definitions and Examples, including Review
Idea: a ring is an “arithmetic system” with two operations, usually written
+ and ×, because the ur-example is the ring of integers. The ring of integers
modulo n is another example, as is the ring of n×n square matrices with integer
entries. There are many more examples.
Definition, in brief. A ring is given by the following data: a set R and two
binary operations, written + and ×, on R. The conditions to be satisfied are:
that (R, +) is an abelian group; that × is associative and distributes over +.
The identity element of the abelian group (R, +) is denoted 0. In this course
“ring” will mean “ring with 1 6= 0”: that means one more datum is given,
namely a specified element, 1 ∈ R, which is different from 0, and there is one
more condition, namely that 1 is an identity for ×.
Definition, in detail. You can find this in your notes for Algebraic Structures 1,
in most of the books mentioned in this course, and on the web. I’ll remind you
in the lecture.
Notation: we write (R; +, ×, 0, 1) if we want to show all the data but usually
we don’t want to do that, so write just R.
Examples 1.1. Examples of rings: (1) Z; (2) Z[i] = {a + bi : a, b ∈ Z}, where
i denotes a square root of −1.
In order to check that these are rings there is no need to check all the details
directly because both are subrings of the ring, C, of complex numbers. So it
is, by the lemma below, enough to check that each of the above sets contains 0
and 1 and is closed under taking negatives and under the operations, + and ×.
Lemma 1.2. If S is a ring and R ⊆ S is a subset then R is a ring, a subring
of S, iff:
• 0, 1 ∈ R;
• r, s ∈ R implies r + s, r × s ∈ R;
• r ∈ R implies −r ∈ R.
That was loosely said: what we should have said was the following. “If
(S; +, ×, 0, 1) is a ring and R ⊆ S then (R; +0 , ×0 , 0, 1) is a ring, where +0 is the
restriction of the operation + to R × R and ×0 is the restriction of × to R × R,
iff (list of conditions as in the lemma)” or, more briefly, “If (S; +, ×, 0, 1) is a
ring and R ⊆ S then R, equipped with the operations inherited from S, is a
ring iff (list of conditions as in the lemma)”.
Anyway, the point is, to check that a subset of a ring S is a subring you
don’t have to check everything: many of the conditions come for free because
they hold everywhere in the ring S.
By the way, don’t get confused by some notation above: × was used both
for the “multiplication” operation in the ring and for the cartesian product of
sets (as in “R × R”). In practice we usually write just rs, or sometimes, r.s for
the multiplication r × s.
2
Examples 1.3. More examples of rings.
Small rings can be presented by giving their “addition and multiplication
tables”. Here are three examples. Can you identify the rings?
+
0
1
2
3
0
0
1
2
3
1
1
2
3
0
2
2
3
0
1
3
3
0
1
2
×
0
1
2
3
0
0
0
0
0
1
0
1
2
3
+
0
1
x
1+x
0
0
1
x
1+x
1
1
0
1+x
x
x
x
1+x
0
1
1+x
1+x
x
1
0
+
0
1
α
1+α
0
0
1
α
1+α
1
1
0
1+α
α
α
α
1+α
0
1
1+α
1+α
α
1
0
2
0
2
0
2
3
0
3
2
1
×
0
1
x
1+x
×
0
1
α
1+α
0
0
0
0
0
0
0
0
0
0
1
0
1
x
1+x
1
0
1
α
1+α
x
0
x
0
x
1+x
0
1+x
x
1
α
0
α
1+α
1
There is an issue here: these tables certainly specify two operations on each
relevant set and it would be easy to pick out the 0 and 1 even if they had not
been identified in name, but it could be rather tedious to check that all the
conditions (associativity and distributivity in particular) for a ring are satisfied.
Of course you could give the tables to a computer to check. We will come back
to these examples and, by recognising what they are, we will be able to avoid
the tedious case-by-case checking of the ring axioms.
Some elementary consequences of the definition are recalled in Exercise 1.4
Exercise* 1.4. Some elementary properties.
If last semester’s course is by now but a hazy memory then you should do
these exercises to bring the basics back into focus. If not then you should be
able to rattle them off. In any case do them. Let R be any ring. Show the
following.
(1) There is just one element z with the property that z + r = r for every r ∈ R.
(2) There is just one element e with the property that er = r for every r ∈ R.
(3) For each element r ∈ R there is just one element r0 such that r + r0 = 0.
Also recall extended associativity and distributivity:
r1P
× r2 × · · · × rn is
Pn
n
unambiguous (no parentheses are needed) and r× i=1 si = i=1 (r×si ). These
are proved by induction (and the first is probably more difficult to formulate
precisely than to prove). Induction is also used to define powers and to prove the
expected things about them: for r ∈ R set r0 = 1 and, inductively, rn+1 = r×rn ,
also define r−n = (r−1 )n .
The characteristic, char(R), of a ring R is the least positive integer n such
that 1 + · · · + 1 = 0 (n 1s), that is, such that n × 1 = 0, that is, such that n = 0.
If, as may well be the case, there is no such positive n, then the characteristic
of R is defined to be 0. For example, Z has characteristic 0, as have the rings Q
and C. On the other hand char(Zn ) = n.
3
1+α
0
1+α
1
α
Lemma 1.5. Suppose that char(R) = n > 0. Then:
(1) nr = 0 for every r ∈ R;
(2) if m is an integer then m1 = 0 iff n|m.
The ring R is a domain if rs = 0 implies r = 0 or s = 0. A non-zero element
r ∈ R is a zero-divisor if there is a non-zero element s ∈ R with rs = 0 or
sr = 0. So a domain is a ring without zero-divisors. (A commutative domain is
also called an integral domain.)
Exercise 1.6 asks you to prove that the characteristic of a domain is either
0 or a prime integer.
Exercise** 1.6. Show that if R is a domain then char(R) = 0 or char(R) is a
prime integer.
Examples 1.7. (1) Z is a domain; (2) Z[i] is a domain - this can be proved
directly or use that C is a domain then apply the next result.
Lemma 1.8. If R is a subring of S and S is a domain then R is a domain.
Examples 1.9. More examples: (3) Q is a domain; (4) the polynomial ring
Q[X] is a domain - to prove this consider leading terms of polynomials. The next
result is the general case (this result and its corollary were proved in Algebraic
Structures I).
Proposition 1.10. Suppose that R is a domain. Then the polynomial ring
R[X] is a domain.
Corollary 1.11. Suppose that R is a domain. Then the ring, R[X1 , . . . , Xn ],
of polynomials in n indeterminates and with coefficients in R, is a domain.
A division ring is a ring in which every non-zero element has a right inverse
and a left inverse: for every r ∈ R there is s ∈ R such that rs = 1 (r is right
invertible) and there is t ∈ R such that tr = 1 (r is left invertible). In this
case, see below, s = t and we write r−1 for this inverse of r and say just that
r is invertible.
A field is a commutative division ring. (Recall that a ring R is said to be
commutative if the multiplication is commutative: rs = sr for all r, s ∈ R.)
Lemma 1.12. If R is a ring and r ∈ R has both a right and a left inverse then
these are equal.
(Just bracket trs in two different ways.)
Beware that an element of a ring can have, for instance, a left inverse without
having a right inverse and, in such a case, there might be more than one left
inverse.
Example 1.13. Let K be your favourite field and let V be the (infinitedimensional) vector space over K with basis v1 , v2 , . . . , vn , . . . . Let R be the
ring of all linear transformations from V to itself, the operations on R being
(pointwise) addition of linear transformations and, for the “multiplication” on
R, use composition of linear transformations (with the convention that sr means
“do r then do s”). Note that an invertible element of R must be an isomorphism
from V to itself. Let r ∈ R be “right-shift”: the linear map which is defined
on the given basis by sending vn to vn+1 . Since this map is not surjective, it is
4
not an isomorphism, hence r is not invertible. But r is left invertible: if s is the
linear map which sends vn+1 back to vn and sends v1 to 0, then sr = 1 (where
1 is, note, the identity map on V ). Let t be the linear map which sends vn+1 to
vn and sends v1 to v1 : then also tr = 1, showing that a left inverse need not be
unique.
Proposition 1.14. Every division ring is a domain.
Exercise 1.15 recalls the condition on n for Zn to be a domain, equivalently
for Zn to be a field.
Exercise 1.15. Show that the following conditions on the integer n ≥ 2 are
equivalent: (i) Zn is a domain; (ii) Zn is a field; (iii) n is a prime (look at your
notes for Algebraic Structures I if you get stuck).
Examples 1.16. More, apart from the obvious ones like Q, R, C, Zp , examples
of fields: (1) if K is a field then so is the ring, K(X), of rational functions
in the variable/indeterminate
X with
√ coefficients in K, more generally so is
√
K(X1 , . . . , Xn ); (2) Q[ 2] = {a + b 2 : a, b ∈ Q}.
Some examples similar to (2) above are in Exercise 1.17.
Exercise** 1.17. Show that the following also are fields: (3) Q[i] (where i2 =
−1); (4) Q[ω] where ω is a primitive cube root of 1 (in this example you should
first figure out the dimension of Z[ω] as a vector space over Q; the dimension is
perhaps not as obvious as it might seem to be).
Tidying up: recall that polynomials are really equivalence classes of polynomial expressions: we consider X 2 + 1 to be the same polynomial as 1 + 0X + X 2
for instance. A similar comment applies to rational functions: these are
expressions of the form p/q where p and q are polynomials but, again, up to
equivalence, for example we identify (X 2 + X)/(X 3 − X) with (X + 1)/(X 2 − 1).
Actually, a more accurate term than “rational function” would be “rational expression”: the distinction between polynomials and polynomial functions also
applies here.
Example 1.18. There are division rings which are not fields, that is, which are
not commutative. The best known is the ring of quaternions: H = {a + bi +
cj + dk : a, b, c, d ∈ R} where i2 = j 2 = k 2 = −1 and ij = k, jk = i, ki = j.
From these relations it follows, for example, that ji = −k 6= k = ij. Notice
that this contains, as a subring, the field, C, of complex numbers. And, in case
you were wondering, “H” stands for “Hamilton”. It was shown in Algebraic
Structures 1 that every non-zero element has an inverse.
A more subtle point is whether the above ring is really what we think it is:
how, for instance, do we know that k 6= −k, even that 0 6= 1? For, certainly
we can write down a presentation of the ring H above by giving generators (as
an R-vectorspace) and relations between these but we should really prove that
there are no unexpected consequences of the relations, more precisely we should
prove that a + bi + cj + dk = a0 + b0 i + c0 j + d0 k iff a = a0 , b = b0 , c = c0 and
d = d0 . One way to do this is to provide a “concrete” representation of the ring,
that is, find some ring S and elements si , sj , sk in S which satisfy the defining
relations of i, j, k and where, for some reason, we know that a + bsi + csj + dsk =
a0 + b0 si + c0 sj + d0 sk iff a = a0 , b = b0 , c = c0 and d = d0 . That will be done later
(Exercise 2.5).
5
√
Exercise* 1.19. Show that Z[ 2] and Z[ω] (ω a primitive cube root of 1) are
rings (check that they are subrings of C). Are they domains? In each case
identify the invertible elements.
We give some more examples of non-commutative rings (i.e. rings which are
not necessarily commutative) below.
Example 1.20. The ring M2 (Z) of 2 × 2 matrices with integer entries is a ring
which is not commutative (exercise: show that this is not commutative). More
generally, if R is a ring, then the ring, Mn (R), of n×n matrices with entries in R
is not commutative if n > 1 (exercise show this), even if R itself is commutative.
(The ring operations in Mn (R) are matrix addition and multiplication; the 0
and 1 are the obvious candidates.)
An element r of a ring R is nilpotent if there is some integer n ≥ 1 with
rn = 0 (and the least such n is the index of nilpotence of r). An element
r ∈ R is idempotent if r2 = r. For example 0 and 1 are idempotent. Exercises
1.21, 1.22, 1.23, 1.24, 1.26, 1.27 are mostly about nilpotent and idempotent
elements.
Exercise** 1.21. Find examples of non-zero nilpotent elements in M2 (Z). Find
examples of idempotent elements in M2 (Z) apart from 0 and 1.
Exercise** 1.22. Prove that if R is a domain then there are no nilpotent elements other than 0 and no idempotent elements other than 0 and 1.
Exercise** 1.23. Find nilpotent elements of M3 (Z) with indices of nilpotence
2 and 3. Is there a nilpotent element of order 4? What is the highest index of
nilpotence of an element of Mn (Z)? Can you prove it?
Exercise* 1.24. When does the ring, Zn , of integers modulo n have non-zero
nilpotent elements? (and what are they?)
