<<

. 11
( 14)



>>

R —R A


GB
A f


Therefore injectivity of 1 — f is equivalent to injectivity of f , and the result follows. ™


10.8.5 Corollary
Every projective module, hence every free module, is ¬‚at.


Proof. By (10.8.3) and (10.8.4), every free module is ¬‚at. Since a projective module is a
direct summand of a free module, it is ¬‚at by (10.8.3). ™


Flat abelian groups can be characterized precisely.
28 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

10.8.6 Theorem
A Z-module is ¬‚at i¬ it is torsion-free.

Proof. Suppose that M is a Z-module that is not torsion-free. Let x ∈ M be a nonzero
element such that nx = 0 for some positive integer n. If f : Z ’ Z is multiplication by n,
then (1 — f ) : M — Z ’ M — Z is given by

y — z ’ y — nz = ny — z

so that (1 — f )(x — 1) = nx — 1 = 0. Since x — 1 corresponds to x under the isomorphism
between M — Z and M , x — 1 = 0, and 1 — f is not injective. Therefore M is not ¬‚at.
The discussion in (10.8.1) shows that in checking ¬‚atness of M , we can restrict to
¬nitely generated submodules of M . [We are examining equations of the form
n
(1 — f )(t) = 0, where t = i=1 xi — yi , xi ∈ M , yi ∈ A.]Thus without loss of gener-
ality, we can assume that M is a ¬nitely generated abelian group. If M is torsion-free,
then by (4.6.5), M is free and therefore ¬‚at by (10.8.5). ™


10.8.7 Corollary
The additive group of rationals Q is a ¬‚at but not projective Z-module.

Proof. Since Q is torsion-free, it is ¬‚at by (10.8.6). If Q were projective, it would be free
by (10.5.5). This is a contradiction (see Section 4.1, Problem 5). ™

Sometimes it is desirable to change the underlying ring of a module; the term base
change is used in these situations.


10.8.8 De¬nitions and Comments
If f : R ’ S is a ring homomorphism and M is an S-module, we can create an R-module
structure on M by rx = f (r)x, r ∈ R, x ∈ M . This is a base change by restriction of
scalars.
If f : R ’ S is a ring homomorphism and M is an R-module, we can make S —R M
into an S-module via

s(s — x) = ss — x, s, s ∈ S, x ∈ M.

This is a base change by extension of scalars. Note that S is an R-module by restriction
of scalars, so the tensor product makes sense. What we are doing is allowing linear
combinations of elements of M with coe¬cients in S. This operation is very common in
algebraic topology.
In the exercises, we will look at the relation between base change and ¬‚atness. There
will also be some problems on ¬nitely generated algebras, so let™s de¬ne these now.
10.9. DIRECT AND INVERSE LIMITS 29

10.8.9 De¬nition
The R-algebra A is ¬nitely generated if there are elements a1 , . . . , an ∈ A such that every
element of A is a polynomial in the ai . Equivalently, the algebra homomorphism from the
polynomial ring R[X1 , . . . , Xn ] ’ A determined by Xi ’ ai , i = 1, . . . , n, is surjective.
Thus A is a quotient of the polynomial ring.
It is important to note that if A is ¬nitely generated as an R-module, then it is ¬nitely
generated as an R-algebra. [If a = r1 a1 + · · · + rn an , then a is certainly a polynomial in
the ai .]

Problems For Section 10.8
1. Give an example of a ¬nitely generated R-algebra that is not ¬nitely generated as an
R-module.
2. Show that R[X] —R R[Y ] ∼ R[X, Y ].
=
3. Show that if A and B are ¬nitely generated R-algebras, so is A —R B.
4. Let f : R ’ S be a ring homomorphism, and let M be an S-module, so that M is an
R-module by restriction of scalars. If S is a ¬‚at R-module and M is a ¬‚at S-module,
show that M is a ¬‚at R-module.
5. Let f : R ’ S be a ring homomorphism, and let M be an R-module, so that S —R M
is an S-module by extension of scalars. If M is a ¬‚at R-module, show that S —R M is
a ¬‚at S-module.
6. Let S be a multiplicative subset of the commutative ring R. Show that for any R-
module M , S ’1 R —R M ∼ S ’1 M via ± : (r/s) — x ’ rx/s with inverse β : x/s ’
=
(1/s) — x.
7. Continuing Problem 6, show that S ’1 R is a ¬‚at R-module.


10.9 Direct and Inverse Limits
If M is the direct sum of R-modules Mi , then R-homomorphisms fi : Mi ’ N can be lifted
uniquely to an R-homomorphism f : M ’ N . The direct limit construction generalizes
this idea. [In category theory, there is a further generalization called the colimit. The
terminology is consistent because the direct sum is the coproduct in the category of
modules.]

10.9.1 Direct Systems
A directed set is a partially ordered set I such that given any i, j ∈ I, there exists k ∈ I
such that i ¤ k and j ¤ k. A typical example is the collection of ¬nite subsets of a set,
ordered by inclusion. If A and B are arbitrary ¬nite subsets, then both A and B are
contained in the ¬nite set A ∪ B.
Now suppose I is a directed set and we have a collection of objects Ai , i ∈ I, in a
category C. Assume that whenever i ¤ j, there is a morphism h(i, j) : Ai ’ Aj . Assume
further that the h(i, j) are compatible in the sense that if i ¤ j ¤ k and we apply h(i, j)
30 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

followed by h(j, k), we get h(i, k). We also require that for each i, h(i, i) is the identity
on Ai . The collection of objects and morphisms is called a direct system. As an example,
take the objects to be the ¬nitely generated submodules of a module, and the morphisms
to be the natural inclusion maps. In this case, the directed set coincides with the set of
objects, and the partial ordering is inclusion.

10.9.2 Direct Limits
Suppose that {Ai , h(i, j), i, j ∈ I} is a direct system. The direct limit of the system will
consist of an object A and morphisms ±i : Ai ’ A. Just as with coproducts, we want to
lift morphisms fj : Aj ’ B to a unique f : A ’ B, that is, f ±j = fj for all j ∈ I. But we
require that the maps ±j be compatible with the h(i, j), in other words, ±j h(i, j) = ±i
whenever i ¤ j. A similar constraint is imposed on the fj , namely, fj h(i, j) = fi , i ¤ j.
Thus the direct limit is an object A along with compatible morphisms ±j : Aj ’ A such
that given compatible morphisms fj : Aj ’ B, there is a unique morphism f : A ’ B
such that f ±j = fj for all j. Figure 10.9.1 summarizes the discussion.

fA’

 ±j
±i
B –d
c dd f
˜˜
fi ˜ dd j
˜
˜ dd
˜
˜˜
G Aj
Ai
h(i,j)

Figure 10.9.1
As in Section 10.2, any two direct limits of a given direct system are isomorphic.
If the ordering on I is the equality relation, then the only element j such that i ¤ j
is i itself. Compatibility is automatic, and the direct limit reduces to a coproduct.
A popular notation for the direct limit is
A = lim Ai .


The direct limit is sometimes called an inductive limit.

10.9.3 Examples
1. A coproduct is a direct limit, as discussed above. In particular, a direct sum of modules
is a direct limit.
2. Any module is the direct limit of its ¬nitely generated submodules. [Use the direct
system indicated in (10.9.1).]
3. The algebraic closure of a ¬eld F can be constructed (informally) by adjoining roots of
all possible polynomials in F [X]; see (3.3.7). This suggests that the algebraic closure
is the direct limit of the collection of all ¬nite extensions of F . This can be proved
with the aid of (3.3.9).
In the category of modules, direct limits always exist.
10.9. DIRECT AND INVERSE LIMITS 31

10.9.4 Theorem
If {Mi , h(i, j), i, j ∈ I} is a direct system of R-modules, then the direct limit of the system
exists.
Proof. Take M to be (•i Mi )/N , with N the submodule of the direct sum generated by
all elements of the form

βj h(i, j)xi ’ βi xi , i ¤ j, xi ∈ Mi (1)

where βi is the inclusion map of Mi into the direct sum. De¬ne ±i : Mi ’ M by

±i xi = βi xi + N.

The ±i are compatible, because

±j h(i, j)xi = βj h(i, j)xi + N = βi xi + N = ±i xi .

