# NOTES ON BASIC HOMOLOGICAL ALGEBRA 0 L M N 0

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1 NOTES ON BASIC HOMOLOGICAL ALGEBRA ANDREW BAKER 1. Chain complexes and their homology Let R be a ring and Mod R the category of right R-modules; a very similar discussion can be had for the category of left R-modules R Mod also makes sense and is left to the reader. Then a sequence of R-module homomorphisms L f M g N is said to be exact if Ker g = Im f. Of course this implies that gf = 0. More generally, a sequence of homomorphisms is exact if for each n, the sequence fn+1 M n is exact. An exact sequence of the form f n Mn 1 f n+1 M n+1 M n f n 1 f n Mn 1 0 L f M g N 0 is called short exact. Such a sequence is split exact if there is a homomorphism r : M L (or equivalently j : N M) so that rf = id L (respectively gj = id N ). L id r g 0 L M N 0 f j id N These equivalent conditions imply that M = L N. Such homomorphisms r and g are said to be retractions, while L and N are said to be retracts of M. A sequence of homomorphisms dn+1 C n is called a chain complex if for each n, (1.1) or equivalently, d n Cn 1 d n d n+1 = 0, Im d n+1 Ker d n. d n 1 An exact or acyclic chain complex is one which each segment d n+1 C n+1 C n d n Cn 1 Date: [28/02/2009]. 1

2 2 ANDREW BAKER is exact. We write (C, d) for such a chain complex and refer to the d n and boundary homomorphisms. We symbolically write d 2 = 0 to indicate that (1.1) holds for all n. For clarity we sometimes (C, d C ) to indicate which boundary is being used. If a chain complex is of finite length finite we often pad it out to a doubly infinite complex by adding in trivial modules and homomorphisms. In particular, if M is a R-module we can view it as the chain complex with M 0 = M and M n = 0 whenever n 0. It is often useful to consider the null complex 0 = ({0}, 0). (C, d) is called bounded below if there is an n 1 such that C n = 0 whenever n < n 1. Similarly, (C, d) is called bounded above if there is an n 2 such that C n = 0 whenever n > n 2. (C, d) is called bounded if it is bounded both below and above. Given a complex (C, d), its homology is defined to be the complex (H (C, d), 0) where H n (C, d) = Ker d n / Im d n+1. The homology of a complex measures its deviation from exactness; in particular, (C, d) is exact if and only if H (C, d) = 0. Notice that there are exact sequences (1.2) 0 Im d n+1 Ker d n H n (C, d) 0, 0 Ker d n C n Im d n 0. Example 1.1. Consider the complex of Z-modules where C n = Z/4, d: Z/4 Z/4; d(t) = 2t. Then Ker d n = 2 Z/4 = Im d n and H n (C, d) = 0, hence (C, d) is acyclic. A homomorphism of chain complexes or chain homomorphism h: (C, d C ) (D, d D ) is a sequence of homomorphisms h n : C n D n for which the following diagram commutes. C n h n d C n C n 1 h n 1 d D D n n D n 1 We often write h: C D when the boundary homomorphisms are clear from the context. A chain homomorphism for which each h n : C n D n is an isomorphism is called a chain isomorphism and admits an inverse chain homomorphism D C consisting of the inverse homomorphisms h 1 n : D C. The category of chain complexes in Mod R, Ch Z (Mod R ), has chain complexes as its objects and chain homomorphisms as its morphisms. Like Mod R, it is an abelian category. Let h: (C, d C ) (D, d D ) be a chain homomorphism. If u Ker d C n we have and if v C n+1, d D n (h n (u)) = h n (d C n (u)) = 0, h n (d C n+1(v)) = d D n+1(h n+1 (v)). Together these allow us to define for each n a well-defined homomorphism h = H n (h): H n (C, d C ) H n (D, d D ); h (u + Im d C n+1) = h n (u) + Im d D n+1. If g : (B, d B ) (C, d C ) is another morphism of chain complexes, it is easy to check that (hg) = h g, while for the identity morphism id: (C, d C ) (C, d C ) we have id = id. Thus each H n is a covariant functor from chain complexes to R-modules.

