## Tag `08BH`

Chapter 70: More on Groupoids in Spaces > Section 70.9: Properties of groups over fields and groupoids on fields

Lemma 70.9.4. In Situation 70.9.2 assume $R$ is a decent space. Then $R$ is a separated algebraic space. In Situation 70.9.1 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 69.6.1 we have to show that $e : S \to G$ is a closed immersion. This follows from Decent Spaces, Lemma 59.13.5.Next, we prove the second 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 69.28.2 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 25.17). Hence $R$ is separated, see Morphisms of Spaces, Lemma 58.4.9. $\square$

The code snippet corresponding to this tag is a part of the file `spaces-more-groupoids.tex` and is located in lines 691–697 (see updates for more information).

```
\begin{lemma}
\label{lemma-group-scheme-over-field-separated}
In Situation \ref{situation-groupoid-on-field}
assume $R$ is a decent space. Then $R$ is a separated algebraic space.
In Situation \ref{situation-group-over-field} assume that
$G$ is a decent algebraic space. Then $G$ is separated algebraic space.
\end{lemma}
\begin{proof}
We first prove the second assertion. By Groupoids in Spaces,
Lemma \ref{spaces-groupoids-lemma-group-scheme-separated}
we have to show that $e : S \to G$ is a closed immersion.
This follows from Decent Spaces, Lemma
\ref{decent-spaces-lemma-finite-residue-field-extension-finite}.
\medskip\noindent
Next, we prove the second 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 \ref{spaces-groupoids-lemma-diagonal}
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 \ref{schemes-section-fibre-products}).
Hence $R$ is separated, see
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-separated-over-separated}.
\end{proof}
```

## Comments (1)

## Add a comment on tag `08BH`

Your email address will not be published.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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.