Given compatible fi : Mi ’ B, we de¬ne f : M ’ B by

f (βi xi + N ) = fi xi ,

the only possible choice. This forces f ±i = fi , provided we show that f is well-de¬ned.
But an element of N of the form (1) is mapped by our proposed f to

fj h(i, j)xi ’ fi xi

which is 0 by compatibility of the fi . Thus f maps everything in N to 0, and the result
follows. ™

10.9.5 Inverse Systems and Inverse Limits
Inverse limits are dual to direct limits. An inverse system is de¬ned as in (10.9.1),
except that if i ¤ j, then h(i, j) maps “backwards” from Aj to Ai . If we apply h(j, k)
followed by h(i, j), we get h(i, k); as before, h(i, i) is the identity on Ai . The inverse
limit of the inverse system {Ai , h(i, j), i, j ∈ I} is an object A along with morphisms
pi : A ’ Ai . As with products, we want to lift morphisms fi : B ’ Ai to a unique
f : B ’ A. There is a compatibility requirement on the pi and fi : if i ¤ j, then
h(i, j)pj = pi , and similarly h(i, j)fj = fi . Thus the inverse limit is an object A along with
compatible morphisms pi : A ’ Ai such that given compatible morphisms fi : B ’ Ai ,
there is a unique morphism f : B ’ A such that pi f = fi for all i. See Figure 10.9.2.

A
y

pj
pi
Bd
dd f
˜˜
fi ˜ dd j
˜
˜˜ dd
 ˜˜ h(i,j) 2
o Aj
Ai
32 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

Figue 10.9.2
As in Section 10.2, the universal mapping property determines the inverse limit up to
isomorphism.
If the ordering on I is the equality relation, the the inverse limit reduces to a product.
In category theory, the limit is a generalization of the inverse limit.
A popular notation for the inverse limit is

A = lim Ai .


The inverse limit is sometimes called a projective limit.
We constructed the direct limit of a family of modules by forming a quotient of the
direct sum. By duality, we expect that the inverse limit involves a submodule of the direct
product.

10.9.6 Theorem
If {Mi , h(i, j), i, j ∈ I} is an inverse system of R-modules, then the inverse limit of the
system exists.
Proof. We take M to be the set of all x = (xi , i ∈ I) in the direct product i Mi such
that h(i, j)xj = xi whenever i ¤ j. Let pi be the restriction to M of the projection on
the ith factor. Then h(i, j)pj x = h(i, j)xj = xi = pi x, so the pi are compatible. Given
compatible fi : N ’ Mi , let f be the product of the fi , that is, f x = (fi x, i ∈ I). By
compatibility, h(i, j)fj x = fi x for i ¤ j, so f maps i Mi into M . By de¬nition of f we
have pi f = fi , and the result follows. ™

10.9.7 Example
Recall from Section 7.9 that a p-adic integer can be represented as a0 + a1 p + a2 p2 + · · · ,
where the ai belong to {0, 1, . . . , p’1}. If we discard all terms after ar’1 pr’1 , r = 1, 2, . . . ,
we get the ring Zpr . These rings form an inverse system; if x ∈ Zps and r ¤ s, we take
h(r, s)x to be the residue of x mod pr . The inverse limit of this system is the ring of
p-adic integers.

Problems For Section 10.9
1. In Theorem (10.9.6), why can™t we say “obviously”, since direct limits exist in the
category of modules, inverse limits must also exist by duality.
2. Show that in the category of modules over a commutative ring, the tensor product
commutes with direct limits. In other words,

lim (M — Ni ) = M — lim Ni
’ ’

assuming that the direct limit of the Ni exists.
3. For each n = 1, 2, . . . , let An be an R-module, with A1 ⊆ A2 ⊆ A3 ⊆ · · · . Take h(i, j),
i ¤ j, to be the inclusion map. What is the direct limit of the An ? (Be more explicit
than in (10.9.4).)
10.9. DIRECT AND INVERSE LIMITS 33

4. Suppose that A, B and C are the direct limits of direct systems {Ai }, {Bi } and {Ci }
of R-modules. Assume that for each i, the sequence

G Bi G Ci
fi gi
Ai
is exact. Give an intuitive argument to suggest that the sequence

GB GC
f g
A
is exact. Thus direct limit is an exact functor.
[A lot of formalism is being suppressed here. We must make the collection of direct
systems into a category, and de¬ne a morphism in that category. This forces compatibility
conditions on the fi and gi : fj hA (i, j) = hB (i, j)fi , gj hB (i, j) = hC (i, j)gi . The direct
limit functor takes a direct system to its direct limit, but we must also specify what it
does to morphisms.]
A possible strategy is to claim that since an element of a direct sum has only ¬nitely
many nonzero components, exactness at B is equivalent to exactness at each Bi . This
is unconvincing because the direct limit is not simply a direct sum, but a quotient of a
direct sum. Suggestions are welcome!
Problems 5 and 6 give some additional properties of direct products.
5. Show that
HomR (•Ai , B) ∼ HomR (Ai , B).
=
i

6. Show that
Bi ) ∼
HomR (A, HomR (A, Bi ).
=
i i

7. If M is a nonzero R-module that is both projective and injective, where R is an integral
domain that is not a ¬eld, show that HomR (M, R) = 0.
8. Let R be an integral domain that is not a ¬eld. If M is an R-module that is both
projective and injective, show that M = 0.


Appendix to Chapter 10
We have seen that an abelian group is injective if and only if it is divisible. In this
appendix we give an explicit characterization of such groups.

A10.1 De¬nitions and Comments
Let G be an abelian group, and T the torsion subgroup of G (the elements of G of
¬nite order). Then G/T is torsion-free, since n(x + T ) = 0 implies nx ∈ T , hence
x ∈ T . If p is a ¬xed prime, the primary component Gp associated with p consists
of all elements whose order is a power of p. Note that Gp is a subgroup of G, for if
pn a = pm b = 0, n ≥ m, then pn (a ’ b) = 0. (We use the fact that G is abelian; for
example, 3(a ’ b) = a ’ b + a ’ b + a ’ b = a + a + a ’ b ’ b ’ b.)
34 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

A10.2 Proposition
The torsion subgroup T is the direct sum of the primary components Gp , p prime.

r
k
Proof. Suppose x has order m = j=1 pj j . If mi = m/pri , then the greatest common
i
divisor of the mi is 1, so there are integers a1 , . . . , ak such that a1 m1 + · · · + ak mk = 1.
k
Thus x = 1x = i=1 ai (mi x), and (by de¬nition of mi ) mi x has order pri and therefore
i
belongs to the primary component Gpi . This proves that G is the sum of the Gp . To show
that the sum is direct, assume 0 = x ∈ Gp © q=p Gq . Then the order of x is a power
of p and also a product of prime factors unequal to p, which is impossible. For example,
if y has order 9 and z has order 125, then 9(125)(y + z) = 0, so the order of y + z is of
the form 3r 5s . ™


A10.3 De¬nitions and Comments
A Pr¨fer group, also called a quasicyclic group and denoted by Z(p∞ ), is a p-primary
u
component of Q/Z, the rationals mod 1. Since every element of Q/Z has ¬nite order, it
follows from (A10.2) that

Z(p∞ ).
Q/Z =
p


Now an element of Q/Z whose order is a power of p must be of the form a/pr + Z for
some integer a and nonnegative integer r. It follows that the elements ar = 1/pr + Z,
r = 1, 2, . . . , generate Z(p∞ ). These elements satisfy the following relations:

pa1 = 0, pa2 = a1 , . . . , par+1 = ar , . . .


A10.4 Proposition
Let H be a group de¬ned by generators b1 , b2 , . . . and relations pb1 = 0, pb2 = b1 , . . . ,
pbr+1 = br , . . . . Then H is isomorphic to Z(p∞ ).

Proof. First note that the relations imply that every element of H is an integer multiple
of some bi . Here is a typical computation:

4b7 + 6b10 + 2b14 = 4(pb8 ) + 6(pb11 ) + 2b14
= · · · = 4(p7 b14 ) + 6(p4 b14 ) + 2b14 = (4p7 + 6p4 + 2)b14 .

By (5.8.5), there is an epimorphism f : H ’ Z(p∞ ), and by the proof of (5.8.5), we can
take f (bi ) = ai for all i. To show that f is injective, suppose f (cbi ) = 0 where c ∈ Z.
Then cf (bi ) = cai = 0, so c/pi ∈ Z, in other words, pi divides c. ( We can reverse
this argument to conclude that f (cbi ) = 0 i¬ pi divides c.) But the relations imply that
pi bi = 0, and since c is a multiple of pi , we have cbi = 0. ™
10.9. DIRECT AND INVERSE LIMITS 35

A10.5 Proposition
Let G be a divisible abelian group. Then its torsion subgroup T is also divisible. Moreover,
G can be written as T • D, where D is torsion-free and divisible.