Exercise** 1.25. If R is a ring and r ∈ R the centraliser of r, denoted C(r), is
the set of elements which commute with r: C(r) = {s ∈ R : rs = sr}. Show that
C(r) is a subring of R. In the ring M2 (Z) find a description for the centraliser
of a general element. Is it always the case that the centraliser of an element is
a commutative ring?
Here is a general construction: if R1 and R2 are rings then their product is
their set-theoretic product R1 × R2 made into a ring by defining the operations
+ and × (don’t confuse “×” and “×”!) by (r1 , r2 ) + (s1 , s2 ) = (r1 + s1 , r2 + s2 )
and (r1 , r2 ) × (s1 , s2 ) = (r1 × s1 , r2 × s2 ) (to be more precise we could write
(r1 +1 s1 , r2 +2 s2 ) to show that the additions refer to operations in different
rings and similarly for multiplication). The element (0, 0) clearly is a, hence
the, zero of this ring and the element (1, 1) is the identity element of R1 × R2 .
Note that the elements (1, 0) and (0, 1) corresponding to the identity elements of the original rings are idempotents of the ring R1 × R2 and they
are orthogonal in the sense that idempotents e1 and e2 are orthogonal if
e1 e2 = 0 = e2 e1 .
Exercise 1.26. Find a pair of (non-trivial) idempotents in the ring M2 (R). What
are the idempotent elements of M2 (R): are there just a few or are they a-dimea-dozen?
(If you think of the elements of M2 (R) as linear transformations of R2 to
itself, with respect to a fixed basis, then one may see that, geometrically, an
idempotent is a projection and that “orthogonal” means just that.)
6
Exercise* 1.27. Let e ∈ R be idempotent. Show that 1 − e is idempotent and
that e and 1 − e are orthogonal.
Example 1.28. Let R be the set of all functions f : [0, 1] −→ R from the unit
interval to the set of reals. Make this a ring by defining + and × to be pointwise
addition and multiplication respectively, for instance f + g is defined to be the
function which takes a ∈ [0, 1] to f (a) + g(a). Exercise 1.29 asks you to check
that this is indeed a ring.
Exercise* 1.29. Let R be the set of all functions f : [0, 1] −→ R from the
unit interval to the set of reals. Make this a ring by defining + and × to be
pointwise addition and multiplication respectively, for instance f + g is defined
to be the function which takes a ∈ [0, 1] to f (a) + g(a). Check that this gives
a commutative ring and identify the 0 and 1 of this ring. (This example also
works if we restrict to functions which are continuous, alternatively functions
which are differentiable: why?)
Exercise 1.30. Let R be the set of all functions f : R −→ R with pointwise
addition for + and take the “multiplication” × to be composition of functions,
with the convention that f g means do g then do f. Show that R is not a ring
(although R satisfies some of the axioms for a ring it does not satisfy them all,
so you should find out what goes wrong).
Exercise 1.31. Could we do something like Exercise 1.29 above but using functions from [0, 1] to [0, 1]?
Example 1.32. Consider the ring, R[X], of polynomials in X with coefficients
from R. Now extend this by a new variable Y which does not commute with
X but, rather, satisfies the relation Y X = XY + 1. We denote this ring, called
the first Weyl algebra, by RhX, Y : Y X = XY + 1i (thus, given R, we are
describing this ring in terms of generators and relations): its elements are “noncommutative polynomials” in X and Y. Every such polynomial can be brought
into a “normal form”: a sum of monomials, each of which has the form X m Y n
for some non-negative integers m, n.
Example 1.33. Let R be any ring
Pand let G be any group. The group ring
RG is the set of formal finite sums ri gi (up to the usual identifications) where
ri ∈ R and gi ∈ G. Addition is formal addition using the rule rg + sg = (r + s)g
and multiplication is based on (rg) × (sh) = (rs)(gh). The identity element is
1e where 1 is the identity of R and e is the identity of G. The zero of R is the
element 0e. (This was described in Algebraic Structures I.)
1.1
Extra exercises
Exercise* 1.34. Is the binary operation on Z (the set of integers) which takes
(a, b) to a − b associative? Is there an identity element for this operation?
Exercise* 1.35. Is the binary operation on P (the set of positive integers) given
by (a, b) 7→ ab associative?
a b
Exercise** 1.36. Let SL2 (Z) denote the set of all matrices of the form
c d
with a, b, c, d ∈ Z and ad − bc = 1. Is this set a ring under matrix addition and
multiplication? Is this set a group under matrix multiplication? What about
7
a b
the set
c d
multiplication?
: a, b, c, d ∈ Z, ad − bc = −1 : is this a group under matrix
Exercise 1.37. Find some subrings of M2 (Z). This exercise becomes more interesting if we don’t insist that subrings contain the identity element so, just for
this exercise, look for subsets containing {0} which are closed under addition,
subtraction and multiplication.
Exercise** 1.38. Suppose that R is a ring. How many solutions in R can there
be to the equation X 2 − 1 = 0?
Exercise** 1.39. Suppose that R is a domain and let a, b ∈ R with a 6= 0. Prove
that the equation ax = b has at most one solution in R. What if, instead of
assuming a 6= 0, we assume b 6= 0: is the conclusion still correct?
√ √
Exercise*
1.40.
√ Let Q[ 2, 3] denote the smallest subring of R which contains
√
3. Show
Q, 2√and √
√ that this consists exactly of the real numbers of the form
a + b 2 + c 3 + d 6 where a, b, c, d ∈ Q. Choose some non-zero elements
of √
this√ring and√find √
their inverses (this is, in fact, a field). Also show that
Q[ 2, 3] = Q[ 2 + 3] (hint: think about what you actually have to do here
- it’s not much).
Exercise 1.41. Let R = Z[i] where i is a square root of −1. Prove that R is a
domain. Give another proof. And another.
Exercise* 1.42. If R1 is a ring with characteristic m and R2 is a ring with
characteristic n what is the characteristic of R1 × R2 ?
Exercise 1.43. Show directly that any ring with exactly three elements must be
a domain.
Exercise 1.44. Let R be a ring. Show that the identity a2 − b2 = (a + b)(a − b)
is true in R iff R is commutative
Exercise** 1.45. Find all idempotent and all nilpotent elements in the ring
Z6 × Z12 .
Exercise 1.46. Suppose that R is a ring such that a2 = a for every a ∈ R. Show
that a + a = 0 for every a ∈ R. Also show that R is commutative.
Exercise 1.47. Let V be a finite-dimensional vector space over a field K and let
R be the ring of linear transformations from V to V (so addition is pointwise
addition and multiplication is composition of transformations). Prove that if
f : V −→ V is in R then f has a left inverse iff f has a right inverse. You may
quote basic results from linear algebra in your proof.
Exercise** 1.48. Let R be the polynomial ring Z8 [X]. Show that the polynomial
1 + 2X is invertible.
Exercise 1.49. Suppose that K is a field with q elements and that G is a group
with n elements. How many elements does the group ring KG have?
Now let K = Z3 be the field with 3 elements and let G be a cyclic group of
order 3, generated by an element a. Write down the general form of an element
of R = KG. Simplify the following elements of R (i.e. write them in “standard
form”):
(i) (1e + 2a + a2 ) + (1e + 2a2 );
(ii) (1e + 2a + a2 ) × (1e + 2a2 );
(iii) (1e + 2a + a2 )3 .
8
2
Isomorphisms, Homomorphisms and Ideals
Isomorphism: the idea is that two rings are isomorphic if they are the same
abstract structure: their addition and multiplication tables “are the same up to
re-arrangement and re-naming of elements”, and an isomorphism is a map which
shows this. For instance, the polynomial rings R[X] and R[Y ] are isomorphic:
the map from the first ring to the second which takes a polynomial in X and
replaces every occurrence of X by Y is an isomorphism. For, clearly this map
preserves (“commutes with”) addition and multiplication and it takes the 0,
respectively 1, of the first ring to the 0, resp. 1, of the second.
Isomorphism: definition. If R and S are rings then an isomorphism from
R to S is a bijection θ : R −→ S such that, for all r, r0 ∈ R we have
θ(r + r0 ) = θ(r) + θ(r0 ) and θ(r × r0 ) = θ(r) × θ(r0 )
(in each case the operation on the left hand side of the equation is an operation
on elements of R and that on the right hand side is an operation on elements
of S). Various things are preserved by θ: see the next lemma.
If θ is an isomorphism from R to S then we write θ : R ' S. We say that R
and S are isomorphic, and write R ' S, if there is an isomorphism from R to
S. The relation of being isomorphic is an equivalence relation on rings (Exercise
2.1).
Exercise 2.1. Show that isomorphism is an equivalence relation on rings. (Of
course, first you’ll need to recall what is meant by an equivalence relation.)
2
Example
2.2. R[X]
] since the map θ : R[X] −→ R[X 2 ] defined by
Pn
Pn ' R[X
i
2i
θ( i=0 ai X ) = i=0 ai X (i.e. the map which substitutes X 2 for X in each
polynomial) is easily seen to be an isomorphism. Here R is any ring. (On the
other hand this operation, if it were regarded as a map from R[X] to R[X],
would not be an isomorphism: why?)
Lemma 2.3. Suppose that θ : R −→ S is an isomorphism. Then:
(1) θ(1) = 1;
(2) θ(0) = 0;
(3) θ(−r) = −θ(r) for every r = R;
(4) r ∈ R is invertible iff θ(r) ∈ S is invertible and, in that case, (θ(r))−1 =
θ(r−1 );
(5) r ∈ R is nilpotent iff θ(r) is nilpotent (and then they have the same index
of nilpotence).
Example 2.4. Let R = C be the field
of complex
numbers and
let S be the ring
1 0
0 −1
of all 2 × 2 matrices of the form a
+b
where a and b are
0 1
1 0
real numbers. Then the map θ from R to S which takes
the complex number
1 0
0 −1
a + bi (a, b ∈ R) to the matrix a
+b
is, one may (and you
0 1
1 0
should) check, an isomorphism.
There is a similar representation of quaternions as matrices with real coefficients, see Exercise 2.5.
1 0
Exercise* 2.5. Show that the ring of 2 × 2 matrices of the form a
+
0 1
9
0 −1
i 0
0 i
b
+c
+d
with a, b, c, d ∈ R (and i a square
1 0
0 −i
i 0
root of −1) is isomorphic to the ring of quaternions.
Example 2.6. All rings with two elements are isomorphic, because the only
elements of such a ring are 0 6= 1 and the map from one two-element ring R
to another two-element ring S which takes 0R to 0S and 1R to 1S is quickly
checked to be an isomorphism. Note that there is a ring with two elements take the ring, Z2 , of integers modulo 2.
It is also the case that, up to isomorphism, there is just one ring with three
elements (Exercise 2.7).
Exercise** 2.7. Show that all rings with three elements are isomorphic.
Example 2.8. There are rings with four elements which are not isomorphic.
In fact, the tables in Examples 1.3 do all give different rings (indeed, there is
even one more).
Homomorphism: this is a weakening of the idea of isomorphism. The requirement that the map commute with the ring structure is kept but the map is not
required to be an isomorphism. That is, the map might collapse some elements
(it need not be injective) and its image need not be the whole of the second ring
(it need not be surjective).
If R and S are rings then a homomorphism from R to S is a map θ :
R −→ S such that, for all r, r0 ∈ R we have θ(r + r0 ) = θ(r) + θ(r0 ) and
θ(r × r0 ) = θ(r) × θ(r0 ). We also require that θ(1) = 1, that is, θ(1R ) = 1S .
Compare the next lemma with that, 2.3, for isomorphisms.
Lemma 2.9. Suppose that θ : R −→ S is a homomorphism. Then:
(i) θ(0) = 0;
(ii) θ(−r) = −θ(r) for every r ∈ R;
(iii) if r ∈ R is invertible then θ(r) ∈ S is invertible and, in that case, (θ(r))−1 =
θ(r−1 );
(iv) if r ∈ R is nilpotent then θ(r) is nilpotent (and the index of nilpotence of
θ(r) is less than or equal to that of r);
(v) the image of θ is a subring of S.
The converse to (iii) above fails: in Exercise 2.10 you are asked to give an
example of an element which is not invertible being sent by a homomorphism
to an element which is invertible.
Exercise** 2.10. Give an example to show that if θ : R −→ S is a homomorphism and if r ∈ R is such that θ(r) is invertible in S, it need not be the case
that r is invertible in R.
Examples 2.11. (1) The map Z −→ Zn defined by taking an integer a to its
congruence class modulo n, [a]n = {b ∈ Z : b ≡ a mod n}, is a homomorphism.
(2) The map θ : Znk −→ Zn which takes [a]nk to [a]n is well-defined and is
a homomorphism of rings. Note that proving well-definedness is necessary since
the map θ was defined in terms of representatives of equivalence classes.
Exercise* 2.12. Prove that there is no homomorphism from Zn to Zl if l is not
an integer factor of n. [Hint, make use of the notion of characteristic]
10
An embedding, or monomorphism, is an injective (i.e. one-to-one) homomorphism.
