The Supplement: 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 Denitions 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 Cn1 called dierentials, 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 nchains, that is, all formal linear combinations with integer coecients 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 dierentials 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 collection 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 distinguish between topological spaces; if the homology groups dier, 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 Denition
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.
Cn
dn
fn
dn
Cn1
G Dn
fn1
G Dn1
We use the same symbol dn to refer to the dierentials in C and D .
[If f : X Y is a continuous map of topological spaces and is a singular n-simplex
in X , then f# ( ) = f 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-dened 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 ) = fn1 dn (z ) = fn1 (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 ) dened 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
and write f
g if there exist homomorphisms hn : Cn Dn+1 such that fn gn =
dn+1 hn + hn1 dn ; see the diagram below.
G Cn1
Cn
yy
hn y
fn gn
yy
yy
|y
v
hn1
G Dn
Dn+1
dn
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 + hn1 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.
A
f
e
d
C
GB
g
GD
(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-dened 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) = g 0 = 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
nn
nn
GB
GE
1 wn
A
1
1
A
1
1
1
1
f
GB
g
GD
xx 0
xx
xx
x9
C
GE
t
GF
e
d
GC
s
GD
G0
h
GF
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
0
GA
d
GC
G0
GC
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 E BDCC .
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 dene the connecting homomorphism : E C by z = x + im d. Symbolically,
= [g 1 e s1 ]
where the brackets indicate that z is the coset of x in C = C/ im d.
We must show that is well-dened. 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
A
f
GB
s
GE
GC
g
GD
t
GF
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 denition of . Thus im s ker . Conversely, assume
z = 0, and let x and y be as in the denition of . Then x = du for some u A, hence
gx = gdu = ef u. But gx = ey by denition of , so y f u ker e = B . Since z = sy by
denition 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 gz = gx +im e. But gx = ey by denition of , so gz = 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 denition
of ) we have z = x + im d. Consequently, x + im d im .
S3. The Long Exact Homology Sequence
S3.1 Denition
We say that
G C
0
f
G D
G E
g
G0
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 ) Hn1 (C )
such that the sequence
g
G Hn+1 (E )
G Hn (C )
f
G Hn (D )
g
G Hn (E )
G Hn1 (C )
f
G
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 )
Cn /Bn (C )
d
0
d
G Zn1 (C )
G En /Bn (E )
G0
d
G Zn1 (E )
G Zn1 (D )
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 Zn1 /Bn1 = Hn1 . The diagram is commutative by the denition
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
0
G Cn
d
0
G Cn1
G Dn
d
G Dn1
G En
G0
d
G En1
G0
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 Zn1 (D )
and dyn = fn1 (xn1 ) for some xn1 Zn1 (C ). Finally, z = xn1 + Bn1 (C ).
S3.4 Naturality
Suppose that we have a commutative diagram of short exact sequences of chain complexes,
as shown below.
0
GC
GD
GE
G0
0
GC
GD
GE
G0
Then there is a corresponding commutative diagram of long exact sequences:
G Hn (C )
G Hn (D )
G Hn (E )
G Hn1 (C )
G
G Hn (C )
G Hn (D )
G Hn (E )
G Hn1 (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 = xn1 + Bn1 (C ), then
(f z ) = f xn1 + Bn1 (C ) = f (xn1 + Bn1 (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 Denitions 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
0
G E0
G E1
G E2 ;
we have an injective resolution if every Ei is injective, . A deleted injective resolution has
the form
0
G E0
y
G E1
G E2
G
M
Exactness at E0 no longer holds because the map E0 E1 need not be injective.
We will use the notation P M for a projective resolution, and M E for an
injective resolution.
S4.2 Proposition
Every module M has a free (hence projective) resolution.
9
Proof. By (4.3.6), M is a homomorphic image of a free module F0 . Let K0 be the kernel
of the map from F0 onto M . In turn, there is a homomorphism with kernel K1 from a
free module F1 onto K0 , and we have the following diagram:
G K1
0
G F1
G K0
G F0
GM
G0
Composing the maps F1 K0 and K0 F0 , we get
G F1
G K1
0
G F0
GM
G0
which is exact. But now we can nd a free module F2 and a with homomorphism kernel K2
mapping F2 onto K1 . The above process can be iterated to produce the desired free
resolution.
