4
$\begingroup$

Let $k$ be a finite field, and let $G$ be the absolute Galois group of $k$, which is isomorphic to $\widehat{\mathbb{Z}}$. Let $\mathcal{C}$ be the category of $G$-modules. Then, we have the following:

For a finite $G$-module $N$, we have $$ Ext^r_{\mathcal{C}}(N, \mathbb{Z}) \simeq H^{r-1}(G,~N^D), $$ where $N^D:=\operatorname{Hom}_\mathbb{Z}(N, \mathbb{Q}/\mathbb{Z})$ is the Pontrjagin dual of $N$, with the natural $G$-action by $(g\phi)(x):=g(\phi(g^{-1}(x)))=\phi(g^{-1}(x))$ for $g\in G$, $x \in N$ and $\phi \in N^D$.

Why is this true?

$\endgroup$
  • $\begingroup$ Note that $H^k(G;M)=\operatorname{Ext}^k_{\mathcal C}(\mathbb Z,M)$ $\endgroup$ – მამუკა ჯიბლაძე Mar 23 '17 at 18:28
  • $\begingroup$ Take the Pontryagin dual of the extension given in the above comment. This is an element of $\operatorname{Ext}^k_{\mathcal{C}}(M^D,\mathbb{Q}/\mathbb{Z})$. Now use the isomorphism to $\operatorname{Ext}^{k+1}_{\mathcal{C}}(M^D,\mathbb{Z})$ in R. van Dobben de Bruyn's answer. Or do I miss something? $\endgroup$ – Chris Wuthrich Mar 24 '17 at 12:15
  • $\begingroup$ @Chris I wonder that $\text{Ext}_\mathcal{C}^k (\mathbb{Z}, M)$ is isomorphic to $\text{Ext}_{\mathcal{C}}^k (M^D, \mathbb{Z}^D)$, by taking Pontryagin duality. I saw a reference that this is true for $k=0, 1$ though $\endgroup$ – user1225 Mar 27 '17 at 5:48
  • $\begingroup$ When $N$ is finite, $\text{Ext}_\mathcal{C}^k (N, \mathbb{Z})=0$ for $k \neq 1$ by Example 1.10 of chapter I of Milne's book on Arithmetic Duality theorem (2nd edition). And as already answered below, $\text{Ext}_\mathcal{C}^1 (N, \mathbb{Z})=\text{Ext}_\mathcal{C}^0 (N, \mathbb{Q}/\mathbb{Z}) = \text{Hom}_{G}(N, \mathbb{Q}/\mathbb{Z})=\text{Hom}_{G}(\mathbb{Z}, N^D)=H^0(G, N^D)$. $\endgroup$ – user1225 Mar 27 '17 at 5:56
6
$\begingroup$

As noted in R. van Dobben de Bruyn's answer, since $N$ is killed by some positive integer we can reformulate the problem as that of constructing natural isomorphisms ${\rm{Ext}}^i_G(N, \mathbf{Q}/\mathbf{Z}) \simeq {\rm{H}}^i(G, N^D)$ for any $i \ge 0$ and any finite discrete $G$-module $N$ (with $G := {\rm{Gal}}(k_s/k)$ for a finite field $k$). We will do this for any field $k$; finiteness is irrelevant.

It is convenient to work in the framework of abelian etale sheaves on ${\rm{Spec}}(k)$. From this viewpoint, a discrete $G$-module $M$ corresponds to an abelian etale sheaf $\mathscr{F}_M$. The local-to-global Ext spectral sequence (which uses that sheaf-Ext into an injective abelian sheaf is flasque and hence acyclic for etale cohomology) is $${\rm{H}}^i(G, \mathscr{E}xt^j_k(\mathscr{F}_N, \mathscr{F}_M)) \Rightarrow {\rm{Ext}}^{i+j}_k(\mathscr{F}_N, \mathscr{F}_M) = {\rm{Ext}}^{i+j}_G(N,M)$$ for any discrete $G$-modules $N$ and $M$.

Now assume $N$ is finite, so $\mathscr{F}_N$ becomes constant over an etale cover of ${\rm{Spec}}(k)$. It is then not hard to show by an erasable $\delta$-functor argument that $\mathscr{E}xt^{\bullet}_k(\mathscr{F}_N,\mathscr{F}_M) = {\rm{Ext}}^{\bullet}_{\rm{Ab}}(N,M)$ as discrete $G$-modules. Hence, the spectral sequence takes the form $${\rm{H}}^i(G, {\rm{Ext}}^j_{\rm{Ab}}(N,M)) \Rightarrow {\rm{Ext}}^{i+j}_G(N,M)$$ for any discrete $G$-module $M$ and finite discrete $G$-module $N$.

Now set $M$ to be the constant $G$-module $\mathbf{Q}/\mathbf{Z}$. Then ${\rm{Hom}}_{\rm{Ab}}(\cdot, M)$ is exact, so ${\rm{Ext}}^{\bullet}_{\rm{Ab}}(\cdot,M)$ vanishes in positive degrees. The spectral sequence with this $M$ therefore degenerates to give isomorphisms $${\rm{H}}^i(G, N^D) \simeq {\rm{Ext}}^i_G(N, \mathbf{Q}/\mathbf{Z})$$ for all $i > 0$, as desired. (To be really useful one should address $\delta$-functoriality in $N$, but this was not requested in the question.)

