Chapter 20

Prove that the example of the standard complex given in \(\$ 1\) is actually a complex, and is exact, so it gives a resolution of \(\mathbf{Z}\). [Hint: To show that the sequence of the standard complex is exact, choose an element \(z \in S\) and define \(h: E^{i} \rightarrow E^{i+1}\) by letting $$ h\left(x_{0} \ldots, x_{i}\right)=\left(z, x_{0}, \ldots, x_{i}\right) . $$ Prove that \(d h+h d=\) id, and that \(d d=0\). Exactness follows at once.]

Let \(G\) be a group. Use \(G\) as the set \(S\) in the standard complex. Define an action of \(G\) on the standard complex \(E\) by letting $$ x\left(x_{0} \ldots, x_{i}\right)=\left(x x_{0} \ldots, x x_{i}\right) . $$ Prove that each \(E_{i}\) is a free module over the group ring \(\mathbf{Z}[G] .\) Thus if we let \(R=\mathbf{Z}[G]\) be the group ring, and consider the category \(\operatorname{Mod}(G)\) of \(G\) -modules, then the standard complex gives a free resolution of \(\mathbf{Z}\) in this category.

The standard complex \(E\) was written in homogeneous form, so the boundary maps have a certain symmetry. There is another complex which exhibits useful features as follows. Let \(F^{i}\) be the free \(\left.\mathbf{Z} \mid G\right]\) -module having for basis \(i\) -tuples (rather than \((i+1)\) -tuples) \(\left(x_{1}, \ldots, x_{i}\right)\). For \(i=0\) we take \(F_{0}=\mathbf{Z}[G]\) itself. Define the boundary operator by the formula $$ \begin{array}{c} d\left(x_{1}, \ldots, x_{i}\right)=x_{1}\left(x_{2}, \ldots, x_{i}\right)+\sum_{j=1}^{i-1}(-1)^{f}\left(x_{1}, \ldots, x_{j} x_{j+1}, \ldots, x_{i}\right) \\ +(-1)^{i+1}\left(x_{1}, \ldots, x_{i}\right) \end{array} $$ Show that \(E=F\) (as complexes of \(G\) -modules) via the association $$ \left(x_{1}, \ldots, x_{i}\right) \mapsto\left(1, x_{1}, x_{1} x_{2}, \ldots, x_{1} x_{2} \cdots x_{i}\right) $$ and that the operator \(d\) given for \(F\) corresponds to the operator \(d\) given for \(E\) under this isomorphism.

If \(A\) is a \(G\) -module, let \(A^{G}\) be the submodule consisting of all elements \(v \in A\) such that \(x v=v\) for all \(x \in G\). Thus \(A^{G}\) has trivial \(G\) -action. (This notation is convenient, but is not the same as for the induced module of Chapter XVIII.) (a) Show that if \(H^{\varphi}(G, A)\) denotes the \(q\) -th homology of the complex \(\mathrm{Hom}_{\alpha}(E, A)\), then \(H^{\circ}(G, A)=A^{G} .\) Thus the left derived functors of \(A \mapsto A^{\circ}\) are the homology groups of the complex \(\mathrm{Hom}_{G}(E, A)\), or for that matter, of the complex \(\operatorname{Hom}(F, A)\), where \(F\) is as in Exercise \(3 .\) (b) Show that the group of 1 -cycles \(Z^{1}(G, A)\) consists of those functions \(f: G \rightarrow A\) satisfying $$ f(x)+x f(y)=f(x y) \text { for all } x, y \in G $$ Show that the subgroup of coboundaries \(B^{1}(G, A)\) consists of those functions \(f\) for which there exists an element \(a \in A\) such that \(f(x)=x a-a .\) The factor group is then \(H^{1}(G, A) .\) See Chapter \(V I\), \(\$ 10\) for the determination of a special case. (c) Show that the group of 2 -cocycles \(Z^{2}(G, A)\) consists of those functions \(f: G \rightarrow A\) satisfying $$ x f(y, z)-f(x y, z)+f(x, y z)-f(x, y)=0 $$ Such 2-cocycles are also called factor sets, and they can be used to describe isomorphism classes of group extensions, as follows.