Example 2.13. (3) There is a string of natural embeddings Z −→ Q −→
R −→ C −→ C[X] −→ C[X, Y ] −→ C(X, Y ). Check that you see what the
embeddings are and why they are embeddings.
Lemma 2.14. If θ : R −→ S and β : S −→ T are homomorphisms (of rings)
then so is the composition βθ : R −→ T .
If θ : R −→ S and β : S −→ T are embeddings then so is the composition
βθ : R −→ T .
If θ : R −→ S and β : S −→ T are homomorphisms and if βθ : R −→ T is an
embedding then θ is an embedding.
Exercise 2.15. Show that if θ : R −→ S and β : S −→ T are homomorphisms
such that the composition βθ : R −→ T is an embedding then it need not be
the case that β is an embedding.
If θ : R −→ S is a homomorphism of rings then the kernel of θ, ker(θ),
is the set, {r ∈ R : θ(r) = 0}, of elements which θ sends to 0S . For instance,
referring to the natural homomorphisms, Z −→ Zn and Z6 −→ Z2 , as in 2.11,
ker(Z −→ Zn ) = {nk : k ∈ Z} and ker(Z6 −→ Z2 ) = {[0]6 , [2]6 , [4]6 }.
Lemma 2.16. If θ : R −→ S is a homomorphism then θ is injective iff ker(θ) =
{0}.
Example 2.17. (4) If r ∈ R then the map from R[X] to R defined by sending
p(X) to p(r) is termed evaluation at r and is√easily checked to be a homomorphism. For instance, the map evaluation at 2 is determined (given that it
is a homomorphism) by sending each
√ constant polynomial s ∈ R[X] to s ∈ R
and sending the polynomial X √
to 2. The kernel of this map is the set of those
polynomials
p(X) such that p( 2) = 0, that is, those polynomials which have
√
(X − 2) as a factor.
Proposition 2.18. If R is any commutative ring and r ∈ R then the map
θr : R[X] −→ R given by θr (p(X)) = p(r) is a homomorphism and ker(θr ) =
{p ∈ R[X] : (X − r)|p}. (The symbol “|” means “divides”.)
Example 2.19. (5) The map from Z[ω], where ω is a primitive cube root of 1,
to Z[ω] given by a + bω + cω 2 7→ a + bω 2 + cω is a well-defined homomorphism
(note that, given that it is a homomorphism, it is determined by the fact that
it fixes each integer and sends ω to ω 2 ). It is an isomorphism, hence has zero
kernel, but it is not the identity map.
An automorphism of a ring is an isomorphism from the ring to itself. The
identity map always is an automorphism and the example just above shows
that there might be√others. In Exercise 2.20 you are asked to find all the
automorphisms of Z[ 2].
√
Exercise**
2.20. Show that the map from Z[ 2] to itself given by sending a +
√
√
b 2 to
√ a−b 2 is an automorphism. Show that there are no more automorphisms
of Z[ 2] apart from the identity map.
Example 2.21. (6) Let R1 × R2 be the product of rings R1 and R2 . The projection map, π1 : R1 × R2 −→ R1 , onto the first coordinate is a homomorphism
and its kernel is {0} × R2 .
11
The injection map ι1 : R1 −→ R1 × R2 which takes r1 to (r1 , 0) is not a
homomorphism, even though it commutes with addition and multiplication, because it fails to send 1 to 1 (if we were dealing with rings which don’t necessarily
have an identity element then it would count as a homomorphism).
Example 2.22. (7) Consider the map from R[X] to R[X] (where R is any ring)
which substitutes X 2 for X, that is, which sends p(X) to p(X 2 ). Then this is a
monomorphism which is not surjective. More generally if q(X) is any polynomial
then substitution of X by q(X), p(X) 7→ p(q(X)), is a homomorphism.
Lemma 2.23. Suppose that θ : R −→ S is a homomorphism. Then ker(θ) is a
subgroup of (R, +). Let r, r0 ∈ R. Then θ(r) = θ(r0 ) iff r − r0 ∈ ker(θ) iff r and
r0 belong to the same coset of ker(θ) in R.
That is, the fibres of a homomorphism θ are the cosets of ker(θ) in the
additive group (R, +). (If f : A −→ B is any map and if b ∈ B then the fibre
of f above b is f −1 b = {a ∈ A : f (a) = b}.)
An ideal of a ring R is a subset I ⊆ R such that:
• 0 ∈ I;
• a, b ∈ I implies a + b ∈ I;
• a ∈ I and r ∈ R implies ar ∈ I and ra ∈ I.
That is, an ideal of R is a subset of R which contains 0, is closed under
addition and is closed under (right and left) multiplication by every element of
the ring (not just under multiplication by elements of I). We write I C R to
mean that I is an ideal of R.
Exercise** 2.24. Give an example of a ring R and a subset H of R which
contains 0 and is closed under addition and multiplication but which is not an
ideal of R.
Example 2.25. If a ∈ R then {r1 as1 + · · · + rn asn : n ≥ 1, ri , si ∈ R} is, you
should check, an ideal which contains a and is the smallest ideal of R containing
a. It is called the principal ideal generated by a and is denoted hai. If R is
commutative then its description simplifies: hai = {ar : r ∈ R}. A principal
ideal is one which can be generated by a single element.
Examples 2.26. (1) In every ring h0i = {0} is the smallest ideal and may be
referred to as the trivial ideal. (2) In every ring h1i = R is the largest ideal
and every other ideal is referred to as a proper ideal. (3) If n ∈ Z then the
principal ideal, hni = {nk : k ∈ Z}, generated by n consists of all multiples of
n. (4) If p ∈ R[X] then hpi is the set of all polynomials with p as a factor.
Note that if I is an ideal of R then a ∈ I implies −a ∈ I, so every ideal also
is closed under subtraction.
The more general notion of right ideal is defined as for ideal but with the
third condition replaced by the weaker condition: a ∈ I and r ∈ R implies
ar ∈ I (and left ideals are defined similarly). Then, if a ∈ R the principal
right ideal generated by a ∈ R is defined to be the set {ar : r ∈ R} and is
denoted aR.
The difference between right ideals, left ideals and (two-sided) ideals is wellillustrated in matrix rings - do the next exercise to see this.
12
Exercise** 2.27. Let R = M2 (Z) be the ring of 2 × 2 matrices with integer
1 0
entries. Let a =
. Compute the right ideal generated by a, the left
0 0
ideal generated by a and the (two-sided) ideal generated by a.
Proposition 2.28. A commutative ring R is a field iff the only ideals of R are
{0} and R.
Exercise 2.29. Show that if R is any ring then R is a division ring iff the only
right ideals of R are {0} and R (and equivalently for left ideals).
[Comment: it is the case (and showing this is a substantial exercise) that
the first Weyl algebra, defined at 1.32, is a (non-commutative) ring which has
no ideals except {0} and R yet which is not a division ring. So the analogue of
2.28 for non-commutative rings must refer to one-sided ideals: the condition on
two-sided ideals is not enough to give a division ring.]
If A ⊆ R is any subset of the ring R then the ideal generated by A is
the set {r1 a1 s1 + · · · + rn an sn : n ≥ 1, ai ∈ A, ri , si ∈ R} and is denoted hAi.
For instance, if A = {a} is a singleton set then hAi, which is also written as
hai, consists of all elements of R which can be written as a sum of terms of the
form ras as r, s range over R. If A = {a, b} then hAi, also written ha, bi, is the
set of all elements of the form r1 as1 + . . . rn asn + t1 bu1 + . . . tm bum where the
ri , si , tj , uj can be any elements of R (and n, m are arbitrary natural numbers).
Exercise 2.30. Show that if A is a subset of a ring R then hAi is the smallest
ideal of R which contains A.
Example/Exercise 2.31. (5) Let R = Z[X] and let I = h2, Xi be the ideal of
Z[X] generated by {2, X}. Prove that I is not a principal ideal. [Hint: suppose,
for a contradiction, that it is, choose a generator, ...]
Exercise** 2.32. The ideal h4, 6i of Z generated by 4 and 6 together is principal
- find a generator for it. Generalise this.
Proposition 2.33. If θ : R −→ S is a homomorphism of rings then ker(θ) is
an ideal of R.
Corollary 2.34. If θ : R −→ S is a homomorphism of rings and R, S both are
fields then θ is a monomorphism.
Later, 3.4, we will prove a kind of converse to 2.33, showing that every ideal
does occur as the kernel of some homomorphism.
We finish the section with some general properties of ideals.
Proposition 2.35. Suppose that I and J are ideals of the ring R. Then I +J =
{a + b : a ∈ I, b ∈ J} (their sum) and I ∩ J (their intersection)
T are ideals. If
{Iλ }λ is any collection of ideals of R then their intersection, λ Iλ , is an ideal.
If R is commutative then (I : J) = {r ∈ R : rJ ⊆ I} is an ideal.
A product of ideals also may be defined: if I, J C R then their product is
defined to be the ideal generated by the set {ab : a ∈ I, b ∈ J} (this set will not
in general be closed under addition, which is why we need to say “generated
by”). Then powers of an ideal may be defined inductively by I n+1 = I n I.
Exercises 2.36, 2.37, 2.38 are concerned with these operations on ideals.
13
Exercise** 2.36. Let R = Z and let I = h3i. What is I 2 ? What is I n ? Let
J = h12i. What is IJ?
Exercise* 2.37. Let I, J be ideals of a ring R. Show that IJ ⊆ I ∩ J. Give an
example to show that this inclusion may be proper.
Exercise** 2.38. Let R = Z. Show that h2i + h5i = Z. Compute h2i ∩ h5i and
the product h2ih5i, writing each of these as a principal ideal, that is, in the form
hni for some integer n. Now replace 2 and 5 by arbitrary integers a and b. What
are hai + hbi, hai ∩ hbi and haihbi? [Hint: unless you’ve guessed the answers
this might be difficult to answer, let alone prove, so try with some other pairs
of integers in place of 2 and 5. In other words, explore until you see what’s
happening.] Compute (h2i : h5i) and (h6i : h8i). What about the general case
(hai : hbi)?
2.1
More exercises
Exercise* 2.39. Find all ring homomorphisms from Z × Z to itself.
√
a 2b
: a, b ∈ Z . Define the
Exercise* 2.40. Let R = Z[ 2] and let S =
b a
√
a 2b
map θ : R −→ S by θ(a + b 2) =
. Prove that θ is an isomorphism.
b a
Exercise** 2.41. Prove that Z6 ' Z2 × Z3 . Is Z10 ' Z2 × Z5 ? Is Z8 ' Z2 × Z4 ?
Justify your answers.
Exercise** 2.42. Prove that if θ : R −→ S is a surjective homomorphism of
rings and R is commutative then S must be commutative. Give an example to
show that the conclusion may be false if θ is not surjective.
Exercise 2.43. Let R be the ring of all infinitely differentiable functions from
the unit interval [0, 1] to R (with pointwise addition and multiplication as the
operations). Let D : R −→ R be the map which takes a function f to its
derivative f 0 . Is D a ring homomorphism?
Exercise* 2.44. Suppose that R is a commutative ring of characteristic 3. Prove
that the map θ : R −→ R defined by θ(r) = r3 is a homomorphism. Suppose
also that R has no non-zero nilpotent elements: prove that θ is injective.
Exercise** 2.45. Show that if I is a right ideal of a ring R and if there is an
element a ∈ I with a right inverse then I = R.
Exercise* 2.46. Let R be a commutative ring and set N (R) = {r ∈ R : ∃n, rn =
0} to be the set of all nilpotent elements of R. Prove that N (R) is an ideal of
R. Compute N (R) for:
(i) R = Z12 ;
(ii) R = Z24;
a b
(iii) R =
: a, b, c ∈ Q .
0 c
Exercise 2.47. Prove that every non-zero ideal of the ring Z[i] contains a nonzero integer.
Exercise* 2.48. Determine whether or not each of the following statements
about ideals in the polynomial ring Q[X, Y ] is true:
(i) hXY + Y 2 , X 2 + XY + Y 2 , X 2 + Y 2 i = hX 2 , XY, Y 2 i;
(ii) hX 2 + XY, XY − Y 2 , X 2 + Y 2 i = hX 2 , XY, Y 2 i.
14
Exercise 2.49. Suppose that I, J are ideals of the commutative ring R such that
I + J = R. Prove that IJ = I ∩ J.
Exercise 2.50. Let R = RhX, Y : Y X = XY + 1i be the first Weyl algebra (as
defined in the notes). Use the defining relation Y X = XY + 1 to write each of
the following elements of R as a sum of monomials of the form X i Y j (i, j ≥ 0):
Y X 2; Y 2X 2; Y X 3; Y X 2 − X 2Y ; Y X 3 − X 3Y .
