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. 10
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t
with ai ∈ I and wi ∈ M . Consequently, x ’ ai wi ∈ N , so x ∈ N + IM . ™
i=1


9.8.4 Nakayama™s Lemma, Version 2
Let N be a submodule of the R-module M , with M/N ¬nitely generated over R. [This
will be satis¬ed if M is ¬nitely generated over R.] If I is a two-sided ideal contained in
J(R), and M = N + IM , then M = N .

Proof. By (9.8.3), I(M/N ) = M/N , so by (9.8.2), M/N = 0, hence M = N . ™

Here is an application of Nakayama™s lemma.

9.8.5 Proposition
Let R be a commutative local ring with maximal ideal J (see (8.5.8)). Let M be a ¬nitely
generated R-module, and let V = M/JM . Then:

(i) V is a ¬nite-dimensional vector space over the residue ¬eld k = R/J.
(ii) If {x1 + JM, . . . , xn + JM } is a basis for V over k, then {x1 , . . . , xn } is a minimal
set of generators for M .
(iii) Any two minimal generating sets for M have the same cardinality.

Proof. (i) Since J annihilates M/JM , it follows from Section 4.2, Problem 6, that V is
a k-module, that is, a vector space over k. Since M is ¬nitely generated over R, V is
¬nite-dimensional over k.
n
(ii) Let N = i=1 Rxi . Since the xi + JM generate V = M/JM , we have
M = N + JM . By (9.8.4), M = N , so the xi generate M . If a proper subset of the xi
were to generate M , then the corresponding subset of the xi + JM would generate V ,
contradicting the assumption that V is n-dimensional.
(iii) A generating set S for M with more than n elements determines a spanning
set for V , which must contain a basis with exactly n elements. By (ii), S cannot be
minimal. ™
9.8. THEOREMS OF HOPKINS-LEVITZKI AND NAKAYAMA 27

Problems For Section 9.8
1. Let a be a nonzero element of the integral domain R. If (at ) = (at+1 ) for some positive
integer t, show that a is invertible.
2. Continuing Problem 1, show that every Artinian integral domain is a ¬eld.
3. If R is a commutative Artinian ring, show that every prime ideal of R is maximal.
4. Let R be a commutative Artinian ring. If S is the collection of all ¬nite intersections
of maximal ideals of R, then S is not empty, hence contains a minimal element
I = I1 © I2 © · · · © In , with the Ij maximal. Show that if P is any maximal ideal of R,
then P must be one of the Ij . Thus R has only ¬nitely many maximal ideals.
5. An R-module is projective if it is a direct summand of a free module. We will study
projective modules in detail in Section 10.5. We bring up the subject now in Prob-
lems 5 and 6 to illustrate a nice application of Nakayama™s lemma.
Let R be a commutative local ring, and let M be a ¬nitely generated projective
module over R, with a minimal set of generators {x1 , . . . , xn } (see (9.8.5)). We can
assume that for some free module F of rank n,

F = M • N.

To justify this, let F be free with basis e1 , . . . , en , and map F onto M via ei ’ xi ,
i = 1, . . . , n. If the kernel of the mapping is K, then we have a short exact sequence

0 ’ K ’ F ’ M ’ 0,

which splits since M is projective. [This detail will be covered in (10.5.3).]

Let J be the maximal ideal of R, and k = R/J the residue ¬eld. Show that

F/JF ∼ M/JM • N/JN.
=



6. Continue from Problem 5 and show that N/JN = 0. It then follows from Nakayama™s
lemma (9.8.2) that N = 0, and therefore M = F . We conclude that a ¬nitely
generated projective module over a commutative local ring is free.
7. We showed in (9.6.6) that there is no distinction between a left and a right-semisimple
ring. This is not the case for Noetherian (or Artinian) rings.

Let X and Y be noncommuting indeterminates, in other words, XY = Y X, and let
Z < X, Y > be the set of all polynomials in X and Y with integer coe¬cients. [Elements
of Z do commute with the indeterminates.] We impose the relations Y 2 = 0 and Y X = 0
to produce the ring R; formally, R = Z < X, Y > /(Y 2 , Y X).
Consider I = Z[X]Y , the set of all polynomials f (X)Y , f (X) ∈ Z[X]. Then I is a
two-sided ideal of R. Show that if I is viewed as a right ideal, it is not ¬nitely generated.
Thus R is not right-Noetherian.

8. Viewed as a left R-module, R = Z[X] • Z[X]Y . Show that R is left-Noetherian.
28 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA

9. Assume the hypothesis of (9.8.5). If {x1 , . . . , xn } is a minimal generating set for M ,
show that {x1 , . . . , xn }, where xi = xi + JM , is a basis for M/JM = V .
10. Continuing Problem 9, suppose that {x1 , . . . , xn } and {y1 , . . . , yn } are minimal gen-
erating sets for M , with yi = j aij xj , aij ∈ R. If A is the matrix of the aij , show
that the determinant of A is a unit in R.
Chapter 10

Introducing Homological
Algebra

Roughly speaking, homological algebra consists of (A) that part of algebra that is funda-
mental in building the foundations of algebraic topology, and (B) areas that arise naturally
in studying (A).


10.1 Categories
We have now encountered many algebraic structures and maps between these structures.
There are ideas that seem to occur regardless of the particular structure under consider-
ation. Category theory focuses on principles that are common to all algebraic systems.

10.1.1 De¬nitions and Comments
A category C consists of objects A, B, C, . . . and morphisms f : A ’ B (where A and B
are objects). If f : A ’ B and g : B ’ C are morphisms, we have a notion of composition,
in other words, there is a morphism gf = g —¦ f : A ’ C, such that the following axioms
are satis¬ed.

(i) Associativity: If f : A ’ B, g : B ’ C, h : C ’ D, then (hg)f = h(gf );
(ii) Identity: For each object A there is a morphism 1A : A ’ A such that for each
morphism f : A ’ B, we have f 1A = 1B f = f .

A remark for those familiar with set theory: For each pair (A, B) of objects, the
collection of morphisms f : A ’ B is required to be a set rather than a proper class.
We have seen many examples:

1. Sets: The objects are sets and the morphisms are functions.
2. Groups: The objects are groups and the morphisms are group homomorphisms.
3. Rings: The objects are rings and the morphisms are ring homomorphisms.

1
2 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

4. Fields: The objects are ¬elds and the morphisms are ¬eld homomorphisms [= ¬eld
monomorphisms; see (3.1.2)].
5. R-mod: The objects are left R-modules and the morphisms are R-module homomor-
phisms. If we use right R-modules, the corresponding category is called mod-R.
6. Top: The objects are topological spaces and the morphisms are continuous maps.
7. Ab: The objects are abelian groups and the the morphisms are homomorphisms from
one abelian group to another.
A morphism f : A ’ B is said to be an isomorphism if there is an inverse morphism
g : B ’ A, that is, gf = 1A and f g = 1B . In Sets, isomorphisms are bijections, and
in Top, isomorphisms are homeomorphisms. For the other examples, an isomorphism is
a bijective homomorphism, as usual.
In the category of sets, a function f is injective i¬ f (x1 ) = f (x2 ) implies x1 = x2 . But
in an abstract category, we don™t have any elements to work with; a morphism f : A ’ B
can be regarded as simply an arrow from A to B. How do we generalize injectivity to an
arbitrary category? We must give a de¬nition that does not depend on elements of a set.
Now in Sets, f is injective i¬ it has a left inverse; equivalently, f is left cancellable, i.e.
if f h1 = f h2 , then h1 = h2 . This is exactly what we need, and a similar idea works for
surjectivity, since f is surjective i¬ f is right cancellable, i.e., h1 f = h2 f implies h1 = h2 .

10.1.2 De¬nitions and Comments
A morphism f is said to be monic if it is left cancellable, epic if it is right cancellable.
In all the categories listed in (10.1.1), a morphism f is monic i¬ f is injective as a
mapping of sets. If f is surjective, then it is epic, but the converse can fail. See Problems 2
and 7“10 for some of the details.
In the category R-mod, the zero module {0} has the property that for any R-
module M , there is a unique module homomorphism from M to {0} and a unique module
homomorphism from {0} to M . Here is a generalization of this idea.