3 BASIC HOMOLOGICAL ALGEBRA 3 A cochain complex is a collection of R-modules C n together with coboundary homomorphisms d n : C n C n+1 for which d n+1 d n = 0; the cohomology of this complex is (H (C, d), 0) where H n (C, d) = Ker d n / Im d n 1. Given a chain complex (C n, d) we can re-index so that C n = C n and form the cochain complex (C n, d); similarly each cochain complex gives rise to a chain complex. We then have H n (C, d) = H n (C, d). We mainly focus on chain complexes, but everything can be reworked for cochain complexes using this correspondence. There are some useful constructions on complexes. Given (C, d), for any integer t we can form (C[t], d[t]) with C[t] n = C n+t, d[t] = ( 1) t d. Then H n (C[t], d[t]) = H n+t (C, d). Viewing an R-module M as a complex concentrated in degree 0, we have a complex M[t] which is a copy of M in degree t. 2. Short exact sequences of chain complexes Given a morphism of chain complexes h: (C, d) (D, d) we may define two new chain complexes Ker h = ((Ker h), d), Im h = ((Im h), d), where (Ker h) n = Ker h: C n D n, (Im h) n = Im h: C n D n. The boundary homomorphisms are the restrictions of d to each of these. Let h: (L, d) (M, d) and k : (M, d) (N, d) be chain homomorphisms and suppose that 0 (L, d) h (M, d) k (N, d) 0 is short exact, i.e., Ker h = 0, Im k = (N, d), Ker k = Im h. It is natural to ask about the relationship between the three homology functors H (L, d), H (M, d) and H (N, d). Example 2.1. Consider the following exact sequence of chain complexes of Z-modules (written vertically). L M N = = 1 0 Z 2 Z Z/2 0 = 0 0 Z Z 0 Z/2 0 =

4 4 ANDREW BAKER On taking homology we obtain H (L ) H (M ) H (N ) = Z/2 = 0 Z/2 0 Z/2 = Z/2 Neither of the rows with = 0, 1 are exact, however there is an isomorphism H 1 (N ) = Z/2 H 0 (L ) = Z/2. Theorem 2.2. There is a long exact sequence of the form H n+1 (N, d) H n (L, d) n+1 h H n (M, d) k H n (N, d) n H n 1 (L, d) and this is natural in the following sense: given a commutative diagram of short exact sequences h 0 (L, d) (M, d) (N, d) 0 p q r 0 (L, d) there is a commutative diagram h (M, d) k k (N, d) 0 H n+1 (N, d) n+1 r H n (L, d) H n+1 (N p, d) n+1 H n (L, d) h H n (M, d) q h H n (M, d) k H n (N, d) n r H n 1 (L, d) k H n (N p, d) n H n 1 (L, d) In Example 2.1, the boundary map 1 induces the isomorphism H 1 (N ) = H 0 (L ) and there is a long exact sequence 0 H 1 (N ) 1 = H 0 (L ) 0 H 0 (M ) = H 0 (N ) 0. For cochain complexes we get a similar result for a short exact sequence of cochain complexes 0 (L, d) h (M, d) k (N, d) 0,

5 but the n are replaced by homomorphisms BASIC HOMOLOGICAL ALGEBRA 5 δ n : H n (N, d) H n+1 (L, d). 3. Chain homotopies and quasi-isomorphisms Given two chain complexes (C, d), (D, d) and chain homomorphisms f, g : C D, a chain homotopy between f and g is a sequence of R-module homomorphisms s n : C n D n+1 for which the following functional equation holds for all n: ds n + s n 1 d = f g. We say that f and g are chain homotopic or just homotopic if there is such a chain homotopy. In particular, f is null homotopic if f 0. Proposition 3.1. is an equivalence relation on the set of chain homomorphisms C D. The equivalences classes of are called chain homotopy classes. Proposition 3.2. Suppose that two chain homomorphisms f, g : C D are chain homotopic. Then the induced homomorphisms f, g : H (C, d) H (D, d) are equal. Proof. Let n Z. Then for [z] H n (C, d) we have dz = 0, hence f(z) g(z) = (ds n + s n 1 d)(z) = ds n (z) + s n 1 (0) = d(s n (z)). Therefore g [z] = [g(z)] = [f(z) d(s n (z))] = [f(z)] = f [z]. The idea behind this definition comes from the geometric mapping cone of a continuous map f : X Y. Here is the algebraic version of this construction. Given a chain homomorphism f : (C, d) (D, d), its (mapping) cone is the complex Cone f with There is an exact sequence (Cone f) n = C n 1 D n, d(u, v) = ( du, dv fu). in which 0 D j (Cone f) q C[ 1] 0 j(v) = (0, v), q(u, v) = u. A particular case of this is the cone on (C, d), Cone C = Cone(C, d) = Cone id C, for which (Cone C) n = C n 1 C n. Proposition 3.3. A chain homomorphism f : C D is null homotopic if and only if there is a factorisation f C D j (Cone C) There is also a closely related notion of (mapping) cylinder Cyl C based on the topological cylinder construction X I for a space X, where I = [0, 1] is the unit interval. A chain homomorphism f : C D is a chain homotopy equivalence if there is a chain homomorphism h: D C for which hf id C and fh id D. Notice that these conditions imply the equations h f = id: H (C, d) H (D, d), f h = id: H (D, d) H (C, d),