Prove that, for every n ∈ N, the ideal generated by X n is equal to R.
Exercise* 2.51. Let R = R[X, Y ] and let S ⊆ R2 be any subset of the real plane.
Set I(S) = {p ∈ R : ∀(r, s) ∈ S, p(r, s) = 0}. Prove that I(S) is an ideal of
R. [Note that this is the kernel of the two-dimensional version of the evaluation
map for polynomials which is defined in the printed notes.]
Exercise** 2.52. Prove that if θ : R −→ S is a surjective homomorphism of
rings and I is a right ideal of R then θ(I) is a right ideal of S. Give an example
to show that the conclusion may be false if θ is not surjective.
Exercise 2.53. Let a be an element of a ring R. Show that the right annihilator
of a, rann(a) = {r ∈ R : ar = 0} is a right ideal of R. If also b ∈ R show
that rann(a + b) ⊇ rann(a) ∩ rann(b). Show that
can be proper.
the inclusion
2 3
Compute rann(a) where R = M2 (Z6 ) and a =
.
0 0
Show that rann(a) need not be a left ideal of R.
Exercise 2.54. Let p be a prime integer and set Z(p) = m
n ∈ Q : p - n}. Prove
that Z(p) is a subring of Z. Show that the only proper non-trivial ideals of Z(p)
are those generated by powers of p.
3
Factor Rings
Given a ring R and an ideal I of R we will define the factor ring (also called
the quotient ring) R/I and the canonical surjective homomorphism π : R −→
R/I with ker(π) = I. Informally, R/I is the “largest ring obtained from R by
collapsing I to 0.” An example of this process which you know is to start with
the ring, Z, of integers, take the ideal, h5i say, generated by 5: the resulting
quotient ring is the ring, Z5 , of integers modulo 5 and the canonical surjection
π is the map which takes an integer to its congruence class modulo 5. We give
the precise general definition now.
Let R be a ring and let I be a proper ideal. Let R/I denote the set of cosets
of I in the additive group (R, +): R/I = {r + I : r ∈ R}. Define operations +
and × on R/I by: (r + I) + (s + I) = (r + s) + I and (r + I) × (s + I) = (r × s) + I
(so we’re adding and multiplying cosets, these being the elements of R/I). As
always, when defining something in terms of representatives, the issue of welldefinedness has to be addressed.
For a number of the results in this section we give the, slighly tedious but
necessary, details here in the notes, so as to allow time for explanation and
illustration in lectures.
Lemma 3.1. The operations + and × on R/I are well-defined. That is, if
r + I = r0 + I and s + I = s0 + I then (r + s) + I = (r0 + s0 ) + I and (r × s) + I =
(r0 × s0 ) + I.
15
Proof. Recall that if H is a subgroup of an abelian group G and if a, b ∈ G then
a + H = b + H iff a − b ∈ H. We apply this with G being the group (R, +) and
H = I.
Since r + I = r0 + I we have r − r0 ∈ I. Since s + I = s0 + I we have
s − s0 ∈ I. Since I is closed under + we deduce that (r − r0 ) + (s − s0 ) ∈ I, that
is (r + s) − (r0 + s0 ) ∈ I, hence (r + s) + I = (r0 + s0 ) + I.
Also, since I is an ideal (r − r0 ) × (s − s0 ) ∈ I, that is, rs − r0 s − rs0 + r0 s0 ∈ I.
Now, rs − r0 s − rs0 + r0 s0 = (rs − r0 s0 ) + 2r0 s0 − r0 s − rs0 = (rs − r0 s0 ) − r0 (s −
s0 ) + (r0 − r)s0 . Since I is an ideal this is in I so, since both s − s0 and r0 − r
belong to I, so does rs − r0 s0 . Hence rs + I = r0 s0 + I, as required.
Lemma 3.2. Let R be a ring and let I be a proper ideal. With the operations
defined above, the set R/I is a ring. The zero element of this ring is the coset
0 + I(= I) and the identity element is 1 + I. Furthermore −(r + I) = (−r) + I
and, if r has an inverse r−1 in R then (r + I) is invertible in R/I with inverse
r−1 + I.
Proof. For instance, to check that the addition we have defined in R/I is
commutative, take r + I, s + I - two typical elements of R/I. By definition
(r + I) + (s + I) = (r + s) + I. Since addition is commutative in R this equals
(s + r) + I which, by definition of + in R/I, equals (s + I) + (r + I), as required.
Similarly for other properties. For instance to check distributivity,
take
elements r + I, s + I, t + I ∈ R/I. Then (r + I) (s + I) + (t + I) = (r +
I) (s + t) + I) (definition of +) = r(s + t) + I (definition of ×) = (rs + rt) + I
(since + is distributive over × in R) = (rs + I) + (rt + I) (definition of +)
= (r + I)(s + I) + (r + I)(t + I) (definition of ×).
To check that 0 + I is the zero element: given r + I ∈ R/I we have (0 + I) +
(r + I) = (0 + r) + I = r + I.
To check that 1 + I is the identity element: given r + I ∈ R/I we have
(1 + I) × (r + I) = (1 × r) + I = r + I.
Also given r + I ∈ R/I we have (r + I) + (−r + I) = (r + −r) + I = 0 + I
so −r + I is the negative of r + I.
Etc. etc.
This ring is the factor ring (or quotient ring) of R by I.
Example 3.3. As stated above, Z/h5i = Z5 .
Theorem 3.4. Let I be a proper ideal of the ring R. The map π : R −→
R/I defined by π(r) = r + I is a surjective ring homomorphism with kernel I
(π is called the canonical surjection, or canonical projection). It is the
“smallest” homomorphism with domain R and kernel I. Indeed, if θ : R −→ S
is a homomorphism with ker(θ) ⊇ I then there is a unique map θ0 : R/I −→ S
with θ0 π = θ. This map θ0 is a homomorphism.
The map θ0 is injective iff ker(θ) = I. If θ is surjective and ker(θ) = I then
0
θ is an isomorphism.
Proof. Since every element of R/I has the form r +I, = π(r)+I for some r ∈ R,
clearly the map π is surjective. We check that it is a homomorphism.
First, π(r + s) = (r + s) + I = (r + I) + (s + I) = π(r) + π(s). Similarly
π(rs) = (rs) + I = (r + I)(s + I) = π(r)π(s). And π(1) = 1 + I which, by the
result above, is the identity of R/I.
16
Next, the kernel of π: ker(π) = {r ∈ R : π(r) = 0R/I }. By the result above
the zero, 0R/I of R/I is the coset I = 0 + I so ker(π) = {r ∈ R : π(r) = 0 + I} =
{r ∈ R : r + I = 0 + I} = {r ∈ R : r ∈ I} = I.
Now suppose that θ : R → S is a homomorphism with ker(θ) ⊇ I. Define
the map θ0 : R/I → S by θ0 (r + I) = θ(r). First we have to show that this
map is well-defined, because the definition used a choice of coset representative.
So suppose r + I = r0 + I; then r − r0 ∈ I so, by assumption, θ(r − r0 ) = 0.
Since θ is a homomorphism this gives θ(r) = θ(r0 ) - so we get the same result
for θ0 (r + I) whatever representative of this coset we use.
0
Then
We have θ0 (r + I) +
we0 have to show
that θ is a homomorphism.
(s + I) = θ (r + s) + I = θ(r + s) = θ(r) + θ(s) = θ0 (r + I) + θ0 (s + I). And
similarly for multiplication.
The homomorphism θ0 is injective iff θ0 (r + I) = 0 implies r + I = 0 + I that
is, iff θ(r) = 0 implies r ∈ I, that is, iff ker(θ) is contained in, hence equals, I,
as required.
Finally, if θ is surjective and ker(θ) = I then it is, by the previous paragraph,
also injective, hence an isomorphism.
This is a place where there are significant conceptual hurdles to get over and
I will spend time on discussion and illustration.
Example 3.5. Take R = Z and I = h6i = 6Z so we have the canonical
surjection π : Z −→ Z6 . Let θ be the canonical projection from Z to Z/h2i = Z2
(which takes a ∈ Z to [a]2 ). Then ker(θ) = h2i ⊇ h6i so, by the theorem, there
is a unique factorisation of θ through π. This is the map Z6 −→ Z2 already seen
in 2.11.
Some notation: if I ⊇ J are ideals then usually I write I ≥ J instead of just
using the subset notation (but it means the same).
Example 3.6. Let R = R[X] and take I = hX 2 + 1i to be the principal ideal
generated by X 2 +1. Let π : R[X] −→ R[X]/hX 2 +1i be the canonical surjection.
Now let θ : R[X] −→ C be the homomorphism which is evaluation at i ∈ C,
that is, θ(p(X)) = p(i) (note that, even though i is not in R this does make
sense and it is a homomorphism). A polynomial p is in the kernel of θ iff it has
X 2 + 1 for a factor, that is, ker(θ) = hX 2 + 1i. By the last part of 3.4 the unique
map, θ0 , from R[X]/hX 2 + 1i to C through which θ factorises, that is, such that
θ = θ0 π, is injective. It is easily checked that θ0 also is surjective and hence is
an isomorphism. Thus R[X]/hX 2 + 1i ' C.
Theorem 3.7. Let I be an ideal of the ring R. Then there is a natural, inclusionpreserving, bijection between the set of ideals of R which contain I and the set
of ideals of the factor ring R/I:
• to an ideal J ≥ I there corresponds πJ = {r + I : r ∈ J} = {π(r) : r ∈ J};
• to an ideal K C R/I there corresponds π −1 K = {r ∈ R : π(r) ∈ K}.
The notation J/I is also used instead of πJ for the image of J in R/I.
Proof. First suppose that J ≥ I is an ideal of R: we check that πJ is an ideal
of R/I.
• Since 0 ∈ J, π(0) ∈ πJ, but π(0) = 0 (more accurately, π(0R ) = 0R/I ), so
0 ∈ πJ.
17
• Suppose that b, b0 ∈ πJ, say b = π(a), b0 = π(a0 ) for some a, a0 ∈ J. Since J
is an ideal a + a0 ∈ J, so π(a + a0 ) ∈ πJ. Since π is a homomorphism (Theorem
3.4), π(a) + π(a0 ) = π(a + a0 ) ∈ πJ, as required.
• Suppose that b ∈ πJ and s ∈ R/I. Say b = π(a) with a ∈ J. Also, π is
surjective so there is r ∈ R with π(r) = s. Since J is an ideal both ar, ra ∈ J.
Hence bs = π(a)π(r) = π(ar) ∈ πJ and similarly sb ∈ πJ, as required.
Thus we have checked the three properties for πJ to be an ideal of R/I.
(You might notice that we did not need to assume that J ≥ I: that assumption
only comes in when we prove the bijection.)
Next, let K be an ideal of R/I and consider its inverse image π −1 K = {a ∈
R : π(a) ∈ K}. We show that π −1 K is an ideal of R and contains I.
• Since 0R/I ∈ K and π(0R ) = 0R/I we have 0R ∈ π −1 K.
• Let a, a0 ∈ π −1 K. So π(a), π(a0 ) ∈ K. Since K is an ideal π(a + a0 ) =
π(a) + π(a0 ) ∈ K. So a + a0 ∈ π −1 K, as required.
• Let a ∈ π −1 K and let r ∈ R. Then π(ar) = π(a)π(r) ∈ K and π(ra) =
π(r)π(a) ∈ K since π(a) ∈ K and K is an ideal. Therefore ar, ra ∈ π −1 K, as
required.
Thus we have shown that π −1 K is an ideal of R.
• Let a ∈ I, then π(a) = 0 ∈ K, so a ∈ π −1 K. Thus π −1 K contains I.
It remains to show that (1) if we start with an ideal J ≥ I, map it across
to R/I to get πJ and then pull that back to R, to get π −1 πJ then we end up
where we started, with J. And, similarly, that, (2) starting with an ideal K
of R/I, pulling back to π −1 K and then mapping across again to get ππ −1 K,
that we end up with K again. The uncommented steps in the arguments which
follow are all direct from the definitions (of πJ and π −1 K etc.).
(1) If a ∈ J then π(a) ∈ πJ so a ∈ π −1 πJ. Thus J ⊆ π −1 πJ. For the converse,
take a ∈ π −1 πJ: then π(a) ∈ πJ, say π(a) = π(a0 ) for some a0 ∈ J. Therefore
π(a−a0 ) = 0 and hence a−a0 ∈ ker(π) = I (by Theorem 3.4). At last we use the
assumption that J ≥ I, to deduce a−a0 ∈ J. Then we have a = a0 +(a−a0 ) ∈ J
(since both a0 and a − a0 are in J). Thus π −1 πJ ⊆ J and hence the two sets
are equal.