$\endgroup$
  • $\begingroup$ Could you explain more specifically your statement that the sheaf Ext is equal to $Ext_{Ab}(N, M)$ as discrete $G$-modules? $\endgroup$ – user1225 Mar 24 '17 at 6:23
  • $\begingroup$ For any scheme $X$, $\mathscr{E}xt^{\bullet}_X(\mathscr{F}, \mathscr{G})$ is sheafification of $U\mapsto {\rm{Ext}}^{\bullet}_U(\mathscr{F}|_U,\mathscr{G}|_U)$ (univ. $\delta$-functor proof). Thus, for $X = {\rm{Spec}}(k)$, the $k_s$-stalk of such sheaf-Ext is direct limit of ${\rm{Ext}}_K^{\bullet}(\mathscr{F}_K,\mathscr{G}_K)$ ($K/k$ finite inside $k_s$), and the natural map from this to ${\rm{Ext}}^{\bullet}_{\rm{Ab}}(\mathscr{F}_{k_s}, \mathscr{G}_{k_s})$ is an isomorphism for lcc $\mathscr{F}$ by reduction to $\mathscr{F}=\mathbf{Z}$ and compatibility of cohomology with limits of schemes. $\endgroup$ – nfdc23 Mar 24 '17 at 9:31
  • $\begingroup$ I'm sure there is a more direct elementary argument; that is just what first came to mind via general cohomological principles and inspired by the usual proof of compatibility of module-Ext with localization when the first argument of module-Ext is finitely presented (for a much wider cohomological context of the same, way beyond just on Spec of fields, see Prop. 4.18 and the associated footnote that are both stated without proof in Chapter I of the book of Freitag & Kiehl on etale cohomology, which focuses there on $n$-torsion sheaves). $\endgroup$ – nfdc23 Mar 24 '17 at 9:39
9
$\begingroup$

Consider the short exact sequence $$0 \to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0.$$ Note that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q) = 0$ for all $i$: it is torsion since $N$ is torsion, but multiplication by $n \in \mathbb Z_{>0}$ is an isomorphism since it is so on $\mathbb Q$. Thus, the sequence above induces isomorphisms $$\operatorname{Ext}^{i-1}_\mathcal C(N,\mathbb Q/\mathbb Z) \stackrel\sim\to \operatorname{Ext}^i_\mathcal C(N,\mathbb Z).$$ Thus, we only have to show that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z) = H^i(G, N^D)$. I will treat the finite group case, and trust that the OP can carry out the limit argument to treat the profinite group $\hat {\mathbb Z}$ (assuming that you are working with discrete [topological] $G$-modules).

Note that if $P$ is a projective $\mathbb Z[G]$-module, then $\operatorname{Hom}_\mathbb Z(P,\mathbb Q/\mathbb Z)$ is an acyclic $\mathbb Z[G]$-module. Indeed, we prove this in three steps:

  1. Let $P = \mathbb Z[G]$. Then we get the co-induced module $M^G(\mathbb Q/\mathbb Z)$, which is acyclic by Shapiro's lemma.

  2. The case of free modules follows by taking sums: $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ turns sums into products, and a product of acyclic modules is acyclic. Indeed (by abstract nonsense): $(-)^G$ has a left adjoint $- \otimes_\mathbb Z \mathbb Z[G]$, hence preserves products (you can also easily prove this by hand). Since products of injectives are injective and products are exact, we can take them out of $H^i(G,-)$ as well.

  3. The general case follows since any projective module is the summand of a free module.

Now consider the composition of the functors $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z) \colon \mathcal C^{\operatorname{op}} \to \mathcal C$ and $(-)^G \colon \mathcal C \to \operatorname{\underline{Ab}}$. The above argument shows that $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ takes projectives to acyclics, hence there is a Grothendieck spectral sequence $$E_2^{pq} = H^p(G,\operatorname{Ext}^q_\mathbb Z(N,\mathbb Q/\mathbb Z)) \Rightarrow \operatorname{Ext}^{p+q}_{\mathbb Z[G]}(N,\mathbb Q/\mathbb Z).$$ But $\mathbb Q/\mathbb Z$ is injective as $\mathbb Z$-module, hence $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z) = 0$ for $i > 0$. Thus, the spectral sequence collapses on the $E_2$ page, and we conclude that $$H^i(G,\operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)) = \operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z).$$ $\square$

Remark. If you don't like spectral sequences, you can also give a more concrete proof of the last part. Indeed, consider a free resolution as $\mathbb Z[G]$-modules $$\ldots \to P_1 \to P_0 \to N.$$ Since $\mathbb Z[G]$ is a free $\mathbb Z$-module, this resolution is also a free resolution as $\mathbb Z$-modules. Thus, it computes both $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z)$ and $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$. More specifically, the former can be computed as the cohomology of the complex $$0 \to \operatorname{Hom}_\mathbb Z(P_0,\mathbb Q/\mathbb Z) \to \operatorname{Hom}_\mathbb Z(P_1,\mathbb Q/\mathbb Z) \to \ldots,\tag{1}\label{Seq 1}$$ and then the latter can be computed as the cohomology of $$0 \to \operatorname{Hom}_{\mathbb Z[G]}(P_0,\mathbb Q/\mathbb Z) \to \operatorname{Hom}_{\mathbb Z[G]}(P_1,\mathbb Q/\mathbb Z) \to \ldots.\tag{2}\label{Seq 2}$$ Now (\ref{Seq 1}) is exact since $\mathbb Q/\mathbb Z$ is an injective $\mathbb Z$-module. By what we argued above, it is actually an acyclic resolution $Q^\bullet$ of $N^D = \operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)$, so we can use it to compute group cohomology. Thus, $H^i(G,N^D)$ is computed as the cohomology of the complex $$0 \to (Q^0)^G \to (Q^1)^G \to (Q^2)^G \to \ldots,$$ which on the other hand computes $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$ by (\ref{Seq 2}). $\square$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.