Proof. If x ∈ T and 0 = n ∈ Z, then for some y ∈ G we have ny = x. Thus in the
torsion-free group G/T we have n(y + T ) = x + T = 0. But then ny ∈ T , so (as in
(A10.1)) y ∈ T and T is divisible, hence injective by (10.6.6). By (10.6.2), the exact
sequence 0 ’ T ’ G ’ G/T ’ 0 splits, so G ∼ T • G/T . Since G/T is torsion-free and
=
divisible (see (10.6.5)), the result follows. ™

We are going to show that an abelian group is divisible i¬ it is a direct sum of copies
of Q (the additive group of rationals) and quasicyclic groups. To show that every divisible
abelian group has this form, it su¬ces, by (A10.2), (A10.5) and the fact that a direct sum
of divisible abelian groups is divisible, to consider only two cases, G torsion-free and G a
p-group.

A10.6 Proposition
If G is a divisible, torsion-free abelian group, then G is isomorphic to a direct sum of
copies of Q.

Proof. The result follows from the observation that G can be regarded as a Q-module,
that is, a vector space over Q; see Section 10.7, Problem 7. ™

For any abelian group G, let G[n] = {x ∈ G : nx = 0}.

A10.7 Proposition
Let G and H be divisible abelian p-groups. Then any isomorphism • of G[p] and H[p]
can be extended to an isomorphism ψ of G and H.

Proof. Our candidate ψ arises from the injectivity of H, as the diagram below indicates.

H —g
yg
gg ψ
gg
gg
G G[p] GG
0

The map from G[p] to G is inclusion, and the map from G[p] to H is the composition of •
and the inclusion from H[p] to H. Suppose that x ∈ G and the order of x is |x| = pn .
We will prove by induction that ψ(x) = 0 implies x = 0. If n = 1, then x ∈ G[p],
so ψ(x) = •(x), and the result follows because • is injective. For the inductive step,
suppose |x| = pn+1 and ψ(x) = 0. Then |px| = pn and ψ(px) = pψ(x) = 0. By induction
hypothesis, px = 0, which contradicts the assumption that x has order pn+1 .
Now we prove by induction that ψ is surjective. Explicitly, if y ∈ H and |y| = pn ,
then y belongs to the image of ψ. If n = 1, then y ∈ H[p] and the result follows because •
36 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

is surjective. If |y| = pn+1 , then pn y ∈ H[p], so for some x ∈ G[p] we have •(x) = pn y.
Since G is divisible, there exists g ∈ G such that pn g = x. Then

pn (y ’ ψ(g)) = pn y ’ ψ(pn g) = pn y ’ ψ(x) = pn y ’ •(x) = 0.

By induction hypothesis, there is an element z ∈ G such that ψ(z) = y ’ ψ(g). Thus
ψ(g + z) = y. ™

A10.8 Theorem
An abelian group G is divisible if and only if G is a direct sum of copies of Q and
quasicyclic groups.

Proof. Suppose that G is such a direct sum. Since Q and Z(p∞ ) are divisible [Z(p∞ ) is a
direct summand of the divisible group Q/Z], and a direct sum of divisible abelian groups
is divisible, G must be divisible. Conversely, assume G divisible. In view of (A10.6) and
the discussion preceding it, we may assume that G is a p-group. But then G[p] is a vector
space over the ¬eld Fp = Z/pZ; the scalar multiplication is given by (n + pZ)g = ng.
Since pg = 0, scalar multiplication is well-de¬ned. If the dimension of G[p] over Fp is d,
let H be the direct sum of d copies of Z(p∞ ). An element of order p in a component of the
direct sum is an integer multiple of 1/p + Z, and consequently H[p] is also a d-dimensional
vector space over Fp . Thus G[p] is isomorphic to H[p], and it follows from (A10.7) that
G is isomorphic to H. ™
Enrichment

Chapters 1“4 form an idealized undergraduate course, written in the style of a graduate
text. To help those seeing abstract algebra for the ¬rst time, I have prepared this section,
which contains advice, explanations and additional examples for each section in the ¬rst
four chapters.


Section 1.1
When we say that the rational numbers form a group under addition, we mean that
rational numbers can be added and subtracted, and the result will inevitably be rational.
Similarly for the integers, the real numbers, and the complex numbers. But the integers
(even the nonzero integers) do not form a group under multiplication. If a is an integer
other than ±1, there is no integer b such that ab = 1. The nonzero rational numbers
form a group under multiplication, as do the nonzero reals and the nonzero complex
numbers. Not only can we add and subtract rationals, we can multiply and divide them
(if the divisor is nonzero). The rational, reals and complex numbers are examples of ¬elds,
which will be studied systematically in Chapter 3.
Here is what the generalized associative law is saying. To compute the product of the
elements a, b, c, d and e, one way is to ¬rst compute bc, then (bc)d, then a((bc)d), and
¬nally [a((bc)d)e]. Another way is (ab), then (cd), then (ab)(cd), and ¬nally ([(ab)(cd)]e).
All procedures give the same result, which can therefore be written as abcde.
Notice that the solution to Problem 6 indicates how to construct a formal proof
of 1.1.4.


Section 1.2
Groups whose descriptions di¬er may turn out to be isomorphic, and we already have an
example from the groups discussed in this section. Consider the dihedral group D6 , with
elements I, R, R2 , F , RF , R2 F . Let S3 be the group of all permutations of {1, 2, 3}. We
claim that D6 and S3 are isomorphic. This can be seen geometrically if we view D6 as a
group of permutations of the vertices of an equilateral triangle. Since D6 has 6 elements
and there are exactly 6 permutations of 3 symbols, we must conclude that D6 and S3
are essentially the same. To display an isomorphism explicitly, let R correspond to the

1
2

permutation (1,2,3) and F to (2,3). Then
I = (1), R = (1, 2, 3), R2 = (1, 3, 2), F = (2, 3), RF = (1, 2), R2 F = (1, 3).
If G is a nonabelian group, then it must have an element of order 3 or more. (For
example, S3 has two elements of order 3.) In other words, if every element of G has order
1 or 2, then G is abelian. To prove this, let a, b ∈ G; we will show that ab = ba. We can
assume with loss of generality that a = 1 and b = 1. But then a2 = 1 and b2 = 1, so that
a is its own inverse, and similarly for b. If ab has order 1, then ab = 1, so a and b are
inverses of each other. By uniqueness of inverses, a = b, hence ab = ba. If ab has order 2,
then abab = 1, so ab = b’1 a’1 = ba.


Section 1.3
Here is another view of cosets that may be helpful. Suppose that a coded message is to
be transmitted, and the message is to be represented by a code word x (an n-dimensional
vector with components in some ¬eld). The allowable code words are solutions of Ax = 0,
where A is an m by n matrix, hence the set H of code words is an abelian group under
componentwise addition, a subgroup of the abelian group G of all n-dimensional vectors.
(In fact, G and H are vector space, but let™s ignore the additional structure.) Transmission
is a¬ected by noise, so that the received vector is of the form z = x + y, where y is another
n-dimensional vector, called the error vector or error pattern vector. Upon receiving z,
we calculate the syndrome s = Az. If s turns out to be the zero vector, we declare that no
error has occurred, and the transmitted word is z. Of course our decision may be incorrect,
but under suitable assumptions about the nature of the noise, our decision procedure will
minimize the probability of making a mistake. Again, let™s ignore this di¬culty and focus
on the algebraic aspects of the problem. We make the following claim:
Two vectors z1 and z2 have the same syndrome if and only if they lie in the same coset
of H in G.
To prove this, observe that Az1 = Az2 i¬ A(z1 ’ z2 ) = 0 i¬ z1 ’ z2 ∈ H i¬ z1 ∈ z2 + H.
(We are working in an abelian group, so we use the additive notation z2 + H rather than
the multiplicative notation z2 H.)
Now suppose that we agree that we are going to correct the error pattern y1 , in other
words, if we receive z = x1 + y1 , where x1 is a code word, we will decode z as x1 . If
we receive z = x1 + y1 , where x1 is another code word, we decode z as x1 . Thus
our procedure corrects y1 regardless of the particular word transmitted. Here is a key
algebraic observation:
If y1 and y2 are distinct vectors that lie in the same coset, it is impossible to correct
both y1 and y2 .
This holds because y1 = y2 + x for some code word x = 0, hence y1 + 0 = y2 + x.
Therefore we cannot distinguish between the following two possibilities:
1. The zero word is transmitted and the error pattern is y1 ;
2. x is transmitted and the error pattern is y2 .
It follows that among all vectors in a given coset, equivalently among all vectors
having the same syndrome, we can choose exactly one as a correctable error pattern. If
3

the underlying ¬eld has only two elements 0 and 1, then (under suitable assumptions) it is
best to choose to correct the pattern of minimum weight, that is, minimum number of 1™s.
In particular, if the coset is the subgroup H itself, then we choose the zero vector. This
agrees with our earlier proposal: if the received vector z has zero syndrome, we decode
z as z itself, thus “correcting” the zero pattern, in other words, declaring that there has
been no error in transmission.
For further discussion and examples, see Information Theory by R. B. Ash, Dover
1991, Chapter 4.