6 6 ANDREW BAKER hence f and h are mutually inverse isomorphisms. Note that chain homotopy equivalences can identify very different complexes. Example 3.4. Consider the complex (C, d) of Z-modules in Example 1.1. Then the chain homomorphism q : (C, d) 0 is a chain homotopy equivalence which is not an chain isomorphism. More generally, a chain homomorphism f : (C, d) (D, d) which induces an isomorphism f : H (C, d) H (D, d) is called a quasi-isomorphism and does not have to be a chain homotopy equivalence. 4. Split complexes A complex (C, d) is called split if for each n Z there is a s n : C n C n+1 such that d = ds n 1 d: C n C n 1. (C, d) is called split exact if it is split and exact. Example 4.1. Suppose that F is a field and (V, d) is a complex of F -vector spaces. Then (V, d) is split. Proof. Consider the exact sequences of (1.2), 0 Im d n+1 Ker d n H n (V, d) 0, 0 Ker d n V n Im d n 0. For each n, choose vector subspaces B n V n, H n Ker d n for which V n = Ker d n B n, d n : B n = Im dn and Ker d n = Im d n+1 H n. Then there is an evident isomorphism V n = B n+1 H n B n under which d corresponds to the linear mapping Now define d : (u, v, w) (w, 0, 0). s n : B n+1 H n B n B n+2 H n+1 B n+1 ; s n(u, v, w) = (0, 0, u). It is easy to see that d s n 1 d = d which implies that the corresponding mapping s n : V n V n+1 satisfies ds n 1 d = d. The arguments used in Example 4.1 apply to any complex of R-modules (C, d). Proposition 4.2. Let (C, d) be a chain complex. (C, d) is split if and only if it has the form with differential C n = B n+1 H n B n d: C n C n+1 ; d(u, v, w) = (w, 0, 0). Furthermore, (C, d) is split exact if and only if H n = 0. Example 4.3. For any complex (C, d), the cone Cone(C, d) = Cone C is split exact, hence H (C, d) = 0.

7 BASIC HOMOLOGICAL ALGEBRA 7 5. Projective and injective resolutions Let R be a ring. A right R-module F is called free if there is a set of elements {b λ : λ Λ} F such that every element x F can be uniquely expressed as x = λ Λ b λ t λ for elements t λ R. We say that the b λ form a basis for F over R. We can make a similar definition for left modules. For example, if n 1, R n = {(t 1,..., t n ) : t 1,..., t n R} is free on the basis consisting of the standard elements e 1 = (1, 0,..., 0),..., e n = (0,..., 0, 1). This works whether we view R n as a left or right module. More generally, if we have a set A, then there is a free module R = {s: A R : s finitely supported}, α A where s finitely supported means that s(α) 0 for only finitely many α A. Notice that for any module M, there is always a surjection F M from some free module. For example, if we take a set of generators m α (α A) for M, then there is a surjection x α m α. α A R M; (x α ) α A An R-module P is called projective if given an exact sequence M f N 0 and a homomorphism p: P N, there is a (not usually unique) homomorphism p: P M for which f p = p. P p p 0 In particular, every free R-module is projective. More generally, we have the following characterisation. Proposition 5.1. An R-module P is projective if and only if it is a retract of a free module. Proof. Suppose that P is projective. Then there is a surjective homomorphism g : F P from a free module F. Consider the diagram of solid arrows M N f P F P id g in which the vertical sequence is exact. By projectivity there is a homomorphism k : P F corresponding to the dotted arrow and making the whole diagram commute. Then g is a retraction and F = P Ker g. 0