10.1.3 De¬nitions and Comments
Let A be an object in a category. If for every object B, there is a unique morphism from
A to B, then A is said to be an initial object. If for every object B there is a unique
morphism from B to A, then A is said to be a terminal object. A zero object is both
initial and terminal.
In the category of sets, there is only one initial object, the empty set. The terminal
objects are singletons {x}, and consequently there are no zero objects. In the category of
groups, the trivial group consisting of the identity alone is a zero object. We are going
to prove that any two initial objects are isomorphic, and similarly for terminal objects.
This will be a good illustration of the duality principle, to be discussed next.

10.1.4 Duality
If C is a category, the opposite or dual category C op has the same objects as C. The
morphisms are those of C with arrows reversed; thus f : A ’ B is a morphism of C op
10.1. CATEGORIES 3

i¬ f : B ’ A is a morphism of C. If the composition gf is permissible in C, then f g is
permissible in C op . To see how the duality principle works, let us ¬rst prove that if A
and B are initial objects of C, then A and B are isomorphic. There is a unique morphism
f : A ’ B and a unique morphism g : B ’ A. But 1A : A ’ A and 1B : B ’ B, and it
follows that gf = 1A and f g = 1B . The point is that we need not give a separate proof
that any two terminal objects are isomorphic. We have just proved the following:
If A and B are objects in a category C, and for every object D of C, there is a unique
morphism from A to D and there is a unique morphism from B to D, then A and B are
isomorphic.
Our statement is completely general; it does not involve the properties of any speci¬c
category. If we go through the entire statement and reverse all the arrows, equivalently,
if we replace C by C op , we get:
If A and B are objects in a category C, and for every object D of C, there is a unique
morphism from D to A and there is a unique morphism from D to B, then A and B are
isomorphic.
In other words, any two terminal objects are isomorphic. If this is unconvincing, just
go through the previous proof, reverse all the arrows, and interchange f g and gf . We say
that initial and terminal objects are dual. Similarly, monic and epic morphisms are dual.
If zero objects exist in a category, then we have zero morphisms as well. If Z is a
zero object and A and B arbitrary objects, there is a unique f : A ’ Z and a unique
g : Z ’ B. The zero morphism from A to B, denoted by 0AB , is de¬ned as gf , and it
is independent of the particular zero object chosen (Problem 3). Note that since a zero
morphism goes through a zero object, it follows that for an arbitrary morphism h, we
have h0 = 0h = 0.

10.1.5 Kernels and Cokernels
If f : A ’ B is an R-module homomorphism, then its kernel is, as we know, {x ∈
A : f (x) = 0}. The cokernel of f is de¬ned as the quotient group B/ im(f ). Thus f
is injective i¬ its kernel is 0, and f is surjective i¬ its cokernel is 0. We will generalize
these notions to an arbitrary category that contains zero objects. The following diagram
indicates the setup for kernels.

GA f GB
i
C –d y ˜b
dd
˜˜
dd g ˜
hd d ˜˜˜ 0
D

We take C to be the kernel of the module homomorphism f , with i the inclusion map. If
f g = 0, then the image of g is contained in the kernel of f , so that g actually maps into C.
Thus there is a unique module homomorphism h : D ’ C such that g = ih; simply take
h(x) = g(x) for all x. The key to the generalization is to think of the kernel as the
morphism i. This is reasonable because C and i essentially encode the same information.
Thus a kernel of the morphism f : A ’ B is a morphism i : C ’ A such that:

(1) f i = 0.
4 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

(2) If g : D ’ A and f g = 0, then there is a unique morphism h : D ’ C such that
g = ih.

Thus any map killed by f can be factored through i.
If we reverse all the arrows in the above diagram and change labels for convenience,
we get an appropriate diagram for cokernels.

GB p GC
f
Ad
dd ˜
˜˜
dd g ˜
0d d2  ˜˜˜˜ h
D

We take p to be the canonical map of B onto the cokernel of f , so that C = B/ im(f ). If
gf = 0, then the image of f is contained in the kernel of g, so by the factor theorem, there
is a unique homomorphism h such that g = hp. In general, a cokernel of a morphism
f : A ’ B is a morphism p : B ’ C such that:

(1 ) pf = 0.
(2 ) If g : B ’ D and gf = 0, then there is a unique morphism h : C ’ D such that
g = hp.

Thus any map that kills f can be factored through p.
Since going from kernels to cokernels simply involves reversing arrows, kernels and
cokernels are dual. Note, however, that in an arbitrary category with 0, kernels and
cokernels need not exist for arbitrary morphisms. But every monic has a kernel and (by
duality) every epic has a cokernel; see Problem 5.

Problems For Section 10.1
1. Show that in any category, the identity and inverse are unique.
2. In the category of rings, the inclusion map i : Z ’ Q is not surjective. Show, however,
that i is epic.
3. Show that the zero morphism 0AB is independent of the particular zero object chosen
in the de¬nition.
4. Show that a kernel must be monic (and by duality, a cokernel must be epic). [In
the de¬nition of kernel, we can assume that i is monic in (1) of (10.1.5), and drop
the uniqueness assumption on h. For i monic forces uniqueness of h, by de¬nition of
monic. Conversely, uniqueness of h forces i to be monic, by Problem 4.]
5. Show that in a category with 0, every monic has a kernel and every epic has a cokernel.
6. Show that if i : C ’ A and j : D ’ A are kernels of f : A ’ B, then C and D are
isomorphic. (By duality, a similar statement holds for cokernels.)
7. Let f : A ’ B be a group homomorphism with kernel K, and assume f not injective,
so that K = {1}. Let g be the inclusion map of K into A. Find a homomorphism h
such that f g = f h but g = h.
10.2. PRODUCTS AND COPRODUCTS 5

8. It follows from Problem 7 that in the category of groups, f monic is equivalent to f
injective as a mapping of sets, and a similar proof works in the category of modules.
Why does the argument fail in the category of rings?
9. Continue from Problem 8 and give a proof that does work in the category of rings.
10. Let f : M ’ N be a module homomorphism with nonzero cokernel, so that f is not
surjective. Show that f is not epic; it follows that epic is equivalent to surjective in
the category of modules.


10.2 Products and Coproducts
We have studied the direct product of groups, rings, and modules. It is natural to try
to generalize the idea to an arbitrary category, and a pro¬table approach is to forget
(temporarily) the algebraic structure and just look at the cartesian product A = i Ai
of a family of sets Ai , i ∈ I. The key property of a product is that if we are given maps
fi from a set S into the factors Ai , we can lift the fi into a single map f : S ’ i Ai .
The commutative diagram below will explain the terminology.

Ai
˜c y
˜
fi ˜
˜ (1)
pi
˜˜
˜
S f GA

In the picture, pi is the projection of A onto the ith factor Ai . If fi (x) = ai , i ∈ I, we
take f (x) = (ai , i ∈ I). It follows that pi —¦ f = fi for all i; this is what we mean by lifting
the fi to f . (Notice that there is only one possible lifting, i.e., f is unique.) If A is the
direct product of groups Ai and the fi are group homomorphisms, then f will also be a
group homomorphism. Similar statements can be made for rings and modules. We can
now give a generalization to an arbitrary category.

10.2.1 De¬nition
A product of objects Ai in a category C is an object A, along with morphisms pi : A ’ Ai ,
with the following universal mapping property. Given any object S of C and morphisms
fi : S ’ Ai , there is a unique morphism f : S ’ A such that pi f = fi for all i.
In a de¬nition via a universal mapping property, we use a condition involving mor-
phisms, along with a uniqueness statement, to specify an object and morphisms associated
with that object. We have already seen this idea in connection with kernels and cokernels
in the previous section, and in the construction of the tensor product in Section 8.7.
Not every category has products (see Problems 1 and 2), but if they do exist, they are
essentially unique. (The technique for proving uniqueness is also essentially unique.)