Section 1.4
Here are some intuitive ideas that may help in visualizing the various isomorphism theo-
rems. In topology, we can turn the real interval [0, 1] into a circle by gluing the endpoints
together, in other words identifying 0 and 1. Something similar is happening when we
form the quotient group G/N where N is a normal subgroup of G. We have identi¬ed
all the elements of N , and since the identity belongs to every subgroup, we can say that
we have set everything in N equal to 1 (or 0 in the abelian case). Formally, (1.4.6) gives
a correspondence between the subgroup of G/N consisting of the identity alone, and the
subgroup N of G.
We have already seen an example of this identi¬cation process. In (1.2.4), we started
with the free group G generated by the symbols R and F , and identi¬ed all sequences
satisfying the relations Rn = I, F 2 = I, and RF = F R’1 (equivalently RF RF = I).
Here we would like to take N to be the subgroup of G generated by Rn , F 2 , and RF RF ,
but N might not be normal. We will get around this technical di¬culty when we discuss
generators and relations in more detail in Section 5.8.


Section 1.5
Direct products provide a good illustration of the use of the ¬rst isomorphism theorem.
Suppose that G = H — K; what can we say about G/H? If (h, k) ∈ G, then (h, k) =
(h, 1K )(1H , k), and intuitively we have identi¬ed (h, 1K ) with the identity (1H , 1K ). What
we have left is (1H , k), and it appears that G/H should be isomorphic to K. To prove
this formally, de¬ne f : G ’ K by f (h, k) = k. Then f is an epimorphism whose kernel
is {(h, 1K ) : h ∈ H}, which can be identi¬ed with H. By the ¬rst isomorphism theorem,
G/H ∼ K.
=


Section 2.1
Here is an interesting ring that will come up in Section 9.5 in connection with group
representation theory. Let G = {x1 , . . . xm } be a ¬nite group, and let R be any ring.
The group ring RG consists of all elements r1 x1 + · · · + rm xm . Addition of elements
is componentwise, just as if the xi were basis vectors of a vector space and the ri were
scalars in a ¬eld. Multiplication in RG is governed by the given multiplication in R, along
4

with linearity. For example,

(r1 x1 + r2 x2 )(s1 x1 + s2 x2 ) = r1 s1 x2 + r1 s2 x1 x2 + r2 s1 x2 x1 + r2 s2 x2 .
1 2

The elements x2 , x1 x2 , x2 x1 , and x2 belong to G, which need not be abelian. The elements
1 2
r1 s1 , r1 s2 , r2 s1 , and r2 s2 belong to R, which is not necessarily commutative. Thus it is
essential to keep track of the order in which elements are written.


Section 2.2
Here is some additional practice with ideals in matrix rings. If I is an ideal of Mn (R),
we will show that I must have the form Mn (I0 ) for some unique ideal I0 of R. [Mn (I0 ) is
the set of all n by n matrices with entries in I0 .]
We note ¬rst that for any matrix A, we have Eij AEkl = ajk Eil . This holds because
Eij A puts row j of A in row i, and AEkl puts column k of A in column l. Thus Eij AEkl
puts ajk in the il position, with zeros elsewhere.
If I is an ideal of Mn (R), let I0 be the set of all entries a11 , where A = (aij ) is a matrix
in I. To verify that I0 is an ideal, observe that (A + B)11 = a11 + b11 , ca11 = (cE11 A)11 ,
and a11 c = (AE11 c)11 . We will show that I = Mn (I0 ).
If A ∈ I, set i = l = 1 in the basic identity involving the elementary matrices Eij
(see the second paragraph above) to get ajk E11 ∈ I. Thus ajk ∈ I0 for all j and k, so
A ∈ Mn (I0 ).
Conversely, let A ∈ Mn (I0 ), so that ail ∈ I0 for all i, l. By de¬nition of I0 , there
is a matrix B ∈ I such that b11 = ail . Take j = k = 1 in the basic identity to get
Ei1 BE1l = b11 Eil = ail Eil . Consequently, ail Eil ∈ I for all i and l. But the sum of the
matrices ail Eil over all i and l is simply A, and we conclude that A ∈ I.
To prove uniqueness, suppose that Mn (I0 ) = Mn (I1 ). If a ∈ I0 , then aE11 ∈ Mn (I0 ) =
Mn (I1 ), so a ∈ I1 . A symmetrical argument completes the proof.


Section 2.3
If a and b are relatively prime integers, then ai and bj are relatively prime for all positive
integers i and j. Here is an analogous result for ideals. Suppose that the ideals I1 and I2
2
of the ring R are relatively prime, so that I1 + I2 = R. Let us prove that I1 and I2 are
relatively prime as well. By the de¬nitions of the sum and product of ideals, we have

R = RR = (I1 + I2 )(I1 + I2 ) = I1 + I1 I2 + I2 I1 + I2 ⊆ I1 + I2 ⊆ R
2 2 2


2 3
so R = I1 + I2 , as asserted. Similarly, we can show that R = I1 + I2 by considering the
2
product of I1 + I2 and I1 + I2 . More generally, an induction argument shows that if

I1 + · · · + In = R,

then for all positive integers m1 , . . . , mn we have

I1 1 + · · · + In n = R.
m m
5

Section 2.4
We have de¬ned prime ideals only when the ring R is commutative, and it is natural to ask
why this restriction is imposed. Suppose that we drop the hypothesis of commutativity,
and de¬ne prime ideals as in (2.4.4). We can then prove that if P is a prime ideal, I and J
are arbitrary ideals, and P ⊇ IJ, then either P ⊇ I or P ⊇ J. [If the conclusion is false,
there are elements a ∈ I \ P and b ∈ J \ P . Then ab ∈ IJ ⊆ P , but a ∈ P and b ∈ P , a
/ /
contradiction.]
If we try to reverse the process and show that a proper ideal P such that P ⊇ IJ
implies P ⊇ I or P ⊇ J must be prime, we run into trouble. If ab belongs to P , then the
principal ideal (ab) is contained in P . We would like to conclude that (a)(b) ⊆ P , so that
(a) ⊆ P or (b) ⊆ P , in other words, a ∈ P or b ∈ P . But (ab) need not equal (a)(b). For
example, to express the product of the element ar ∈ (a) and the element sb ∈ (b) as a
multiple of ab, we must invoke commutativity.
An explicit example: Let P be the zero ideal in the ring Mn (R) of n by n matrices over
a division ring R (see Section 2.2, exercises). Since Mn (R) has no nontrivial two-sided
ideals, P ⊇ IJ implies P ⊇ I or P ⊇ J. But ab ∈ P does not imply a ∈ P or b ∈ P ,
because the product of two nonzero matrices can be zero.
This example illustrates another source of di¬culty. The zero ideal P is maximal,
but Mn (R)/P is not a division ring. Thus we cannot generalize (2.4.3) by dropping
commutativity and replacing “¬eld” by “division ring”. [If R/M is a division ring, it does
follow that M is a maximal ideal; the proof given in (2.4.3) works.]


Section 2.5
Let™s have a brief look at polynomials in more than one variable; we will have much
more to say in Chapter 8. For example, a polynomial f (X, Y, Z) in 3 variables is a sum of
monomials; a monomial is of the form aX i Y j Z k where a belongs to the underlying ring R.
The degree of such a monomial is i + j + k, and the degree of f is the maximum monomial
degree. Formally, we can de¬ne R[X, Y ] as (R[X])[Y ], R[X, Y, Z] as (R[X, Y ])[Z], etc.
Let f be a polynomial of degree n in F [X, Y ], where F is a ¬eld. There are many cases
in which f has in¬nitely many roots in F . For example, consider f (X, Y ) = X + Y over
the reals. The problem is that there is no direct extension of the division algorithm (2.5.1)
to polynomials in several variables. The study of solutions to polynomial equations in
more than one variable leads to algebraic geometry, which will be introduced in Chapter 8.


Section 2.6
We have shown in (2.6.8) that every principal ideal domain is a unique factorization
domain. Here is an example of a UFD that is not a PID. Let R = Z[X], which will be
shown to be a UFD in (2.9.6). Let I be the maximal ideal < 2, X > (see (2.4.8)). If I is
principal, then I consists of all multiples of a polynomial f (X) with integer coe¬cients.
Since 2 ∈ I, we must be able to multiply f (X) = a0 + a1 X + · · · + an X n by a polynomial
g(X) = b0 + b1 X + · · · + bm X m and produce 2. There is no way to do this unless f (X) = 1
or 2. But if f (X) = 1 then I = R, a contradiction (a maximal ideal must be proper).
6

Thus f (X) = 2, but we must also be able to multiply f (X) by some polynomial in Z[X]
to produce X. This is impossible, and we conclude that I is not principal.
A faster proof that Z[X] is not a PID is as follows. In (2.4.8) we showed that < 2 >
and < X > are prime ideals that are not maximal, and the result follows from (2.6.9).
On the other hand, the ¬rst method produced an explicit example of an ideal that is not
principal.


