The Stacks project

Lemma 70.12.6. Let $k$ be a field. Let $X$ be an algebraic space over $k$. Let $\overline{k}$ be a (possibly infinite) Galois extension of $k$. Let $V \subset X_{\overline{k}}$ be a quasi-compact open. Then

  1. there exists a finite subextension $k \subset k' \subset \overline{k}$ and a quasi-compact open $V' \subset X_{k'}$ such that $V = (V')_{\overline{k}}$,

  2. there exists an open subgroup $H \subset \text{Gal}(\overline{k}/k)$ such that $\sigma (V) = V$ for all $\sigma \in H$.

Proof. Choose a scheme $U$ and a surjective ├ętale morphism $U \to X$. Choose a quasi-compact open $W \subset U_{\overline{k}}$ whose image in $X_{\overline{k}}$ is $V$. This is possible because $|U_{\overline{k}}| \to |X_{\overline{k}}|$ is continuous and because $|U_{\overline{k}}|$ has a basis of quasi-compact opens. We can apply Varieties, Lemma 33.7.9 to $W \subset U_{\overline{k}}$ to obtain the lemma. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 70.12: Geometrically connected algebraic spaces

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A15. Beware of the difference between the letter 'O' and the digit '0'.