8 8 ANDREW BAKER Now suppose that for some module Q, P Q is a free module with basis e α (α A) say. Suppose there is a diagram P p in which the vertical sequence is exact. Then we can extend p to the homomorphism p : P Q N given by p (x, y) = p(x). Then the elements p (e α ) N can be lifted to elements u α M for which f(u α ) = p (e α ), and there is a unique homomorphism p : P Q M for which p (e α ) = u α. Now restricting this to P P Q gives the required p satisfying f p = p. An exact complex (5.1) F k F 1 F 0 M 0 is called a resolution of M, and denoted F M 0. Here we view M as the ( 1)-term and 0 as the ( 2)-term. If each F k is free (resp. projective) over R then it is called a free (resp. projective) resolution of M. Proposition 5.2. Let M be an R-module. (a) There is a free (hence projective) resolution F M 0. (b) If f : M M is a homomorphism and P M 0, P M 0 are projective resolutions, then there is a chain homomorphism between these which is f in degree 1. Furthermore, any two such chain homomorphisms are homotopic. (c) If P M 0 and Q M 0 are two projective resolutions, then they are chain homotopy equivalent by a chain map which is the identity in degree 1. An R-module J is called injective if given an exact sequence 0 K g L and a homomorphism q : K J, there is a (not usually unique) homomorphism q : L J for which qg = q. 0 K L g q It is easy to see that if J is injective, then Hom R (, J) is right exact. Injective modules are less easy to characterise than projective modules. Let R be a commutative integral domain. Then an R-module M is said to be divisible if for every non-zero r R, the multiplication by r homomorphism r : M M is surjective. For example, if the fraction field of R is F, then for any vector space over F is an R-module and is divisible. Another example is the quotient R-module F/R. Proposition 5.3. Let R be a pid. Then an R-module is injective if and only if it is divisible. Furthermore, for every R-module M, there is an injective homomorphism j : M J with J injective. Thus the injective Z-modules (= abelian groups) are the divisible groups. Any torsion-free abelian group can be viewed as a subgroup of a Q-vector space. On the other hand the torsion group Z/n is isomorphic to the subgroup of Q/Z consisting of all elements of order dividing n. M N q 0 J f

9 BASIC HOMOLOGICAL ALGEBRA 9 Now let R be any ring and let M be a right R-module. We may view M as a Z-module and choose and injection of Z-modules i: M I. Now consider Hom Z (R, I) which is right R-module with the action of r R on f Hom Z (R, I) given by Then there is a homomorphism of R-modules (f r)(x) = f(rx). θ : M Hom Z (R, I); θ(m)(x) = i(mx) (x R). Since i(m) = 0 implies m = 0, this is an injection. In fact, Hom Z (R, I) is an injective R-module since for any right R-module N, there is a natural isomorphism Hom R (N, Hom Z (R, I)) = Hom Z (N, I). An exact cochain complex 0 M J in which each J n is injective is called an injective resolution of M. Proposition 5.4. Let M be an R-module. (a) There is an injective resolution 0 M J. (b) If f : M M is a homomorphism and 0 M J, 0 M K are injective resolutions, then there is a chain homomorphism between these which is f in degree 1. Furthermore, any two such chain homomorphisms are homotopic. (c) If 0 M J and 0 M K are two injective resolutions, then they are chain homotopy equivalent by a chain map which is the identity in degree Hom and Ext Let M, N be two right R-modules. Then we define Hom R (M, N) = {h : M N : h is a homomorphism of R-modules}. If R is commutative then Hom R (M, N) is also an R-module, otherwise it may only be an abelian group. If f : M M and g : N N are homomorphisms of R-modules, then there are functions f : Hom R (M, N) Hom R (M, N); f h = h f, g : Hom R (M, N) Hom R (M, N ); g h = g h. These are group homomorphisms and homomorphisms of R-modules if R is commutative. Proposition 6.1. Let M, N be right R-modules. (a) Given a short exact sequence of R-modules the sequence f 1 f 2 0 M 1 M2 M3 0, 0 Hom R (M 3, N) f 2 Hom R (M 2, N) f 1 Hom R (M 1, N) is exact. (b) Given a short exact sequence of R-modules the sequence is exact. g 1 g 2 0 N 1 N2 N3 0, 0 Hom R (M, N 1 ) g 1 Hom R (M, N 2 ) g 2 Hom R (M, N 3 )