Section 2.7
It may be useful to look at the Gaussian integers in more detail, and identify the primes
in Z[i]. To avoid confusion, we will call a prime in Z[i] a Gaussian prime and a prime
in Z a rational prime. Anticipating some terminology from algebraic number theory, we
de¬ne the norm of the Gaussian integer a + bi as N (a + bi) = a2 + b2 . We will outline
a sequence of results that determine exactly which Gaussian integers are prime. We use
Greek letters for members of Z[i] and roman letters for ordinary integers.

1. ± is a unit in Z[i] i¬ N (±) = 1. Thus the only Gaussian units are ±1 and ±i.

[If ±β = 1, then 1 = N (1) = N (±)N (β), so both N (±) and N (β) must be 1.]
Let n be a positive integer.

2. If n is a Gaussian prime, then n is a rational prime not expressible as the sum of two
squares.

[n is a rational prime because any factorization in Z is also a factorization in Z[i]. If
n = x2 + y 2 = (x + iy)(x ’ iy), then either x + iy or x ’ iy is a unit. By (1), n = 1, a
contradiction.]
Now assume that n is a rational prime but not a Gaussian prime.

3. If ± = a + bi is a nontrivial factor of n, then gcd(a, b) = 1.

[If the greatest common divisor d is greater than 1, then d = n. Thus ± divides n and n
divides ±, so n and ± are associates, a contradiction.]

4. n is a sum of two squares.

[Let n = (a + bi)(c + di); since n is real we have ad + bc = 0, so a divides bc. By (3), a and
b are relatively prime, so a divides c, say c = ka. Then b(ka) = bc = ’ad, so d = ’bk.
Thus n = ac ’ bd = ka2 + kb2 = k(a2 + b2 ). But a + bi is a nontrivial factor of n, so
a2 + b2 = N (a + bi) > 1. Since n is a rational prime, we must have k = 1 and n = a2 + b2 .]
By the above results, we have:

5. If n is a positive integer, then n is a Gaussian prime if and only if n is a rational prime
not expressible as the sum of two squares.

Now assume that ± = a + bi is a Gaussian integer with both a and b nonzero. (The cases
in which a or b is 0 are covered by (1) and (5).)

6. If N (±) is a rational prime, then ± is a Gaussian prime.
7

[If ± = βγ where β and γ are not units, then N (±) = N (β)N (γ), where N (β) and N (γ)
are greater than 1, contradicting the hypothesis.]
Now assume that ± is a Gaussian prime.
7. If N (±) = hk is a nontrivial factorization, so that h > 1 and k > 1, then ± divides
either h or k. If, say, ± divides h, then so does its complex conjugate ±.
[We have N (±) = a2 + b2 = (a + bi)(a ’ bi) = ±± = hk. Since ± divides the product hk,
it must divide one of the factors. If ±β = h, take complex conjugates to conclude that
±β = h.]
8. N (±) is a rational prime.
[If not, then we can assume by (7) that ± and ± divide h. If ± and ± are not associates,
then N (±) = ±± divides h, so hk divides h and therefore k = 1, a contradiction. If ± and
its conjugate are associates, then one is ±i times the other. The only way this can happen
is if ± = γ(1 ± i) where γ is a unit. But then N (±) = N (γ)N (1 ± i) = N (1 ± i) = 2, a
rational prime.]
By the above results, we have:
9. If ± = a + bi with a = 0, b = 0, then ± is a Gaussian prime if and only if N (±) is a
rational prime.
The assertions (5) and (9) give a complete description of the Gaussian primes, except
that it would be nice to know when a rational prime p can be expressed as the sum of two
squares. We have 2 = 12 + 12 , so 2 is not a Gaussian prime, in fact 2 = (1 + i)(1 ’ i). If
p is an odd prime, then p is a sum of two squares i¬ p ≡ 1 mod 4, as we will prove at the
beginning of Chapter 7. Thus we may restate (5) as follows: 10. If n is a positive integer,
then n is a Gaussian prime i¬ n is a rational prime congruent to 3 mod 4.
[Note that a number congruent to 0 or 2 mod 4 must be even.]


Section 2.8
Suppose that R is an integral domain with quotient ¬eld F , and g is a ring homomorphism
from R to an integral domain R . We can then regard g as mapping R into the quotient
¬eld F of R . It is natural to try to extend g to a homomorphism g : F ’ F . If a, b ∈ R
with b = 0, then a = b(a/b), so we must have g(a) = g(b)g(a/b). Thus if an extension
exists, it must be given by
g(a/b) = g(a)[g(b)]’1 .
For this to make sense, we must have g(b) = 0 whenever b = 0, in other words, g is a
monomorphism. [Note that if x, y ∈ R and g(x) = g(y), then g(x’y) = 0, hence x’y = 0,
so x = y.] We will see in (3.1.2) that any homomorphism of ¬elds is a monomorphism, so
this condition is automatically satis¬ed. We can establish the existence of g by de¬ning it
as above and then showing that it is a well-de¬ned ring homomorphism. This has already
been done in Problem 8. We are in the general situation described in Problem 7, with S
taken as the set of nonzero elements of R. We must check that g(s) is a unit in F for
every s ∈ S, but this holds because g(s) is a nonzero element of F .
8

Section 2.9
Here is another useful result relating factorization over an integral domain to factorization
over the quotient ¬eld. Suppose that f is a monic polynomial with integer coe¬cients,
and that f can be factored as gh, where g and h are monic polynomials with rational
coe¬cients. Then g and h must have integer coe¬cients. More generally, let D be a
unique factorization domain with quotient ¬eld F , and let f be a monic polynomial in
D[X]. If f = gh, with g, h ∈ F [X], then g and h must belong to D[X].
To prove this, we invoke the basic proposition (2.9.2) to produce a nonzero » ∈ F
such that »g ∈ D[X] and »’1 h ∈ D[X]. But g and h are monic, so » = 1 and the result
follows.
Let f be a cubic polynomial in F [X]. If f is reducible, it must have a linear factor and
hence a root in F . We can check this easily if F is a ¬nite ¬eld; just try all possibilities. A
¬nite check also su¬ces when F = Q, by the rational root test (Section 2.9, Problem 1).
If g is a linear factor of f , then f /g = h is quadratic. We can factor h as above, and in
addition the quadratic formula is available if square roots can be extracted in F . In other
words, if a ∈ F , then b2 = a for some b ∈ F .


Section 3.1
All results in this section are basic and should be studied carefully. You probably have
some experience with polynomials over the rational numbers, so let™s do an example with
a rather di¬erent ¬‚avor. Let F = F2 be the ¬eld with two elements 0 and 1, and let
f ∈ F [X] be the polynomial X 2 + X + 1. Note that f is irreducible over F , because if f
were factorable, it would have a linear factor and hence a root in F . This is impossible,
as f (0) = f (1) = 1 = 0. If we adjoin a root ± of f to produce an extension F (±), we
know that f is the minimal polynomial of ± over F , and that F (±) consists of all elements
b0 + b1 ±, with b0 and b1 in F . Since b0 and b1 take on values 0 and 1, we have constructed
a ¬eld F (±) with 4 elements. Moreover, all nonzero elements of F (±) can be expressed
as powers of ±, as follows:
±0 = 1, ±1 = ±, ±2 = ’± ’ 1 = 1 + ±. (The last equality follows because 1 + 1 = 0
in F .)
This is a typical computation involving ¬nite ¬elds, which will be studied in detail in
Chapter 6.


Section 3.2
We found in Problem 3 that a splitting ¬eld for X 4 ’ 2 has degree 8 over Q. If we make a
seemingly small change and consider f (X) = X 4 ’ 1, the results are quite di¬erent. The
roots of f are 1, i, ’1 and ’i. Thus Q(i) is the desired splitting ¬eld, and it has degree 2
over Q because the minimal polynomial of i over Q has degree 2.
A general problem suggested by this example is to describe a splitting ¬eld for X n ’ 1
over Q for an arbitrary positive integer n. The splitting ¬eld is Q(ω), where ω is a
primitive nth root of unity, for example, ω = ei2π/n . We will see in Section 6.5 that the
degree of Q(ω) over Q is •(n), where • is the Euler phi function.
9

Section 3.3
In Problem 8 we used the existence of an algebraic closure of F to show that any set of
nonconstant polynomials in F [X] has a splitting ¬eld over F . Conversely, if we suppose
that it is possible to ¬nd a splitting ¬eld K for an arbitrary family of polynomials over
the ¬eld F , then the existence of an algebraic closure of F can be established quickly.
Thus let K be a splitting ¬eld for the collection of all polynomials in F [X], and let C
be the algebraic closure of F in K (see (3.3.4)). Then by de¬nition, C is an algebraic
extension of F and every nonconstant polynomial in F [X] splits over C. By (3.3.6), C is
an algebraic closure of F .


