Homological Algebra  Problem Set 3


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1 Homological Algebra  Problem Set 3 Problem 1. Let R be a ring. (1) Show that an Rmodule M is projective if and only if there exists a module M and a free module F such that F = M M (we say M is a summand of F). (2) Show that the Z/6Zmodules Z/3Z and Z/2Z are examples of projective modules which are not free. Solution: (1) Assume M is projective. Choose a surjection f : F M where F is a free Rmodule. By the projectivity of M, the short exact sequence 0 ker(f) F f M 0 splits so that F = M ker(f). Conversely, assume M is a summand of F such that F = M M. Wehaveaprojectionmapπ M : F M andaninclusionmapi M : M F. Given an epic f : A B and a morphism g : M B, we first obtain a morphism g π M : F B which can be lifted to a morphism r : F A, since the free module F is projective. Then r i M : M A is a lift of g showing that M is projective. (2) We have Z/6Z = Z/2Z Z/3Z and therefore, by (1), both summands are projective Z/6Zmodules. Problem 2. Let N be a left Rmodule. Show that the functor R N : mod R Ab,M M R N from the category of right Rmodules to the category of abelian group is right exact. Solution: This is a direct consequence of the adjunction Hom Ab (M R N,A) = Hom mod R (M,Hom Ab (N,A)) and the general fact that left adjoints are right exact. But we haven t discussed adjunctions yet, so we give a direct proof. Let 0 A f B g C 0 (1) be a short exact sequence of right Rmodules. We have to show that A R N f N B R N g N C R N 0 is exact. The surjectivity of g N is immediate. It remains to show im(f N) = ker(g N). Since we certainly have im(f N) ker(g N), we obtain an induced surjective morphism ψ : B R N/im(f N) C R N.
2 The exactness condition im(f N) = ker(g N) is equivalent to this morphism being an isomorphism. We construct an inverse morphism by first defining a map of sets C N B R N/im(f N),(c,x) c x where c denotes a preimage of c under g. Note that this map is welldefined due to the exactness of (1). It is further bilinear and hence, by the universal property of the tensor product, induces a morphism of abelian groups ϕ : C R N B R N/im(f N). It is immediate to verify that ϕ and ψ are inverse to one another. Problem 3. A left Rmodule N is called flat if the functor R N is exact. (1) Let N be a left Rmodule. Show that the following are equivalent: (2) Let (a) N is flat. (b) Tor R i (M,N) = 0 for every right Rmodule M and every i > 0. (c) Tor R 1(M,N) = 0 for every right Rmodule M. 0 A B C 0 be a short exact sequence of left Rmodules and assume that B and C are flat. Show that A is flat. (3) Let k be a field and let R = k[x,y] denote the polynomial ring. Show that the ideal I = (x,y) is an example of a torsionfree Rmodule which is not flat. Solution: (1) (a) (b): Assume N is flat and let M be a right Rmodule. We first claim that R N preserves exactness, i.e., it preserves exact sequences of the form A f B g C. Given such a sequence we obtain the diagram ker(f) im(g) f g A B C im(f) coker(g) where the three diagonal sequences are short exact. Using the fact that R N preserves those short exact sequences, it is immediate that it preserves exactness of the original sequence. To compute Tor R (M,N), we choose a free resolution P M. Since the complex P is exact in positive degrees, and R N preserves exactness, the complex P R N is exact in positive degrees. Therefore, we obtain Tor R i (M,N) = 0 for i > 0. 2
3 (b) (c) is clear. (c) (a) is an immediate consequence of the long exact sequence in Tor. (2) Let F be a free R module on a set X, i.e., F = X R. Then the functor F R is given by A F R A = A. X Applying it to a short exact sequence of left Rmodules, we obtain the sequence 0 A A A 0 0 X A X A X A 0 which, as we see by explicit element theoretic verification, is again short exact. Therefore, the functor F R is exact. (Using that any projective module is a summand of a free module, we can in fact see that P R is exact for any projective Rmodule, but we will not need this for this problem). Assume now that B,C are flat left Rmodules and we are given a short exact sequence 0 A B C 0. We show that A is flat. By (1), it suffices to show that, for every right Rmodule M, the higher Tor groups Tor i (M,A) = 0 for i > 0. Choose a free resolution F M of M. The resulting sequence 0 F R A F R B F R C 0 isexact,sinceeveryrowisgivenbytensoringwithafreemodule. Passingtotheassociated long exact sequence, we obtain, using that Tor i (M,B) = Tor i (M,C) = 0 for i > 0, that Tor i (M,A) = 0 for i > 0. (3) Applying R I to the short exact sequence yields the sequence 0 I R k 0 0 I R I I k 2 0 which is not exact on the left: For example the nonzero element x y y x of I R I maps to xy yx = 0 in I. Also note, that this implies Tor R 1(k,I) 0. Problem 4. Let R be a commutative ring. Note that, in this case, there is no difference between left and right Rmodules. Further, given two Rmodules M and N, we can define a natural Rmodule structure on M R N by letting r.