Lemma 40.14.10. Let $(U, R, s, t, c)$ be a groupoid scheme. Let $u \in U$. Assume

1. $s, t$ are finite morphisms,

2. $U$ is separated and locally Noetherian,

3. $\dim (\mathcal{O}_{U, u'}) \leq 1$ for every point $u'$ in the orbit of $u$.

Then $u$ is contained in an $R$-invariant affine open of $U$.

Proof. The $R$-orbit of $u$ is finite. By conditions (2) and (3) it is contained in an affine open $U'$ of $U$, see Varieties, Proposition 33.42.7. Then $t(s^{-1}(U \setminus U'))$ is an $R$-invariant closed subset of $U$ which does not contain $u$. Thus $U \setminus t(s^{-1}(U \setminus U'))$ is an $R$-invariant open of $U'$ containing $u$. Replacing $U$ by this open we may assume $U$ is quasi-affine.

By Lemma 40.14.6 we may replace $U$ by its reduction and assume $U$ is reduced. This means $R$-invariant subschemes $W' \subset W \subset U$ of Lemma 40.14.2 are equal $W' = W$. As $U = t(s^{-1}(\overline{W}))$ some point $u'$ of the $R$-orbit of $u$ is contained in $\overline{W}$ and by Lemma 40.14.6 we may replace $U$ by $\overline{W}$ and $u$ by $u'$. Hence we may assume there is a dense open $R$-invariant subscheme $W \subset U$ such that the morphisms $s_ W, t_ W$ of the restriction $(W, R_ W, s_ W, t_ W, c_ W)$ are finite locally free.

If $u \in W$ then we are done by Groupoids, Lemma 39.24.1 (because $W$ is quasi-affine so any finite set of points of $W$ is contained in an affine open, see Properties, Lemma 28.29.5). Thus we assume $u \not\in W$ and hence none of the points of the orbit of $u$ is in $W$. Let $\xi \in U$ be a point with a nontrivial specialization to a point $u'$ in the orbit of $u$. Since there are no specializations among the points in the orbit of $u$ (Lemma 40.14.8) we see that $\xi$ is not in the orbit. By assumption (3) we see that $\xi$ is a generic point of $U$ and hence $\xi \in W$. As $U$ is Noetherian there are finitely many of these points $\xi _1, \ldots , \xi _ m \in W$. Because $s_ W, t_ W$ are flat the orbit of each $\xi _ j$ consists of generic points of irreducible components of $W$ (and hence $U$).

Let $j : U \to \mathop{\mathrm{Spec}}(A)$ be an immersion of $U$ into an affine scheme (this is possible as $U$ is quasi-affine). Let $J \subset A$ be an ideal such that $V(J) \cap j(W) = \emptyset$ and $V(J) \cup j(W)$ is closed. Apply Lemma 40.14.7 to the groupoid scheme $(W, R_ W, s_ W, t_ W, c_ W)$, the morphism $j|_ W : W \to \mathop{\mathrm{Spec}}(A)$, the points $\xi _ j$, and the ideal $J$ to find an $f \in J$ such that $(j|_ W)^{-1}D(f)$ is an $R_ W$-invariant affine open containing $\xi _ j$ for all $j$. Since $f \in J$ we see that $j^{-1}D(f) \subset W$, i.e., $j^{-1}D(f)$ is an $R$-invariant affine open of $U$ contained in $W$ containing all $\xi _ j$.

Let $Z$ be the reduced induced closed subscheme structure on

$U \setminus j^{-1}D(f) = j^{-1}V(f).$

Then $Z$ is set theoretically $R$-invariant (but it may not be scheme theoretically $R$-invariant). Let $(Z, R_ Z, s_ Z, t_ Z, c_ Z)$ be the restriction of $R$ to $Z$. Since $Z \to U$ is finite, it follows that $s_ Z$ and $t_ Z$ are finite. Since $u \in Z$ the orbit of $u$ is in $Z$ and agrees with the $R_ Z$-orbit of $u$ viewed as a point of $Z$. Since $\dim (\mathcal{O}_{U, u'}) \leq 1$ and since $\xi _ j \not\in Z$ for all $j$, we see that $\dim (\mathcal{O}_{Z, u'}) \leq 0$ for all $u'$ in the orbit of $u$. In other words, the $R_ Z$-orbit of $u$ consists of generic points of irreducible components of $Z$.