10.2.2 Proposition
If (A, pi , i ∈ I) and (B, qi , i ∈ I) are products of the objects Ai , then A and B are
isomorphic.
6 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

Proof. We use the above diagram (1) with S = B and fi = qi to get a morphism f : B ’ A
such that pi f = qi for all i. We use the diagram with S = A, A replaced by B, pi replaced
by qi , and fi = pi , to get a morphism h : A ’ B such that qi h = pi . Thus

pi f h = q i h = p i and qi hf = pi f = qi .

But

pi 1A = pi and qi 1B = qi

and it follows from the uniqueness condition in (10.2.1) that f h = 1A and hf = 1B .
Formally, we are using the diagram two more times, once with S = A and fi = pi , and
once with S = B, A replaced by B, pi replaced by qi , and fi = qi . Thus A and B are
isomorphic. ™
The discussion of diagram (1) indicates that in the categories of groups, abelian groups,
rings, and R-modules, products coincide with direct products. But a category can have
products that have no connection with a cartesian product of sets; see Problems 1 and 2.
Also, in the category of torsion abelian groups (torsion means that every element has
¬nite order), products exist but do not coincide with direct products; see Problem 5.
The dual of a product is a coproduct, and to apply duality, all we need to do is reverse
all the arrows in (1). The following diagram results.

Mj
}
}
fj }
}} (2)
ij
}}
˜} 
o
NfM

We have changed the notation because it is now pro¬table to think about modules. Sup-
pose that M is the direct sum of the submodules Mj , and ij is the inclusion map of Mj
into M . If the fj are module homomorphisms out of the factors Mj and into a module
N , the fj can be lifted to a single map f . If xj1 ∈ Mj1 , . . . , xjr ∈ Mjr , we take

f (xj1 + · · · + xjr) = fj1 (xj1 ) + · · · + fjr (xjr ).

Lifting means that f —¦ij = fj for all j. We can now give the general de¬nition of coproduct.

10.2.3 De¬nition
A coproduct of objects Mj in a category C is an object M , along with morphisms
ij : Mj ’ M , with the following universal mapping property. Given any object N of C
and morphisms fj : Mj ’ N , there is a unique morphism f : M ’ N such that f ij = fj
for all j
Exactly as in (10.2.2), any two coproducts of a given collection objects are isomorphic.
The discussion of diagram (2) shows that in the category of R-modules, the coproduct
is the direct sum, which is isomorphic to the direct product if there are only ¬nitely many
factors. In the category of sets, the coproduct is the disjoint union. To explain what
10.2. PRODUCTS AND COPRODUCTS 7

this means, suppose we have sets Aj , j ∈ J. We can disjointize the Aj by replacing Aj
by Aj = {(x, j) : x ∈ Aj }. The coproduct is A = j∈J Aj , with morphisms ij : Aj ’ A
given by ij (aj ) = (aj , j). If for each j we have fj : Aj ’ B, we de¬ne f : A ’ B by
f (aj , j) = fj (aj ).
The coproduct in the category of groups will be considered in the exercises.


Problems For Section 10.2
1. Let S be a preordered set, that is, there is a re¬‚exive and transitive relation ¤ on S.
Then S can be regarded as a category whose objects are the elements of S. If x ¤ y,
there is a unique morphism from x to y, and if x ¤ y, there are no morphisms from
x to y. Re¬‚exivity implies that there is an identity morphism on x, and transitivity
implies that associativity holds. Show that a product of the objects xi , if it exists,
must be a greatest lower bound of the xi . The greatest lower bound will be unique
(not just essentially unique) if S is a partially ordered set, so that ¤ is antisymmetric.
2. Continuing Problem 1, do products always exist?
3. Continuing Problem 2, what can be said about coproducts?
4. If A is an abelian group, let T (A) be the set of torsion elements of A. Show that T (A)
is a subgroup of A.
5. Show that in the category of torsion abelian groups, the product of groups Ai is
T ( Ai ), the subgroup of torsion elements of the direct product.
6. Assume that we have a collection of groups Gi , pairwise disjoint except for a common
identity 1. The free product of the Gi (notation —i Gi ) consists of all words (¬nite
sequences) a1 · · · an where the aj belong to distinct groups. Multiplication is by con-
catenation with cancellation. For example, with the subscript j indicating membership
in Gj ,

(a1 a2 a3 a4 )(b4 b2 b6 b1 b3 ) = a1 a2 a3 (a4 b4 )b2 b6 b1 b3

and if b4 = a’1 , this becomes a1 a2 a3 b2 b6 b1 b3 . The empty word is the identity, and
4
inverses are calculated in the usual way, as with free groups (Section 5.8). In fact a
free group on S is a free product of in¬nite cyclic groups, one for each element of S.
Show that in the category of groups, the coproduct of the Gi is the free product.
7. Suppose that products exist in the category of ¬nite cyclic groups, and suppose that
the cyclic group C with generator a is the product of the cyclic groups C1 and C2 with
generators a1 and a2 respectively. Show that the projections p1 and p2 associated with
the product of C1 and C2 are surjective.
8. By Problem 7, we may assume without loss of generality that pi (a) = ai , i = 1, 2.
Show that for some positive integer n, na1 = a1 and na2 = 0. [Take f1 : C1 ’ C1 to
be the identity map, and let f2 : C1 ’ C2 be the zero map (using additive notation).
Lift f1 and f2 to f : C1 ’ C.]
9. Exhibit groups C1 and C2 that can have no product in the category of ¬nite cyclic
groups.
8 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

10.3 Functors
We will introduce this fundamental concept with a concrete example. Let HomR (M, N )
be the set of R-module homomorphisms from M to N . As pointed out at the beginning
of Section 4.4, HomR (M, N ) is an abelian group. It will also be an R-module if R is a
commutative ring, but not in general. We are going to look at HomR (M, N ) as a function
of N , with M ¬xed.

10.3.1 The Functor HomR (M, )
We are going to construct a mapping from the category of R-modules to the category of
abelian groups. Since a category consists of both objects and morphisms, our map will
have two parts:
(i) Associate with each R-module N the abelian group HomR (M, N ).
(ii) Associate with each R-module homomorphism h : N ’ P a homomorphism h— from
the abelian group HomR (M, N ) to the abelian group HomR (M, P ). The following
diagram suggests how h— should be de¬ned.

GN GP
f h
M
Take

h— (f ) = hf.

Note that if h is the identity on N , then h— is the identity on HomR (M, N ).
Now suppose we have the following situation:

GN GP GQ
f g h
M

Then (hg)— (f ) = (hg)f = h(gf ) = h— (g— (f )), so that

(hg)— = h— g— (= h— —¦ g— ).

To summarize, we have a mapping F called a functor that takes an object A in a category C
to an object F (A) in a category D; F also takes a morphism h : A ’ B in C to a morphism
h— = F (h) : F (A) ’ F (B) in D. The key feature of F is the functorial property:

F (hg) = F (h)F (g) and F (1A ) = 1F (A) .

Thus a functor may be regarded as a homomorphism of categories.

10.3.2 The Functor HomR ( , N )
We now look at HomR (M, N ) as a function of M , with N ¬xed. Here is an appropriate
diagram:

GL GM GN
g f
h
K
10.3. FUNCTORS 9

If M is an R-module, we take F (M ) to be the abelian group HomR (M, N ). If h : L ’ M
is an R-module homomorphism, we take h— = F (h) to be a homomorphism from the
abelian group HomR (M, N ) to the abelian group HomR (L, N ), given by

h— (f ) = f h.

It follows that

(hg)— (f ) = f (hg) = (f h)g = g — (f h) = g — (h— (f ))

hence

(hg)— = g — h— ,

and if h is the identity on M , then h— is the identity on HomR (M, N ).
Thus F does not quite obey the functorial property; we have F (hg) = F (g)F (h)
instead of F (hg) = F (h)F (g). However, F is a legal functor on the opposite category
of R-mod. In the literature, HomR ( , N ) is frequently referred to as a contravariant
functor on the original category R-mod, and HomR (M, ) as a covariant functor on
R-mod.
If we replace the category of R-modules by an arbitrary category, we can still de¬ne
functors (called hom functors) as in (10.3.1) and (10.3.2). But we must replace the
category of abelian groups by the category of sets.


