Lemma 70.12.7. Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. Let $X$ be an algebraic space over $k$. Let $\overline{T} \subset |X_{\overline{k}}|$ have the following properties

$\overline{T}$ is a closed subset of $|X_{\overline{k}}|$,

for every $\sigma \in \text{Gal}(\overline{k}/k)$ we have $\sigma (\overline{T}) = \overline{T}$.

Then there exists a closed subset $T \subset |X|$ whose inverse image in $|X_{k'}|$ is $\overline{T}$.

**Proof.**
Let $T \subset |X|$ be the image of $\overline{T}$. Since $|X_{\overline{k}}| \to |X|$ is surjective, the statement means that $T$ is closed and that its inverse image is $\overline{T}$. Choose a scheme $U$ and a surjective étale morphism $U \to X$. By the case of schemes (see Varieties, Lemma 33.7.10) there exists a closed subset $T' \subset |U|$ whose inverse image in $|U_{\overline{k}}|$ is the inverse image of $\overline{T}$. Since $|U_{\overline{k}}| \to |X_{\overline{k}}|$ is surjective, we see that $T'$ is the inverse image of $T$ via $|U| \to |X|$. By our construction of the topology on $|X|$ this means that $T$ is closed. In the same manner one sees that $\overline{T}$ is the inverse image of $T$.
$\square$

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