Let $I \subset A$ be an ideal such that $V(I) \cap j(U) =\emptyset$ and $V(I) \cup j(U)$ is closed. Apply Lemma 40.14.7 to the groupoid scheme $(Z, R_ Z, s_ Z, t_ Z, c_ Z)$, the restriction $j|_ Z$, the ideal $I$, and the point $u \in Z$ to obtain $h \in I$ such that $j^{-1}D(h) \cap Z$ is an $R_ Z$-invariant open affine containing $u$.

Consider the $R_ W$-invariant (Groupoids, Lemma 39.23.2) function

$g = \text{Norm}_{s_ W}(t_ W^\sharp (j^\sharp (h)|_ W)) \in \Gamma (W, \mathcal{O}_ W)$

(In the following we only need the restriction of $g$ to $j^{-1}D(f)$ and in this case the norm is along a finite locally free morphism of affines.) We claim that

$V = (W_ g \cap j^{-1}D(f)) \cup (j^{-1}D(h) \cap Z)$

is an $R$-invariant affine open of $U$ which finishes the proof of the lemma. It is set theoretically $R$-invariant by construction. As $V$ is a constuctible set, to see that it is open it suffices to show it is closed under generalization in $U$ (Topology, Lemma 5.19.10 or the more general Topology, Lemma 5.23.6). Since $W_ g \cap j^{-1}D(f)$ is open in $U$, it suffices to consider a specialization $u_1 \leadsto u_2$ of $U$ with $u_2 \in j^{-1}D(h) \cap Z$. This means that $h$ is nonzero in $j(u_2)$ and $u_2 \in Z$. If $u_1 \in Z$, then $j(u_1) \leadsto j(u_2)$ and since $h$ is nonzero in $j(u_2)$ it is nonzero in $j(u_1)$ which implies $u_1 \in V$. If $u_1 \not\in Z$ and also not in $W_ g \cap j^{-1}D(f)$, then $u_1 \in W$, $u_1 \not\in W_ g$ because the complement of $Z = j^{-1}V(f)$ is contained in $W \cap j^{-1}D(f)$. Hence there exists a point $r_1 \in R$ with $s(r_1) = u_1$ such that $h$ is zero in $t(r_1)$. Since $s$ is finite we can find a specialization $r_1 \leadsto r_2$ with $s(r_2) = u_2$. However, then we conclude that $h$ is zero in $u'_2 = t(r_2)$ which contradicts the fact that $j^{-1}D(h) \cap Z$ is $R$-invariant and $u_2$ is in it. Thus $V$ is open.

Observe that $V \subset j^{-1}D(h)$ for our function $h \in I$. Thus we obtain an immersion

$j' : V \longrightarrow \mathop{\mathrm{Spec}}(A_ h)$

Let $f' \in A_ h$ be the image of $f$. Then $(j')^{-1}D(f')$ is the principal open determined by $g$ in the affine open $j^{-1}D(f)$ of $U$. Hence $(j')^{-1}D(f)$ is affine. Finally, $j'(V) \cap V(f') = j'(j^{-1}D(h) \cap Z)$ is closed in $\mathop{\mathrm{Spec}}(A_ h/(f')) = \mathop{\mathrm{Spec}}((A/f)_ h) = D(h) \cap V(f)$ by our choice of $h \in I$ and the ideal $I$. Hence we can apply Lemma 40.14.9 to conclude that $V$ is affine as claimed above. $\square$

Comment #6718 by 羽山籍真 on

In the second last paragraph, and the last second line, "we conclude that f is zero...". I think here $f$should be $h$, no? Because $u_2'$ is in the orbit of $u_2\in j^{-1}D(h)\cap Z$, if $f$ is replaced by $h$ then this can lead to the desired contradiction.

Comment #6914 by on

Thanks very much! See changes here.

To everybody: if you want your name in your own language added to the list of contributors, please either email me or leave another comment with your name in the form "name in Latin alphabet (name in other characters)".

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).