(m n) = (rm) n. We define the tensor product X R Y of chain complexes of Rmodules X and Y to be given by (X R Y) n = X i R Y j i+j=n and differential obtained by linearly extending the formula d(x i y j ) = d X (x i ) y j + ( 1) i x i d Y (y j ). 3
4 (1) For x R, we define the complex K(x) = 1 R x 0 R, where 0 and 1 indicate the degrees of the terms. Let X be a complex of Rmodules and assume that, for every n, multiplication by x is injective on the homology module H n (X). Show that, for every n, we have H n (X R K(x)) = H n (X)/xH n (X). (2) Let R be the polynomial ring k[x 1,x 2,...,x n ] over a field k. Show that the complex K(x 1 ) R K(x 2 ) R R K(x n ) is a free resolution of the residue field k = R/(x 1,x 2,...,x n ). Compute the Tor groups Tor R (k,k). Solution: (1) We have (K(x) X) n = Xn X n 1 and the differential is given by the formula d(y n,y n 1 ) = (dy n +xy n 1, dy n 1 ). We compute the homology in degree n. Suppose (y n,y n 1 ) is a cycle, then y n 1 is a cycle in X n 1 and dy n = xy n 1. In other words, xy n 1 is a boundary and hence zero in homology. By our assumption, this implies that y n 1 is zero in homology and hence a boundary. We may therefore choose z n X n such that dz n = y n 1. Therefore, the homology class represented by (y n,y n 1 ) can always be represented by a cycle of the form (y n,0). A cycle of this form is a boundary if and only if y n = dz n+1 +xz n with dz n = 0. But this is simply saying H n (X R K(x)) = H n (X)/xH n (X). (2) Let R be a ring, A collection {x 1,...,x n } of elements of R is said to be a regular sequence in R if x d is not a zerodivisor in k = R / (x 1,...,x d 1 ). For a regular sequence, we shall prove by induction that is a free resolution of R / (x 1,...,x n ). K(x 1 ) R... R K(x n ) Now it is clear that H i (K(x 1 )) is R / (x 1 ) in degree 0, and 0 in all other degrees. Suppose we have shown that H n (K(x 1 ) R... R K(x d )) is R / (x 1,...,x d ) in degree 0, and 0 in all other degrees. Then x d+1 acts injectively on H n (K(x 1 ) R... R K(x d )) so by part (1) H n (K(x 1 ) R... R K(x d ) R K(x d+1 )) = H n (K(x 1 ) R... R K(x d )) / x d+1 H n (K(x 1 ) R... R K(x d )) We get 4
5 { / R (x1,...,x H i (K(x 1 ) R... R K(x d )) = d+1 ) i = 0 0 i 0 So we do indeed have a free resolution of k. Now, for n complexes X 1,...,X n we have The differentials are (X 1... X n ) m = i i n=m X 1 i 1 R... R X n i n d(y i1... y in ) = d X1 (y i1 ) y i2... y in +( 1) i 1 y i1 d(y i2... y in ) In particular =... = d X1 (y i1 ) y i2... y in Tensoring with k gives a complex with + ( 1) i i j y i1... x j d Xj+1 (y ij+1 ) y ij+2... y in (K(x 1 ) R... R K(x d )) m = R (n m) (K(x 1 ) R... R K(x d ) R k) m = k (n m) However, since the nonzero differentials of each complex K(x i ) are given by multiplicationbyx i, andx i k = 0foralli, thedifferentialsinthecomplexk(x 1 ) R... R K(x d ) R k are all zero. Tor i R(k,k) = H i (K(x 1 ) R... R K(x d ) R k) = (K(x 1 ) R... R K(x d ) R k) i { = R (n i) 0 i n 0 i > n Problem 5. Let A be an abelian category, and let f : X Y be a map of chain complexes in A. We define the mapping cone of f to be the chain complex given by cone(f) n = X n 1 Y n with differential d = ( d X,d Y f). Show that we have natural maps of complexes Y cone(f) and cone(f) X[ 1]. Here we define X[k], k Z, to be chain complex given by X[k] n = X k+n and differential ( 1) k d X. Show that the resulting diagram induces a long exact sequence X Y cone(f) X[ 1] Y[ 1]... H n (X) H n (Y) H n (cone(f)) H n 1 (X) H n 1 (Y)... in A. Deduce that f is a quasiisomorphism if and only if cone(f) is exact. 5
6 Solution: We simply note that there is a short exact sequence of complexes 0 Y cone(f) X[ 1] 0 which induces the claimed long exact sequence in homology. The only thing to verify is that the connecting homomorphisms are induced by the morphism f, but this follows from the usual diagram chase. 6
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