TCC Homological Algebra: Assignment #3 (Solutions)


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1 TCC Homological Algebra: Assignment #3 (Solutions) David Loeffler, 30th November 2016 This is the third of 4 problem sheets. Solutions should be submitted to me (via any appropriate method) by noon on 30th November. This problem sheet will be marked out of a total of 25; the number of marks available for each question is indicated. Note that rings are assumed to be unital (i.e. having a multiplicative identity element 1), ring homomorphisms are assumed to map 1 to 1, and modules are left modules, unless otherwise stated. 1. [2 points] Let C be an abelian category with enough injectives, and X an object of C. Show that X is injective if and only if it is Facyclic for every leftexact functor F from C to an abelian category. Solution: The if direction was discussed in lectures, so it suffices to prove only if. Let A Obj C and consider the leftexact functor F = Hom(A, ) : C Ab. By assumption, X is Facyclic, which is to say that we have Ext i (A, X) = 0 for all A and all i 1. In other words, the functors Ext i (, X) are the zero functors for i 1. By another result from lectures this implies that X is injective. 2. [2 points] Let 0 X Y Z 0 be a short exact sequence of abelian groups, and I X, I Y, I Z injective resolutions of X, Y, Z. Is it necessarily the case that the maps in the short exact sequence lift to a short exact sequence of complexes 0 I X I Y I Z 0? Give a proof or counterexample as appropriate. Solution: No, such liftings do not necessarily exist. Consider the case where all three of X, Y, Z are zero. We may take I X and I Z to be the zero resolution. Then the assertion of the question would imply that any injective resolution of the zero object must be zero, which is clearly not true (take e.g. the 2term complex Q id Q). 3. Let S : 0 X α Y β Z 0 be a short exact sequence in an abelian category C. We say the sequence S is split if there is an isomorphism Y = X Z compatible with the maps X Y and Y Z. (a) [1 point] Show that the following are equivalent: (i) The exact sequence S is split. (ii) There exists a morphism π : Y X such that π α = id X. (iii) There exists a morphism φ : Z Y such that β φ = id Z. Solution: It is clear that (i) implies both (ii) and (iii), since we may take π and φ to be the projection (resp. inclusion) maps in the definition of the direct sum. 1
2 If (ii) holds, consider the morphism t = id Y (α π) : Y Y. Then t α = 0, so t factors through Z: that is, we have t = φ β for some φ : Z Y. We compute that β φ β = β (id Y α π) = β. Since β has zero cokernel it is an epimorphism (rightcancellative): the equality β φ β = β = id Z β implies β φ = id Z. Thus (ii) (iii), and exactly the same argument in reverse shows that (iii) (ii). Now suppose (ii) holds. Then (iii) holds also, and we may assume without loss of generality that φ has the form constructed above, so that α π + φ β = id Y. I claim that π φ = 0. Since β is rightcancellative, it suffices to show that π φ β = 0; but we know φ β = id Y α π. As π α = id X we have π (id Y α π) = 0. Thus Y fits into the star diagram characterising X Z, X id with both diagonal compositions zero. X π α Y φ β Z id Z, (b) [1 point] Show that if either X is injective, or Z is projective, then S must be split. Solution: Suppose X is injective. Then id X must extend to a morphism Y X, so S satisfies (ii) above. Similarly if Z is projective, then id Z must extend to a morphism Z Y, so S satisfies (iii). (c) [2 points] Assume C is RMod for some ring R. Prove the following converse to (b): if X is such that every short exact sequence starting with X is split, then X is an injective object. Solution: Since RMod has enough injectives, there exists a short exact sequence 0 X I I/X 0 where I is an injective object. By hypothesis this must split, so X is a direct summand of an injective object. Let A B be a monomorphism in C and φ : A X a morphism. Composing φ with X I we obtain a homomorphism A I, which must lift to B I because of the injectivity of I. Composing this extension with the projection map I X gives a homomorphism B X which also extends φ. Thus X must itself be injective. 4. [3 points] Let k be a field, and let R be the ring k[x, Y]/(X 2, XY, Y 2 ). Let I = (X, Y) be the unique maximal ideal of R, so that R/I = k. Find a projective resolution of R/I as an Rmodule, and hence compute the groups Ext i R (R/I, R). (Hint: Do not expect your projective resolution to have only finitely many terms!) Solution: It is clear how to find a surjection from a projective object P 0 onto R/I, namely the canonical surjection from R itself. The kernel of this map is I, so we need to find a projective P 1 which surjects onto I. Since we have two generators of I, we take P 1 = R R with the map given by (X Y). ( ) What is the kernel of this map? If we write a generic element of R 2 a + bx + cy as a + b X + c with Y a, b, c, a, b, c k, then the condition that this maps to 0 is simply that ax + a Y = 0, i.e. that a = 0 and a = 0. Thus the kernel is simply I 2, and we end up taking P 2 = R 4, P 3 = R 8, etc. 2
