39.19 Invariant subschemes
In this section we discuss briefly the notion of an invariant subscheme.
Definition 39.19.1. Let $(U, R, s, t, c)$ be a groupoid scheme over the base scheme $S$.
A subset $W \subset U$ is set-theoretically $R$-invariant if $t(s^{-1}(W)) \subset W$.
An open $W \subset U$ is $R$-invariant if $t(s^{-1}(W)) \subset W$.
A closed subscheme $Z \subset U$ is called $R$-invariant if $t^{-1}(Z) = s^{-1}(Z)$. Here we use the scheme theoretic inverse image, see Schemes, Definition 26.17.7.
A monomorphism of schemes $T \to U$ is $R$-invariant if $T \times _{U, t} R = R \times _{s, U} T$ as schemes over $R$.
For subsets and open subschemes $W \subset U$ the $R$-invariance is also equivalent to requiring that $s^{-1}(W) = t^{-1}(W)$ as subsets of $R$. If $W \subset U$ is an $R$-equivariant open subscheme then the restriction of $R$ to $W$ is just $R_ W = s^{-1}(W) = t^{-1}(W)$. Similarly, if $Z \subset U$ is an $R$-invariant closed subscheme, then the restriction of $R$ to $Z$ is just $R_ Z = s^{-1}(Z) = t^{-1}(Z)$.
Lemma 39.19.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$.
For any subset $W \subset U$ the subset $t(s^{-1}(W))$ is set-theoretically $R$-invariant.
If $s$ and $t$ are open, then for every open $W \subset U$ the open $t(s^{-1}(W))$ is an $R$-invariant open subscheme.
If $s$ and $t$ are open and quasi-compact, then $U$ has an open covering consisting of $R$-invariant quasi-compact open subschemes.
Proof.
Part (1) follows from Lemmas 39.3.4 and 39.13.2, namely, $t(s^{-1}(W))$ is the set of points of $U$ equivalent to a point of $W$. Next, assume $s$ and $t$ open and $W \subset U$ open. Since $s$ is open the set $W' = t(s^{-1}(W))$ is an open subset of $U$. Finally, assume that $s$, $t$ are both open and quasi-compact. Then, if $W \subset U$ is a quasi-compact open, then also $W' = t(s^{-1}(W))$ is a quasi-compact open, and invariant by the discussion above. Letting $W$ range over all affine opens of $U$ we see (3).
$\square$
Lemma 39.19.3. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $s$ and $t$ quasi-compact and flat and $U$ quasi-separated. Let $W \subset U$ be quasi-compact open. Then $t(s^{-1}(W))$ is an intersection of a nonempty family of quasi-compact open subsets of $U$.
Proof.
Note that $s^{-1}(W)$ is quasi-compact open in $R$. As a continuous map $t$ maps the quasi-compact subset $s^{-1}(W)$ to a quasi-compact subset $t(s^{-1}(W))$. As $t$ is flat and $s^{-1}(W)$ is closed under generalization, so is $t(s^{-1}(W))$, see (Morphisms, Lemma 29.25.9 and Topology, Lemma 5.19.6). Pick a quasi-compact open $W' \subset U$ containing $t(s^{-1}(W))$. By Properties, Lemma 28.2.4 we see that $W'$ is a spectral space (here we use that $U$ is quasi-separated). Then the lemma follows from Topology, Lemma 5.24.7 applied to $t(s^{-1}(W)) \subset W'$.
$\square$
Lemma 39.19.4. Assumptions and notation as in Lemma 39.19.3. There exists an $R$-invariant open $V \subset U$ and a quasi-compact open $W'$ such that $W \subset V \subset W' \subset U$.
Proof.
Set $E = t(s^{-1}(W))$. Recall that $E$ is set-theoretically $R$-invariant (Lemma 39.19.2). By Lemma 39.19.3 there exists a quasi-compact open $W'$ containing $E$. Let $Z = U \setminus W'$ and consider $T = t(s^{-1}(Z))$. Observe that $Z \subset T$ and that $E \cap T = \emptyset $ because $s^{-1}(E) = t^{-1}(E)$ is disjoint from $s^{-1}(Z)$. Since $T$ is the image of the closed subset $s^{-1}(Z) \subset R$ under the quasi-compact morphism $t : R \to U$ we see that any point $\xi $ in the closure $\overline{T}$ is the specialization of a point of $T$, see Morphisms, Lemma 29.6.5 (and Morphisms, Lemma 29.6.3 to see that the scheme theoretic image is the closure of the image). Say $\xi ' \leadsto \xi $ with $\xi ' \in T$. Suppose that $r \in R$ and $s(r) = \xi $. Since $s$ is flat we can find a specialization $r' \leadsto r$ in $R$ such that $s(r') = \xi '$ (Morphisms, Lemma 29.25.9). Then $t(r') \leadsto t(r)$. We conclude that $t(r') \in T$ as $T$ is set-theoretically invariant by Lemma 39.19.2. Thus $\overline{T}$ is a set-theoretically $R$-invariant closed subset and $V = U \setminus \overline{T}$ is the open we are looking for. It is contained in $W'$ which finishes the proof.
$\square$
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