The Functors M —R —R N
10.3.3 and
To avoid technical complications, we consider tensor products of modules over a commu-
tative ring R. First we discuss the tensor functor T = M —R The relevant diagram is
given below.

GP GQ
g f
N

If N is an R-module, we take T (N ) = M —R N . If g : N ’ P is an R-module homomor-
phism, we set T (g) = 1M — g : M —R N ’ M —R P , where 1M is the identity mapping
on M [Recall that (1M — g)(x — y) = x — g(y).] Then

T (f g) = 1M — f g = (1M — f )(1M — g) = T (f )T (g)

and

T (1N ) = 1T (N )

so T is a functor from R-mod to R-mod.
The functor S = —R N is de¬ned in a symmetrical way. If M is an R-module, then
S(M ) = M —R N , and if f : L ’ M is an R-module homomorphism, then S(f ) : L—R N ’
M —R N is given by S(f ) = f — 1N .
10 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

10.3.4 Natural Transformations
Again we will introduce this idea with an explicit example. The diagram below summarizes
the data.

G G(A)
tA
F (A)
(1)
Ff Gf
 
G G(B)
F (B) tB


We start with abelian groups A and B and a homomorphism f : A ’ B. We apply
the forgetful functor, also called the underlying functor U. This is a fancy way of saying
that we forget the algebraic structure and regard A and B simply as sets, and f as
a mapping between sets. Now we apply the free abelian group functor F to produce
F (A) = FU(A), the free abelian group with A as basis (and similarly for F (B)). Thus
F (A) is the direct sum of copies of Z, one copy for each a ∈ A. The elements of F (A)
can be represented as a n(a)x(a), n(a) ∈ Z, where x(a) is the member of the direct
sum that is 1 in the ath position and 0 elsewhere. [Similarly, we represent elements of B
as b n(b)y(b).] The mapping f determines a homomorphism F f : F (A) ’ F (B), via
n(a)x(a) ’ n(a)y(f (a)).
Now let G be the identity functor, so that G(A) = A, G(B) = B, Gf = f . We de¬ne
an abelian group homomorphism tA : F (A) ’ A by tA ( n(a)x(a)) = n(a)a, and
similarly we de¬ne tB ( n(b)y(b)) = n(b)b. (Remember that we began with abelian
groups A and B.) The diagram (1) is then commutative, because

f tA (x(a)) = f (a) and tB [(F f )(x(a))] = tB (y(f (a)) = f (a).

To summarize, we have two functors, F and G from the category C to the category D.
(In this case, C = D = the category of abelian groups.) For all objects A, B ∈ C and
morphisms f : A ’ B, we have morphisms tA : F (A) ’ G(A) and tB : F (B) ’ G(B)
such that the diagram (1) is commutative. We say that t is a natural transformation from
F to G. If for every object C ∈ C, tC is an isomorphism (not the case in this example), t
is said to be a natural equivalence.
The key intuitive point is that the process of going from F (A) to G(A) is “natural”
in the sense that as we move from an object A to an object B, the essential features of
the process remains the same.

Problems For Section 10.3
1. Let F : S ’ T , where S and T are preordered sets. If we regard S and T as categories,
as in Section 10.2, Problem 1, what property must F have in order to be a functor?
2. A group may be regarded as a category with a single object 0, with a morphism for
each element g ∈ G. The composition of two morphisms is the morphism associated
with the product of the elements. If F : G ’ H is a function from a group G to a
group H, and we regard G and H as categories, what property must F have in order
to be a functor?
10.3. FUNCTORS 11

3. We now look at one of the examples that provided the original motivation for the
concept of a natural transformation. We work in the category of vector spaces (over a
given ¬eld) and linear transformations. If V is a vector space, let V — be the dual space,
that is, the space of linear maps from V to the ¬eld of scalars, and let V —— be the dual
of V — . If v ∈ V , let v ∈ V —— be de¬ned by v(f ) = f (v), f ∈ V — . The mapping from v
to v is a linear transformation, and in fact an isomorphism if V is ¬nite-dimensional.
Now suppose that f : V ’ W and g : W ’ X are linear transformations. De¬ne
f — : W — ’ V — by f — (±) = ±f , ± ∈ W — . Show that (gf )— = f — g — .
4. The double dual functor takes a vector space V into its double dual V —— , and takes a
linear transformation f : V ’ W to f —— : V —— ’ W —— , where f —— (v —— ) = v —— f — . Show
that the double dual functor is indeed a functor.
5. Now consider the following diagram.

G V ——
tV
V
f ——
f


G W ——
W tW


We take tV (v) = v, and similarly for tW . Show that the diagram is commutative, so
that t is a natural transformation from the identity functor to the double dual functor.
In the ¬nite-dimensional case, we say that there is a natural isomorphism between a
vector space and its double dual. “Natural” means coordinate-free in the sense that it
is not necessary to choose a speci¬c basis. In contrast, the isomorphism of V and its
single dual V — is not natural.
6. We say that D is a subcategory of C if the objects of D are also objects of C, and
similarly for morphisms (and composition of morphisms). The subcategory D is full if
every C-morphism f : A ’ B, where A and B are objects of D (the key point) is also
a D-morphism. Show that the category of groups is a full subcategory of the category
of monoids.
7. A functor F : C ’ D induces a map from C-morphisms to D-morphisms; f : A ’ B
is mapped to F f : F A ’ F B. If this map is injective for all objects A, B of C, we
say that F is faithful. If the map is surjective for all objects A, B of C, we say that F
is full.

(a) The forgetful functor from groups to sets assigns to each group its underlying
set, and to each group homomorphism its associated map of sets. Is the forgetful
functor faithful? full?
(b) We can form the product C — D of two arbitrary categories; objects in the product
are pairs (A, A ) of objects, with A ∈ C and A ∈ D. A morphism from (A, A ) to
(B, B ) is a pair (f, g), where f : A ’ B and g : A ’ B . The projection functor
from C — D to C takes (A, A ) to A and (f, g) to f . Is the projection functor
faithful? full?
12 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

10.4 Exact Functors
10.4.1 De¬nitions and Comments
We are going to investigate the behavior of the hom and tensor functors when presented
with an exact sequence. We will be working in the categories of modules and abelian
groups, but exactness properties can be studied in the more general setting of abelian
categories, which we now describe very informally.
In any category C, let HomC (A, B) (called a “hom set”) be the set of morphisms in C
from A to B. [As remarked in (10.1.1), the formal de¬nition of a category requires that
HomC (A, B) be a set for all objects A and B. The collection of all objects of C is a class
but need not be a set.] For C to be an abelian category, the following conditions must be
satis¬ed.

1. Each hom set is an abelian group.
2. The distributive laws f (g + h) = f g + f h, (f + g)h = f h + gh hold.
3. C has a zero object.
4. Every ¬nite set of objects has a product and a coproduct. (The existence of ¬nite
coproducts can be deduced from the existence of ¬nite products, along with the re-
quirements listed so far.)
5. Every morphism has a kernel and a cokernel.
6. Every monic is the kernel of its cokernel.
7. Every epic is the cokernel of its kernel.
8. Every morphism can be factored as an epic followed by a monic.

Exactness of functors can be formalized in an abelian category, but we are going to
return to familiar ground by assuming that each category that we encounter is R-mod
for some R. When R = Z, we have the category of abelian groups.

10.4.2 Left Exactness of HomR (M, )
Suppose that we have a short exact sequence

GA GB GC G0
f g
(1)
0

We apply the covariant hom functor F = HomR (M, ) to the sequence, dropping the last
term on the right. We will show that the sequence

G FA G FB G FC
Ff Fg
(2)
0

is exact. A functor that behaves in this manner is said to be left exact.
We must show that the transformed sequence is exact at F A and F B. We do this in
three steps.