3 To compute the Ext groups, we must compute the cohomology of the Hom cochain complex Hom R (P, R). One sees easily that Hom R (R, R) is isomorphic to (the underlying additive group of) R via φ φ(1), so the nth term of the complex is R 2n, with differentials given by ( f 1,..., f 2 n) (X f 1, Y f 1, X f 2, Y f 2,..., X f 2 n, Y f 2 n). Hence the 0cohomology is simply { f R : i f = 0 i I}, which is just I itself; and for n 1, the cohomology is the direct sum of 2 n 1 copies of the quotient group I I {X f, Y f : f R} = k 3. So Ext i (R/I, R) is isomorphic as an abelian group to k 2 if i = 0, and to k 3 2n 1 for n 1. [In fact, since R is commutative the Ext groups are naturally Rmodules, and are isomorphic to the direct sum of this many copies of R/I this is easy to see since I must annihilate Ext i (R/I, anything). The fact that Ext i (R/I, R) is nonzero for arbitrarily large i shows that R is not a Gorenstein ring.] 5. [2 points] Let R be a commutative ring, and A, B, C any three Rmodules, with A projective. Show that Ext i R (A R B, C) = Hom R (A, Ext i R (B, C)) for every i 1. (You may assume the statement is true for i = 0). Solution: By definition Ext i (A B, C) is the cohomology of the complex Hom(A B, I ) where I is an injective resolution of C. Using the case i = 0 this is the same as Hom(A, Hom(B, I )). Since A is projective, Hom(A, ) is an exact functor, so commutes with the formation of cohomology: H i (Hom(A, Hom(B, I ))) = Hom(A, H i (Hom(B, I ))) = Hom(A, Ext i (B, C)). 6. Let G be a group. Recall the bar resolution X n (G) of the trivial Z[G]module Z, discussed in lectures. (a) [2 points] Write down the differential d 1 : X 2 (G) X 1 (G) explicitly. Use this and the formula for d 0 given in lectures to show that for any Gmodule M we have H 1 (G, M) = Z 1 (G, M)/B 1 (G, M) where Z 1 (G, M) = {functions σ : G M such that σ(gh) = σ(g) + gσ(h)} and B 1 (G, M) = {functions such that σ(g) = gm m for some m M}. Solution: By definition X 2 (G) is spanned by formal symbols (g, h) with g, h G. By definition, we have d((g, h)) = d 0 d 1 + d 2 where d 0 = g (h), d 1 = (gh), and d 2 = (g). Dually C i (G, M) is the Z[G]module homs X i (G) M, which are uniquely determined by the values on the basis elements (g 1,..., g i ). So the map d 1 : C 1 (G, M) C 2 (G, M) sends a function σ : G M to the function τ : G G M mapping (g, h) to the value of σ on d 1 ((g, h)); that is, τ(g, h) = gσ(h) σ(gh) + σ(g). This is 0 if and only if σ(gh) = σ(g) + gσ(h) for all g, h G, giving the formula for Z 1 (G, M). Similarly, the formula d 0 ((g)) = (g 1) ( ) from lectures (where denotes the unique element of G 0 ) shows that the map C 0 (G, M) C 1 (G, M) sends σ : G 0 M to τ : G 1 M defined by τ(g) = (g 1)σ( ), which gives the formula for B 1 (G, M). 3
4 (b) [2 points] Let M be a Gmodule. An extension of Z by M is a short exact sequence of Gmodules 0 M E Z 0, for some E; two such extensions are equivalent if there is a morphism between the two short exact sequences which is the identity on M and on Z. Show that there is a bijection between H 1 (G, M) and the equivalence classes of extensions of Z by M. Solution: If σ Z 1 (G, M), we may construct an extension of Z by M by defining E = M Z as an abelian group, and letting g G map (0, 1) M Z to (σ(g), 1). The cocycle condition σ(gh) = σ(g) + gσ(h) is exactly the condition required for this to be a group action. If two cocycles σ, σ lead to isomorphic extensions, then the isomorphism must send (0, 1) E to some element of the form (m, 1) E, and this shows that σ = σ + d 0 (m), so σ and σ differ by a coboundary. Conversely, if σ and σ differ by a coboundary d 0 (m), then translating by m gives an isomorphism between the two extensions. It remains to see that every extension arises in this way; but if E is an extension, then it must split as a short exact sequence of abelian groups (since Z is a projective object) and hence we may choose an isomorphism of groups E = M Z; let e be the element of E corresponding to 1 Z. If g G, then ge e maps to 0 in Z, so it must lie in M and this gives a map G M which is the required cocycle. (Alternatively, one can check that the cohomology class associated to E is the image of 1 Z under the coboundary map H 0 (G, Z) H 1 (G, M) in the cohomology long exact sequence.) 