Section 3.4
Let™s have another look at Example 3.4.8 with p = √ to get some additional practice with
2
separability and inseparability. We have seen that t is not separable over F , in fact it is

purely inseparable because its minimal polynomial X 2 ’ t can be written as (X ’ t)2 .

But if we adjoin a cube root of t, the resulting element 3 t is separable over F , because
X 3 ’ t has nonzero derivative, √ equivalently does not belong to F [X 2 ]√(see 3.4.3).

3 6
√ √ also that adjoining t and t is equivalent to6 adjoining t, in other words,
Notice √ √ √ √
F ( t, 3 t) = F ( 6 t). √ see this, ¬rst observe that if ± = t, then t = ±3 and 3 t = ±2 .
To √
On the other hand, ( t/ 3 t)6 = t.
It is possible for an element ± to be both separable and purely inseparable over F , but
it happens if and only if ± belongs to F . The minimal polynomial of ± over F must have
only one distinct root and no repeated roots, so min(±, F ) = X ’ ±. But the minimal
polynomial has coe¬cients in F (by de¬nition), and the result follows.


Section 3.5
Suppose we wish to ¬nd the Galois group of the extension E/F , where E = F (±). Assume
that ± is algebraic over F with minimal polynomial f , and that f has n distinct roots
±1 = ±, ±2 , . . . , ±n in some splitting ¬eld. If σ ∈ Gal(E/F ), then σ permutes the roots
of f by (3.5.1). Given any two roots ±i and ±j , i = j, we can ¬nd an F -isomorphism
that carries ±i into ±j ; see (3.2.3). Do not jump to the conclusion that all permutations
are allowable, and therefore Gal(E/F ) is isomorphic to Sn . For example, we may not be
able to simultaneously carry ±1 into ±2 and ±3 into ±4 . Another di¬culty is that the
F -isomorphism carrying ±i into ±j need not be an F -automorphism of E. This suggests
that normality of the extension is a key property. If E/F is the non-normal extension of
Example (3.5.10), the only allowable permutation is the identity.


Section 4.1
Finitely generated algebras over a commutative ring R frequently appear in applications
to algebraic number theory and algebraic geometry. We say that A is a ¬nitely generated
R-algebra if there are ¬nitely many elements x1 , . . . , xn in A such that every element of
10

A is a polynomial f (x1 , . . . , xn ) with coe¬cients in R. Equivalently, A is a homomorphic
image of the polynomial ring R[X1 , . . . , Xn ]. The homomorphism is determined explicitly
by mapping Xi to xi , i = 1, . . . , n. The polynomial f (X1 , . . . , Xn ) is then mapped to
f (x1 , . . . , xn ).
If every element is not just a polynomial in the xi but a linear combination of the xi
with coe¬cients in R, then A is a ¬nitely generated module over R. To see the di¬erence
clearly, look at the polynomial ring R[X], which is a ¬nitely generated R algebra. (In the
above discussion we can take n = 1 and x1 = X.) But if f1 , . . . , fn are polynomials in
R[X] and the maximum degree of the fi is m, there is no way to take linear combinations
of the fi and produce a polynomial of degree greater than m. Thus R[X] is a not a ¬nitely
generated R-module.


Section 4.2
Here is some practice working with quotient modules. Let N be a submodule of the
R-module M , and let π be the canonical map from M onto M/N , taking x ∈ M to
x + N ∈ M/N . Suppose that N1 and N2 are submodules of M satisfying

(a) N1 ¤ N2 ;
(b) N1 © N = N2 © N ;
(c) π(N1 ) = π(N2 ).

Then N1 = N2 .
To prove this, let x ∈ N2 . Hypothesis (c) says that (N1 + N )/N = (N2 + N )/N ; we
don™t write Ni /N, i = 1, 2, because N is not necessarily a submodule of N1 or N2 . Thus
x + N ∈ (N2 + N )/N = (N1 + N )/N , so x + N = y + N for some y ∈ N1 . By (a), y ∈ N2 ,
hence x ’ y ∈ N2 © N = N1 © N by (b). Therefore x ’ y and y both belong to N1 , and
consequently so does x. We have shown that N2 ¤ N1 , and in view of hypothesis (a), we
are ¬nished.


Section 4.3
If M is a free R-module with basis S = (xi ), then an arbitrary function f from S to
an arbitrary R-module N has a unique extension to an R-homomorphism f : M ’ N ;
see (4.3.6).
This property characterizes free modules, in other words, if M is an R-module with a
subset S satisfying the above property, then M is free with basis S. To see this, build a
free module M with basis S = (yi ) having the same cardinality as S. For example, we
can take M to be the direct sum of copies of R, as many copies as there are elements
of S. De¬ne f : S ’ S ⊆ M by f (xi ) = yi , and let f be the unique extension of f to
an R-homomorphism from M to M . Similarly, de¬ne g : S ’ S ⊆ M by g(yi ) = xi ,
and let g be the unique extension of g to an R-homomorphism from M to M . Note
that g exists and is unique because M is free. Now g —¦ f is the identity on S, so by
uniqueness of extensions from S to M , g —¦ f is the identity on M . Similarly, f —¦ g is the
11

identity on M . Thus M and M are not only isomorphic, but the isomorphism we have
constructed carries S into S . It follows that M is free with basis S.
This is an illustration of the characterization of an algebraic object by a universal
mapping property. We will see other examples in Chapter 10.


Section 4.4
Here is some practice in decoding abstract presentations. An R-module can be de¬ned as
a representation of R in an endomorphism ring of an abelian group M . What does this
mean?
First of all, for each r ∈ R, we have an endomorphism fr of the abelian group M , given
by fr (x) = rx, x ∈ M . To say that fr is an endomorphism is to say that r(x+y) = rx+ry,
x, y ∈ M , r ∈ R.
Second, the mapping r ’ fr is a ring homomorphism from R to EndR (M ). (Such
a mapping is called a representation of R in EndR (M ).) This says that fr+s (x) =
fr (x) + fs (x), frs (x) = fr (fs (x)), and f1 (x) = x. In other words, (r + s)x = rx + sx,
(rs)x = r(sx), and 1x = x.
Thus we have found a fancy way to write the module axioms. If you are already
comfortable with the informal view of a module as a “vector space over a ring”, you are
less likely to be thrown o¬ stride by the abstraction.


Section 4.5
The technique given in Problems 1“3 for ¬nding new bases and generators is worth em-
phasizing. We start with a matrix A to be reduced to Smith normal form. The equations
U = AX give the generators U of the submodule K in terms of the basis X of the free
module M . The steps in the Smith calculation are of two types:
1. Premultiplication by an elementary row matrix R. This corresponds to changing gen-
erators via V = RU .
2. Postmultiplication by an elementary column matrix C. This corresponds to changing
bases via Y = C ’1 X.
Suppose that the elementary row matrices appearing in the calculation are R1 , . . . , Rs ,
in that order, and the elementary column matrices are C1 , . . . , Ct , in that order. Then
the matrices Q and P are given by
P ’1 = C1 C2 · · · Ct
Q = Rs · · · R 2 R1 ,
’1 ’1 ’1
hence P = Ct · · · C2 C1 . The ¬nal basis for M is Y = P X, and the ¬nal generating
set for K is V = QU = SY , where S = QAP ’1 is the Smith normal form (see 4.4.2).


Section 4.6
Here is a result that is used in algebraic number theory. Let G be a free abelian group
of rank n, and H a subgroup of G. By the simultaneous basis theorem, there is a basis
12

y1 , . . . yn of G and there are positive integers a1 , . . . ar , r ¤ n, such that ai divides ai+1
for all i, and a1 y1 , . . . , ar yr is a basis for H. We claim that the abelian group G/H is
¬nite if and only if r = n, and in this case, the size of G/H is |G/H| = a1 a2 · · · ar .
To see this, look at the proof of (4.6.3) with Rn replaced by G and K by H. The
argument shows that G/H is the direct sum of cyclic groups Z/Zai , i = 1, . . . , n, with
ai = 0 for r < i ¤ n. In other words, G/H is the direct sum of r ¬nite cyclic groups (of
order a1 , . . . , ar respectively) and n ’ r copies of Z. The result follows.
Now assume that r = n, and let x1 , . . . , xn and z1 , . . . , zn be arbitrary bases for G
and H respectively. Then each zi is a linear combination of the xi with integer coe¬cients;
in matrix form, z = Ax. We claim that |G/H| is the absolute value of the determinant
of A. To verify this, ¬rst look at the special case xi = yi and zi = ai yi , i = 1, . . . , n.
Then A is a diagonal matrix with entries ai , and the result follows. But the special case
implies the general result, because any matrix corresponding to a change of basis of G
or H is unimodular, in other words, has determinant ±1. (See Section 4.4, Problem 1.)


