The Stacks project

Proof. Let $\{ Z_ r\} $ be the stratification of $U$ given by the Fitting ideals of the finite type quasi-coherent modules $s_*\mathcal{O}_ R$. See Divisors, Lemma 31.9.6. Since the identity $e : U \to R$ is a section to $s$ we see that $s_*\mathcal{O}_ R$ contains $\mathcal{O}_ S$ as a direct summand. Hence $U = Z_{-1} = Z_0$ (details omitted). Since formation of Fitting ideals commutes with base change (More on Algebra, Lemma 15.8.4) we find that $s^{-1}(Z_ r)$ corresponds to the $r$th Fitting ideal of $\text{pr}_{1, *}\mathcal{O}_{R \times _{s, U, t} R}$ because the lower right square of diagram (40.3.0.1) is cartesian. Using the fact that the lower left square is also cartesian we conclude that $s^{-1}(Z_ r) = t^{-1}(Z_ r)$, in other words $Z_ r$ is $R$-invariant. The morphism $s^{-1}(Z_{r - 1}) \setminus s^{-1}(Z_ r) \to Z_{r - 1} \setminus Z_ r$ is finite locally free of rank $r$ because the module $s_*\mathcal{O}_ R$ pulls back to a finite locally free module of rank $r$ on $Z_{r - 1} \setminus Z_ r$ by Divisors, Lemma 31.9.6. $\square$

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

1. $W_1, \ldots , W_ n$ are disjoint,

2. $W_ r$ and $W_ r'$ are $R$-invariant,

3. $U \setminus Z_ n \subset \bigcup _{r \leq n} t(s^{-1}(\overline{W_ r}))$ set theoretically,

4. $W_ r$ is a thickening of $W'_ r$,

5. 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). $\square$

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. $\square$

Proof. Let $W' \subset W \subset U$ be as in Lemma 40.14.2. By Lemma 40.14.3 we get $u_ j \in W$ and that $W' \to W$ is a thickening of finite presentation. By Limits, Lemma 32.11.3 it suffices to find an $R$-invariant affine open subscheme $V'$ of $W'$ containing $u_ j$ (because then we can let $V \subset W$ be the corresponding open subscheme which will be affine). Thus we may replace $(U, R, s, t, c)$ by the restriction $(W', R', s', t', c')$ to $W'$. In other words, we may assume we have a groupoid scheme $(U, R, s, t, c)$ whose morphisms $s$ and $t$ are finite locally free. By Properties, Lemma 28.29.1 we can find an affine open containing the union of the orbits of $u_1, \ldots , u_ m$. Finally, we can apply Groupoids, Lemma 39.24.1 to conclude. $\square$

Proof. Let $\{ u_{j1}, \ldots , u_{jn_ j}\} $ be the orbit of $u_ j$. Let $W' \subset W \subset U$ be as in Lemma 40.14.2. Since $U = t(s^{-1}(\overline{W}))$ we see that at least one $u_{ji} \in \overline{W}$. Since $u_{ji}$ is a generic point of an irreducible component and $U$ locally Noetherian, this implies that $u_{ji} \in W$. Since $W$ is $R$-invariant, we conclude that $u_ j \in W$ and in fact the whole orbit is contained in $W$. By Cohomology of Schemes, Lemma 30.13.3 it suffices to find an $R$-invariant affine open subscheme $V'$ of $W'$ containing $u_1, \ldots , u_ m$ (because then we can let $V \subset W$ be the corresponding open subscheme which will be affine). Thus we may replace $(U, R, s, t, c)$ by the restriction $(W', R', s', t', c')$ to $W'$. In other words, we may assume we have a groupoid scheme $(U, R, s, t, c)$ whose morphisms $s$ and $t$ are finite locally free. By Properties, Lemma 28.29.1 we can find an affine open containing $\{ u_{ij}\} $ (a locally Noetherian scheme is quasi-separated by Properties, Lemma 28.5.4). Finally, we can apply Groupoids, Lemma 39.24.1 to conclude. $\square$

Proof. Let $W' \subset U'$ be an $R'$-invariant affine open. Set $\tilde R = U' \times _{g, U, t} R$ with maps $\text{pr}_0 : \tilde R \to U'$ and $h = s \circ \text{pr}_1 : \tilde R \to U$. Observe that $\text{pr}_0$ and $h$ are integral. It follows that $\tilde W = \text{pr}_0^{-1}(W')$ is affine. Since $W'$ is $R'$-invariant, the image $W = h(\tilde W)$ is set theoretically $R$-invariant and $\tilde W = h^{-1}(W)$ set theoretically (details omitted). Thus, if we can show that $W$ is open, then $W$ is a scheme and the morphism $\tilde W \to W$ is integral surjective which implies that $W$ is affine by Limits, Proposition 32.11.2. However, our assumption on orbits meeting $U'$ implies that $h : \tilde R \to U$ is surjective. Since an integral surjective morphism is submersive (Topology, Lemma 5.6.5 and Morphisms, Lemma 29.44.7) it follows that $W$ is open. $\square$

