The Stacks project

Lemma 79.15.6 (Existence of splitting). In Situation 79.15.3 there exists an algebraic space $U'$, an étale morphism $U' \to U$, and a point $u' : \mathop{\mathrm{Spec}}(\kappa (u)) \to U'$ lying over $u : \mathop{\mathrm{Spec}}(\kappa (u)) \to U$ such that the restriction $R' = R|_{U'}$ of $R$ to $U'$ is split over $u'$.

Proof. Let $f : (U', Z_{univ}, s', t', c') \to (U, R, s, t, c)$ be as constructed in Lemma 79.14.1. Recall that $R' = R \times _{(U \times _ S U)} (U' \times _ S U')$. Thus we get a morphism $(f, t', s') : Z_{univ} \to R'$ of groupoids in algebraic spaces

\[ (U', Z_{univ}, s', t', c') \to (U', R', s', t', c') \]

(by abuse of notation we indicate the morphisms in the two groupoids by the same symbols). Now, as $Z_{univ} \subset R \times _{s, U, g} U'$ is open and $R' \to R \times _{s, U, g} U'$ is étale (as a base change of $U' \to U$) we see that $Z_{univ} \to R'$ is an open immersion. By construction the morphisms $s', t' : Z_{univ} \to U'$ are finite. It remains to find the point $u'$ of $U'$.

We think of $u$ as a morphism $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$ as in the statement of the lemma. Set $F_ u = R \times _{s, U} \mathop{\mathrm{Spec}}(\kappa (u))$. Let $G_ u \subset F_ u$ be the scheme theoretic fibre of $G \to U$ over $u$. By assumption $G_ u$ is finite and $F_ u \to \mathop{\mathrm{Spec}}(\kappa (u))$ is quasi-finite at each point of $G_ u$ by assumption. Hence we can find a decomposition into open and closed subschemes

\[ F_ u = Z_ u \amalg Rest \]

for some scheme $Z_ u$ finite over $\kappa (u)$ whose support is $G_ u$. Note that $e(u) \in Z_ u$. Hence by the construction of $U'$ in Section 79.14 $(u, Z_ u)$ defines a $\mathop{\mathrm{Spec}}(\kappa (u))$-valued point $u'$ of $U'$.

We still have to show that the set $\{ g' \in |G'| : g'\text{ maps to }u'\} $ is contained in $|Z_{univ}|$. Pick any point $g'$ in this set and represent it by a morphism $z' : \mathop{\mathrm{Spec}}(k) \to G'$. Denote $z : \mathop{\mathrm{Spec}}(k) \to G$ the composition of $z'$ with the map $G' \to G$. Clearly, $z$ defines a point of $G_ u$. In fact, let us write $\tilde u : \mathop{\mathrm{Spec}}(k) \to u \to U$ for the corresponding map to $u$ or $U$. Consider the triple

\[ (\tilde u, Z_ u \times _{u, \tilde u} \mathop{\mathrm{Spec}}(k), z) \]

where $Z_ u$ is as above. This defines a $\mathop{\mathrm{Spec}}(k)$-valued point of $Z_{univ}$ whose image via $s', t'$ in $U'$ is $u'$ and whose image via $Z_{univ} \to R'$ is the point $z'$ (because the image in $R$ is $z$). This finishes the proof. $\square$

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 0DTC. Beware of the difference between the letter 'O' and the digit '0'.