Section 59.56 (03QW): Galois action on stalks

59.56 Galois action on stalks

In this section we define an action of the absolute Galois group of a residue field of a point $s$ of $S$ on the stalk functor at any geometric point lying over $s$ .

Galois action on stalks. Let $S$ be a scheme. Let $\overline{s}$ be a geometric point of $S$ . Let $\sigma \in \text{Aut}(\kappa (\overline{s})/\kappa (s))$ . Define an action of $\sigma$ on the stalk $\mathcal{F}_{\overline{s}}$ of a sheaf $\mathcal{F}$ as follows

59.56.0.1

$\begin{equation} \label{etale-cohomology-equation-galois-action} \begin{matrix} \mathcal{F}_{\overline{s}} & \longrightarrow & \mathcal{F}_{\overline{s}} \\ (U, \overline{u}, t) & \longmapsto & (U, \overline{u} \circ \mathop{\mathrm{Spec}}(\sigma ), t). \end{matrix} \end{equation}$

where we use the description of elements of the stalk in terms of triples as in the discussion following Definition 59.29.6. This is a left action, since if $\sigma _ i \in \text{Aut}(\kappa (\overline{s})/\kappa (s))$ then

$\begin{align*} \sigma _1 \cdot (\sigma _2 \cdot (U, \overline{u}, t)) & = \sigma _1 \cdot (U, \overline{u} \circ \mathop{\mathrm{Spec}}(\sigma _2), t) \\ & = (U, \overline{u} \circ \mathop{\mathrm{Spec}}(\sigma _2) \circ \mathop{\mathrm{Spec}}(\sigma _1), t) \\ & = (U, \overline{u} \circ \mathop{\mathrm{Spec}}(\sigma _1 \circ \sigma _2), t) \\ & = (\sigma _1 \circ \sigma _2) \cdot (U, \overline{u}, t) \end{align*}$

It is clear that this action is functorial in the sheaf $\mathcal{F}$ . We note that we could have defined this action by referring directly to Remark 59.29.8.

Definition 59.56.1. Let $S$ be a scheme. Let $\overline{s}$ be a geometric point lying over the point $s$ of $S$ . Let $\kappa (s) \subset \kappa (s)^{sep} \subset \kappa (\overline{s})$ denote the separable algebraic closure of $\kappa (s)$ in the algebraically closed field $\kappa (\overline{s})$ .

In this situation the absolute Galois group of $\kappa (s)$ is $\text{Gal}(\kappa (s)^{sep}/\kappa (s))$ . It is sometimes denoted $\text{Gal}_{\kappa (s)}$ .
The geometric point $\overline{s}$ is called algebraic if $\kappa (s) \subset \kappa (\overline{s})$ is an algebraic closure of $\kappa (s)$ .

Example 59.56.2. The geometric point $\mathop{\mathrm{Spec}}(\mathbf{C}) \to \mathop{\mathrm{Spec}}(\mathbf{Q})$ is not algebraic.

Let $\kappa (s) \subset \kappa (s)^{sep} \subset \kappa (\overline{s})$ be as in the definition. Note that as $\kappa (\overline{s})$ is algebraically closed the map

$\text{Aut}(\kappa (\overline{s})/\kappa (s)) \longrightarrow \text{Gal}(\kappa (s)^{sep}/\kappa (s)) = \text{Gal}_{\kappa (s)}$

is surjective. Suppose $(U, \overline{u})$ is an étale neighbourhood of $\overline{s}$ , and say $\overline{u}$ lies over the point $u$ of $U$ . Since $U \to S$ is étale, the residue field extension $\kappa (u)/\kappa (s)$ is finite separable. This implies the following

If $\sigma \in \text{Aut}(\kappa (\overline{s})/\kappa (s)^{sep})$ then $\sigma$ acts trivially on $\mathcal{F}_{\overline{s}}$ .
More precisely, the action of $\text{Aut}(\kappa (\overline{s})/\kappa (s))$ determines and is determined by an action of the absolute Galois group $\text{Gal}_{\kappa (s)}$ on $\mathcal{F}_{\overline{s}}$ .
Given $(U, \overline{u}, t)$ representing an element $\xi$ of $\mathcal{F}_{\overline{s}}$ any element of $\text{Gal}(\kappa (s)^{sep}/K)$ acts trivially, where $\kappa (s) \subset K \subset \kappa (s)^{sep}$ is the image of $\overline{u}^\sharp : \kappa (u) \to \kappa (\overline{s})$ .

Altogether we see that $\mathcal{F}_{\overline{s}}$ becomes a $\text{Gal}_{\kappa (s)}$ -set (see Fundamental Groups, Definition 58.2.1). Hence we may think of the stalk functor as a functor

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

and from now on we usually do think about the stalk functor in this way.

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. $\square$

Remark 59.56.4. Another way to state the conclusion of Theorem 59.56.3 and Fundamental Groups, Lemma 58.2.2 is to say that every sheaf on $\mathop{\mathrm{Spec}}(K)_{\acute{e}tale}$ is representable by a scheme $X$ étale over $\mathop{\mathrm{Spec}}(K)$ . This does not mean that every sheaf is representable in the sense of Sites, Definition 7.12.3. The reason is that in our construction of $\mathop{\mathrm{Spec}}(K)_{\acute{e}tale}$ we chose a sufficiently large set of schemes étale over $\mathop{\mathrm{Spec}}(K)$ , whereas sheaves on $\mathop{\mathrm{Spec}}(K)_{\acute{e}tale}$ form a proper class.

