# Galois cohomology of finite fields

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?

• Note that $H^k(G;M)=\operatorname{Ext}^k_{\mathcal C}(\mathbb Z,M)$ – მამუკა ჯიბლაძე Mar 23 '17 at 18:28
• 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? – Chris Wuthrich Mar 24 '17 at 12:15
• @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 – user1225 Mar 27 '17 at 5:48
• 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, $\text{Ext}_\mathcal{C}^1 (N, \mathbb{Z})=\text{Ext}_\mathcal{C}^0 (N, \mathbb{Q}/\mathbb{Z}) = \text{Hom}_{G}(N, \mathbb{Q}/\mathbb{Z})=\text{Hom}_{G}(\mathbb{Z}, N^D)=H^0(G, N^D)$. – user1225 Mar 27 '17 at 5:56

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.)

• Could you explain more specifically your statement that the sheaf Ext is equal to $Ext_{Ab}(N, M)$ as discrete $G$-modules? – user1225 Mar 24 '17 at 6:23
• 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. – nfdc23 Mar 24 '17 at 9:31
• 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). – nfdc23 Mar 24 '17 at 9:39

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}). $\square$