10 10 ANDREW BAKER These results show that Hom R (, N) and Hom R (M, ) are left exact. If P is projective, then Hom R (P, ) is right exact, since for given a short exact sequence 0 L M N 0, every homomorphism P N lifts to a homomorphism P M. Now suppose that P M 0 is a resolution of M by projective modules (for example, each P n could be free). This means that the complex P M 0 is exact. We can forget the M and just look at P 0, whose homology is just M in degree 0. Then for any N we can form the cochain complex Hom R (P, N) whose n-th term is Hom R (P n, N). The n-th cohomology group of this is Ext n R(M, N) = H n (Hom R (P, N)). It turns out that this is independent of the choice of projective resolution of M. Notice also that Ext 0 R(M, N) = Hom R (M, N). Now let 0 M J be an injective resolution of M. Then we may form the cochain complex Hom R (M, J ) whose n-term is Hom R (M, J n ). The n-th cohomology group of this complex is rext n R(M, N) = H n (Hom R (M, J )). It turns out that this is independent of the choice of injective resolution of N. Notice also that rext 0 R(M, N) = Hom R (M, N). Proposition 6.2. For R-modules M, N there is a natural isomorphism rext n R(M, N) = Ext n R(M, N). Thus Ext n R (, ) is a balanced functor of its two variables. The main properties of Ext R are summarised in the following result. Let R 0 be the centre of R, i.e., the subring R 0 = {r R : rs = sr for all s R}. Theorem 6.3. Let M and N be right R-modules. (a) For each n 0, Ext n R (M, ) is a covariant functor Mod R Mod R0. (b) For each n 0, Ext n R (, N) is a contravariant functor Mod R Mod R0. (c) For short exact sequences 0 N 1 N 2 N 3 0, 0 M 1 M 2 M 3 0, there are natural long exact sequences 0 Ext 0 R (M, N 1) Ext 0 R (M, N 2) Ext 0 R (M, N 3) Ext n 1 R (M, N 1) Ext n R (M, N 1) Ext n R (M, N 2) Ext n R (M, N 3) Ext n+1 R (M, N 1)

11 BASIC HOMOLOGICAL ALGEBRA 11 and 0 Ext 0 R (M 3, N) Ext 0 R (M 2, N) Ext 0 R (M 1, N) Ext n 1 R (M 1, N) Ext n R (M 3, N) Ext n R (M 2, N) Ext n R (M 1, N) Ext n+1 R (M 3, N) 7. Tensor products and Tor Let M be any right R-module and N be left R-module. Then we can form the tensor product M R N which is an abelian group. If R is commutative, M R N is also an R-module. The definition involves forming the free Z-module F (M, N) with basis consisting of all the pairs (m, n) where m M and n N. Then M R N = F (M, N)/S(M, N), where S(M, N) F (M, N) is the subgroup generated by all the elements of form (m 1 + m 2, n) (m 1, n) (m 2, n), (m, n 1 + n 2 ) (m, n 1 ) (m, n 2 ), (mr, n) (m, rn), where m, m 1, m 2 M, n, n 1, n 2 N, r R. We usually denote the coset of (m, n) by m n; such elements generate the group M R N. The tensor product M R N has an important universal property which characterizes it up to isomorphism. Write q : M N M R N for the quotient function. Let f : M N V be a function into an abelian group V which satisfies f(m 1 + m 2, n) = f(m 1, n) + f(m 2, n), f(m, n 1 + n 2 ) = f(m, n 1 ) + f(m, n 2 ), f(mr, n) = f(m, rn) for all m, m 1, m 2 M, n, n 1, n 2 N, r R, there is a unique homomorphism f : M R N V for which f = f q. M N q M R N f V! f Proposition 7.1. If f : M 1 M 2 and g : N 1 N 2 are homomorphisms of R-modules, there is a group homomorphism f g : M 1 R N 1 M 2 R N 2 for which f g(m n) = f(m) g(n). Proposition 7.2. Given a short exact sequence of left R-modules there is an exact sequence g 1 g 2 0 N 1 N2 N3 0, 1 g 1 1 g 2 M R N 1 M R N 2 M R N 3 0.