Lemma 59.56.5. Assumptions and notations as in Theorem 59.56.3. There is a functorial bijection

$\Gamma (S, \mathcal{F}) = (\mathcal{F}_{\overline{s}})^ G$

Proof. We can prove this using formal arguments and the result of Theorem 59.56.3 as follows. Given a sheaf $\mathcal{F}$ corresponding to the $G$ -set $M = \mathcal{F}_{\overline{s}}$ we have

$\begin{eqnarray*} \Gamma (S, \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (S_{\acute{e}tale})}(h_{\mathop{\mathrm{Spec}}(K)}, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{G\textit{-Sets}}(\{ *\} , M) \\ & = & M^ G \end{eqnarray*}$

Here the first identification is explained in Sites, Sections 7.2 and 7.12, the second results from Theorem 59.56.3 and the third is clear. We will also give a direct proof ¹.

Suppose that $t \in \Gamma (S, \mathcal{F})$ is a global section. Then the triple $(S, \overline{s}, t)$ defines an element of $\mathcal{F}_{\overline{s}}$ which is clearly invariant under the action of $G$ . Conversely, suppose that $(U, \overline{u}, t)$ defines an element of $\mathcal{F}_{\overline{s}}$ which is invariant. Then we may shrink $U$ and assume $U = \mathop{\mathrm{Spec}}(L)$ for some finite separable field extension of $K$ , see Proposition 59.26.2. In this case the map $\mathcal{F}(U) \to \mathcal{F}_{\overline{s}}$ is injective, because for any morphism of étale neighbourhoods $(U', \overline{u}') \to (U, \overline{u})$ the restriction map $\mathcal{F}(U) \to \mathcal{F}(U')$ is injective since $U' \to U$ is a covering of $S_{\acute{e}tale}$ . After enlarging $L$ a bit we may assume $K \subset L$ is a finite Galois extension. At this point we use that

$\mathop{\mathrm{Spec}}(L) \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(L) = \coprod \nolimits _{\sigma \in \text{Gal}(L/K)} \mathop{\mathrm{Spec}}(L)$

where the maps $\mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(L \otimes _ K L)$ come from the ring maps $a \otimes b \mapsto a\sigma (b)$ . Hence we see that the condition that $(U, \overline{u}, t)$ is invariant under all of $G$ implies that $t \in \mathcal{F}(\mathop{\mathrm{Spec}}(L))$ maps to the same element of $\mathcal{F}(\mathop{\mathrm{Spec}}(L) \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(L))$ via restriction by either projection (this uses the injectivity mentioned above; details omitted). Hence the sheaf condition of $\mathcal{F}$ for the étale covering $\{ \mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(K)\}$ kicks in and we conclude that $t$ comes from a unique section of $\mathcal{F}$ over $\mathop{\mathrm{Spec}}(K)$ . $\square$

Remark 59.56.6. Let $S$ be a scheme and let $\overline{s} : \mathop{\mathrm{Spec}}(k) \to S$ be a geometric point of $S$ . By definition this means that $k$ is algebraically closed. In particular the absolute Galois group of $k$ is trivial. Hence by Theorem 59.56.3 the category of sheaves on $\mathop{\mathrm{Spec}}(k)_{\acute{e}tale}$ is equivalent to the category of sets. The equivalence is given by taking sections over $\mathop{\mathrm{Spec}}(k)$ . This finally provides us with an alternative definition of the stalk functor. Namely, the functor

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

is isomorphic to the functor

$\mathop{\mathit{Sh}}\nolimits (S_{\acute{e}tale}) \longrightarrow \mathop{\mathit{Sh}}\nolimits (\mathop{\mathrm{Spec}}(k)_{\acute{e}tale}) = \textit{Sets}, \quad \mathcal{F} \longmapsto \overline{s}^*\mathcal{F}$

To prove this rigorously one can use Lemma 59.36.2 part (3) with $f = \overline{s}$ . Moreover, having said this the general case of Lemma 59.36.2 part (3) follows from functoriality of pullbacks.

[1] For the doubting Thomases out there.

Comments (2)

Comment #415 by Rex on December 29, 2013 at 00:07

(second attempt)

Concerning "...where the maps \Spec(L \otimes_K L) \to \Spec(L) come from the ring maps $a \otimes b \mapsto a\sigma(b)$ ."

Shouldn't the ring maps be going in the other direction i.e. L \longrightarrow L \otimes_K L via $a \mapsto a \otimes \sigma(a)$ ?

Comment #416 by Johan on December 30, 2013 at 20:34

Thanks for pointing out the mistake. I fixed the direction of the arrow of schemes, in other words, I wrote $\text{Spec}(L) \to \text{Spec}(L \otimes L)$ . The reason is that the ring map $L \to L \otimes L$ does not have a simple formula. See here.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$ ). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

The Stacks project

59.56 Galois action on stalks

Comments (2)

Post a comment