Section 4.7
Here is some extra practice in diagram chasing. The diagram below is commutative with
exact rows.

GB GC G0
f g
A
u v
t
  
GB GC G0
A
f g


If t and u are isomorphisms, we will show that v is also an isomorphism. (The hypothesis
on t can be weakened to surjectivity.)
Let c ∈ C ; then c = g b for some b ∈ B . Since u is surjective, g b = g ub for some
b ∈ B. By commutativity, g ub = vgb, which proves that v is surjective.
Now assume vc = 0. Since g is surjective, c = gb for some b ∈ B. By commutativity,
vgb = g ub = 0. Thus ub ∈ ker g = im f , so ub = f a for some a ∈ A . Since t is
surjective, f a = f ta for some a ∈ A. By commutativity, f ta = uf a. We now have
ub = uf a, so b ’ f a ∈ ker u, hence b = f a because u is injective. Consequently,

c = gb = gf a = 0

which proves that v is injective.
Supplement: The Long Exact
Homology Sequence and
Applications

S1. Chain Complexes
In the supplement, we will develop some of the building blocks for algebraic topology.
As we go along, we will make brief comments [in brackets] indicating the connection
between the algebraic machinery and the topological setting, but for best results here,
please consult a text or attend lectures on algebraic topology.


S1.1 De¬nitions and Comments
A chain complex (or simply a complex ) C— is a family of R-modules Cn , n ∈ Z, along
with R-homomorphisms dn : Cn ’ Cn’1 called di¬erentials, satisfying dn dn+1 = 0 for
all n. A chain complex with only ¬nitely many Cn ™s is allowed; it can always be extended
with the aid of zero modules and zero maps. [In topology, Cn is the abelian group of n-
chains, that is, all formal linear combinations with integer coe¬cients of n-simplices in a
topological space X. The map dn is the boundary operator, which assigns to an n-simplex
an n ’ 1-chain that represents the oriented boundary of the simplex.]
The kernel of dn is written Zn (C— ) or just Zn ; elements of Zn are called cycles in
dimension n. The image of dn+1 is written Bn (C— ) or just Bn ; elements of Bn are called
boundaries in dimension n. Since the composition of two successive di¬erentials is 0, it
follows that Bn ⊆ Zn . The quotient Zn /Bn is written Hn (C— ) or just Hn ; it is called the
nth homology module (or homology group if the underlying ring R is Z).
[The key idea of algebraic topology is the association of an algebraic object, the col-
lection of homology groups Hn (X), to a topological space X. If two spaces X and Y
are homeomorphic, in fact if they merely have the same homotopy type, then Hn (X)
and Hn (Y ) are isomorphic for all n. Thus the homology groups can be used to distin-
guish between topological spaces; if the homology groups di¬er, the spaces cannot be
homeomorphic.]
Note that any exact sequence is a complex, since the composition of successive maps
is 0.

1
2

S1.2 De¬nition
A chain map f : C— ’ D— from a chain complex C— to a chain complex D— is a collection
of module homomorphisms fn : Cn ’ Dn , such that for all n, the following diagram is
commutative.

G Dn
fn
Cn
dn dn
 
G Dn’1
Cn’1
fn’1


We use the same symbol dn to refer to the di¬erentials in C— and D— .
[If f : X ’ Y is a continuous map of topological spaces and σ is a singular n-simplex
in X, then f# (σ) = f —¦σ is a singular n-simplex in Y , and f# extends to a homomorphism
of n-chains. If we assemble the f# ™s for n = 0, 1, . . . , the result is a chain map.]

S1.3 Proposition
A chain map f takes cycles to cycles and boundaries to boundaries. Consequently, the
map zn + Bn (C— ) ’ fn (zn ) + Bn (D— ) is a well-de¬ned homomorphism from Hn (C— ) to
Hn (D— ). It is denoted by Hn (f ).

Proof. If z ∈ Zn (C— ), then since f is a chain map, dn fn (z) = fn’1 dn (z) = fn’1 (0) = 0.
Therefore fn (z) ∈ Zn (D— ). If b ∈ Bn (C— ), then dn+1 c = b for some c ∈ Cn+1 . Then
fn (b) = fn (dn+1 c) = dn+1 fn+1 c, so fn (b) ∈ Bn (D— ). ™

S1.4 The Homology Functors
We can create a category whose objects are chain complexes and whose morphisms are
chain maps. The composition gf of two chain maps f : C— ’ D— and g : D— ’ E— is the
collection of homomorphisms gn fn , n ∈ Z. For any n, we associate with the chain complex
C— its nth homology module Hn (C— ), and we associate with the chain map f : C— ’ D—
the map Hn (f ) : Hn (C— ) ’ Hn (D— ) de¬ned in (S1.3). Since Hn (gf ) = Hn (g)Hn (f ) and
Hn (1C— ) is the identity on Hn (C— ), Hn is a functor, called the nth homology functor.

S1.5 Chain Homotopy
Let f and g be chain maps from C— to D— . We say that f and g are chain homotopic
g if there exist homomorphisms hn : Cn ’ Dn+1 such that fn ’ gn =
and write f
dn+1 hn + hn’1 dn ; see the diagram below.

G Cn’1
dn
Cn
y
hn yy
yy fn ’gn
yy
|y 
G Dn v
hn’1
Dn+1
dn+1
3

[If f and g are homotopic maps from a topological space X to a topological space Y ,
then the maps f# and g# (see the discussion in (S1.2)) are chain homotopic,]


S1.6 Proposition
If f and g are chain homotopic, then Hn (f ) = Hn (g).

Proof. Let z ∈ Zn (C— ). Then

fn (z) ’ gn (z) = (dn+1 hn + hn’1 dn )z ∈ Bn (D— )

since dn z = 0. Thus fn (z) + Bn (D— ) = gn (z) + Bn (D— ), in other words, Hn (f ) =
Hn (g). ™


S2. The Snake Lemma
We isolate the main ingredient of the long exact homology sequence. After an elaborate
diagram chase, a homomorphism between two modules is constructed. The domain and
codomain of the homomorphism are far apart in the diagram, and the arrow joining them
tends to wiggle like a serpent. First, a result about kernels and cokernels of module
homomorphisms.


S2.1 Lemma
Assume that the diagram below is commutative.

GB
f
A
e
d
 
GD
C g




(i) f induces a homomorphism on kernels, that is, f (ker d) ⊆ ker e.
(ii) g induces a homomorphism on cokernels, that is, the map y + im d ’ g(y) + im e,
y ∈ C, is a well-de¬ned homomorphism from coker d to coker e.
(iii) If f is injective, so is the map induced by f , and if g is surjective, so is the map
induced by g.

Proof. (i) If x ∈ A and d(x) = 0, then ef (x) = gd(x) = g0 = 0.
(ii) If y ∈ im d, then y = dx for some x ∈ A. Thus gy = gdx = ef x ∈ im e. Since g is
a homomorphism, the induced map is also.
(iii) The ¬rst statement holds because the map induced by f is simply a restriction.
The second statement follows from the form of the map induced by g. ™
4

Now refer to our snake diagram, Figure S2.1. Initially, we are given only the second
and third rows (ABE0 and 0CDF), along with the maps d, e and h. Commutativity of the
squares ABDC and BEFD is assumed, along with exactness of the rows. The diagram is
now enlarged as follows. Take A =ker d, and let the map from A to A be inclusion. Take
C = coker d, and let the map from C to C be canonical. Augment columns 2 and 3 in
a similar fashion. Let A ’ B be the map induced by f on kernels, and let C ’ D be
the map induced by g on cokernels. Similarly, add B ’ E and D ’ F . The enlarged
diagram is commutative by (S2.1), and it has exact columns by construction.

n h‚ ‚
nn ‚‚
n ‚‚
nn ‚‚
wn n GB GE
1 A
1
1
  
1 GB GE G0
f s
A
1
1 e
d h
1   
x GC GD GF
g t
xx0
xx
xx   
9 GD GF
C

Figure S2.1


S2.2 Lemma
The ¬rst and fourth rows of the enlarged snake diagram are exact.

