## 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 $\kappa (\overline{s})$.

1. In this situation the absolute Galois group of $\kappa (s)$ is $\text{Gal}(\kappa (s)^{sep}/\kappa (s))$. It is sometimes denoted $\text{Gal}_{\kappa (s)}$.

2. 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

1. If $\sigma \in \text{Aut}(\kappa (\overline{s})/\kappa (s)^{sep})$ then $\sigma$ acts trivially on $\mathcal{F}_{\overline{s}}$.

2. 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 $\mathcal{F}_{\overline{s}}$.

3. 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 proof1.

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.

 For the doubting Thomases out there.

Comment #415 by Rex on

(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 on

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).