The Stacks project

Proof. Pick $i \in I$. Take $U_ i \subset S_ i$ an affine open. Denote $U_{i'} = f_{i'i}^{-1}(U_ i)$ and $U = f_ i^{-1}(U_ i)$. Here $f_{i'i} : S_{i'} \to S_ i$ is the transition morphism and $f_ i : S \to S_ i$ is the projection. By Lemma 32.2.2 we have $U = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_ i$. Suppose we can show that $U_{set} = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{i', set}$. Then the lemma follows by a simple argument using an affine covering of $S_ i$. Hence we may assume all $S_ i$ and $S$ affine. This reduces us to the algebra question considered in the next paragraph.

Suppose given a system of rings $(A_ i, \varphi _{ii'})$ over $I$. Set $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$ with canonical maps $\varphi _ i : A_ i \to A$. Then

Namely, suppose that we are given primes $\mathfrak p_ i \subset A_ i$ such that $\mathfrak p_ i = \varphi _{ii'}^{-1}(\mathfrak p_{i'})$ for all $i' \geq i$. Then we simply set

It is clear that this is an ideal and has the property that $\varphi _ i^{-1}(\mathfrak p) = \mathfrak p_ i$. Then it follows easily that it is a prime ideal as well. $\square$

Proof. We will use the criterion of Topology, Lemma 5.14.3. We have seen that $S_{set} = \mathop{\mathrm{lim}}\nolimits _ i S_{i, set}$ in Lemma 32.4.1. The maps $f_ i : S \to S_ i$ are morphisms of schemes hence continuous. Thus $f_ i^{-1}(U_ i)$ is open for each open $U_ i \subset S_ i$. Finally, let $s \in S$ and let $s \in V \subset S$ be an open neighbourhood. Choose $0 \in I$ and choose an affine open neighbourhood $U_0 \subset S_0$ of the image of $s$. Then $f_0^{-1}(U_0) = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}(U_0)$, see Lemma 32.2.2. Then $f_0^{-1}(U_0)$ and $f_{i0}^{-1}(U_0)$ are affine and

either by the proof of Lemma 32.2.2 or by Lemma 32.2.1. Choose $a \in \mathcal{O}_ S(f_0^{-1}(U_0))$ such that $s \in D(a) \subset V$. This is possible because the principal opens form a basis for the topology on the affine scheme $f_0^{-1}(U_0)$. Then we can pick an $i \geq 0$ and $a_ i \in \mathcal{O}_{S_ i}(f_{i0}^{-1}(U_0))$ mapping to $a$. It follows that $D(a_ i) \subset f_{i0}^{-1}(U_0) \subset S_ i$ is an open subset whose inverse image in $S$ is $D(a)$. This finishes the proof. $\square$

Proof. Choose $0 \in I$. Note that $I$ is nonempty as the limit is directed. Choose an affine open covering $S_0 = \bigcup _{j = 1, \ldots , m} U_ j$. Since $I$ is directed there exists a $j \in \{ 1, \ldots , m\} $ such that $f_{i0}^{-1}(U_ j) \not= \emptyset $ for all $i \geq 0$. Hence $\mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}(U_ j)$ is not empty since a directed colimit of nonzero rings is nonzero (because $1 \not= 0$). As $\mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}(U_ j)$ is an open subscheme of the limit we win. $\square$