Let \(W\) be a group and \(A\) a normal subgroup, written multiplicatively. Let \(G=W / A\) be the factor group. Let \(F: G \rightarrow W\) be a choice of coset representatives. Define $$ f(x, y)=F(x) F(y) F(x y)^{-1} . $$ (a) Prove that \(f\) is \(A\) -valued, and that \(f: G \times G \rightarrow A\) is a 2-cocycle. (b) Given a group \(G\) and an abelian group \(A\), we view an extension \(W\) as an exact sequence $$ 1 \rightarrow A \rightarrow W \rightarrow G \rightarrow 1 $$ Show that if two such extensions are isomorphic then the 2-cocycles associated to these extensions as in (a) define the same class in \(H^{\prime}(G, A)\). (c) Prove that the map which we obtained above from isomorphism classes of group extensions to \(H^{2}(G, A)\) is a bijection.

Let \(\lambda: G^{\prime} \rightarrow G\) be a group homomorphism. Then \(\lambda\) gives rise to an exact functor $$ \Phi_{A}: \operatorname{Mod}(G) \rightarrow \operatorname{Mod}\left(G^{\prime}\right) $$ because every \(G\) -module can be viewed as a \(G^{\prime}\) -module by defining the operation of \(\sigma^{\prime} \in G^{\prime}\) to be \(\sigma^{\prime} a=\lambda\left(\sigma^{\prime}\right) a .\) Thus we obtain a cohomology functor \(H^{\sigma} \circ \Phi_{\lambda}\). Let \(G^{\prime}\) be a subgroup of \(G .\) In dimension 0 , we have a morphism of functors \(\lambda^{*}: H_{G}^{0} \rightarrow H_{\sigma}^{0} \circ \Phi_{4}\) given by the inclusion \(A^{G} \hookrightarrow A^{G}=\Phi_{\lambda}(A)^{G^{*}}\). (a) Show that there is a unique morphism of \(\delta\) -functors $$ \lambda^{*}: H_{G} \rightarrow H_{G} \circ \Phi_{\lambda} $$ which has the above effect on \(H_{G}^{0}\). We have the following important special cases. Restriction. Let \(H\) be a subgroup of \(G\). Let \(A\) be a \(G\) -module. A function from \(G\) into \(A\) restricts to a function from \(H\) into \(A\). In this way, we get a natural homomorphism called the restriction $$ \text { res: } H^{q}(G, A) \rightarrow H^{q}(H, A) $$ Inflation. Suppose that \(H\) is normal in \(G\). Let \(A^{H}\) be the subgroup of \(A\) consisting of those elements fixed by \(H\). Then it is immediately verified that \(A^{H}\) is stable under \(G\), and so is a \(G / H\) -module. The inclusion \(A^{H} \subset \rightarrow A\) induces a homomorphism $$ H_{G}^{q}(u)=u_{q}: H^{q}\left(G, A^{H}\right) \rightarrow H^{q}(A) $$ Define the inflation $$ \inf _{G / H}^{H}: H^{\nabla}\left(G / H, A^{H}\right) \rightarrow H^{\top}(G, A) $$ as the composite of the functorial morphism \(H^{\circ}\left(G / H, A^{H}\right) \rightarrow H^{\varphi}\left(G, A^{H}\right)\) followed by the induced homomorphism \(u_{q}=H_{G}^{q}(u)\) as above. In dimension 0, the inflation gives the identity \(\left(A^{H}\right)^{G / H}=A^{G}\). (b) Show that the inflation can be expressed on the standard cochain complex by the natural map which to a function of \(G / H\) in \(A^{H}\) associates a function of \(G\) into \(A^{H} \subset A\). (c) Prove that the following sequence is exact. $$ 0 \rightarrow H^{\prime}\left(G / H, A^{H}\right) \stackrel{\text { inf }}{\rightarrow} H^{1}(G, A) \stackrel{\text { res }}{\rightarrow} H^{\prime}(H, A) $$ (d) Describe how one gets an operation of \(G\) on the cohomology functor \(H_{G}\) "by conjugation" and functoriality. (c) In (c), show that the image of restriction on the right actually lies in \(H^{\prime}(H, A)^{G}\) (the fixed subgroup under \(\left.G\right)\). Remark. There is an analogous result for higher cohomology groups, whose proof needs a spectral sequence of Hochschild-Serre. See [La 96]. Chapter VI, \(\$ 2\), Theorem \(2 .\) It is actually this version for \(H^{2}\) which is applied to \(H^{2}\left(G, K^{*}\right)\), when \(K\) is a Galois extension, and is used in class field theory [ArT 67].