Proof. Let $u_1, \ldots , u_ m \in V' \subset V \subset U$ be as in Lemma 40.14.4. Since $U \setminus V$ is closed in $U$, $j$ an immersion, and $V(I) \cup j(U)$ is closed in $\mathop{\mathrm{Spec}}(A)$, we can find an ideal $J \subset I$ such that $V(J) = V(I) \cup j(U \setminus V)$. For example we can take the ideal of elements of $I$ which vanish on $j(U \setminus V)$. Thus we can replace $(U, R, s, t, c)$, $j : U \to \mathop{\mathrm{Spec}}(A)$, and $I$ by $(V', R', s', t', c')$, $j|_{V'} : V' \to \mathop{\mathrm{Spec}}(A)$, and $J$. In other words, we may assume that $U$ is affine and that $s$ and $t$ are finite locally free. Take any $f \in I$ which does not vanish at all the points in the $R$-orbits of $u_1, \ldots , u_ m$ (Algebra, Lemma 10.15.2). Consider

Since $f \in I$ and since $V(I) \cup j(U)$ is closed we see that $U \cap D(f) \to D(f)$ is a closed immersion. Hence $f^ ng$ is the image of an element $h \in I$ for some $n > 0$. We claim that $h$ works. Namely, we have seen in Groupoids, Lemma 39.23.2 that $g$ is an $R$-invariant function, hence $D(g) \subset U$ is $R$-invariant. Since $f$ does not vanish on the orbit of $u_ j$, the function $g$ does not vanish at $u_ j$. Moreover, we have $V(g) \supset V(j^\sharp (f))$ and hence $j^{-1}D(h) = D(g)$. $\square$

Proof. Let $r \in R$ with $s(r) = u$ and $t(r) = u'$. If $u \leadsto u'$ then we can find a nontrivial specialization $r \leadsto r'$ with $s(r') = u'$, see Schemes, Lemma 26.19.8. Set $u'' = t(r')$. Note that $u'' \not= u'$ as there are no specializations in the fibres of a finite morphism. Hence we can continue and find a nontrivial specialization $r' \leadsto r''$ with $s(r'') = u''$, etc. This shows that the orbit of $u$ contains an infinite sequence $u \leadsto u' \leadsto u'' \leadsto \ldots $ of specializations which is nonsense as the orbit $t(s^{-1}(\{ u\} ))$ is finite. $\square$

Proof. This follows from Morphisms, Lemma 29.11.14 but we will also give a direct proof. Let $A' = \Gamma (V, \mathcal{O}_ V)$. Then $j' : V \to \mathop{\mathrm{Spec}}(A')$ is a quasi-compact open immersion, see Properties, Lemma 28.18.4. Let $f' \in A'$ be the image of $f$. Then $(j')^{-1}D(f') = j^{-1}D(f)$ is affine. On the other hand, $j'(V) \cap V(f')$ is a subscheme of $\mathop{\mathrm{Spec}}(A')$ which maps isomorphically to the closed subscheme $j(V) \cap V(f)$ of $\mathop{\mathrm{Spec}}(A)$. Hence it is closed in $\mathop{\mathrm{Spec}}(A')$ for example by Schemes, Lemma 26.21.11. Thus we may replace $A$ by $A'$ and assume that $j$ is an open immersion and $A = \Gamma (V, \mathcal{O}_ V)$.

In this case we claim that $j(V) = \mathop{\mathrm{Spec}}(A)$ which finishes the proof. If not, then we can find a principal affine open $D(g) \subset \mathop{\mathrm{Spec}}(A)$ which meets the complement and avoids the closed subset $j(V) \cap V(f)$. Note that $j$ maps $j^{-1}D(f)$ isomorphically onto $D(f)$, see Properties, Lemma 28.18.3. Hence $D(g)$ meets $V(f)$. On the other hand, $j^{-1}D(g)$ is a principal open of the affine open $j^{-1}D(f)$ hence affine. Hence by Properties, Lemma 28.18.3 again we see that $D(g)$ is isomorphic to $j^{-1}D(g) \subset j^{-1}D(f)$ which implies that $D(g) \subset D(f)$. This contradiction finishes the proof. $\square$

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

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

(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

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

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. $\square$