The Stacks project

Lemma 33.7.18. Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. There is an action

\[ \text{Gal}(\overline{k}/k)^{opp} \times \pi _0(X_{\overline{k}}) \longrightarrow \pi _0(X_{\overline{k}}) \]

with the following properties:

  1. An element $\overline{T} \in \pi _0(X_{\overline{k}})$ is fixed by the action if and only if there exists a connected component $T \subset X$, which is geometrically connected over $k$, such that $T_{\overline{k}} = \overline{T}$.

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

    \[ \xymatrix{ \text{Gal}(\overline{k}'/k') \times \pi _0(X_{\overline{k}'}) \ar[r] \ar[d] & \pi _0(X_{\overline{k}'}) \ar[d] \\ \text{Gal}(\overline{k}/k) \times \pi _0(X_{\overline{k}}) \ar[r] & \pi _0(X_{\overline{k}}) } \]

    is commutative (where the right vertical arrow is a bijection according to Lemma 33.7.6).

Proof. The action ( of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$ induces an action on its connected components. Connected 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 connected over $k$, see Lemma 33.7.7. To see that $T$ is a connected component of $X$, suppose that $T \subset T'$, $T \not= T'$ where $T'$ is a connected component of $X$. In this case $T'_{k'}$ strictly contains $\overline{T}$ and hence is disconnected. By Lemma 33.7.12 this means that $T'$ is disconnected! Contradiction.

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

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