Let \(G\) be a group, \(B\) an abelian group and \(M_{G}(B)=M(G, B)\) the set of mappings from \(G\) into \(B\). For \(x \in G\) and \(f \in M(G, B)\) define \(([x] f)(y)=f(y x)\). (a) Show that \(B \mapsto M_{G}(B)\) is a covariant, additive, exact functor from \(\operatorname{Mod}(\mathbf{Z})\) (category of abelian groups) into \(\operatorname{Mod}(G)\). (b) Let \(G^{\prime}\) be a subgroup of \(G\) and \(G=\bigcup_{x_{j}} G^{\prime}\) a coset decomposition. For \(f \in M(G, B)\) let \(f_{j}\) be the function in \(M\left(G^{\prime}, B\right)\) such that \(f_{j}(y)=f\left(x_{j} y\right)\). Show that the map $$ f \mapsto \prod_{j} f_{j} $$ is a \(G^{\prime}\) -isomorphism from \(M(G, B)\) to \(\prod_{j} M\left(G^{\prime}, B\right)\).

For each \(G\) -module \(A \in \operatorname{Mod}(G)\), define \(\varepsilon_{A}: A \rightarrow M(G, A)\) by the condition \(\varepsilon_{A}(a)=\) the function \(f_{a}\) such that \(f_{x}(\sigma)=\sigma a\) for \(\sigma \in G\). Show that \(a \mapsto f_{a}\) is a \(G\) -module embedding, and that the exact sequence $$ 0 \rightarrow A \stackrel{E}{\rightarrow} M(G, A) \rightarrow X_{A}=\text { coker } \varepsilon_{A} \rightarrow 0 $$ splits over \(\mathbf{Z}\). (In fact, the map \(f \mapsto f(e)\) splits the left side arrow.)

Let \(G\) be a group and \(S\) a subgroup. Show that the bifunctors \((A, B) \mapsto \operatorname{Hom}_{G}\left(A, M_{G}^{S}(B)\right)\) and \((A, B) \mapsto \operatorname{Hom}_{S}(A, B)\) on \(\operatorname{Mod}(G) \times \operatorname{Mod}(S)\) with value in \(\operatorname{Mod}(\mathbf{Z})\) are isomorphic. The isomorphism is given by the maps \(\varphi \mapsto\left(a \mapsto g_{a}\right)\), for \(\varphi \in \mathrm{Hom}_{5}(A, B)\), where \(g_{a}(\sigma)=\varphi(\sigma a), g_{a} \in M_{G}^{S}(B) .\) The inverse mapping is given by $$ f \mapsto f(1) \text { with } f \in \operatorname{Hom}_{G}\left(A, M_{G}^{S}(B)\right) . $$ Recall that \(M_{G}^{s}(B)\) was defined in Chapter \(X V I I I, \$ 7\) for the induced representation. Basically you should already know the above isomorphism.

Let \(G\) be a group and \(S\) a subgroup. Show that the map $$ H^{?}\left(G, M_{G}^{5}(B)\right) \rightarrow H^{4}(S, B) \text { for } B \in \operatorname{Mod}(S) $$ obtained by composing the restriction res? with the \(S\) -homomorphism \(f \mapsto f(1)\), is an isomorphism for \(q>0 .\) [Hint: Use the uniqueness theorem for cohomology functors.]