(2) If b ∈ K then, since π is surjective, there is r ∈ R with π(r) = b. So
r ∈ π −1 K. Then b = π(r) ∈ ππ −1 K. Thus K ⊆ ππ −1 K. For the converse,
take b ∈ ππ −1 K. So there is r ∈ π −1 K such that b = π(r). But the fact that
r ∈ π −1 K says that π(r) ∈ K. Thus b ∈ K. Therefore ππ −1 K ⊆ K and so the
two sets are equal.
Example 3.8. The ideals of Z6 = Z/h6i are in bijection with the ideals of Z
which contain h6i and these are Z, h2i, h3i, h6i, giving four ideals of the quotient
ring Z6 (you should write them down explicitly).
There are some exercises (Exercises 3.9, 3.10) around this.
Exercise** 3.9. Determine the ideals of Z24 and match these up with the ideals
of Z which contain h24i.
Exercise 3.10. In the situation of 3.7 there is a similar correspondence for right
(respectively left) ideals J containing I and right (resp. left) ideals of R/I (but
I itself should still be a two-sided ideal in order that R/I be a ring). Check
this.
18
An ideal I of a ring R is maximal if it is proper (i.e. I 6= R) and if there
is no ideal between it and R, more formally, if any ideal J with I ≤ J ≤ R is
equal to I or to R.
Corollary 3.11. If R is a commutative ring then an ideal I C R is maximal
iff the quotient ring R/I is a field.
Theorem 3.12. If I ≤ J are ideals of R, so J/I is an ideal of R/I, then
(R/I)/(J/I) ' R/J.
Exercise 3.14 illustrates this.
Example 3.13. We have h6i ≤ h2i and the theorem says that Z/h6i / h2i/h6i
is isomorphic to Z/h2i (this is already essentially in 2.11, 3.5).
Exercise* 3.14. Describe explicitly the isomorphism between Z/h10i / h5i/h10i
and Z5 .
Exercise 3.15. An ideal I of a commutative ring R is prime if whenever r, s ∈ R
and rs ∈ I then either r ∈ I or s ∈ I. What are the prime ideals of Z? Prove
that an ideal I of the commutative ring R is prime iff the factor ring R/I is a
domain.
3.1
More exercises
Exercise* 3.16. Show that the factor ring Z[i]/h1 + 3ii is isomorphic to Z10 .
Exercise 3.17. Prove that if K is a field then K[X, Y ]/hXY i is isomorphic to
the subring of the product K[X]×K[Y ] consisting of all pairs (p(X), q(Y )) with
the same constant term (i.e. such that p(0) = q(0)).
Exercise 3.18. Let R = K[X] where K is a field. Let I = hX(X − 1)i. Define a
map θ from R to S = K × K as follows. Given p ∈ R, write p = q · X(X − 1) + r
where q, r ∈ R and r is linear, so r has the form λ + µX for some λ, µ ∈ K.
Then define θ(p) = (λ, λ + µ). Prove that θ is a homomorphism. Prove that θ
induces an isomorphism from R/I to S.
4
4.1
Polynomial Rings and Factorisation
Division of polynomials
Throughout this subsection K is a field and polynomials will be in a single variable X, thus members ofP
the polynomial ring K[X]. Recall that the canonical
n
form of f ∈ K[X] is f = i=0 ai X i with ai ∈ K. If an 6= 0 then the degree of
f is n, deg(f ) = n.
Theorem 4.1. (Division Theorem for Polynomials) Let K be a field and take
f, g ∈ K[X] with g 6= 0. Then there are (unique) q, r ∈ K[X] with
f = qg + r and deg(r) < deg(g).
(q is the quotient and r the remainder when f is divided by g)
(This should remind you of the Division Theorem for integers. So should the
proof.)
19
Proof. Let A = {f − gh : h ∈ K[X]} be the set of all polynomials which can be
obtained by subtracting polynomial multiples of g from f . This set is non-empty
since it contains, for example, f = f − g · 0.
Choose a polynomial, say r, in A of least degree, so r = f − qg for some
q ∈ K[X]. Rearranging gives f = qg + r, so it has to be shown than deg(r) <
deg(g). If it were not, then (think about it) some multiple of g could be used to
reduce the degree of r: say deg(r − hg) < deg(g) for some h ∈ K[X]. But then
r − hg = (f − qg) − hg = f − (q + h)g would be in A and would have smaller
degree than r, contradicting the choice of r.
It remains to show uniqueness. If we had qg − r = f = q 0 g − r0 with both
deg(r), deg(r0 ) < deg(g) then (q − q 0 )g = r − r0 . Because deg(r0 − r) < deg(g)
this can happen only if both sides are 0, that is, only if q = q 0 and r = r0 , as
required.
An element a ∈ K is a root (or zero) of f ∈ K[X] if f (a) = 0.
Corollary 4.2. Let K be a field, let f ∈ K[X] and let a ∈ K. Then a is a root
of f iff X − a is a factor of f .
Proof. If X −a is a factor of f then certainly f (a) = 0. For the converse, suppose
that f (a) = 0. Use the Division Theorem with g = X −a to deduce that there is
an equation f = q · (X − a) + r for some q, r ∈ K[X] with deg(r) < deg(X − a).
Since X − a has degree 1 it must be that r is a constant. So, substituting a
for X we obtain f (a) = 0 + r. In particular, if f (a) = 0 then r = 0 and hence
f = q · (X − a) is indeed a multiple of X − a.
The greatest common divisor (or highest common factor) of polynomials f, g is a polynomial d such that d|f , d|g and, if h is any polynomial
dividing both f and g then h divides d. Write d = gcd(f, g). This polynomial
is defined only up to non-zero scalar multiple so, if we want a unique gcd then
we can insist that d be monic (have leading coefficient 1).
Division Algorithm This is entirely analogous to the Division Algorithm
(Euclid’s Algorithm) for integers. The key observation is that if f = qg + r
then gcd(f, g) = gcd(g, r) (because, by that equation and its rearrangement
f − qg = r, the common divisors of f and g are the common divisors of g and
r).
So, given f, g ∈ K[X] with g 6= 0, apply the Division Theorem to obtain
f = q0 g + r1 with deg(r1 ) < deg(g).
Then apply it to divide r1 into g, obtaining:
g = q1 r1 + r2 with deg(r2 ) < deg(r1 )
et cetera
r1 = q2 r2 + r3 with deg(r3 ) < deg(r2 )
...
rn−2 = qn−1 rn−1 + rn with deg(rn ) < deg(rn−1 )
and, because the degrees of the remainders are strictly decreasing, eventually
a remainder of 0 is reached:
rn−1 = qn rn .
Then gcd(f, g) = gcd(g, r1 ) = gcd(r1 , r2 ) = · · · = gcd(rn−1 , rn ) = rn (the
last since rn divides rn−1 ).
So the gcd of f and g is the last non-zero remainder in this algorithm.
20
Also, tracking back through the equations allows the expression of the gcd,
rn , as a linear combination of f and g; “linear” in the polynomial sense, that
is, there are polynomials k, l ∈ K[X] such that gcd(f, g) = kf + lg. And from
this follows the next corollary.
Corollary 4.3. Let K be a field and take f, g ∈ K[X]. Then the ideal generated
by f together with g equals the ideal generated by their greatest common divisor:
hf, gi = hgcd(f, g)i.
Corollary 4.4. Let K be a field. Then every ideal of the polynomial ring K[X]
is principal (i.e. can be generated by a single polynomial).
Proof. It follows from the corollary above, and induction, that every ideal generated by finitely many polynomials can be generated by a single polynomial
(their greatest common divisor) and then one could quote the Hilbert Basis
Theorem which says that every ideal in K[X] is finitely generated. But that’s
a bit heavy-handed and the idea in the proof of the Division Algorithm can be
used: namely, given an ideal, choose a polynomial in it of least degree (ignoring
the zero polynomial), and show that this generates the ideal (details left as an
exercise).
4.2
Factorisation
Throughout this section the ring R is assumed to be commutative.
Most of the examples that we use will be polynomial rings but we set these in a
more general context. Although we state various definitions and results about
this general context we will, in practice, treat the general case rather lightly.
In particular, when it comes to revising this subsection for the examination,
concentrate on what was actually covered in lectures.
An element r ∈ R is irreducible if r is not invertible and if, whenever r = st
either s or t is invertible. For example, the irreducible elements of Z are the
prime (positive and negative) integers.
Exercise* 4.5. What are the irreducible elements of the polynomial ring R[X]?
Elements r, s ∈ R are associated if s = ur for some invertible element
u ∈ R. For instance two integers r, s are associated iff r = ±s.
Exercise 4.6. Show that the relation “associated” is an equivalence relation on
any commutative ring R.
Exercise 4.7. What are the association equivalence classes in the polynomial
ring R[X]?
A commutative domain R is said to be a unique factorisation domain if every
non-zero, non-invertible element of R has an essentially unique factorisation as
a product of irreducible elements. More formally: the commutative domain
R is a unique factorisation domain, or UFD for short, if for every nonzero r ∈ R which is not invertible there are irreducible elements r1 , . . . , rk ∈ R
such that r = r1 × · · · × rk (existence of irreducible factorisation) and, if also
r = s1 × · · · × sl where the sj are irreducible elements of R, then k = l and
there is a permutation σ of {1, . . . , k} such that, for each i, sσ(i) is associated
to ri (uniqueness of irreducible factorisation).
21
Example 4.8. Z is, as you know, a unique factorisation domain. For instance
take r = −24. Then −24 = (−2)×2×(−2)×(−3) is one irreducible factorisation.
There are others, such as −24 = 3 × (−2) × 2 × 2 but you can surely see how to
permute the factors so that they match up as associated elements.
√
Example 4.9. The ring Z[ −5], though
a commutative
domain, is not a UFD.
√
√
For instance 21 = 3 × 7 = (1 + 2 −5)(1 − 2 −5) are, one may check, two
factorisations into irreducibles but, again one may check, the irreducible factors
do not match up into associated pairs.
Example 4.10. By default, every field is a UFD.
A principal ideal domain, or PID for short, is a commutative domain in
which every ideal is principal (that is, is generated by some single element).
Example 4.11. The ring Z is a PID since every ideal has the form hni for
some integer n. If K is a field then K[X] is a PID (every ideal consists of all
the multiples of a particular polynomial). In each case the proof makes use of
the division theorem.
It is a general result that every PID is a UFD so it follows that if K is a field
then the polynomial ring K[X] is a UFD. A much harder result to prove is that
K[X1 , . . . , Xn ] is a UFD: we won’t prove it in the course but it is, in fact, true
that if R is a UFD then so is R[X] hence, by induction, so is R[X1 , . . . , Xn ] (for
a proof see, e.g., Fraleigh, Section 45).
Example 4.12. For an example of a ring which is a UFD but not a PID take
either Z[X] or K[X, Y ] where K is a field. It is not difficult to show that these
are not PIDs (cf. Example 2.31). The fact that they are UFDs follows from the,
hard, general result mentioned earlier, that if R is a UFD then so is R[X].
Remark 4.13. Almost all the material that we present in this course is from
the foundations of the subject and dates back around 100 years (or more).
But this is a good place to mention a somewhat more recent result (from the
late 1950s) of M. Auslander and D. Buchsbaum which says that regular local
rings are UFDs: regular local rings are commutative rings which are of central
importance in algebraic geometry.
Remark 4.14. Another (this time, rather old) result which we should mention
is the Hilbert Basis Theorem (named after D. Hilbert) which states that if K is
a field then every ideal in a polynomial ring K[X1 , . . . , Xn ] is finitely generated
(i.e. has a finite set of generators).
Exercise* 4.15. Let K be a field and consider the polynomial ring
R = K[X1 , . . . , Xn , . . . (n ∈ P)] in infinitely many generators. Find an ideal I
of R which is not finitely generated. Can you prove that your choice for I isn’t
finitely generated?
There is a variety of tests for irreducibility in polynomial rings. Some examples will be given in lectures.
22
4.3
More exercises
Exercise** 4.16. Find a generator, h, for the ideal hf, gi of Q[X], where f =
X 3 − X 2 − X + 1, g = X 5 + X 2 − X − 1. Write h in the form af + bg for suitable
a, b ∈ Q[X].
Exercise* 4.17. Find a generator, h, for the ideal hf, gi of C[X], where f =
X 3 − iX 2 + 2X − 2i, g = X 2 + 1. Write h in the form af + bg for suitable
a, b ∈ Q[X].
Exercise 4.18. Find a generator, k, for the ideal hf, g, hi of Q[X], where f =
25
2
X 3 + 4X 2 + 2X − 3, g = X 4 + 3X 3 − X 2 − 3X, h = X 4 − X 3 − 15
2 X + 2 X − 3.
Write k in the form af + bg + ch for suitable a, b, c ∈ Q[X].
Exercise 4.19. For which integers n does the polynomial X 2 + X + 1 divide
X 4 + 3X 3 + X 2 + 6X + 10 in the polynomial ring Zn [X].
