Lemma 40.14.3. In Lemma 40.14.2 assume in addition that $s$ and $t$ are of finite presentation. Then
the morphism $W' \to W$ is of finite presentation, and
if $u \in U$ is a point whose $R$-orbit consists of generic points of irreducible components of $U$, then $u \in W$.
Proof. In this case the stratification $U = Z_0 \supset Z_1 \supset Z_2 \supset \ldots $ of Lemma 40.14.1 is given by closed immersions $Z_ k \to U$ of finite presentation, see Divisors, Lemma 31.9.6. Part (1) follows immediately from this as $W' \to W$ is locally given by intersecting the open $W$ by $Z_ r$. To see part (2) let $\{ u_1, \ldots , u_ n\} $ be the orbit of $u$. Since the closed subschemes $Z_ k$ are $R$-invariant and $\bigcap Z_ k = \emptyset $, we find an $k$ such that $u_ i \in Z_ k$ and $u_ i \not\in Z_{k + 1}$ for all $i$. The image of $Z_ k \to U$ and $Z_{k + 1} \to U$ is locally constructible (Morphisms, Theorem 29.22.3). Since $u_ i \in U$ is a generic point of an irreducible component of $U$, there exists an open neighbourhood $U_ i$ of $u_ i$ which is contained in $Z_ k \setminus Z_{k + 1}$ set theoretically (Properties, Lemma 28.2.2). In the proof of Lemma 40.14.2 we have constructed $W$ as a disjoint union $\coprod W_ r$ with $W_ r \subset Z_{r - 1} \setminus Z_ r$ such that $U = \bigcup t(s^{-1}(\overline{W_ r}))$. As $\{ u_1, \ldots , u_ n\} $ is an $R$-orbit we see that $u \in t(s^{-1}(\overline{W_ r}))$ implies $u_ i \in \overline{W_ r}$ for some $i$ which implies $U_ i \cap W_ r \not= \emptyset $ which implies $r = k$. Thus we conclude that $u$ is in
as desired. $\square$
Your email address will not be published.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)