**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.
$\square$

## Comments (2)

Comment #6718 by 羽山籍真 on

Comment #6914 by Johan on