Exercise** 4.20. Consider the field Z2 with two elements. Write down all the
monic polynomials of degree 2 in R = Z2 [X]. For each of these, determine
whether it is reducible or irreducible. Now do the same for monic polynomials
of degree 3.
Exercise** 4.21. Prove that the polynomial 7X 3 − 6X 2 + 2X − 1 ∈ Z[X] is
irreducible. Then prove this using a different method.
Exercise** 4.22. Prove that the polynomial X 5 + X 2 + X − 1 ∈ Z[X] is irreducible.
Exercise* 4.23. Show that if K is a finite field then there is a polynomial in
K[X] which has no root in K.
Exercise 4.24. Show that the only invertible elements of the ring Z[i] are ±1,
±i.
5
Constructing Roots for Polynomials
You know that some (non-constant) polynomials with rational coefficients, for
example X 2 − 2, don’t have a rational root. That particular polynomial does
have a solution in the field R which extends Q. For some other polynomials,
for example, X 2 + 2, there is still no root in R but we could extend further, to
C. In fact, we don’t have to go to such large field extensions. In this section we
will see how, given a field K and a polynomial with coefficients in K, we can
produce a root of that polynomial in a “minimal” field extension of K.
That is, given a field K and p ∈ K[X], we will produce a field, L, extending
K (i.e. into which K embeds) and containing a root for p. Our construction will
be such that L is generated (as a ring) by K and a root of p; so the construction
is an economical one.
Example 5.1. Take p = X 2 − 2 ∈ Q[X]. For√an extension field containing a
root we could take R or, less extravagantly, Q[ 2].
Example 5.2. Take p = X 2 + 1 ∈ R[X]: then for L we may take C (and that
is the smallest possible choice since it is generated as a ring, so certainly as a
field, by R and a root of p).
23
Notice that, to produce a root for a polynomial p, it is enough to produce a
root of one of its irreducible factors. Therefore we may concentrate on the case
of an irreducible polynomial.
Theorem 5.3. (Kronecker’s Theorem) Let K be a field and let f ∈ K[X]
be irreducible of degree n. Then the canonical homomorphism π : K[X] −→
K[X]/hf i induces an embedding ι : K −→ K[X]/hf i of K into L = K[X]/hf i
and L is a field. Also α = π(X) ∈ L is a root of f . The dimension of L as a
vector space over K is n, with {1, α, α2 , . . . , αn−1 } being a basis of L over K,
so every element of L has a unique representation of the form an−1 αn−1 + · · · +
a1 α + a0 with an−1 , . . . , a1 , a0 ∈ K.
(Note that in the latter part of the statement we have identified K with its image
in L.)
The last part of the theorem says that L, which we also write as K[α], is
minimal, not just as a field but even as a ring, given that we have added a root
of f .
Examples 5.4. (1) K = R, f = X 2 + 1. The degree of f is 2, so every
element of R[X]/hX 2 + 1i has the form a1 i + a0 with a1 , a0 ∈ R, where i,
instead of α, has (for obvious reasons) been used as notation for π(X). Clearly
L = R[X]/hX 2 + 1i ' C.
(2)√K = Q, f = X 2 − 2. Then it’s easy to check that L = Q[X]/hX 2 − 2i '
Q[ 2].
Corollary 5.5. Let f ∈ K[X] be irreducible, where K is a field, and let π, ι
and α be as in the statement of Kronecker’s Theorem. Let θ : K −→ L0 be an
extension field of K and suppose that L0 contains a root, β, of f . Then there
is a homomorphism ρ : L = K[X]/hf i −→ L0 extending θ (in the sense that
ρι = θ) and taking α to β. (So there is a copy of L sitting between θ(K) and
L0 .)
2
0
Example 5.6.
√ Take K = Q, f = X − 2 and L = R. The three fields involved
are: Q ≤ Q[ 2] ≤ R.
Example 5.7. (3) K = Q, f = X 4 +1. First we will check that f is irreducible.
That done, since deg(f ) = 4 the canonical form for elements of L = Q[X]/hX 4 +
1i is a3 α3 +a2 α2 +a1 α+a0 with the ai rational and where α is as in the theorem.
Of course we can identify α with
√ a primitive fourth root of −1 in C and see that
Q[X]/hX 4 + 1i ' Q[(1 + i)/ 2]. This is continued in Exercise 5.8.
√
√
Exercise 5.8. Show that the ring Q[(1+i)/ 2] (see Example 5.7) equals√
Q[i, 2],
by which is meant the smallest subfield
which contains Q, i and 2. Also
√ of C √
show that this is equal to {a + bi + c 2 + di 2; a, b, c, d ∈ Q}).
Exercise* 5.9. (quite long but worth doing) Show that the polynomial f =
X 4√
+1 factorises completely (i.e. into linear factors) in the extension field Q[(1+
i)/ 2] constructed above: thus, adding a single root of f gives all the roots.
By way of contrast, consider g = X 3 − 2 ∈ Q[X]. Check that the field
L = Q[α] constructed in the theorem may be regarded as Q[21/3 ] but that the
other two roots of g are not in this field. Find a field L0 , of degree (i.e. dimension
as a vector space) 6 over Q, over which g does split into linear factors.
Now check that the field L could have been taken to be Q[η] where η is one
of the properly complex roots of g and find the factorisation of g as (X − η)g 0
24
where g 0 is a quadratic polynomial with coefficients in Q[η]. This emphasises
that the three roots of g are, abstractly, equivalent (precisely, Q[21/3 ] ' Q[η]).
If you want to understand better what’s happening here, pursue the topic
called Galois Theory.
In Kronecker’s Theorem the polynomial f is assumed to be irreducible: what
happens if we drop that assumption?
Example 5.10. Let K be any field and let f = X(X −1). Then K[X]/hX(X −
1)i is not a field. To see this, let α = π(X) where, as usual, π : K[X] −→
K[X]/hX(X − 1)i is the canonical projection. Then neither α nor α − 1 is zero
(since neither is a multiple of f ) but their product α(α − 1) is 0 = π(X(X − 1)).
So the ring K[X]/hX(X − 1)i is not even a domain.
Examples 5.11. We will consider some finite field extensions of F2 , F3 , F5 ,
including the degree 2 extension of F2 . In particular we’ll use X 3 +X+1 ∈ Z2 [X],
X 2 + X + 2 ∈ Z3 [X] and look at a quadratic extension of Z5 .
Example 5.12. Let p be an odd prime. Then −1 is a quadratic residue modulo
p (that is, a square modulo p) iff p has the form 4k + 1 for some integer k, so
Fp [X 2 + 1] is a proper extension of Fp iff p ∼
= 3 mod 4.
In any case, for any prime p, there always will be an irreducible polynomial
in Fp [X] of the form X 2 + aX + b (just by counting) and hence some proper
extension of Fp of degree 2.
The remainder of this section is purely for your edification: likely we will
not have time to cover it and it will not be examined. So: purely for interest!
A field K is algebraically closed if every non-constant polynomial with
coefficients from K has a solution from K.
Exercise 5.13. Prove that the following are equivalent for a field K: (i) K
is algebraically closed; (ii) every irreducible polynomial over K is linear; (iii)
every non-constant polynomial with coefficients from K has a factorisation as a
product of linear factors.
Examples 5.14. Neither Q nor R is algebraically closed but C is (this is the
so-called Fundamental Theorem of Algebra).
If K ≤ L is an extension of fields then an element λ ∈ L is said to be
algebraic over K if it is a root of some non-zero polynomial with coefficients
from K. For example every element in C is algebraic over R, indeed is a root of a
quadratic polynomial with coefficients in R (exercise: describe such a polynomial
explicitly, given a complex number λ = a + bi).
On the other hand C is not algebraic over Q. There is a fairly short “counting” argument to prove this. From that argument it follows that many (indeed
“most”) elements of C are transcendental (not algebraic) over Q. It is, however, surprisingly difficult to prove that a particular real or complex number is
transcendental over Q. For instance neither π nor e is a root of any (non-zero)
polynomial with rational coefficients but neither of these facts is easy to prove
(Hermite in 1873 showed it for e and Lindemann in 1882 for π). And, although
one might guess that π + e and eπ are transcendental over Q, this has never
been proved. (It is not difficult to show that at least one of them must be,
probably both are, but no-one knows. See, e.g., the article on mathworld about
transcendental numbers.)
25
e containing K which is
Theorem 5.15. Let K be a field. There is a field K
e is algebraic over K. This
algebraically closed and such that every element of K
field is unique up to isomorphism (over K) and is called the algebraic closure
of K.
Example 5.16. The algebraic closure of Q is not the field of complex numbers.
In fact it is much smaller: if you trace through the proof of the above result with
e is countable, hence is much
K being Q then you can see that the end result, Q,
smaller than C. This is a fundamental mathematical object which, nevertheless,
you are unlikely to have encountered before.
5.1
More exercises
Exercise** 5.17. Let F4 be the field with 4 elements (so F4 ' Z2 [X]/hX 2 + X +
1i), say F4 = {0, 1, α, α + 1} where α2 + α + 1 = 0.
(i) Show that the polynomial p = X 2 + αX + (α + 1) ∈ F4 [X] is reducible.
(ii) Show that q = X 2 + αX + 1 ∈ F4 [X] is irreducible.
(iii) Let F16 = F4 [X]/hX 2 + αX + 1i be the corresponding extension field of F4 ,
and let β be a root of X 2 + αX + 1 in F16 , so 1, β is a basis of F16 over F4 .
Compute β 3 and β −1 in terms of this basis.
Write down a basis of F16 over F2 .
Exercise 5.18. Show that the map from R[X, Y ]/hX 2 +1, Y 2 +1i to C×C which
takes a polynomial p(X, Y ) to (p(i, i), p(i, −i)) is an isomorphism.
√
√
Exercise 5.19. Show that each of 7 + 21/3 and 3 + −5 is algebraic over Q.
That is, for each, find a non-zero polynomial with rational coefficients of which
it is a root.
Exercise 5.20. Prove that Z2 [X]/hX 3 + X + 1i is a field and that Z3 [X]/hX 3 +
X +1i is not. [Hint: this is not a long computation, it’s a case of seeing what the
point is, quoting a result from the notes, and doing a very small computation.]
Part II
Groups
6
Groups, Isomorphisms and Group Actions
Recall that a group is given by the following data:
• a set G;
• a binary operation, “multiplication”, denote it · say, on G;
• a unary operation, “inverse”, denote it (−)−1 say, on G;
• a specific element, the identity element, denote it e, in G;
such that:
∗ the operation · is associative, a · (b · c) = (a · b) · c for all a, b, c ∈ G;
∗ a · a−1 = e = a−1 · a for all a ∈ G;
∗ a · e = a = e · a for all a ∈ G.
In practice one usually gives just the set G together with an associative
operation · on G and then checks existence of an identity and existence of
inverses. Also, in practice, we write ab for a · b and just G for (G, ·).
26
Recall how to draw up the group table for a (small) group G: arrange the
elements of G along the top and left-hand side and, at the entry in the a-labelled
row and b-labelled column, place the product ab.
Recall that the order of an element a ∈ G, we write o(a), is the least k ≥ 1
such that ak = e and is ∞ if there is no such k. For instance the identity
element of G is the only one of its elements with order 1. Elements of order 2
sometimes are called involutions.
The order of the group G is quite a different notion and means simply the
number of elements in G: we denote this by |G|.
Example 6.1. (1) Sn - the symmetric group on n letters, that is, the group of
all permutations of n objects. Recall that every permutation may be written,
in an essentially unique way, as a product of disjoint cycles. The order of an
element is obtained by inspection from its cycle decomposition (and that is easy
to compute).
Exercise** 6.2. Draw up the group table for S3 . List, giving each as a product of
disjoint cycles, all the permutations in S4 . Determine the order of each element
of S4 .
Recall that if (G, ·) is a group, with identity element e, then a subset H of
G forms a subgroup if:
• a, b ∈ H implies a · b ∈ H;
• a ∈ H implies a−1 ∈ H;
• e ∈ H.
If these conditions are satisfied then H, with the restriction of “·” to H forms
a group in its own right. The trivial group, {e} has just one subgroup (namely
itself). Every other group G has at least two subgroups: the whole group G (all
other subgroups are proper) and the trivial subgroup {e}. We write H ≤ G
when H is a subgroup of G.
If a ∈ G then the cyclic subgroup generated by a is the set of all, positive
and negative (and 0), powers hai = {ak : k ∈ Z} and this is a subgroup of G (if
the order of a is n < ∞ then only positive powers are needed since a−1 = an−1 ).
A group is said to be cyclic if it can be generated by a (suitable) single element.
