Theorem 59.56.3 (03QT)—The Stacks project

Theorem 59.56.3. Let $S = \mathop{\mathrm{Spec}}(K)$ with $K$ a field. Let $\overline{s}$ be a geometric point of $S$ . Let $G = \text{Gal}_{\kappa (s)}$ denote the absolute Galois group. Taking stalks induces an equivalence of categories

$\mathop{\mathit{Sh}}\nolimits (S_{\acute{e}tale}) \longrightarrow G\textit{-Sets}, \quad \mathcal{F} \longmapsto \mathcal{F}_{\overline{s}}.$

Proof. Let us construct the inverse to this functor. In Fundamental Groups, Lemma 58.2.2 we have seen that given a $G$ -set $M$ there exists an étale morphism $X \to \mathop{\mathrm{Spec}}(K)$ such that $\mathop{\mathrm{Mor}}\nolimits _ K(\mathop{\mathrm{Spec}}(K^{sep}), X)$ is isomorphic to $M$ as a $G$ -set. Consider the sheaf $\mathcal{F}$ on $\mathop{\mathrm{Spec}}(K)_{\acute{e}tale}$ defined by the rule $U \mapsto \mathop{\mathrm{Mor}}\nolimits _ K(U, X)$ . This is a sheaf as the étale topology is subcanonical. Then we see that $\mathcal{F}_{\overline{s}} = \mathop{\mathrm{Mor}}\nolimits _ K(\mathop{\mathrm{Spec}}(K^{sep}), X) = M$ as $G$ -sets (details omitted). This gives the inverse of the functor and we win. $\square$

Comments (2)

Comment #888 by Konrad Voelkel on August 07, 2014 at 20:37

The formulation "the functor above" threw me off, since I had to look up broader context to figure out that the functor "above" is just the functor written down in the Theorem.

Comment #905 by Johan on August 10, 2014 at 13:09

OK, I changed the wording of the theorem, see here.

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2 comment(s) on Section 59.56: Galois action on stalks

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