Lemma 77.9.4. In Situation 77.9.2 assume $R$ is a decent space. Then $R$ is a separated algebraic space. In Situation 77.9.1 assume that $G$ is a decent algebraic space. Then $G$ is separated algebraic space.
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 76.28.2 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 65.4.9. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.