Recall that if H is a subgroup of G and a ∈ G then the corresponding left
coset is aH = {ah : h ∈ H} and the right coset of a with respect to H is
Ha = {ha : h ∈ H}. The sets H, aH and Ha all have the same number of
elements and the, say left, cosets partition G into disjoint pieces. The number of
distinct left cosets of H in G is the index of H in G (and equals the number of
distinct right cosets of H in G). This is written [G : H]. Lagrange’s theorem
says that |G| = |H|[G : H].
Exercise** 6.3. Give an example of a group G, a subgroup H and an element
a ∈ G such that the left and right cosets aH and Ha are not equal.
Lemma 6.4. Let H be a subgroup of the group G and let a, b ∈ G.
(i) aH = H iff a ∈ H iff Ha = H
(ii) aH = bH iff a−1 b ∈ H iff b−1 a ∈ H.
A group G is abelian (or commutative) if ab = ba for all a, b ∈ G.
27
Example 6.5. (3) If (R; +, ×, 0, 1) is a ring then (R, +) is an abelian group.
The most basic example is perhaps (Z, +). Note that every non-zero element of
this group has infinite order.
Exercise* 6.6. Write down the group tables, where the operation is addition,
for Z4 and also for Z5 . Do the same for the field F4 with 4 elements (see Section
5).
Example 6.7. (4) If (R; +, ×, 0, 1) is a ring then the set, U (R) of invertible
(with respect to multiplication) elements of R, with the operation × forms a
group (exercise: prove this), called the group of units of R.
Exercise** 6.8. Find U (R) and draw up the group table for each of the following
rings R: Z; Z5 ; Z8 ; F4 .
Exercise 6.9. Let R be the ring M2 (Z) of 2 × 2 matrices with integer entries.
Find some elements in the group of units of R and
compute
their orders.
a b
Now replace Z by R. What is a criterion for
∈ U (M2 (R)? Given
c d
n can you find an element of order n in U (M2 (R))?
Example 6.10. (5) Groups of symmetries of geometric figures (composition
being the operation) is another good source of groups.
Exercise** 6.11. Denote by Dn the group of symmetries of a regular n-gon.
Show that the dihedral group Dn has 2n elements, of which n are rotations
and n are reflections. Draw up the group table for D3 .
We say that two groups G and H are isomorphic if they have the same
shape as abstract structures. Precisely, an isomorphism from G to H is a
bijection θ : G −→ H such that θ(ab) = θ(a)θ(b) for all a, b ∈ G. Note the
similarity with the notion of isomorphism for rings (this is, if anything, simpler).
Similar comments apply: for example in the equation above, the operation on
the left hand side is the multiplication in G whereas that on the right is the
multiplication in H. The next lemma is analogous to 2.3.
Lemma 6.12. Suppose that θ : G −→ H is an isomorphism between groups.
Then:
(1) θ(eG ) = eH ;
(2) θ(a−1 ) = (θ(a))−1 ;
(3) o(θ(a)) = o(a) for every a ∈ G;
(4) |G| = |H|.
We say that G and H are isomorphic, and write G ' H, if there is an isomorphism from G to H. As with rings, this is an equivalence relation (exercise).
To say that G and H are isomorphic is to say that they are the same group
“up to re-naming elements”. In terms of the group tables, this means that, reordering elements if necessary, the group tables are the same up to re-naming
elements (θ is the “renaming function”).
Exercise** 6.13. Show that the symmetric group S3 and the dihedral group
D3 are isomorphic by bringing their group tables to the same form. Define an
isomorphism from S3 to D3 . Now define another isomorphism from S3 to D3 .
28
Suppose that G and H are groups. The direct product, G × H, of G and
H is the group with underlying set G × H = {(g, h); g ∈ G, h ∈ H} and with
operation defined by (g, h) · (g 0 , h0 ) = (gg 0 , hh0 ) (i.e. (g ·G g 0 , h ·H h0 )).
Exercise** 6.14. Check that G × H is a group. Prove that G × H ' H × G
(note that, to do this, you should produce an isomorphism θ).
Exercise** 6.15. Draw up the multiplication table for the group of symmetries
of a rectangle which is not a square. Show that this group is isomorphic to
Z2 × Z2 (where addition is the operation).
Exercise 6.16. Show that the group of rotations of a cube is isomorphic to S4 .
How many symmetries of a cube are there? Show, by considering the relation
between an octahedron and a cube, that an octahedron has the same group of
symmetries.
6.1
The sign of a permutation
Recall that Sn denotes the symmetric group on n objects: the set of all (n!)
permutations of, say, {1, 2, . . . , n} with the operation being composition of permutations. We use the convention that if σ, τ ∈ Sn then στ means “do τ then do
σ” (so the usual order for composition of functions). Thus (1 2)(2 3) = (1 2 3).
To every permutation σ ∈ Sn we associate a matrixPσ (a permutation ma- 1
2
...
n
trix) as follows. Write σ using “two-row notation”, as
.
σ(1) σ(2) . . . σ(n)
Take the n × n identity matrix In and rearrange its columns according to the
second row: so the σ(1)-th column of In becomes the first column of Pσ ; the
σ(2)-th column of In becomes the second column of Pσ , et cetera. We define the
sign, sgn(σ), of σ to be the determinant of Pσ . Therefore sgn(σ) = ±1 since,
recall, switching two columns of a matrix multiplies its determinant by −1 and
since Pσ can be obtained from In (which has determinant 1) by a sequence of
such switches. This shows the following.
Lemma 6.17. If a permutation σ is written as a product of k transpositions
then sgn(σ) = (−1)k .
For instance
take 
(1 2), (2 3) in S3. Then the
corresponding matrices are:

0 1 0
1 0 0
P(1 2) =  1 0 0  and P(2 3) =  0 0 1 . Note that P(1 2) P(2 3) =
0 0 1
0 1 0

0 0 1
 1 0 0  = P(2 3 1) = P(1 2)(2 3) . This is a special case of the next result.
0 1 0
Lemma 6.18. If σ, τ ∈ Sn then Pσ Pτ = Pστ .
Proof. (Sketch) Consider the matrices Pσ , Pτ which are to be multiplied. The
1 in the first row of Pτ is in the τ −1 (1)-th column (that’s where the first column
of In was moved to). The 1 in the first column of Pσ is in the σ(1)-th row (the
σ(1)-th column is the one that was moved into the 1-st position). When we
multiply to obtain Pσ Pτ the 1 in the first column of the product is going to be
obtained from this product, of the σ(1)-th row of Pσ with the τ −1 (1)-th column
of Pτ , so we end up with a 1 in the (σ(1), τ −1 (1)) position. Similarly for all
29
i = 1, . . . , n. So the non-zero entries (all 1) of Pσ Pτ are in the (σ(i), τ −1 (i))
positions. Hence the permutation, ρ say, represented by Pσ Pτ puts the σ(i)-th
column in the τ −1 (i)-th place. That is (think about it) ρ takes τ −1 (i) to σ(i).
Therefore (again, think about it) ρτ −1 = σ. Therefore (multiply by τ on the
right) ρ = στ . So Pστ does represent the permutation στ .
Corollary 6.19. If σ, τ ∈ Sn then sgn(στ ) = sgn(σ)sgn(τ ).
Proof. sgn(στ ) = det(Pστ ) = det(Pσ Pτ ) (by the lemma above) = det(Pσ )det(Pτ ) =
sgn(σ)sgn(τ ).
Corollary 6.20. For any permutation σ, sgn(σ) = sgn(σ −1 ).
Lemma 6.21. Any cycle of length t can be written as the product of t − 1
transpositions.
(A proof will be given but you might like to try it for yourself first.)
Corollary 6.22. If σ is a cycle of length t then sgn(σ) = (−1)t−1 .
So, if a permutation is written as a product of disjoint cycles then its sign
can be calculated immediately.
A permutation is said to be even if its sign is +1 and is odd if its sign is
−1. So, for instance, the identity permutation is even, every transposition is
odd, any product of two transpositions is even, every 3-cycle is even, in general
a cycle of even length is odd and vice versa.
Let An denote the set of even permutations in Sn : An = {σ ∈ Sn : sgn(σ) =
1}. This is the alternating group (on n symbols).
Proposition 6.23. An is a subgroup of Sn and is of index 2 in Sn .
Proof. The identity permutation is even, the inverse of any even permutation
is even (by 6.20) and a product of even permutations is even (by 6.19) so An is
a subgroup. If σ, τ are in Sn but not in An then both are odd, so the product
σ −1 τ is even, that is σ −1 τ ∈ An , hence the cosets σAn and τ An are equal. Thus
all elements of Sn \ An form a single coset and so there are just two cosets of
An in Sn (namely the set, An of even permutations and the set, Sn \ An , of odd
permutations).
Example 6.24. (2) An denotes the subgroup of Sn consisting of the even
permutations (recall the division of the elements of Sn into the even and odd
permutations and, as an exercise, prove that An is a subgroup). The sign
of a permutation, even and odd permutations will be discussed on a separate
handout.
Exercise* 6.25. Draw up some of the group table for A4 .
Exercise 6.26. Show that the group of symmetries of a tetrahedron is S4 .
Show that the subgroup of those symmetries which are realisable as rotations
(i.e. without going into a mirror 3-space) is isomorphic to A4 . (In each case,
use reasoning rather than try to draw up the (rather large) group table.)
30
6.2
Group actions
Let G be a group. A G-set is a set, X, together with an action of G on X, that is,
a map α : G × X −→ X which satisfies α(g, α(h, x)) = α(gh, x) and α(e, x) = x
where g, h ∈ G and x ∈ X. Normally we write g · x or just gx for α(g, x)
(generalising the notation for the operation in a group), so then the conditions
become: g(hx) = (gh)x and ex = x (perhaps more readily comprehensible).
Thus, for every element g ∈ G we have a function x 7→ gx from X to X (which
can be though of as the action of g on X) such that composition of functions
corresponds to multiplication in G and the identity element of G corresponds
to the identity function on X.
Example 6.27. G acting on itself: the group multiplication in G is a function
G × G −→ G satisfying the condition for being a G-action.
Exercise* 6.28. Show that, in the action of G on itself described in Example
6.27, the orbit of each element is the whole group and the stabiliser of each
element is the trivial subgroup.
Example 6.29. If G is a group of “symmetries” of a set or object X (e.g. the
group of symmetries of a geometric object or the group of permutations of a
set) then the natural action of G on this set or object is a G-action (in the case
of symmetries of a polygon or polyhedron, think of the action on, say, the set
of vertices to get an action on an actual set X).
Suppose that G is a group and X is a G-set (so that means a set with a
specified G-action). Let x ∈ X. The orbit of x is {gx : g ∈ G}. This is written
Gx. If Gx = {x} then x is a fixed point of the G-action. The stabiliser of x,
stab(x), is {g ∈ G : gx = x}. So x is a fixed point iff Stab(x) = G.
Exercise** 6.30. Compute some orbits and stabilisers in geometric examples
and also for the action of the symmetric group Sn on {1, . . . , n}. For example,
compute the stabiliser, in the group of symmetries of a cube, of a vertex of the
cube. Also consider the action of this group of symmetries on the set of edges
of the cube and compute the stabiliser of an edge. Compute the orbit, under
the group of symmetries, of each vertex of a square-based regular (as far as it
can be) pyramid. Compute the stabiliser of 4 under the natural action of S4
on {1, 2, 3, 4}. Compute the stabiliser of the set {1, 2} under the action of S5
on two-element subsets of {1, . . . , 5} which is induced by the action of S5 on
{1, . . . , 5}.
Theorem 6.31. (Orbit-Stabiliser Theorem) Suppose that G is a finite group
and that X is a G-set. Let x ∈ X. Then its stabiliser, Stab(x), is a subgroup of
G and the index of Stab(x) in G is equal to the number of elements in the orbit
of x. Hence |G| = |Gx| |Stab(x)|.
Exercise** 6.32. Check the statement of the Orbit-Stabiliser Theorem in the
examples you looked at in Exercise 6.30.
If G is a group and g ∈ G then conjugation by g is the map, (−)g , from G
to itself defined by a 7→ g −1 ag, the notation ag is used as a slight abbreviation
for g −1 ag. This map will be the identity map from G to G iff g belongs to
the centre, ZG = {b ∈ G : ba = ab for all b ∈ G}, of G (exercise). The
centraliser, of a ∈ G is defined to be C(a) = {g ∈ G : ag = ga} = {g ∈ G :
ag = a}
31
Lemma 6.33. If G is a group and g ∈ G then conjugation by g is an automorphism of G (that is, an isomorphism from G to G). The action of G on G given
by α(g, a) = gag −1 is a G-action. The stabiliser of a ∈ G under this action is
its centraliser C(a).
The orbit of a under this action, i.e. the set of elements of the form g −1 ag
as g varies over G, is the set of conjugates of a.