Let \(G\) be a group. Let \(\varepsilon: \mathbf{Z}[G] \rightarrow \mathbf{Z}\) be the homomorphism such that \(\varepsilon\left(\sum n(x) x\right)=\) \(\sum n(x) .\) Let \(I_{G}\) be its kernel. Prove that \(I_{G}\) is an ideal of \(\mathbf{Z}[G]\) and that there is an isomorphism of functors (on the category of groups) $$ G / G^{\varepsilon}=I_{G} / I_{G}^{2}, \quad \text { by } \quad x G^{c} \mapsto(x-1)+I_{G}^{2} . $$

Let \(A \in \operatorname{Mod}(G)\) and \(\alpha \in H^{\prime}(G, A) .\) Let \(\\{a(x)\\}_{x \in G}\) be a standard 1 -cocycle representing \(\alpha\). Show that there exists a \(G\) -homomorphism \(f: I_{G} \rightarrow A\) such that \(f(x-1)=a(x)\). so \(f \in\left(\mathrm{Hom}\left(I_{G}, A\right)\right)^{6}\). Show that the sequence $$ 0 \rightarrow A=\operatorname{Hom}(\mathbf{Z}, A) \rightarrow \operatorname{Hom}(\mathbf{Z}[G], A) \rightarrow \operatorname{Hom}\left(I_{G}, A\right) \rightarrow 0 $$ is exact, and that if \(\delta\) is the coboundary for the cohomology sequence, then \(\delta(f)=-\alpha .\)

(a) Show that a projective object in \(\operatorname{Mod}(G)\) is \(G\) -regular. (b) Let \(R\) be a commutative ring and let \(A\) be in \(\operatorname{Mod}_{R}(G)\) (the category of \((G, R)\) modules). Show that \(A\) is \(R[G]\) -projective if and only if \(A\) is \(R\) -projective and \(R[G]\) -regular, meaning that id \(_{A}=T_{G}(u)\) for some \(R\) -homomorphism \(u: A \rightarrow A\).

Let \(G\) be a finite cyclic group of order \(n\). Let \(\sigma\) be a generator of \(G\). Let \(K^{i}=\mathbf{Z}[G]\) for \(i>0 .\) Let \(\varepsilon: K^{0} \rightarrow \mathbf{Z}\) be the augmentation as before. For \(i\) odd \(\geq 1\), let \(d^{i}: K^{i} \rightarrow K^{i-1}\) be multiplication by \(1-\sigma .\) For \(i\) even \(\geq 2\), let \(d^{i}\) be multiplication by \(1+\sigma+\cdots+\sigma^{n-1}\). Prove that \(K\) is a resolution of \(\mathbf{Z}\). Conclude that: For i odd: \(H^{\prime}(G, A)=A^{G} / T_{G} A\) where \(T_{G}: a \mapsto\left(1+\sigma+\cdots+\sigma^{n-1}\right) a\); For i even \(\geq 2: H^{\prime}(G, A)=A_{T} /(1-\sigma) A\), where \(A_{T}\) is the kernel of \(T_{G}\) in \(A .\)

Let \(G\) be a finite group. Show that there exists a \(\delta\) -functor \(\mathbf{H}\) from \(\operatorname{Mod}(G)\) to Mod \((\mathbf{Z})\) such that: (1) \(\mathbf{H}^{0}\) is (isomorphic to) the functor \(A \mapsto A^{G} / T_{G} A\). (2) \(\mathbf{H}^{q}(A)=0\) if \(A\) is injective and \(q>0\), and \(\mathrm{H}^{q}(A)=0\) if \(A\) is projective and \(q\) is arbitrary. (3) \(\mathbf{H}\) is erased by \(G\) -regular modules. In particular, \(\mathbf{H}\) is erased by \(M_{G}\). The \(\delta\) -functor of Exercise 17 is called the special cohomology functor. It differs from the other one only in dimension \(0 .\)

