The Stacks project

Lemma 79.15.11. In Situation 79.15.2 assume in addition that $s, t$ are flat and locally of finite presentation and that $U$ is affine. Then there exists an affine scheme $U'$, an étale morphism $U' \to U$, and a point $u' \in U'$ lying over $u$ with $\kappa (u) = \kappa (u')$ such that the restriction $R' = R|_{U'}$ of $R$ to $U'$ is strongly split over $u'$.

Proof. Let $U' \to U$ and $u' \in U'$ be the separated étale morphism of schemes we found in Lemma 79.15.8. Let $P \subset R'$ be the strong splitting of $R'$ over $u'$. By More on Groupoids, Lemma 40.9.1 the morphisms $s', t' : R' \to U'$ are flat and locally of finite presentation. They are finite by assumption. Hence $s', t'$ are finite locally free, see Morphisms, Lemma 29.48.2. In particular $t(s^{-1}(u'))$ is a finite set of points $\{ u'_1, u'_2, \ldots , u'_ n\} $ of $U'$. Choose a quasi-compact open $W \subset U'$ containing each $u'_ i$. As $U$ is affine the morphism $W \to U$ is quasi-compact (see Schemes, Lemma 26.19.2). The morphism $W \to U$ is also locally quasi-finite (see Morphisms, Lemma 29.36.6) and separated. Hence by More on Morphisms, Lemma 37.43.2 (a version of Zariski's Main Theorem) we conclude that $W$ is quasi-affine. By Properties, Lemma 28.29.5 we see that $\{ u'_1, \ldots , u'_ n\} $ are contained in an affine open of $U'$. Thus we may apply Groupoids, Lemma 39.24.1 to conclude that there exists an affine $P$-invariant open $U'' \subset U'$ which contains $u'$.

To finish the proof denote $R'' = R|_{U''}$ the restriction of $R$ to $U''$. This is the same as the restriction of $R'$ to $U''$. As $P \subset R'$ is an open and closed subscheme, so is $P|_{U''} \subset R''$. By construction the open subscheme $U'' \subset U'$ is $P$-invariant which means that $P|_{U''} = (s'|_ P)^{-1}(U'') = (t'|_ P)^{-1}(U'')$ (see discussion in Groupoids, Section 39.19) so the restrictions of $s''$ and $t''$ to $P|_{U''}$ are still finite. The sub groupoid scheme $P|_{U''}$ is still a strong splitting of $R''$ over $u''$; above we verified (a), (b) and (c) holds as $\{ r' \in R': t'(r') = u', s'(r') = u'\} = \{ r'' \in R'': t''(r'') = u', s''(r'') = u'\} $ trivially. The lemma is proved. $\square$

Comments (0)

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Your email address will not be published.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 04RZ. Beware of the difference between the letter 'O' and the digit '0'.