Specifying a module by generators and relations (see (4.6.6) for abelian groups) involves nding an appropriate F0 and K0 , as in the rst step of the above iterative process.
Thus a projective resolution may be regarded as a generalization of a specication by generators and relations.
Injective resolutions can be handled by dualizing the proof of (S4.2).
S4.3 Proposition
Every module M has an injective resolution.
Proof. By (10.7.4), M can be embedded in an injective module E0 . Let C0 be the cokernel
of M E0 , and map E0 canonically onto C0 . Embed C0 in an injective module E1 , and
let C1 be the cokernel of the embedding map. We have the following diagram:
GM
0
G E0
G C0
G E1
G C1
G0
Composing E0 C0 and C0 E1 , we have
GM
0
G E0
G E1
G C1
G0
which is exact. Iterate to produce the desired injective resolution.
S5. Derived Functors
S5.1 Left Derived Functors
Suppose that F is a right exact functor from modules to modules. (In general, the
domain and codomain of F can be abelian categories, but the example we have in mind is
M R .) Given a short exact sequence 0 A B C 0, we form deleted projective
resolutions PA A, PB B , PC C . It is shown in texts on homological algebra
that it is possible to dene chain maps to produce a short exact sequence of complexes
as shown below.
0
GA
y
GB
y
GC
y
G0
0
G PA
G PB
G PC
G0
10
The functor F will preserve exactness in the diagram, except at the top row, where we only
have F A F B F C 0 exact. But remember that we are using deleted resolutions,
so that the rst row is suppressed. The left derived functors of F are dened by taking
the homology of the complex F (P ), that is,
(Ln F )(A) = Hn [F (PA )].
The word left is used because the Ln F are computed using left resolutions. It can
be shown that up to natural equivalence, the derived functors are independent of the
particular projective resolutions chosen. By (S3.2), we have the following long exact
sequence:
G (Ln F )(A)
G (Ln F )(B )
G (Ln F )(C )
G (Ln1 F )(A)
G
S5.2 Right Derived Functors
Suppose now that F is a left exact functor from modules to modules, e.g., HomR (M, ).
We can dualize the discussion in (S5.1) by reversing the vertical arrows in the commutative diagram of complexes, and replacing projective resolutions such as PA by injective
resolutions EA . The right derived functors of F are dened by taking the homology of
F (E ). In other words,
(Rn F )(A) = H n [F (EA )]
where the superscript n indicates that we are using right resolutions and the indices are
increasing as we move away from the starting point. By (S3.2), we have the following
long exact sequence:
G
(Rn F )(A)
G (Rn F )(B )
G (Rn F )(C )
G (Rn+1 F )(A)
G
S5.3 Lemma
(L0 F )(A) F (A) (R0 F )(A)
=
=
Proof. This is a good illustration of the advantage of deleted resolutions. If P A, we
have the following diagram:
F (P1 )
G F (P0 )
G0
F (A)
0
The kernel of F (P0 ) 0 is F (P0 ), so the 0th homology module (L0 F )(A) is F (P0 ) mod
the image of F (P1 ) F (P0 ) [=the kernel of F (P0 ) F (A).] By the rst isomorphism
11
theorem and the right exactness of F , (L0 F )(A) F (A). To establish the other isomor=
phism, we switch to injective resolutions and reverse arrows:
0
G F (E0 )
y
G F (E1 )
F (A)
y
0
The kernel of F (E0 ) F (E1 ) is isomorphic to F (A) by left exactness of F , and the image
of 0 F (E0 ) is 0. Thus (R0 F )(A) F (A).
=
S5.4 Lemma
If A is projective, then (Ln F )(A) = 0 for every n > 0; if A is injective, then (Rn F )(A) = 0
for every n > 0.
Proof. If A is projective [resp. injective], then 0 A A 0 is a projective [resp.
injective] resolution of A. Switching to a deleted resolution, we have 0 A 0 in each
case, and the result follows.