If H, H 0 are subgroups of a group G then we say that H and H 0 are conjugate if there is an element g ∈ G such that g −1 Hg = H 0 , where g −1 Hg =
{g −1 hg : h ∈ H}. Note that, if this is the case, then the map h 7→ g −1 hg is an
isomorphism from H to H 0 .
Exercise 6.34. Show that being conjugate is an equivalence relation on the set
of subgroups of G.
Proposition 6.35. Suppose that X is a G-set. Let x, y ∈ X lie in the same orbit
of G, say gx = y. Then g −1 Stab(y)g = Stab(x), so the map, θg , from Stab(y)
to Stab(x) which takes h ∈ Stab(y) to g −1 hg is an isomorphism Stab(y) '
Stab(x). In particular Stab(x) and Stab(y) are conjugate subgroups of G.
(This applies, for example if G is the group of symmetries of a geometric
figure X and if x and y are two vertices of X which lie in the same orbit of G:
their stabilisers will be conjugate subgroups of G.)
Example 6.36. (6) If (−) is any mathematical object then an automorphism
of (−) is a structure-preserving bijection from (−) to itself. The set, Aut(−)
of automorphisms of (−) forms a group under composition (the composition of
two automorphisms is an automorphism, the inverse of an automorphism is an
automorphism, the identity map is an automorphism). Below are some specific
examples.
Example 6.37. Let K be a field and let p ∈ K[X] be an irreducible polynomial.
Let L = K[α] be the minimal field containing a root of p, as constructed in 5.3.
A K-automorphism of L is an automorphism, θ, of the field L which fixes
each element of K. It is easy (exercise) to show that θ(α) must be a root of
p and that θ is completely determined by the value θ(α). It is somewhat less
easy (a more extended exercise) to show that, if β ∈ L is any root of p then
mapping α to β and fixing K pointwise determines a K-automorphism of L.
Thus the K-automorphisms of L are in bijection with the roots of p which lie
in L. Earlier examples, 2.20 and 5.9, show that there may be just one or more
than one such, depending on the polynomial chosen.
Example 6.38. Let K be a field and consider the polynomial ring R = K[X1 , . . . , Xn ].
There is a natural action of the symmetric group Sn on R: if σ ∈ Sn and
p(X1 , . . . , Xn ) ∈ R then σp, more usually written pσ , is the polynomial p(Xσ(1) , . . . , Xσ(n) )
obtained by permuting the variables according to σ. The set of fixed points is
usually denoted K[X1 , . . . , Xn ]Sn and is the set of all symmetric polynomials
(those polynomials unchanged by every permutation of the variables)
Exercise** 6.39. Consider the natural action of S3 on K[X1 , X2 , X3 ]. Compute
the orbit and stabiliser of each of the following polynomials: X1 , X1 + X2 ,
X1 + X2 + X3 , X1 X32 + X3 X12 , X1 X22 + X2 X32 + X3 X12 , X1 + X22 + X33 .
32
6.3
More exercises
Exercise 6.40. Let R = M2 (Z) be the ring of 2 × 2 matrices with entries from
the ring Z of integers. Determine the condition on a matrix to be in the group
of units of this ring and give some examples of units.
Exercise* 6.41. Let R = M2 (Z4 ) be the ring of 2 × 2 matrices with entries from
the ring Z4 of integers modulo 4. Find some elements of the group of units of
this ring. Is this group U (R) of units of R abelian?
What is a criterion for a matrix in M2 (Z8 ) to be invertible?
Exercise 6.42. Let R = M2 (F4 ) be the ring of 2 × 2 matrices with entries from
the field F4 with 4 elements. Find some elements of the group of units of this
ring. How many elements are in the group U (R) of units of R? How many of
these have determinant 1?
Exercise* 6.43. Identify the groups of symmetries of the following geometric
figures: (i) a triangle; (ii) a square; (iii) a rectangle which is not a square; (iv)
a hexagon; (v) a tetrahedron; (vi) a cube. Why is the group of symmetries of a
cube the same as that of a regular octahedron? (Or, for that matter, why is the
group of symmetries of a dodecahedron the same as that of an icosahedron?)
Exercise 6.44. Let X be a square-based pyramid with all sides of equal length.
Find the group G of symmetries of X and compute the stabiliser of each vertex.
7
Homomorphisms of Groups, Normal Subgroups
and Factor Groups
The idea of a homomorphism is, for groups as for rings, that of a structurepreserving map. A homomorphism from the group G to the group H is a
map θ : G −→ H such that θ(ab) = θ(a)θ(b) for all a, b ∈ G.
We have a lemma completely analogous to 2.9.
Lemma 7.1. Suppose that θ : G −→ H is a homomorphism between groups.
Then:
(1) θ(eG ) = eH ;
(2) θ(a−1 ) = (θ(a))−1 for every a ∈ G;
(3) o(θ(a)) ≤ o(a) for every a ∈ G;
(4) the image of θ, im(θ) = {θ(a) : a ∈ G} is a subgroup of H.
The kernel of a homomorphism θ : G −→ H of groups is the set of elements
which map to the identity of H: ker(θ) = {a ∈ G : θ(a) = e}. A subgroup N
of G is said to be normal if it is stable as a set under conjugation, that is,
if a−1 N a = N for every a ∈ G (this is not to say that every element of N is
fixed under conjugation - the elements of N might well be permuted around,
but they won’t be sent outside N ). We write N C G to indicate that N is a
normal subgroup of G.
The following lemma slightly simplifies checking that a subgroup is normal.
Lemma 7.2. Suppose that N is a subgroup of the group G. If a−1 N a ⊆ N for
every a ∈ G then N is a normal subgroup of G.
33
Proof. Let a ∈ G; we have to show that N ⊆ a−1 N a. So let n ∈ N . By
assumption (a−1 )−1 na−1 = n0 for some n0 ∈ N , that is, ana−1 = n0 , so an =
n0 a, hence n = a−1 n0 a ∈ a−1 N a, as required.
Note that if G is finite then the proof is even easier: because conjugation by
a is a bijection, the number of elements in a−1 N a is the same as the number of
elements in N so, if one is a subset of the other, then they must be equal.
The “manipulating sets” argument is also short (and valid): if a−1 N a ⊆ N
then multiply on the left by a and on the right by a−1 to get aa−1 N aa−1 ⊆
aN a−1 , that is, N ⊆ aN a−1 which, noting that it’s for all a and so we can
replace a by a−1 , gives what we want.
Lemma 7.3. The following conditions are equivalent for a subgroup N of the
group G.
(i) N is a normal subgroup of G;
(ii) aN = N a for every a ∈ G;
(ii) every right coset of N in G is also a left coset of N in G.
Proposition 7.4. Suppose that θ : G −→ H is a homomorphism of groups.
Then ker(θ) is a normal subgroup of G.
In Exercises 7.5 and 7.6 you are asked to prove that the centre of G is
a normal subgroup and also to prove that any subgroup of index 2 must be
normal.
Exercise** 7.5. Show that the centre of G is a normal subgroup.
Exercise** 7.6. Prove that any subgroup of index 2 is normal.
Thus normal subgroups of groups are analogous to ideals of rings. And the
analogy does not stop there. Let N be a normal subgroup of the group G.
Consider the set, written G/N, of cosets (right or left, it doesn’t matter by
the lemma above) of N in G. We make this set into a group by defining the
product of two cosets: aN · bN = (ab)N. As usual, because the definition uses
particular representatives, we have to show that this is well-defined, but that is
straightforward enough. We refer to G/N as the factor group of G by N (or
as the quotient of G by N ).
See Exercise 7.7 for an example of what goes wrong if you try to do this
construction with a non-normal subgroup.
Exercise* 7.7. If H is a subgroup of G which is not normal then trying to
define a product as above won’t work. For instance, let G = S3 and let H be
the (cyclic) subgroup generated by the transposition (1 2). Compute the left and
right cosets of H in G to see that H is not normal in G. Show that the product of
the cosets (1 3)H and (2 3)H is not well-defined (note that (1 3)H = (1 3 2)H).
Theorem 7.8. Let N be a normal subgroup of the group G. Defining a product
on the set, G/N, of cosets of N in G by (aN )(bN ) = (ab)N gives a well-defined
structure of a group to G/N. There is a canonical projection π : G −→ G/N
defined by π(a) = aN and this is a surjective homomorphism of groups, with
kernel N. Any group homomorphism θ : G −→ H with ker(θ) ≥ N factors
uniquely through π.
34
Examples 7.9. (1) Consider the map from Sn to the group {−1, +1} (with
multiplication as the operation) given by assigning to a permutation its sign.
This is a homomorphism with kernel the alternating group, An .
(2) If R is any ring and I is an ideal of R then the canonical surjection onto
the factor ring R → R/I is, forgetting the multiplication, a homomorphism of
additive groups, with kernel I. For instance, consider Z6 → Z6 /([0]6 , [2]6 , [4]6 ) '
Z2 .
As for rings there is a correspondence between (normal) subgroups of the
quotient G/N and (normal) subgroups of G which contain N .
Theorem 7.10. Suppose that N is a normal subgroup of the group G. Then
there is a bijection between subgroups H of G which contain N and subgroups
K of the quotient group G/N given by:
H 7→ H/N = π(H);
K 7→ π −1 K = {a ∈ G : π(a) ∈ K}.
Furthermore H is normal in G iff H/N is normal in G/N, in which case the
quotient (G/N )/(H/N ) is isomorphic to G/H.
If H1 and H2 are subgroups of a group G then we define H1 H2 to be the
subgroup of G generated by H1 and H2 : this is the smallest subgroup containing H1 and H2 and is more explicitly defined as the set of all products of the
form h11 h21 h12 h22 . . . h1k h2k for some k, where hij ∈ Hi . That is (think about
it), form all possible products of elements from the set H1 ∪ H2 . If one of these
groups is normal then this becomes much simpler to describe.
Exercises** 7.11. (1) Let G = S4 , H1 = h(1 2)i, H2 = h(2 3)i, H3 = h(3 4)i.
Compute H1 H2 and H1 H3 .
(2) Find the subgroup of D6 (the group of symmetries of a regular hexagon)
generated by H1 = hρi where ρ is rotation about the centre by 2π/3 and H2 =
hσi where σ is one of the reflections.
(3) (additive notation) Find the subgroup of Z generated by 6 and 4 (i.e. by the
cyclic subgroups h6i and h4i).
Lemma 7.12. Suppose that H is a subgroup of the group G and that N is a
normal subgroup. Then HN = {hn : h ∈ H, n ∈ N } = {nh : n ∈ N, h ∈ H}.
Exercise* 7.13. In Exercise 7.11(2) above show that H1 (the subgroup generated
by rotation through 2π/3 is normal in D6 and verify the statement of Lemma
7.12 in this case, that is, show that H1 H2 can be described as in the Lemma.
Proposition 7.14. Suppose that H is a subgroup of the group G and that N is
a normal subgroup. Then N is a normal subgroup of HN , H ∩ N is a normal
subgroup of H and HN/N ' H/(H ∩ N ).
7.1
More exercises
Exercise* 7.15. Find all the group homomorphisms from Z12 to itself. Which
of these are automorphisms?
Exercise 7.16. Show that if G is a group
T and if {Ni }i∈I is any set of normal
subgroups of G then their intersection, i Ni , is a normal subgroup of G. Deduce
that if A is any subset of G then there is a smallest normal subgroup of G which
contains A (this is referred to as the normal subgroup of G generated by A).
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Show that the normal subgroup generated by A is the set of all elements
which are products of elements of the form g −1 a g with g ∈ G, a ∈ A and
= ±1 (allowing the identity element in as being the “empty product”).
Exercise 7.17. Suppose that N is a subgroup of the group G and that N is
abelian. Suppose also that N 0 is a subgroup of G which is conjugate to N .
Prove that N 0 is abelian.
Exercise 7.18. (i) Let α be an automorphism of the group G. Suppose that
N is a normal subgroup of G and let N 0 = α(N ). Prove that N 0 is a normal
subgroup of G.
(ii) Suppose now only that N and N 0 are subgroups of G with N normal in
G and N 0 isomorphic to N . Show, by giving an appropriate example, that N 0
need not be normal in G.
Exercise** 7.19. Let H be a subgroup of the group G. Show that the set
NG (H) = {g ∈ G : g −1 Hg = H} is a subgroup of G, the normaliser of H in
G. Show that NG (H) contains H and that H is a normal subgroup of NG (H).
Show that NG (H) is the largest subgroup of G in which H is normal (in the
sense that if H1 is a subgroup of G which strictly contains NG (H) then H is
not normal in H1 ).
Exercise** 7.20. Let G be the group S(3) and let H be the subgroup generated
by the transposition (1 2). Find the normaliser of H in G. Find the normal
subgroup of G generated by H. Repeat the exercise using the subgroup, H 0 ,
generated by the 3-cycle (1 2 3).
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