7. [2 points] Let G be a group and M a k[g]module, where k is a field of characteristic 0. Show that for any normal subgroup H G of finite index, the restriction map is an isomorphism. res G/H : H 1 (G, M) H 1 (H, M) G/H Solution: From the inflationrestriction exact sequence (the fiveterm exact sequence of lowdegree terms arising from the Hochschild Serre spectral sequence), we have H 1 (G/H, M H ) H 1 (G, M) res G/H H 1 (H, M) G/H H 2 (G/H, M H ). So it suffices to show that H 1 (G/H, M H ) = H 2 (G/H, M H ) = 0. I claim that for any finite group Γ, and k a field of characteristic 0 (or any finite characteristic not dividing Γ ), the cohomology functors H i (Γ, ) vanish on the category of k[γ]modules. The map a γ [γ] 1 Γ ( a γ )( [γ]) γ γ is a projection from k[γ] to a direct summand isomorphic to the trivial representation k; thus k is a projective object in k[γ]mod, so the functors Ext i k[γ] (k, ) are zero for i [3 points] Let G = C 2 = {1, σ}. (a) Show that... Z[G] σ 1 Z[G] σ+1 Z[G] σ 1 Z[G] is a projective resolution of the trivial module, as a Z[G]module. 4
5 Solution: This is obviously a complex since (σ + 1)(σ 1) = σ 2 1 = 1 1 = 0. So we must check it has the correct cohomology. A generic element of Z[G] looks like a + bσ for a, b Z, and we have (σ ± 1)(a + bσ) = (a ± b)(σ ± 1). Thus the image of multiplication by σ ± 1 is Z (σ ± 1), and its kernel is {a + bσ : a ± b = 0} = Z (σ 1), which proves that the cohomology in all degrees = 0 is trivial. Meanwhile, the image of the last map is exactly the kernel of the surjection Z[G] Z, a + bσ a + b. (b) Hence compute the cohomology groups of i. Z with the trivial Gaction; Solution: We must compute the cohomology of the complex of homomorphisms from the above resolution to Z, which is Z 0 Z 2 Z 0 Z... So the cohomology is Z in degree 0, trivial in odd degrees, and Z/2Z in positive even degrees. ii. Z with the generator σ acting as 1. Solution: Now we obtain the complex Z 2 Z 0 Z 2 Z... so the cohomology is 0 in even degrees (including degree 0) and Z/2Z in odd degrees. 9. [3 points] Let G be the infinite dihedral group, which has two generators a, b satisfying b 2 = 1 and bab = a 1. Let H be the normal subgroup generated by a. Write down the E 2 sheet of the Hochschild Serre spectral sequence for the cohomology of the trivial Gmodule Z in terms of cohomology of H and G/H, and hence show that the groups H k (G, Z) (for trivial Gaction on Z) are Z for k = 0, trivial for k odd, and have order 4 for k even. (You may assume that the generator of G/H = C 2 acts on H 1 (H, Z) as 1.) Solution: Recall that the Hochschild Serre spectral sequence has E pq 2 = H p (G/H, H q (H, )). From 8.b.i above, the q = 0 row of the E 2 page starts with a Z at (p, q) = (0, 0), and then alternates 0 s and Z/2Z s. From 8.b.ii, the q = 1 row alternates starting with a zero term at (0, 1). So the E 2 page looks like q 1 0 Z/2Z 0 Z/2Z 0 Z 0 Z/2Z p All differentials (the dotted arrows) have either source or target zero; so they are the zero maps, and the E 3 page is the same as the E 2 page. Once we get to E 3 and beyond, the differentials will 5
6 land outside the 2 strip in which the nonzero entries lie, so all differentials are zero from the E 2 page onwards and the E page is the same as the E 2 one. Hence H 0 (G, Z) = Z (duh); H k (G, Z) is zero for odd k; and for positive even k, the group H k (G, Z) has a 2step filtration with sub and quotient both isomorphic to Z/2Z, so it must be either Z/4Z or (Z/2Z) 2, and in particular it has order 4. (In fact the evendegree cohomology groups are (Z/2Z) 2 ; compare Exercise of Weibel, which is the analogous computation for group homology H n (G, Z).) 10. [Not assessed] Find an pair of δfunctors S = (S n ) n 0 and T = (T n ) n 0 such that S 0 = T 0 but S is not isomorphic to T. Solution: One (rather silly) source of examples is as follows. If F 0, F 1,... is any sequence of exact functors, then one can regard (F n ) n 0 as a δfunctor with all differentials δ n being zero, and it is clear that F 0 does not in any way determine the rest of the F i. For instance, one can take C = Ab, S to be the zero functor, and T to be the functor given by this construction with F i = 0 for i = 1 and F 1 the identity functor. 6
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