Let \(\mathbf{H}=\mathbf{H}_{G}\) be the special cohomology functor for a finite group \(G\). Show that: $$ \begin{array}{l} \mathbf{H}^{0}\left(I_{G}\right)=0 ; \mathbf{H}^{0}(Z) \sim \mathbf{H}^{1}(I)=\mathbf{Z} / n \mathbf{Z} \text { where } n=\\#(G) ; \\ \mathbf{H}^{\circ}(Q / Z)=\mathbf{H}^{1}(Z)=\mathbf{H}^{2}(I)=0 \\ \mathbf{H}^{\prime}(Q / Z)=\mathbf{H}^{2}(Z)=\mathbf{H}^{3}(I) \approx G^{\wedge}=\operatorname{Hom}(G, \mathbf{Q} / \mathbf{Z}) \text { by definition. } \end{array} $$

(a) Show that if an abelian group \(T\) is injective in the category of abelian groups, then It is divisible. (b) Let \(A\) be a principal ent?e ring. Define the notion of divisibility by elements of \(A\) for modules in a manner analogous to that for abelian groups. Show that an \(A\) module is injective if and only if it is \(A\) -divisible. [The proof for \(Z\) should work in exactly the same way.]

Let \(S\) be a multiplicative subset of the commutative Noetherian ring \(A\). If \(I\) is an injective \(A\) -module, show that \(S^{-1} I\) is an injective \(S^{-1} A\) -module.

(a) Show that a direct sum of projective modules is projective. (b) Show that a direct product of injective modules is injective.

Show that a factor module, direct summand, direct product, and direct sum of divisible modules are divisible.

Let \(Q\) be a module over a commutative ring \(A\). Assume that for every left ideal \(J\) of A, every homomorphism \(\varphi: J \rightarrow Q\) can be extended to a homomorphism of \(A\) into Q. Show that \(Q\) is injective. [Hint: Given \(M^{\prime} \subset M\) and \(f: M^{\prime} \rightarrow Q\). let \(x_{0} \in M\) and \(x_{0} \notin M^{\prime}\). Let \(J\) be the left ideal of elements \(a \in A\) such that \(a x_{0} \in M\) : Let \(\varphi(a)=f\left(a x_{0}\right)\) and extend \(\varphi\) to \(A\), as can be done by hypothesis. Then show that one can extend \(f\) to \(M\) by the formula $$ f\left(x^{\prime}+b x_{0}\right)=f\left(x^{\prime}\right)+\varphi(b) $$ for \(x^{\prime} \in M\) and \(b \in A\). Then use Zorn's lemma. This is the same pattern of proof as the proof of Lemma \(4.2 .1\)

Let $$ 0 \rightarrow I_{1} \rightarrow I_{2} \rightarrow I_{3} \rightarrow 0 $$ be an exact sequence of modules. Assume that \(I_{1}, I_{2}\) are injective. (a) Show that the sequence splits. (b) Show that \(I_{3}\) is injective. (c) If \(I\) is injective and \(I=M \oplus N\), show that \(M\) is injective.

Let \(A\) be a commutative ring. Let \(E\) be an \(A\) -module, and let \(E^{\wedge}=\operatorname{Hom}_{\mathbf{Z}}(E, \mathbf{Q} / \mathbf{Z})\) be the dual module, Prove the following statements. (a) A sequence $$ 0 \rightarrow N \rightarrow M \rightarrow E \rightarrow 0 $$ is exact if and only if the dual sequence $$ 0 \rightarrow E^{\wedge} \rightarrow M^{\wedge} \rightarrow N^{\wedge} \rightarrow 0 $$ is exact. (b) Let \(F\) be flat and \(I\) injective in the category of \(A\) -modules. Show that \(\mathrm{Hom}_{A}(F, I)\) is injective. (c) \(E\) is flat if and only if \(E^{\wedge}\) is injective.

