Lemma 40.14.2. Let $(U, R, s, t, c)$ be a groupoid scheme over a scheme $S$. Assume $s, t$ are finite. There exists an open subscheme $W \subset U$ and a closed subscheme $W' \subset W$ such that
$W$ and $W'$ are $R$-invariant,
$U = t(s^{-1}(\overline{W}))$ set theoretically,
$W$ is a thickening of $W'$, and
the maps $s'$, $t'$ of the restriction $(W', R', s', t', c')$ are finite locally free.
Proof. Consider the stratification $U = Z_0 \supset Z_1 \supset Z_2 \supset \ldots $ of Lemma 40.14.1.
We will construct disjoint unions $W = \coprod _{r \geq 1} W_ r$ and $W' = \coprod _{r \geq 1} W'_ r$ with each $W'_ r \to W_ r$ a thickening of $R$-invariant subschemes of $U$ such that the morphisms $s_ r', t_ r'$ of the restrictions $(W_ r', R_ r', s_ r', t_ r', c_ r')$ are finite locally free of rank $r$. To begin we set $W_1 = W'_1 = U \setminus Z_1$. This is an $R$-invariant open subscheme of $U$, it is true that $W_0$ is a thickening of $W'_0$, and the maps $s_1'$, $t_1'$ of the restriction $(W_1', R_1', s_1', t_1', c_1')$ are isomorphisms, i.e., finite locally free of rank $1$. Moreover, every point of $U \setminus Z_1$ is in $t(s^{-1}(\overline{W_1}))$.
Assume we have found subschemes $W'_ r \subset W_ r \subset U$ for $r \leq n$ such that
$W_1, \ldots , W_ n$ are disjoint,
$W_ r$ and $W_ r'$ are $R$-invariant,
$U \setminus Z_ n \subset \bigcup _{r \leq n} t(s^{-1}(\overline{W_ r}))$ set theoretically,
$W_ r$ is a thickening of $W'_ r$,
the maps $s_ r'$, $t_ r'$ of the restriction $(W_ r', R_ r', s_ r', t_ r', c_ r')$ are finite locally free of rank $r$.
Then we set
set theoretically and
scheme theoretically. Then $W_{n + 1}$ is an $R$-invariant open subscheme of $U$ because $Z_{n + 1} \setminus \overline{U \setminus Z_{n + 1}}$ is open in $U$ and $\overline{U \setminus Z_{n + 1}}$ is contained in the closed subset $\bigcup \nolimits _{r \leq n} t(s^{-1}(\overline{W_ r}))$ we are removing by property (3) and the fact that $t$ is a closed morphism. It is clear that $W'_{n + 1}$ is a closed subscheme of $W_{n + 1}$ with the same underlying topological space. Finally, properties (1), (2) and (3) are clear and property (5) follows from Lemma 40.14.1.
By Lemma 40.14.1 we have $\bigcap Z_ r = \emptyset $. Hence every point of $U$ is contained in $U \setminus Z_ n$ for some $n$. Thus we see that $U = \bigcup _{r \geq 1} t(s^{-1}(\overline{W_ r}))$ set theoretically and we see that (2) holds. Thus $W' \subset W$ satisfy (1), (2), (3), and (4). $\square$
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.
Comments (0)