(a) F f is monic.
10.4. EXACT FUNCTORS 13

Suppose that (F f )(±) = f ± = 0. Since f is monic (by exactness of the sequence (1)),
± = 0 and the result follows.
(b) im F f ⊆ ker F g.
If β ∈ im F f , then β = f ± for some ± ∈ HomR (M, A). By exactness of (1), im f ⊆
ker g, so gβ = gf ± = 0± = 0. Thus β ∈ ker g.
(c) ker F g ⊆ im F f .
If β ∈ ker F g, then gβ = 0, with β ∈ HomR (M, B). Thus if y ∈ M , then β(y) ∈
ker g = im f , so β(y) = f (x) for some x = ±(y) ∈ A. Note that x is unique since f is
monic, and ± ∈ HomR (M, A). Thus β = f ± ∈ im F f . ™

10.4.3 Left Exactness of HomR ( , N )
The contravariant hom functor G = HomR ( , N ) is a functor on the opposite category, so
before applying it to the sequence (1), we must reverse all the arrows. Thus left-exactness
of G means that the sequence

G GC G GB G GA
Gg Gf
(3)
0
is exact. Again we have three steps.
(a) Gg is monic.
If (Gg)± = ±g = 0, then ± = 0 since g is epic.
(b) im Gg ⊆ ker Gf .
If β ∈ im Gg, then β = ±g for some ± ∈ HomR (C, N ). Thus (Gf )β = βf = ±gf = 0,
so β ∈ ker Gf .
(c) ker Gf ⊆ im Gg.
Let β ∈ HomR (B, N ) with β ∈ ker Gf , that is, βf = 0. If y ∈ C, then since g is epic,
we have y = g(x) for some x ∈ B. If g(x1 ) = g(x2 ), then x1 ’ x2 ∈ ker g = im f , hence
x1 ’ x2 = f (z) for some z ∈ A. Therefore β(x1 ) ’ β(x2 ) = β(f (z)) = 0, so it makes sense
to de¬ne ±(y) = β(x). Then ± ∈ HomR (C, N ) and ±g = β, that is, (Gg)± = β. ™

—R N
10.4.4 Right Exactness of the Functors M—R and
If we apply the functor H = M —R to the exact sequence

GA GB GC G0
f g
0
[see (1) of (10.4.2)], we will show that the sequence

G HB G HC G0
Hf Hg
(4)
HA
is exact. A similar result holds for —R N . A functor that behaves in this way is said to
be right exact. Once again, there are three items to prove.
14 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

(i) Hg is epic.

An element of M — C is of the form t = i xi — yi with xi ∈ M and yi ∈ C.
Since g is epic, there exists zi ∈ B such that g(zi ) = yi . Thus (1 — g)( i xi — zi ) =
i xi — g(zi ) = t.

(ii) im Hf ⊆ ker Hg.

This is a brief computation: (1 — g)(1 — f ) = 1 — gf = 1 — 0 = 0.

(iii) ker Hg ⊆ im Hf .

By (ii), the kernel of 1 — g contains L = im(1 — f ), so by the factor theorem, there is a
homomorphism g : (M —R B)/L ’ M —R C such that g(m — b + L) = m — g(b), m ∈ M ,
b ∈ B.
Let π be the canonical map of M —R B onto (M —R B)/L. Then gπ(m — b) =
g(m — b + L) = m — g(b), so

gπ = 1 — g.

If we can show that g is an isomorphism, then

ker(1 — g) = ker(gπ) = ker π = L = im(1 — f )

and we are ¬nished. To show that g is an isomorphism, we will display its inverse. First
let h be the bilinear map from M — C to (M —R B)/L given by h(m, c) = m — b + L, where
g(b) = c. [Such a b exists because g is epic. If g(b) = g(b ) = c, then b ’ b ∈ ker g = im f ,
so b ’ b = f (a) for some a ∈ A. Then m — b ’ m — b = m — f (a) = (1 — f )(m — a) ∈ L,
and h is well-de¬ned.] By the universal mapping property of the tensor product, there is
a homomorphism h : M —R C ’ (M —R B)/L such that

h(m — c) = h(m, c) = m — b + L, where g(b) = c.

But g : (M —R B)/L ’ M —R C and

g(m — b + L) = m — g(b) = m — c.

Thus h is the inverse of g. ™


10.4.5 De¬nition
A functor that is both left and right exact is said to be exact. Thus an exact functor is
one that maps exact sequences to exact sequences. We have already seen one example,
the localization functor (Section 8.5, Problems 4 and 5).
If we ask under what conditions the hom and tensor functors become exact, we are
led to the study of projective, injective and ¬‚at modules, to be considered later in the
chapter.
10.5. PROJECTIVE MODULES 15

Problems For Section 10.4
In Problems 1“3, we consider the exact sequence (1) of (10.4.2) with R = Z, so that we
are in the category of abelian groups. Take A = Z, B = Q, the additive group of rational
numbers, and C = Q/Z, the additive group of rationals mod 1. Let f be inclusion, and g
the canonical map. We apply the functor F = HomR (M, ) with M = Z2 . [We will omit
the subscript R when R = Z, and simply refer to Hom(M, ).]
1. Show that Hom(Z2 , Q) = 0.
2. Show that Hom(Z2 , Q/Z) = 0.
3. Show that Hom(Z2 , ) is not right exact.
In Problems 4 and 5, we apply the functor G = Hom( , N ) to the above exact
sequence, with N = Z.
4. Show that Hom(Q, Z) = 0.
5. Show that Hom( , Z) is not right exact.
Finally, in Problem 6 we apply the functor H = M — to the above exact sequence,
with M = Z2 .
6. Show that Z2 — (and similarly — Z2 ) is not left exact.
7. Refer to the sequences (1) and (2) of (10.4.2). If (2) is exact for all possible R-modules
M , show that (1) is exact.
8. State an analogous result for the sequence (3) of (10.4.3), and indicate how the result
is proved.


10.5 Projective Modules
Projective modules are direct summands of free modules, and are therefore images of
natural projections. Free modules are projective, and projective modules are sometimes
but not always free. There are many equivalent ways to describe projective modules, and
we must choose one of them as the de¬nition. In the diagram below and the de¬nition
to follow, all maps are R-module homomorphisms. The bottom row is exact, that is, g is
surjective.
P
}}
}f
h}
}}
˜}} 
GN G0
M g


10.5.1 De¬nition
The R-module P is projective if given f : P ’ N , and g : M ’ N surjective, there
exists h : P ’ M (not necessarily unique) such that the diagram is commutative, that is,
f = gh. We sometimes say that we have “lifted” f to h.
The de¬nition may look obscure, but the condition described is a familiar property of
free modules.
16 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

10.5.2 Proposition
Every free module is projective.
Proof. Let S be a basis for the free module P . By (4.3.6), f is determined by its behavior
on basis elements s ∈ S. Since g is surjective, there exists a ∈ M such that g(a) = f (s).
Take h(s) = a and extend by linearity from S to P . Since f = gh on S, the same must
be true on all of P . ™
Here is the list of equivalences.

10.5.3 Theorem
The following conditions on the R-module P are equivalent.
(1) P is projective.
(2) The functor HomR (P, ) is exact.
Every short exact sequence 0 ’ M ’ N ’ P ’ 0 splits.
(3)
(4) P is a direct summand of a free module.
Proof. (1) is equivalent to (2). In view of the left exactness of F = HomR (P, ) (see
(10.4.2)), (2) says that if g : M ’ N is surjective, then so is F g : F M ’ F N . But F g
maps h : P ’ M to gh : P ’ N , so what we must prove is that for an arbitrary morphism
f : P ’ N , there exists h : P ’ M such that gh = f . This is precisely the de¬nition of
projectivity of P .
(2) implies (3). Let 0 ’ M ’ N ’ P ’ 0 be a short exact sequence, with g : N ’ P
(necessarily surjective). Since P is projective, we have the following diagram.