12 12 ANDREW BAKER Because of this property, we say that M R ( ) is right exact. Obviously it would be helpful to understand Ker 1 g 1 which measures the deviation from left exactness of M R ( ). Every M admits a projective resolution P M 0, and given a left R-module N, we can form a new complex P R N 0, where the boundary maps are obtained by tensoring those of P with the identity on N an taking P 0 R N 0 rather than using the original map P 0 M. Here P n R N is in degree n. We define It is easy to see that Tor R n (M, N) = H n (P R N). Tor R 0 (M, N) = M R N. Of course we could also form a projective resolution of N, tensor it with M and then take homology. Theorem 7.3. Tor R has the following properties. i) Tor R (M, N) can be computed by using projective resolutions of either variable and the answers agree up to isomorphism. ii) Given R-module homomorphisms f : M 1 M 2 and g : N 1 N 2 there are homomorphisms (f g) = f g : Tor R n (M 1, N 1 ) Tor R n (M 2, N 2 ) generalizing f g : M 1 R N 1 M 2 R N 2. iii) For a projective right (resp. left) R-module P (resp. Q) and n > 0, we have Tor R n (P, N) = 0 = Tor R n (M, Q). iv) Associated to a short exact sequence of right R-modules there is a long exact sequence 0 M 1 M 2 M 3 0 Tor R n+1(m 3, N) Tor R n (M 1, N) Tor R n (M 2, N) Tor R n (M 3, N) Tor R n 1(M 1, N) M 1 R N M 2 R N M 3 R N 0 and associated to a short exact sequence of left R-modules 0 N 1 N 2 N 3 0

13 there is a long exact sequence BASIC HOMOLOGICAL ALGEBRA 13 Tor R n+1(m, N 3 ) Tor R n (M, N 1 ) Tor R n (M, N 2 ) Tor R n (M, N 3 ) Tor R n 1(M, N 1 ) M R N 1 M R N 2 M R N 3 0 Corollary 7.4. Let Q be a left R-module for which Tor R n (M, Q) = 0 for all n > 0 and M. Then for any exact complex (C, d), the complex (C R Q, d 1) is exact, and H n (C R Q, d 1) = H n (C, d) R Q. An R-module M for which Tor R n (M, N) = 0 for all n > 0 and left R-module N is called flat. Given a module M, it is always possible to find a resolution F M 0 for which each F k is flat (for example a projective resolution). Proposition 7.5. If F M 0 is a flat resolution, then Tor R n (M, N) = H n (F R N, d 1). 8. The Künneth Theorem Suppose that (C, d) is a chain complex of flat (for example projective) right R-modules. For any left R-module N we have another chain complex (C R N, d 1) with homology H (C R N, d 1). We would like to understand the connection between this homology and H (C, d) R N. Begin by taking a flat resolution of N, F N 0. For each n the complex C n R F C n R N 0 is still exact since C n is flat. The double complex C R F has two compatible families of boundaries, namely the horizontal ones coming from the boundaries maps d tensored with the identity, d 1, and the vertical ones coming from the identity tensored with the boundary maps δ of F, 1 δ. We can take the two types of homology in different orders to obtain H v (H h (C R F )) = H v = Tor R (H (C, d), N), H h (H v (C R F )) = H h (C R N)) = H (C R N, d 1). In general, the precise relationship between these two involves a spectral sequence, however there are situations where the relationship is more direct. Suppose that N has a flat resolution of the form 0 F 1 F 0 N 0. This will always happen when R = Z or any (commutative) pid and for semi-simple rings. Then for any right R-module M and n > 1, Tor R n (M, N) = 0. Now consider what happens when we tensor C with such a resolution. We obtain a short exact sequence of chain complexes 0 C R F 1 C R F 0 C R N 0