Proof. This is an instructive diagram chase, showing many standard patterns. Induced
maps will be denoted by an overbar, and we ¬rst prove exactness at B . If x ∈ A = ker
d and y = f x = f x, then sy = sf x = 0, so y ∈ker s. On the other hand, if y ∈ B ⊆ B
and sy = sy = 0, then y = f x for some x ∈ A. Thus 0 = ey = ef x = gdx, and since g is
injective, dx = 0. Therefore y = f x with x ∈ A , and y ∈im f .
Now to prove exactness at D , let x ∈ C. Then t(gx + im e) = tgx + im h = 0 by
exactness of the third row, so im g ⊆ker t. Conversely, if y ∈ D and t(y + im e) =
ty + im h = 0, then ty = hz for some z ∈ E. Since s is surjective, z = sx for some x ∈ B.
Now

ty = hz = hsx = tex

so y ’ ex ∈ ker t = im g, say y ’ ex = gw, w ∈ C. Therefore

y + im e = g(w + im d)

and y + im e ∈ im g. ™
5

S2.3 Remark
Sometimes an even bigger snake diagram is given, with column 1 assumed to be an exact
sequence

GA GA GC GC G0
d
0

and similarly for columns 2 and 3. This is nothing new, because by replacing modules
by isomorphic copies we can assume that A is the kernel of d, C is the cokernel of d,
A ’ A is inclusion, and C ’ C is canonical.

S2.4 The Connecting Homomorphism
We will now connect E to C in the snake diagram while preserving exactness. The idea
is to zig-zag through the diagram along the path E EBDCC .
Let z ∈ E ⊆ E; Since s is surjective, there exists y ∈ B such that z = sy. Then
tey = hsy = hz = 0 since E = ker h. Thus ey ∈ ker t = im g, so ey = gx for some x ∈ C.
We de¬ne the connecting homomorphism ‚ : E ’ C by ‚z = x + im d. Symbolically,

‚ = [g ’1 —¦ e —¦ s’1 ]

where the brackets indicate that ‚z is the coset of x in C = C/ im d.
We must show that ‚ is well-de¬ned. Suppose that y is another element of B with
sy = z. Then y ’ y ∈ ker s = im f , so y ’ y = f u for some u ∈ A. Thus e(y ’ y ) =
ef u = gdu. Now we know from the above computation that ey = gx for some x ∈ C, and
similarly ey = gx for some x ∈ C. Therefore g(x ’ x ) = gdu, so x ’ x ’ du ∈ ker g.
Since g is injective, x ’ x = du, so x + im d = x + im d. Thus ‚z is independent of the
choice of the representatives y and x. Since every map in the diagram is a homomorphism,
so is ‚.

S2.5 Snake Lemma
The sequence

GB GE GC GD GF
f g
s ‚ t
A
is exact.
Proof. In view of (S2.2), we need only show exactness at E and C . If z = sy, y ∈ B =
ker e, then ey = 0, so ‚z = 0 by de¬nition of ‚. Thus im s ⊆ ker ‚. Conversely, assume
‚z = 0, and let x and y be as in the de¬nition of ‚. Then x = du for some u ∈ A, hence
gx = gdu = ef u. But gx = ey by de¬nition of ‚, so y ’ f u ∈ ker e = B . Since z = sy by
de¬nition of ‚, we have

z = s(y ’ f u + f u) = s(y ’ f u) ∈ im s.

To show exactness at C , consider an element ‚z in the image of ‚. Then ‚z = x + im d ,
so g‚z = gx+im e. But gx = ey by de¬nition of ‚, so g‚z = 0 and ‚z ∈ ker g. Conversely,
6

suppose x ∈ C and g(x + im d) = gx + im e = 0. Then gx = ey for some y ∈ B. If z = sy,
then hsy = tey = tgx = 0 by exactness of the third row. Thus z ∈ E and (by de¬nition
of ‚) we have ‚z = x + im d. Consequently, x + im d ∈ im ‚. ™


S3. The Long Exact Homology Sequence
S3.1 De¬nition
We say that

G C— G D— G E— G0
f g
0
where f and g are chain maps, is a short exact sequence of chain complexes if for each n, the
corresponding sequence formed by the component maps fn : Cn ’ Dn and gn : Dn ’ En ,
is short exact. We will construct connecting homomorphisms ‚n : Hn (E— ) ’ Hn’1 (C— )
such that the sequence

G Hn+1 (E— ) G Hn (C— ) G Hn (D— ) G Hn (E— ) G Hn’1 (C— ) G···
g f g f
‚ ‚
···

is exact. [We have taken some liberties with the notation. In the second diagram, f
stands for the map induced by fn on homology, namely, Hn (f ); similarly for g.] The
second diagram is the long exact homology sequence, and the result may be summarized
as follows.

S3.2 Theorem
A short exact sequence of chain complexes induces a long exact sequence of homology
modules.
Proof. This is a double application of the snake lemma. The main ingredient is the
following snake diagram.
G Dn /Bn (D— ) G En /Bn (E— ) G0
Cn /Bn (C— )
d d d
  
G Zn’1 (C— ) G Zn’1 (D— ) G Zn’1 (E— )
0
The horizontal maps are derived from the chain maps f and g, and the vertical maps are
given by d(xn + Bn ) = dxn . The kernel of a vertical map is {xn + Bn : xn ∈ Zn } = Hn ,
and the cokernel is Zn’1 /Bn’1 = Hn’1 . The diagram is commutative by the de¬nition
of a chain map. But in order to apply the snake lemma, we must verify that the rows
are exact, and this involves another application of the snake lemma. The appropriate
diagram is
G Cn G Dn G En G0
0
d d d
  
G Cn’1 G Dn’1 G En’1 G0
0
7

where the horizontal maps are again derived from f and g. The exactness of the rows of
the ¬rst diagram follows from (S2.2) and part (iii) of (S2.1), shifting indices from n to
n ± 1 as needed. ™

S3.3 The connecting homomorphism explicitly
If z ∈ Hn (E— ), then z = zn + Bn (E— ) for some zn ∈ Zn (E— ). We apply (S2.4) to compute
‚z. We have zn + Bn (E— ) = gn (yn + Bn (D— )) for some yn ∈ Dn . Then dyn ∈ Zn’1 (D— )
and dyn = fn’1 (xn’1 ) for some xn’1 ∈ Zn’1 (C— ). Finally, ‚z = xn’1 + Bn’1 (C— ).

S3.4 Naturality
Suppose that we have a commutative diagram of short exact sequences of chain complexes,
as shown below.
GC GD GE G0
0

  
GC GD GE G0
0

Then there is a corresponding commutative diagram of long exact sequences:

G Hn (C— ) G Hn (D— ) G Hn (E— ) G Hn’1 (C— ) G ···
‚ ‚
···

  
G Hn (C— ) G Hn (D— ) G Hn (E— ) G Hn’1 (C— ) G ···
‚ ‚
···

Proof. The homology functor, indeed any functor, preserves commutative diagrams, so
the two squares on the left commute. For the third square, an informal argument may
help to illuminate the idea. Trace through the explicit construction of ‚ in (S3.3), and
let f be the vertical chain map in the commutative diagram of short exact sequences. The
¬rst step in the process is

zn + Bn (E— ) ’ yn + Bn (D— ).

By commutativity,

f zn + Bn (E— ) ’ f yn + Bn (D— ).

Continuing in this fashion, we ¬nd that if ‚z = xn’1 + Bn’1 (C— ), then

‚(f z) = f xn’1 + Bn’1 (C— ) = f (xn’1 + Bn’1 (C— )) = f (‚z). ™



A formal proof can be found in “An Introduction to Algebraic Topology” by J. Rotman,
page 95.
8

S4. Projective and Injective Resolutions
The functors Tor and Ext are developed with the aid of projective and injective resolutions
of a module, and we will now examine these constructions.


S4.1 De¬nitions and Comments
A left resolution of a module M is an exact sequence

G P2 G P1 G P0 GM G0
···

A left resolution is a projective resolution if every Pi is projective, a free resolution if
every Pi is free. By the ¬rst isomorphism theorem, M is isomorphic to the cokernel of the
map P1 ’ P0 , so in a sense no information is lost if M is removed. A deleted projective
resolution is of the form

G P2 G P1 G P0 G0
···


M

and the deleted version turns out to be more convenient in computations. Notice that in
a deleted projective resolution, exactness at P0 no longer holds because the map P1 ’ P0
need not be surjective. Resolutions with only ¬nitely many Pn ™s are allowed, provided
that the module on the extreme left is 0. The sequence can then be extended via zero
modules and zero maps.
Dually, a right resolution of M is an exact sequence

GM G E0 G E1 G E2 · · · ;
0

we have an injective resolution if every Ei is injective, . A deleted injective resolution has
the form

G E0 G E1 G E2 G ···
0 y


M

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