P
}}
}1
h}
}}
˜}} 
GP G0
M g

Thus there exists h : P ’ N such that gh = 1P , which means that the exact sequence
splits (see (4.7.1)).
(3) implies (4). By (4.3.6), P is a quotient of a free module, so there is an exact
sequence 0 ’ M ’ N ’ P ’ 0 with N free. By (3), the sequence splits, so by (4.7.4),
P is a direct summand of N .
(4) implies (1). Let P be a direct summand of the free module F , and let π be the
natural projection of F on P ; see the diagram below.
GP
π
F
f
h
 
GN G0
M g

Given f : P ’ N , we have f π : F ’ N , so by (10.5.2) there exists h : F ’ M such that
f π = gh. If h is the restriction of h to P , then f = gh . ™
10.5. PROJECTIVE MODULES 17

10.5.4 Corollary
The direct sum P = •Pj is projective if and only if each Pj is projective.

Proof. If P is a direct summand of a free module, so is each Pj , and therefore the Pj
are projective by (4) of (10.5.3). Conversely, assume that each Pj is projective. Let
f : P ’ N and g : M ’ N , with g surjective. If ij is the inclusion map of Pj into P ,
then f ij : Pj ’ N can be lifted to hj : Pj ’ M such that f ij = ghj . By the universal
mapping property of direct sum (Section 10.2), there is a morphism h : P ’ M such that
hij = hj for all j. Thus f ij = ghij for every j, and it follows from the uniqueness part of
the universal mapping property that f = gh. ™

If we are searching for projective modules that are not free, the following result tells
us where not to look.

10.5.5 Theorem
A module M over a principal ideal domain R is projective if and only if it is free.

Proof. By (10.5.2), free implies projective. If M is projective, then by (4) of (10.5.3), M
is a direct summand of a free module. In particular, M is a submodule of a free module,
hence is free by (4.6.2) and the discussion following it. ™

10.5.6 Examples
1. A vector space over a ¬eld k is a free k-module, hence is projective.
2. A ¬nite abelian group G is not a projective Z-module, because it is not free. [If g ∈ G
and n = |G|, then ng = 0, so g can never be part of a basis.]
3. If p and q are distinct primes, then R = Zpq = Zp • Zq . We claim that Zp and Zq
are projective but not free R-modules. (As in Example 2, they are not projective
Z-modules.) This follows from (4) of (10.5.3) and the fact that any ring R is a free
R-module (with basis {1}).

Problems For Section 10.5
In Problems 1“5, we are going to prove the projective basis lemma, which states that an R-
module P is projective if and only if there are elements xi ∈ P (i ∈ I) and homomorphisms
fi : P ’ R such that for every x ∈ P , fi (x) = 0 for all but ¬nitely many i and

x= fi (x)xi .
i

The set of xi ™s is referred to as the projective basis.

1. To prove the “only if” part, let P be a direct summand of the free module F with basis
{ei }. Take f to be the inclusion map of P into F , and π the natural projection of F
onto P . Show how to de¬ne the fi and xi so that the desired results are obtained.
18 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

2. To prove the “if” part, let F be a free module with basis {ei , i ∈ I}, and de¬ne
π : F ’ P by π(ei ) = xi . De¬ne f : P ’ F by f (x) = i fi (x)ei . Show that πf is the
identity on P .
3. Continuing Problem 2, show that P is projective.
4. Assume that P is ¬nitely generated by n elements. If Rn is the direct sum of n copies
of R, show that P is projective i¬ P is a direct summand of Rn .
5. Continuing Problem 4, show that if P is projective and generated by n elements, then
P has a projective basis with n elements.
6. Show that a module P is projective i¬ P is a direct summand of every module of
which it is a quotient. In other words, if P ∼ M/N , then P is isomorphic to a direct
=
summand of M .
7. In the de¬nition (10.5.1) of a projective module, give an explicit example to show that
the mapping h need not be unique.


10.6 Injective Modules
If we reverse all arrows in the mapping diagram that de¬nes a projective module, we
obtain the dual notion, an injective module. In the diagram below, the top row is exact,
that is, f is injective.

GN GM
f
0
}
}}
}}
g
}
 ˜}} h
E

10.6.1 De¬nition
The R-module E is injective if given g : N ’ E, and f : N ’ M injective, there exists
h : M ’ E (not necessarily unique) such that g = hf . We sometimes say that we have
“lifted” g to h.
As with projectives, there are several equivalent ways to characterize an injective
module.

10.6.2 Theorem
The following conditions on the R-module E are equivalent.

(1) E is injective.
(2) The functor HomR ( , E) is exact.
(3) Every exact sequence 0 ’ E ’ M ’ N ’ 0 splits.

Proof. (1) is equivalent to (2). Refer to (3) of (10.4.3), (1) of (10.4.2) and the de¬nition
of the contravariant hom functor in (10.3.2) to see what (2) says. We are to show that
if f : N ’ M is injective, then f — : HomR (M, E) ’ HomR (N, E) is surjective. But
10.6. INJECTIVE MODULES 19

f — (h) = hf , so given g : N ’ E, we must produce h : M ’ E such that g = hf . This is
precisely the de¬nition of injectivity.
(2) implies (3). Let 0 ’ E ’ M ’ N ’ 0 be a short exact sequence, with f : E ’ M
(necessarily injective). Since E is an injective module, we have the following diagram:

GE GM
f
0
}}
}}
}} g
1
 ˜}}
E
Thus there exists g : M ’ E such that gf = 1E , which means that the exact sequence
splits.
(3) implies (1). Given g : N ’ E, and f : N ’ M injective, we form the pushout of f
and g, which is a commutative square as indicated in the diagram below.

GM
f
N
g g


GQ
E
f

Detailed properties of pushouts are developed in the exercises. For the present proof, all
we need to know is that since f is injective, so is f . Thus the sequence

GE GQ G Q/ im f G0
f
0

is exact. By (3), there exists h : Q ’ E such that hf = 1E . We now have hg : M ’ E
with hg f = hf g = 1E g = g, proving that E is injective. ™
We proved in (10.5.4) that a direct sum of modules is projective i¬ each component
is projective. The dual result holds for injectives.

10.6.3 Proposition
A direct product j Ej of modules is injective i¬ each Ej is injective. Consequently, a
¬nite direct sum is injective i¬ each summand is injective.
Proof. If f : N ’ M is injective and g : N ’ i Ei , let gi = pi g, where pi is the projection
of the direct product on Ei . Then ¬nding h : M ’ i Ei such that g = hf is equivalent
to ¬nding, for each i, a morphism hi : M ’ Ei such that gi = hi f . [If pi g = hi f = pi hf
for every i, then g = hf by the uniqueness part of the universal mapping property for
products.] The last assertion holds because the direct sum of ¬nitely many modules
coincides with the direct product. ™
In checking whether an R-module E is injective, we are given g : N ’ E, and
f : N ’ M , with f injective, and we must lift g to h : M ’ E with g = hf . The
next result drastically reduces the collection of maps f and g that must be examined. We
may take M = R and restrict N to a left ideal I of R, with f the inclusion map.
20 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

10.6.4 Baer™s Criterion
The R-module E is injective if and only if every R-homomorphism f : I ’ E, where I is
a left ideal of R, can be extended to an R-homomorphism h : R ’ E.

Proof. The “only if” part follows from the above discussion, so assume that we are given
g : N ’ E and f : N ’ M , where (without loss of generality) f is an inclusion map.
We must extend g to h : M ’ E. A standard Zorn™s lemma argument yields a maximal
extension g0 in the sense that the domain M0 of g0 cannot be enlarged. [The partial
ordering is (g1 , D1 ) ¤ (g2 , D2 ) i¬ D1 ⊆ D2 and g1 = g2 on D1 .] If M0 = M , we are
¬nished, so assume x ∈ M \ M0 . Let I be the left ideal {r ∈ R : rx ∈ M0 }, and de¬ne
h0 : I ’ E by h0 (r) = g0 (rx). By hypothesis, h0 can be extended to h0 : R ’ E. Let
M1 = M0 + Rx and de¬ne h1 : M1 ’ E by

h1 (x0 + rx) = g0 (x0 ) + rh0 (1).