S5.5 Denitions and Comments
If F is the right exact functor M R , the left derived functor Ln F is called TorR (M, ).
n
If F is the left exact functor HomR (M, ), the right derived functor Rn F is called
Extn (M, ). It can be shown that the Ext functors can also be computed using proR
jective resolutions and the contravariant hom functor. Specically,
Extn (M, N ) = [Rn HomR ( , N )](M ).
R
A switch from injective to projective resolutions is a simplication, because projective
resolutions are easier to nd in practice.
The next three results sharpen Lemma S5.4. [The ring R is assumed xed, and we
write R simply as . Similarly, we drop the R in TorR and ExtR . When discussing Tor,
we assume R commutative.]
S5.6 Proposition
If M is an R-module, the following conditions are equivalent.
(i) M is at;
(ii) Torn (M, N ) = 0 for all n 1 and all modules N ;
(iii) Tor1 (M, N ) = 0 for all modules N .
12
Proof. (i) implies (ii): Let P N be a projective resolution of N . Since M
exact functor (see (10.8.1)), the sequence
is an
M P1 M P0 M N 0
is exact. Switching to a deleted resolution, we have exactness up to M P1 but not at
M P0 . Since the homology modules derived from an exact sequence are 0, the result
follows.
(ii) implies (iii): Take n = 1.
(iii) implies (i): If 0 A B C 0 is a short exact sequence, then by (S5.1), we
have the following long exact sequence:
Tor1 (M, C ) Tor0 (M, A) Tor0 (M, B ) Tor0 (M, C ) 0.
By hypothesis, Tor1 (M, C ) = 0, so by (S5.3),
0M AM B M C 0
is exact, and therefore M is at.
S5.7 Proposition
If M is an R-module, the following conditions are equivalent.
(i) M is projective;
(ii) Extn (M, N ) = 0 for all n 1 and all modules N ;
(iii) Ext1 (M, N ) = 0 for all modules N .
Proof. (i) implies (ii): By (S5.4) and (S5.5), Extn (M, N ) = [Extn ( , N )](M ) = 0 for
n 1.
(ii) implies (iii): Take n = 1.
(iii) implies (i): Let 0 A B M 0 be a short exact sequence. If N is any
module, then using projective resolutions and the contravariant hom functor to construct
Ext, as in (S5.5), we get the following long exact sequence:
0 Ext0 (M, N ) Ext0 (B, N ) Ext0 (A, N ) Ext1 (M, N )
By (iii) and (S5.3),
0 Hom(M, N ) Hom(B, N ) Hom(A, N ) 0
is exact. Take N = A and let g be the map from A to B . Then the map g from Hom(B, A)
to Hom(A, A) is surjective. But 1A Hom(A, A), so there is a homomorphism f : B A
such that g (f ) = f g = 1A . Therefore the sequence 0 A B M 0 splits, so by
(10.5.3), M is projective.
13
S5.8 Corollary
If N is an R-module, the following conditions are equivalent.
(a) N is injective;
(b) Extn (M, N ) = 0 for all n 1 and all modules M ;
(c) Ext1 (M, N ) = 0 for all modules M .
Proof. Simply saying duality may be unconvincing, so lets give some details. For (a)
implies (b), we have [Extn (M, )](N ) = 0. For (c) implies (a), note that the exact
sequence 0 N A B 0 induces the exact sequence
0 Ext0 (M, N ) Ext0 (M, A) Ext0 (M, B ) 0.
Replace Ext0 (M, N ) by Hom(M, N ), and similarly for the other terms. Then take M = B
and proceed exactly as in (S5.7).
S6. Some Properties of Ext and Tor
We will compute Extn (A, B ) and TorR (A, B ) in several interesting cases.
R
n
S6.1 Example
We will calculate Extn (Zm , B ) for an arbitrary abelian group B . To ease the notational
Z
burden slightly, we will omit the subscript Z in Ext and Hom, and use = (most of the
time) when we really mean We have the following projective resolution of Zm :
=.
