Lemma 39.24.1. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $s$, $t$ are finite locally free. Let $u \in U$ be a point such that $t(s^{-1}(\{ u\} ))$ is contained in an affine open of $U$. Then there exists an $R$-invariant affine open neighbourhood of $u$ in $U$.

Proof. Since $s$ is finite locally free it has finite fibres. Hence $t(s^{-1}(\{ u\} )) = \{ u_1, \ldots , u_ n\}$ is a finite set. Note that $u \in \{ u_1, \ldots , u_ n\}$. Let $W \subset U$ be an affine open containing $\{ u_1, \ldots , u_ n\}$, in particular $u \in W$. Consider $Z = R \setminus s^{-1}(W) \cap t^{-1}(W)$. This is a closed subset of $R$. The image $t(Z)$ is a closed subset of $U$ which can be loosely described as the set of points of $U$ which are $R$-equivalent to a point of $U \setminus W$. Hence $W' = U \setminus t(Z)$ is an $R$-invariant, open subscheme of $U$ contained in $W$, and $\{ u_1, \ldots , u_ n\} \subset W'$. Picture

$\{ u_1, \ldots , u_ n\} \subset W' \subset W \subset U.$

Let $f \in \Gamma (W, \mathcal{O}_ W)$ be an element such that $\{ u_1, \ldots , u_ n\} \subset D(f) \subset W'$. Such an $f$ exists by Algebra, Lemma 10.15.2. By our choice of $W'$ we have $s^{-1}(W') \subset t^{-1}(W)$, and hence we get a diagram

$\xymatrix{ s^{-1}(W') \ar[d]_ s \ar[r]_-t & W \\ W' }$

The vertical arrow is finite locally free by assumption. Set

$g = \text{Norm}_ s(t^\sharp f) \in \Gamma (W', \mathcal{O}_{W'})$

By construction $g$ is a function on $W'$ which is nonzero in $u$, as $t^\sharp (f)$ is nonzero in each of the points of $R$ lying over $u$, since $f$ is nonzero in $u_1, \ldots , u_ n$. Similarly, $D(g) \subset W'$ is equal to the set of points $w$ such that $f$ is not zero in any of the points equivalent to $w$. This means that $D(g)$ is an $R$-invariant affine open of $W'$. The final picture is

$\{ u_1, \ldots , u_ n\} \subset D(g) \subset D(f) \subset W' \subset W \subset U$

and hence we win. $\square$

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).