14 14 ANDREW BAKER and on taking homology, an associated long exact sequence as in Theorem 2.2. H n+1 (C R N) n+1 H n (C R F 1 ) H n (C R F 0 ) H n (C R N) n H n 1 (C R F 1 ) As F 0 and F 1 are flat, in each case Corollary 7.4 gives so our long exact sequence becomes H n (C R F i ) = H n (C ) R F i, H n+1 (C R N) n+1 H n (C ) R F 1 H n (C ) R F 0 H n (C R N) The segment n H n 1 (C ) R F 1 H n (C ) R F 1 H n (C ) R F 0 is part of the complex used to compute Tor R (H n (C ), N) and so we have the sequence 0 Tor R 1 (H n (C ), N) H n (C ) R F 1 H n (C ) R F 0 Tor R 0 (H n (C ), N) 0 in which Tor R 0 (H n (C ), N) = H n (C ) R N. For n : H n (C R N) H n 1 (C R F 1 ) we find that Ker n = Tor R 0 (H n (C ), N), Im n = Tor R 1 (H n 1 (C ), N). Theorem 8.1 (Künneth Theorem). Let (C, d) be a chain complex of flat right R-modules and N a left R-module. For each n there is an exact sequence 0 H n (C ) R N H (C R N) Tor R 1 (H n 1 (C ), N) Some examples of resolutions and calculations 9.1. Regular sequences. Let R be a commutative ring and suppose that u R is not a zero-divisor. Then there is an exact sequence of R-modules 0 R u R R/(u) 0. This is a free resolution of R/(u) with F 0 = F 1 = R and all other terms zero. For any R-module N we have the complex whose cohomology is Ext R (R/(u), N). Hence 0 Hom R (P 0, N) u Hom R (P 1, N) 0 Similarly, Ext 0 R(R/(u), N) = Ker(N u N), Tor R 0 (R/(u), N) = R/(u) R N = N/uN, Ext 1 R(R/(u), N) = N/uN. Tor R 1 (R/(u), N) = Ker(N u N).

15 BASIC HOMOLOGICAL ALGEBRA 15 This can be generalised. Again let R be a commutative ring. Suppose that u 1,..., u n R is a regular sequence, i.e., for each k = 1,..., n, the residue class u k + (u 1,..., u k 1 ) R/(u 1,..., u k 1 ) is not a zero divisor. Now form the graded R-algebra where e k has degree 1 and K(u 1,..., u n ) = Λ R (e 1,..., e n ), e k e l = ( 1) lk e l e k. This has a differential d which drops degree by 1 and is a derivation, i.e., for x Λ R (e 1,..., e n ) r and y Λ R (e 1,..., e n ) s it satisfies and also Then there is an exact sequence of R-modules d(xy) = (dx)y + ( 1) r xdy, de r = u r. d Λ R (e 1,..., e n ) s d ΛR (e 1,..., e n ) s 1 d d Λ R (e 1,..., e n ) 1 d ΛR (e 1,..., e n ) 0 R/(u 1,..., u n ) 0. This is a free resolution of R/(u 1,..., u n ) and Λ R (e 1,..., e n ) is called the Koszul resolution. Then Ext R(R/(u 1,..., u n ), N) = H (Hom R (Λ R (e 1,..., e n ), N)) and Tor R (R/(u 1,..., u n ), N) = H (Λ R (e 1,..., e n ) R N). For example, if N = R/(u 1,..., u n ), then Tor R (R/(u 1,..., u n ), R/(u 1,..., u n )) = H (Λ R (e 1,..., e n ) R R/(u 1,..., u n )). Here the boundary is trivial so Tor R (R/(u 1,..., u n ), R/(u 1,..., u n )) = Λ R/(u1,...,u n )(e 1,..., e n ) Exterior algebras. Let k be a commutative ring and let R = Λ k (ε), so ε 2 = 0. Sending ε to 0 gives a ring homomorphism R k and using this k becomes an R-module. There is a free resolution of k, ε ε ε ε Rγ s Rγs 1 Rγ0 k 0, ε where Rγ s Rγs 1 is the obvious multiplication by ε map. Books on homological algebra Here are some textbooks that cover homological algebra from rather different times and perspectives. Anyone seriously learning about the subject as it now exists would probably find [3] the most sensible book to use since it covers topics such as derived categories that are absent from the others. However, it is certainly much more sophisticated than [2]. Although older, [1] is still a good source for traditional parts of the subject and contains a number of specialist topics that are not as well treated in other texts. There are many other books that aim to cover the whole or part of the subject, and algebraic topology, algebraic geometry, and algebra books tend to cover at least the basic notions. References [1] H. Cartan & S. Eilenberg, Homological Algebra, Princeton University Press (1956). [2] J. J. Rotman, An introduction to Homological Algebra, Academic Press (1979). [3] C. A. Weibel, An Introduction to Homological Algebra, Cambridge University Press (1994).

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