Lemma 70.12.4. Let $k \subset k'$ be an extension of fields. Let $X$ be an algebraic space over $k$. Assume $k$ separably algebraically closed. Then the morphism $X_{k'} \to X$ induces a bijection of connected components. In particular, $X$ is geometrically connected over $k$ if and only if $X$ is connected.

**Proof.**
Since $k$ is separably algebraically closed we see that $k'$ is geometrically connected over $k$, see Algebra, Lemma 10.47.4. Hence $Z = \mathop{\mathrm{Spec}}(k')$ is geometrically connected over $k$ by Varieties, Lemma 33.7.5. Since $X_{k'} = Z \times _ k X$ the result is a special case of Lemma 70.12.3.
$\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: