Lemma 33.8.12. Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. There is an action

$\text{Gal}(\overline{k}/k)^{opp} \times \text{IrredComp}(X_{\overline{k}}) \longrightarrow \text{IrredComp}(X_{\overline{k}})$

with the following properties:

1. An element $\overline{T} \in \text{IrredComp}(X_{\overline{k}})$ is fixed by the action if and only if there exists an irreducible component $T \subset X$, which is geometrically irreducible over $k$, such that $T_{\overline{k}} = \overline{T}$.

2. For any field extension $k'/k$ with separable algebraic closure $\overline{k}'$ the diagram

$\xymatrix{ \text{Gal}(\overline{k}'/k') \times \text{IrredComp}(X_{\overline{k}'}) \ar[r] \ar[d] & \text{IrredComp}(X_{\overline{k}'}) \ar[d] \\ \text{Gal}(\overline{k}/k) \times \text{IrredComp}(X_{\overline{k}}) \ar[r] & \text{IrredComp}(X_{\overline{k}}) }$

is commutative (where the right vertical arrow is a bijection according to Lemma 33.8.7).

Proof. The action (33.7.8.1) of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$ induces an action on its irreducible components. Irreducible components are always closed (Topology, Lemma 5.7.3). Hence if $\overline{T}$ is as in (1), then by Lemma 33.7.10 there exists a closed subset $T \subset X$ such that $\overline{T} = T_{\overline{k}}$. Note that $T$ is geometrically irreducible over $k$, see Lemma 33.8.8. To see that $T$ is an irreducible component of $X$, suppose that $T \subset T'$, $T \not= T'$ where $T'$ is an irreducible component of $X$. Let $\overline{\eta }$ be the generic point of $\overline{T}$. It maps to the generic point $\eta$ of $T$. Then the generic point $\xi \in T'$ specializes to $\eta$. As $X_{\overline{k}} \to X$ is flat there exists a point $\overline{\xi } \in X_{\overline{k}}$ which maps to $\xi$ and specializes to $\overline{\eta }$. It follows that the closure of the singleton $\{ \overline{\xi }\}$ is an irreducible closed subset of $X_{\overline{\xi }}$ which strictly contains $\overline{T}$. This is the desired contradiction.

We omit the proof of the functoriality in (2). $\square$

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