GZ
0
m
GZ
G0
G Zm
where the m over the arrow indicates multiplication by m. Switching to a deleted resolution and applying the contravariant hom functor, we get
0
G Hom(Z, B )
y
m
G Hom(Z, B )
G0
Hom(Zm , B )
But by (9.4.1), we have
HomR (R, B ) B
=
(1)
and the above diagram becomes
0
GB
m
GB
G0
(2)
By (S5.3), Ext0 (Zm , B ) = Hom(Zm , B ). Now a homomorphism f from Zm to B is
determined by f (1), and f (m) = mf (1) = 0. If B (m) is the set of all elements of B
14
that are annihilated by m, that is, B (m) = {x B : mx = 0}, then the map of B (m) to
Hom(Zm , B ) given by x f where f (1) = x, is an isomorphism. Thus
Ext0 (Zm , B ) = B (m).
It follows from (2) that
Extn (Zm , B ) = 0, n 2
and
Ext1 (Zm , B ) = ker(B 0)/ im(B B ) = B/mB.
The computation for n 2 is a special case of a more general result.
S6.2 Proposition
Extn (A, B ) = 0 for all n 2 and all abelian groups A and B .
Z
Proof. If B is embedded in an injective module E , we have the exact sequence
0 B E E/B 0.
This is an injective resolution of B since E/B is divisible, hence injective; see (10.6.5)
and (10.6.6). Applying the functor Hom(A, ) = HomZ (A, ) and switching to a deleted
resolution, we get the sequence
0
G Hom(A, E )
y
G Hom(A, E/B )
G0
Hom(A, B )
whose homology is 0 for all n 2.
S6.3 Lemma
Ext0 (Z, B ) = HomZ (Z, B ) = B and Ext1 (Z, B ) = 0.
Z
Z
Proof. The rst equality follows from (S5.3) and the second from (1) of (S6.1). Since Z
is projective, the last statement follows from (S5.7).
S6.4 Example
We will compute TorZ (Zm , B ) for an arbitrary abelian group B . As before, we drop the
n
superscript Z and write = for We use the same projective resolution of Zm as in (S6.1),
=.
15
and apply the functor
before. Thus
B . Since R R B B by (8.7.6), we reach diagram (2) as
=
Torn (Zm , B ) = 0, n 2;
Tor1 (Zm , B ) = ker(B B ) = {x B : mx = 0} = B (m);
Tor0 (Zm , B ) = Zm B = B/mB.
[To verify the last equality, use the universal mapping property of the tensor product to
produce a map of Zm B to B/mB such that n x n(x + mB ). The inverse of this
map is x + mB 1 x.]
The result for n 2 generalizes as in (S6.2):
S6.5 Proposition
TorZ (A, B ) = 0 for all n 2 and all abelian groups A and B .
n
Proof. B is the homomorphic image of a free module F . If K is the kernel of the homomorphism, then the exact sequence 0 K F B 0 is a free resolution of B .
[K is a submodule of a free module over a PID, hence is free.] Switching to a deleted
resolution and applying the tensor functor, we get a four term sequence as in (S6.2), and
the homology must be 0 for n 2.
S6.6 Lemma
Tor1 (Z, B ) = Tor1 (A, Z) = 0; Tor0 (Z, B ) = Z B = B .
Proof. The rst two equalities follow from (S5.6) since Z is at. The other two equalities
follow from (S5.3) and (8.7.6).
S6.7 Finitely generated abelian groups
We will show how to compute Extn (A, B ) and Torn (A, B ) for arbitrary nitely generated
abelian groups A and B . By (4.6.3), A and B can be expressed as a nite direct sum of
cyclic groups. Now Tor commutes with direct sums:
Torn (A, r=1 Bj ) = r=1 Torn (A, Bj ).
j
j
[The point is that if Pj is a projective resolution of Bj , then the direct sum of the Pj
is a projective resolution of j Bj , by (10.5.4). Since the tensor functor is additive on
direct sums, by (8.8.6(b)), the Tor functor will be additive as well. Similar results hold
when the direct sum is in the rst coordinate, and when Tor is replaced by Ext. (We use
(10.6.3) and Problems 5 and 6 of Section 10.9, and note that the direct product and the
direct sum are isomorphic when there are only nitely many factors.)]