To show that h1 is well de¬ned, assume x0 + rx = y0 + sx, with x0 , y0 ∈ M0 and r, s ∈ R.
Then (r ’ s)x = y0 ’ x0 ∈ M0 , so r ’ s ∈ I. Using the fact that h0 extends h0 , we have

g0 (y0 ’ x0 ) = g0 ((r ’ s)x) = h0 (r ’ s) = h0 (r ’ s) = (r ’ s)h0 (1)

and consequently, g0 (x0 ) + rh0 (1) = g0 (y0 ) + sh0 (1) and h1 is well de¬ned. If x0 ∈ M0 ,
take r = 0 to get h1 (x0 ) = g0 (x0 ), so h1 is an extension of g0 to M1 ⊃ M0 , contradicting
the maximality of g0 . We conclude that M0 = M . ™

Since free modules are projective, we can immediately produce many examples of
projective modules. The primary source of injective modules lies a bit below the surface.


10.6.5 De¬nitions and Comments
Let R be an integral domain. The R-module M is divisible if each y ∈ M can be divided
by any nonzero element r ∈ R, that is, there exists x ∈ M such that rx = y. For
example, the additive group of rational numbers is a divisible abelian group, as is Q/Z,
the rationals mod 1. The quotient ¬eld of any integral domain (regarded as an abelian
group) is divisible. A cyclic group of ¬nite order n > 1 can never be divisible, since it is
not possible to divide by n. The group of integers Z is not divisible since the only possible
divisors of an arbitrary integer are ±1. It follows that a nontrivial ¬nitely generated
abelian group, a direct sum of cyclic groups by (4.6.3), is not divisible.
It follows from the de¬nition that a homomorphic image of a divisible module is
divisible, hence a quotient or a direct summand of a divisible module is divisible. Also, a
direct sum of modules is divisible i¬ each component is divisible.


10.6.6 Proposition
If R is any integral domain, then an injective R-module is divisible. If R is a PID, then
an R-module is injective if and only if it is divisible.
10.6. INJECTIVE MODULES 21

Proof. Assume E is injective, and let y ∈ E, r ∈ R, r = 0. Let I be the ideal Rr,
and de¬ne an R-homomorphism f : I ’ E by f (tr) = ty. If tr = 0, then since R is
an integral domain, t = 0 and f is well de¬ned. By (10.6.4), f has an extension to an
R-homomorphism h : R ’ E. Thus

y = f (r) = h(r) = h(r1) = rh(1)

so division by r is possible and E is divisible. Conversely, assume that R is a PID and
E is divisible. Let f : I ’ E, where I is an ideal of R. Since R is a PID, I = Rr for
some r ∈ R. We have no trouble extending the zero mapping, so assume r = 0. Since E
is divisible, there exists x ∈ E such that rx = f (r). De¬ne h : R ’ E by h(t) = tx. If
t ∈ R, then

h(tr) = trx = tf (r) = f (tr)

so h extends f , proving E injective. ™

Problems For Section 10.6
We now describe the construction of the pushout of two module homomorphisms f : A ’ C
and g : A ’ B; refer to Figure˜10.6.1. Take

D = (B • C)/W, where W = {(g(a), ’f (a)) : a ∈ A},

and

g (c) = (0, c) + W, f (b) = (b, 0) + W.

In Problems 1“6, we study the properties of this construction.

GC
f
A
g g
 
GD
f g
B dd
dd
d
hd d2 
CE
f


Figure 10.6.1


1. Show that the pushout square ACDB is commutative, that is, f g = g f .
2. Suppose we have another commutative pushout square ACEB with maps f : B ’ E
and g : C ’ E, as indicated in Figure 10.6.1. De¬ne h : D ’ E by

h((b, c) + W ) = g (c) + f (b).

Show that h is well de¬ned.
22 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

3. Show that h makes the diagram commutative, that is, hg = g and hf = f .
4. Show that if h : D ’ E makes the diagram commutative, then h = h.
The requirements stated in Problems 1, 3 and 4 can be used to de¬ne the pushout
via a universal mapping property. The technique of (10.2.2) shows that the pushout
object D is unique up to isomorphism.
5. If f is injective, show that f is also injective. By symmetry, the same is true for g
and g .
6. If f is surjective, show that f is surjective. By symmetry, the same is true for g
and g .
Problems 7“10 refer to the dual construction, the pullback, de¬ned as follows (see
Figure 10.6.2). Given f : A ’ B and g : C ’ B, take
D = {(a, c) ∈ A • C : f (a) = g(c)}
and
g (a, c) = a, f (a, c) = c.

Ed
dd g
dd
h
dd
2 4
GA
D
f g
f
f
) 
GB
C g

Figure 10.6.2

7. Show that the pullback square DABC is commutative, that is, f g = gf .
8. If we have another commutative pullback square EABC with maps f : E ’ C
and g : E ’ A, show that there is a unique h : E ’ D that makes the diagram
commutative, that is, g h = g and f h = f .
9. If f is injective, show that f is injective. By symmetry, the same is true for g and g .
10. If f is surjective, show that f is surjective. By symmetry, the same is true for g
and g .
11. Let R be an integral domain with quotient ¬eld Q, and let f be an R-homomorphism
from an ideal I of R to Q. Show that f (x)/x is constant for all nonzero x ∈ I.
12. Continuing Problem 11, show that Q is an injective R-module.


10.7 Embedding into an Injective Module
We know that every module is a quotient of a projective (in fact free) module. In this
section we prove the more di¬cult dual statement that every module can be embedded
in an injective module. (To see that quotients and submodules are dual, reverse all the
arrows in a short exact sequence.) First, we consider abelian groups.
10.7. EMBEDDING INTO AN INJECTIVE MODULE 23

10.7.1 Proposition
Every abelian group can be embedded in a divisible abelian group.
Proof. If A is an abelian group, then A is a quotient of a free abelian group F , say
A ∼ F/B. Now F is a direct sum of copies of Z, hence F can be embedded in a direct
=
sum D of copies of Q, the additive group of rationals. It follows that F/B can be embedded
in D/B; the embedding is just the inclusion map. By (10.6.5), D/B is divisible, and the
result follows. ™

10.7.2 Comments
In (10.7.1), we used Q as a standard divisible abelian group. It would be very desirable
to have a canonical injective R-module. First, we consider H = HomZ (R, A), the set of
all abelian group homomorphisms from the additive group of the ring R to the abelian
group A. If we are careful, we can make this set into a left R-module. The abelian group
structure of H presents no di¬culties, but we must also de¬ne scalar multiplication. If
f ∈ H and s ∈ R, we set

(sf )(r) = f (rs), r ∈ R.

Checking the module properties is routine except for associativity:

((ts)f )(r) = f (rts), and (t(sf ))(r) = (sf )(rt) = f (rts)

so (ts)f = t(sf ). Notice that sf is an abelian group homomorphism, not an R-module
homomorphism.
Now if A is an R-module and B an abelian group, we claim that

HomZ (A, B) ∼ HomR (A, HomZ (R, B)), (1)
=

equivalently,

HomZ (R —R A, B) ∼ HomR (A, HomZ (R, B)). (2)
=

This is a special case of adjoint associativity:
If S MR , R N , S P , then

HomS (M —R N, P ) ∼ HomR (N, HomS (M, P )). (3)
=

Thus if F is the functor M —R and G is the functor HomS (M, ), then

HomS (F N, P ) ∼ HomR (N, GP ) (4)
=

which is reminiscent of the adjoint of a linear operator on an inner product space. We
say that F and G are adjoint functors, with F left adjoint to G and G right adjoint
to F . [There is a technical naturality condition that is added to the de¬nition of adjoint
functors, but we will not pursue this since the only adjoints we will consider are hom and
tensor.]
24 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

