Lemma 76.9.4. In Situation 76.9.2 assume $R$ is a decent space. Then $R$ is a separated algebraic space. In Situation 76.9.1 assume that $G$ is a decent algebraic space. Then $G$ is separated algebraic space.

**Proof.**
We first prove the second assertion. By Groupoids in Spaces, Lemma 75.6.1 we have to show that $e : S \to G$ is a closed immersion. This follows from Decent Spaces, Lemma 65.14.5.

Next, we prove the first assertion. To do this we may replace $B$ by $S$. By the paragraph above the stabilizer group scheme $G \to U$ is separated. By Groupoids in Spaces, Lemma 75.28.2 the morphism $j = (t, s) : R \to U \times _ S U$ is separated. As $U$ is the spectrum of a field the scheme $U \times _ S U$ is affine (by the construction of fibre products in Schemes, Section 26.17). Hence $R$ is separated, see Morphisms of Spaces, Lemma 64.4.9. $\square$

## 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Your email address will not be published.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 (2)

Comment #2802 by Evan Warner on

Comment #2905 by Johan on