Let \(M, N\) be modules over a ring. By an extension of \(M\) by \(N\) we mean an exact sequence \(\left(^{*}\right)\) $$ 0 \rightarrow N \rightarrow E \rightarrow M \rightarrow 0 $$ We shall now define a map from such extensions to \(\mathrm{Ext}^{1}(M, N)\). Let \(P\) be projective. with a surjective homomorphism onto \(M\), so we get an exact sequence \((* *)\) $$ 0 \rightarrow K \stackrel{w}{\rightarrow} P \stackrel{p}{\rightarrow} M \rightarrow 0 $$ where \(K\) is defined to be the kernel. Since \(P\) is projective, there exists a homomorphism \(u: P \rightarrow E\), and depending on \(u\) a unique homomorphism \(v: K \rightarrow N\) making the diagram commutative: On the other hand, we have the exact sequence \((* * *) \quad 0 \rightarrow \operatorname{Hom}(M, N) \rightarrow \operatorname{Hom}(P, N) \rightarrow \operatorname{Hom}(K, N) \rightarrow\) Ext \(^{1}(M, N) \rightarrow 0\), with the last term on the right being equal to 0 because Ext \(^{\prime}(P, N)=0 .\) To the extension \(\left(^{*}\right)\) we associate the image of \(v\) in \(\operatorname{Ext}^{1}(M, N)\). Prove that this association is a bijection between isomorphism classes of extensions (i.e. isomorphism classes of exact sequences as in \(\left.\left({ }^{*}\right)\right)\), and \(\operatorname{Ext}^{1}(M, N)\). [Hint: Construct an inverse as follows. Given an element \(e\) of Ext ' \((M, N)\), using an exact sequence \((* *)\), there is some element \(v \in \operatorname{Hom}(K, N)\) which maps on \(e\) in \((* * *)\). Let \(E\) be the push-out of \(v\) and \(w\). In other words, let \(J\) be the submodule of \(N \oplus P\) consisting of all elements \((v(x),-w(x))\) with \(x \in K\), and let \(E=(N \oplus P) / J .\) Show that the map \(y \mapsto(y, 0)\) mod \(J\) gives an injection of \(N\) into \(E\). Show that the map \(N \oplus P \rightarrow M\) vanishes on \(J\), and so gives a surjective homomorphism \(E \rightarrow M \rightarrow 0\). Thus we obtain an exact sequence \(\left(^{*}\right) ;\) that is, an extension of \(M\) by \(N\). Thus to each element of Ext \(^{1}(M, N)\) we have associated an isomorphism class of extensions of \(M\) by \(N\). Show that the maps we have defined are inverse to each other between isomorphism classes of extensions and elements of Ext \(\left.{ }^{\prime}(M, N) .\right]\)

Let \(R\) be a principal entire ring. Let \(a \in R .\) For every \(R\) -module \(N\), prove: (a) Ext \(^{\prime}(R / a R, N)=N / a N\). (b) For \(b \in R\) we have Ext \(^{1}(R / a R, R / b R)=R /(a, b)\), where \((a, b)\) is the \(\mathrm{g.c.d}\) of \(a\) and \(b\), assuming \(a b \neq 0\).

Let \(K=\bigoplus K_{b}\) and \(L=\bigoplus L_{4}\) be two complexes indexed by the integers, and with boundary maps lower indices by 1. Define \(K \otimes L\) to be the direct sum of the modules \((K \otimes L)_{n}\), where $$ (K \otimes L)_{n}=\bigoplus_{p+q=n} K_{p} \otimes L_{q} $$ Show that there exist unique homomorphisms $$ d=d_{n}:(K \otimes L)_{n} \rightarrow(K \otimes L)_{n-1} $$ such that $$ d(x \otimes y)=d(x) \otimes y+(-1)^{\prime} x \otimes d(y) . $$ Show that \(K \otimes L\) with these homomorphisms is a complex, that is \(d \circ d=0\).

Let \(K, L\) be double complexes. We write \(K_{i}\) and \(L_{i}\) for the ordinary column complexes of \(K\) and \(L\) respectively. Let \(\varphi: K \rightarrow L\) be a homomorphism of double complexes. Assume that each homomorphism $$ \varphi_{i}: K_{i} \rightarrow L_{i} $$ is a homology isomorphism. (a) Prove that \(\operatorname{Tot}(\varphi): \operatorname{Tot}(K) \rightarrow \operatorname{Tot}(L)\) is a homology isomorphism. (If you want to see this worked out, cf. [FuL 85]. Chapter \(\mathrm{V}\), Lemma \(5.4 .\) ) (b) Prove Theorem \(9.8\) using (a) instead of spectral sequences.