Thus to complete the computation, we need to know Ext(A, B ) and Tor(A, B ) when
A = Z or Zm and B = Z or Zn . We have already done most of the work. By (S6.2) and
16
(S6.5), Extn and Torn are identically 0 for n 2. By (S6.1),
Ext0 (Zm , Z) = Z(m) = {x Z : mx = 0} = 0;
Ext0 (Zm , Zn ) = Zn (m) = {x Zn : mx = 0} = Zd
where d is the greatest common divisor of m and n. [For example, Z12 (8) = {0, 3, 6, 9}
=
Z4 . The point is that the product of two integers is their greatest common divisor times
their least common multiple.] By (S6.3),
Ext0 (Z, Z) = Hom(Z, Z) = Z; Ext0 (Z, Zn ) = Zn .
By (S6.1),
Ext1 (Zm , Z) = Z/mZ = Zm ;
Ext1 (Zm , Zn ) = Zn /mZn = Zd
as above. By (S5.7),
Ext1 (Z, Z) = Ext1 (Z, Zn ) = 0
By (S6.4),
Tor1 (Zm , Z) = Tor1 (Z, Zm ) = Z(m) = 0;
Tor1 (Zm , Zn ) = Zn (m) = Zd .
By (8.7.6) and (S6.4),
Tor0 (Z, Z) = Z;
Tor0 (Zm , Z) = Z/mZ = Zm ;
Tor0 (Zm , Zn ) = Zn /mZn = Zd .
Notice that Tor1 (A, B ) is a torsion group for all nitely generated abelian groups A and B .
This is a partial explanation of the term Tor. The Ext functor arises in the study of
group extensions.
S7. Base Change in the Tensor Product
Let M be an A-module, and suppose that we have a ring homomorphism from A to B (all
rings are assumed commutative). Then B A M becomes a B module (hence an A-module)
via b(b m) = bb m. This is an example of base change, as discussed in (10.8.8). We
examine some frequently occurring cases. First, consider B = A/I , where I is an ideal
of A.
S7.1 Proposition
(A/I ) A M M/IM .
=
17
Proof. Apply the (right exact) tensor functor to the exact sequence of A-modules
0 I A A/I 0
to get the exact sequence
I A M A A M (A/I ) A M 0.
Recall from (8.7.6) that A A M is isomorphic to M via a m am. By the rst
isomorphism theorem, (A/I ) A M is isomorphic to M mod the image of the map from
I A M to M . This image is the collection of all nite sums
ai mi with ai I and
mi M , which is IM .
Now consider B = S 1 A, where S is a multiplicative subset of A.
S7.2 Proposition
(S 1 A) A M S 1 M .
=
Proof. The map from S 1 A M to S 1 M given by (a/s, m) am/s is A-bilinear, so by
the universal mapping property of the tensor product, there is a linear map : S 1 A A
M S 1 M such that ((a/s) m) = am/s. The inverse map is given by (m/s) =
(1/s) m. To show that is well-dened, suppose that m/s = m /s . Then for some
t S we have ts m = tsm . Thus
1/s m = ts /tss m = 1/tss ts m = 1/tss tsm = 1/s m .
Now followed by takes a/s m to am/s and then to 1/s am = a/s m. On the other
hand, followed by takes m/s to 1/s m and then to m/s. Consequently, and are
inverses of each other and yield the desired isomorphism of S 1 A A M and S 1 M .
Finally, we look at B = A[X ].
S7.3 Proposition
[X ] A M M [X ]
=
where the elements of M [X ] are of the form a0 m0 + a1 Xm1 + a2 X 2 m2 + + an X n mn ,
ai A, mi M , n = 0, 1, . . . .
Proof. This is very similar to (S7.2). In this case, the map from A[X ] A M to M [X ]
takes f (X ) m to f (X )m, and the map from M [X ] to A[X ] A M takes X i m to X i m.
Here, there is no need to show that is well-dened.