Before giving the formal proof of (3), we will argue intuitively. The left side of the
equation describes all biadditive, R-balanced maps from M — N to P . If (x, y) ’ f (x, y),
x ∈ M , y ∈ N , is such a map, then f determines a map g from N to HomS (M, P ),
namely, g(y)(x) = f (x, y). This is a variation of the familiar fact that a bilinear map
amounts to a family of linear maps. Now if s ∈ S, then g(y)(sx) = f (sx, y), which need
not equal sf (x, y), but if we factor f through the tensor product M —R N , we can then
pull out the s. Thus g(y) ∈ HomS (M, P ). Moreover, g is an R-module homomorphism,
because g(ry)(x) = f (x, ry), and we can factor out the r by the same reasoning as above.
Since g determines f , the correspondence between f and g is an isomorphism of abelian
groups.
To prove (3), let f : M —R N ’ P be an S-homomorphism. If y ∈ N , de¬ne
fy : M ’ P by fy (x) = f (x — y), and de¬ne ψ(f ) : N ’ HomS (M, P ) by y ’ fy .
[HomS (M, P ) is a left R-module by Problem 1.]
(a) ψ(f ) is an R-homomorphism:
ψ(f )(y1 + y2 ) = fy1 +y2 = fy1 + fy2 = ψ(f )(y1 ) + ψ(f )(y2 );
ψ(f )(ry) = fry = rfy = (rψ(f ))(y).
[fry (x) = f (x — ry) and (rfy )(x) = fy (xr) = f (xr — y).]
(b) ψ is an abelian group homomorphism:
We have fy (x) + gy (x) = f (x — y) + g(x — y) = (f + g)(x — y) = (f + g)y (x) so
ψ(f + g) = ψ(f ) + ψ(g).
(c) ψ is injective:
If ψ(f ) = 0, then fy = 0 for all y ∈ N , so f (x — y) = 0 for all x ∈ M and y ∈ N .
Thus f is the zero map.
(d) If g ∈ HomR (N, HomS (M, P )), de¬ne •g : M — N ’ P by •g (x, y) = g(y)(x). Then
•g is biadditive and R-balanced:
By de¬nition, •g is additive in each coordinate, and [see Problem 1] •g (xr, y) =
g(y)(xr) = (rg(y))(x) = g(ry)(x) = •g (x, ry).
(e) ψ is surjective:
By (d), there is a unique S-homomorphism β(g) : M —R N ’ P such that β(g)(x—y) =
•g (x, y) = g(y)(x), x ∈ M , y ∈ N . It follows that ψ(β(g)) = g, because ψ(β(g))(y) =
β(g)y , where β(g)y (x) = β(g)(x — y) = g(y)(x). Thus β(g)y = g(y) for all y in N , so ψβ
is the identity and ψ is surjective.
It follows that ψ is an abelian group isomorphism. This completes the proof of (3).
Another adjointness result, which can be justi¬ed by similar reasoning, is that if NR ,
R MS , PS , then

HomS (N —R M, P ) ∼ HomR (N, HomS (M, P )) (5)
=

—R M and G = HomS (M,
which says that F = ) are adjoint functors.

10.7.3 Proposition
If E is a divisible abelian group, then HomZ (R, E) is an injective left R-module.
10.7. EMBEDDING INTO AN INJECTIVE MODULE 25

Proof. By (10.6.2), we must prove that HomR ( , HomZ (R, E)) is exact. As in the proof
of (1) implies (2) in (10.6.2), if 0 ’ N ’ M is exact, we must show that

HomR (M, HomZ (R, E)) ’ HomR (N, HomZ (R, E)) ’ 0

is exact. By (1) of (10.7.2), this is equivalent to showing that

HomZ (M, E) ’ HomZ (N, E) ’ 0

is exact. [As indicated in the informal discussion in (10.7.2), this replacement is allowable
because a bilinear map can be regarded as a family of linear maps. A formal proof would
invoke the naturality condition referred to in (10.7.2).] Since E is an injective Z-module,
the result now follows from (10.6.2). ™

We can now prove the main result.

10.7.4 Theorem
If M is an arbitrary left R-module, then M can be embedded in an injective left R-module.

Proof. If we regard M as an abelian group, then by (10.7.1), we can assume that M is a
subset of the divisible abelian group E. We will embed M in the injective left R-module
N = HomZ (R, E) (see (10.7.3)). If m ∈ M , de¬ne f (m) : R ’ E by f (m)(r) = rm.
Then f : M ’ N , and we claim that f is an injective R-module homomorphism. If
f (m1 ) = f (m2 ), then rm1 = rm2 for every r ∈ R, and we take r = 1 to conclude
that m1 = m2 , proving injectivity. To check that f is an R-homomorphism, note that if
r, s ∈ R and m ∈ M , then

f (sm)(r) = rsm and (sf (m))(r) = f (m)(rs) = rsm


by de¬nition of scalar multiplication in the R-module N ; see (10.7.2).

It can be shown that every module M has an injective hull, that is, there is a smallest
injective module containing M .

Problems For Section 10.7
1. If R MS and R N , show that HomR (M, N ) is a left S-module via

(sf )(m) = f (ms).

2. If R MS and NS , show that HomS (M, N ) is a right R-module via

(f r)(m) = f (rm).

3. If MR and S NR , show that HomR (M, N ) is a left S-module via

(sf )(m) = sf (m).
26 CHAPTER 10. INTRODUCING HOMOLOGICAL ALGEBRA

4. If S M and S NR , show that HomS (M, N ) is a right R-module via

(f r)(m) = f (m)r.

5. A useful mnemonic device for remembering the result of Problem 1 is that since M
and N are left R-modules, we write the function f on the right of its argument. The
result is m(sf ) = (ms)f , a form of associativity. Give similar devices for Problems 2, 3
and 4.

Note also that in Problem 1, M is a right S-module, but HomR (M, N ) is a left S-
module. The reversal might be expected, because the hom functor is contravariant in its
¬rst argument. A similar situation occurs in Problem 2, but in Problems 3 and 4 there is
no reversal. Again, this might be anticipated because the hom functor is covariant in its
second argument.
6. Let R be an integral domain with quotient ¬eld Q. If M is a vector space over Q, show
that M is a divisible R-module.
7. Conversely, if M is a torsion-free divisible R-module, show that M is a vector space
over Q.
8. If R is an integral domain that is not a ¬eld, and Q is the quotient ¬eld of R, show
that HomR (Q, R) = 0.


10.8 Flat Modules
10.8.1 De¬nitions and Comments
We have seen that an R-module M is projective i¬ its covariant hom functor is exact,
and M is injective i¬ its contravariant hom functor is exact. It is natural to investigate
the exactness of the tensor functor M —R , and as before we avoid complications by
assuming all rings commutative. We say that M is ¬‚at if M —R is exact. Since the
tensor functor is right exact by (10.4.4), an equivalent statement is that if f : A ’ B is
an injective R-module homomorphism, then

1—f: M —A’M —B

is injective. In fact it su¬ces to consider only R-modules A and B that are ¬nitely
generated. This can be deduced from properties of direct limits to be considered in the
next section. [Any module is the direct limit of its ¬nitely generated submodules (10.9.3,
Example 2). The tensor product commutes with direct limits (Section 10.9, Problem 2).
The direct limit is an exact functor (Section 10.9, Problem 4).] A proof that does not
involve direct limits can also be given; see Rotman, “An Introduction to Homological
Algebra”, page 86.

10.8.2 Example
Since Z2 —Z is not exact (Section 10.4, Problem 6), Z2 is not a ¬‚at Z-module.
The next result is the analog for ¬‚at modules of property (10.5.4) of projective modules.
10.8. FLAT MODULES 27

10.8.3 Proposition
The direct sum •i Mi is ¬‚at if and only if each Mi is ¬‚at.


Proof. Let f : A ’ B be an injective R-homomorphism. In view of (8.8.6(b)), investigat-
ing the ¬‚atness of the direct sum amounts to analyzing the injectivity of the mapping

g : •i (Mi — A) ’ •i (Mi — B)

given by

xi1 — a1 + · · · + xin — an ’ xi1 — f (a1 ) + · · · + xin — f (an ).

The map g will be injective if and only if all component maps xi — ai ’ xi — f (ai ) are
injective. This says that the direct sum is ¬‚at i¬ each component is ¬‚at. ™


We now examine the relation between projectivity and ¬‚atness.


10.8.4 Proposition
R is a ¬‚at R-module.


Proof. If f : A ’ B is injective, we must show that (1—f ) : R—R A ’ R—R B is injective.
But by (8.7.6), R —R M ∼ M via r — x ’ rx. Thus the following diagram is commutative.
=

G R —R B
1—f

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