Proof. Choose $0 \in I$ and an affine open covering $S_0 = \bigcup _{j \in J} U_{0, j}$. For $i \geq 0$ let $U_{i, j} = f_{i, 0}^{-1}(U_{0, j})$ and set $U_ j = f_0^{-1}(U_{0, j})$. Here $f_{i'i} : S_{i'} \to S_ i$ is the transition morphism and $f_ i : S \to S_ i$ is the projection. For $j \in J$ the following are equivalent: (a) $s \in U_ j$, (b) $s_0 \in U_{0, j}$, (c) $s_ i \in U_{i, j}$ for all $i \geq 0$. Let $J' \subset J$ be the set of indices for which (a), (b), (c) are true. Then $\overline{\{ s\} } = \bigcup _{j \in J'} (\overline{\{ s\} } \cap U_ j)$ and similarly for $\overline{\{ s_ i\} }$ for $i \geq 0$. Note that $\overline{\{ s\} } \cap U_ j$ is the closure of the set $\{ s\} $ in the topological space $U_ j$. Similarly for $\overline{\{ s_ i\} } \cap U_{i, j}$ for $i \geq 0$. Hence it suffices to prove the lemma in the case $S$ and $S_ i$ affine for all $i$. This reduces us to the algebra question considered in the next paragraph.

Suppose given a system of rings $(A_ i, \varphi _{ii'})$ over $I$. Set $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$ with canonical maps $\varphi _ i : A_ i \to A$. Let $\mathfrak p \subset A$ be a prime and set $\mathfrak p_ i = \varphi _ i^{-1}(\mathfrak p)$. Then

This follows from Lemma 32.4.1 because $A/\mathfrak p = \mathop{\mathrm{colim}}\nolimits A_ i/\mathfrak p_ i$. This equality of rings also shows the final statement about reduced induced scheme structures holds true. The equality $\kappa (\mathfrak p) = \mathop{\mathrm{colim}}\nolimits \kappa (\mathfrak p_ i)$ follows from the statement as well. $\square$

Proof. Part (1) is a special case of Lemma 32.4.1.

Part (2) is a special case of Lemma 32.4.2.

Part (3) is a special case of Lemma 32.4.4. $\square$

Proof. Write $\mathcal{A}_ j = f_{i0, *} \mathcal{O}_{S_ i}$. This is a quasi-coherent sheaf of $\mathcal{O}_{S_0}$-algebras (see Morphisms, Lemma 29.11.5) and $S_ i$ is the relative spectrum of $\mathcal{A}_ i$ over $S_0$. In the proof of Lemma 32.2.2 we constructed $S$ as the relative spectrum of $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathcal{A}_ i$ over $S_0$. Set

and

Then we have $f_{i0, *} \mathcal{F}_ i = \mathcal{M}_ i$ and $f_{0, *}\mathcal{F} = \mathcal{M}$. Since $\mathcal{A}$ is the colimit of the sheaves $\mathcal{A}_ i$ and since tensor product commutes with directed colimits, we conclude that $\mathcal{M} = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathcal{M}_ i$. Since $S_0$ is quasi-compact and quasi-separated we see that

see Sheaves, Lemma 6.29.1 and Topology, Lemma 5.27.1 for the middle equality. $\square$

Proof. Let $Z_ i \subset S_ i$ also denote the reduced closed subscheme associated to $Z_ i$, see Schemes, Definition 26.12.5. A closed immersion is affine, and a composition of affine morphisms is affine (see Morphisms, Lemmas 29.11.9 and 29.11.7), and hence $Z_{i'} \to S_ i$ is affine when $i' \geq i$. We conclude that the morphism $f_{i'i} : Z_{i'} \to Z_ i$ is affine by Morphisms, Lemma 29.11.11. Each of the schemes $Z_ i$ is quasi-compact as a closed subscheme of a quasi-compact scheme. Hence we may apply Lemma 32.4.3 to see that $Z = \mathop{\mathrm{lim}}\nolimits _ i Z_ i$ is nonempty. Since there is a canonical morphism $Z \to S$ we win. $\square$

Proof. By Lemma 32.2.3 we see that $T \times _{S_ i} S = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} T \times _{S_ i} S_{i'}$. Hence the result follows from Lemma 32.4.3. $\square$

Proof. Writing $S_ i$ as a finite union of open affine subschemes reduces the question to the case that $S_ i$ is affine and $E$ is constructible, see Lemma 32.2.2 and Properties, Lemma 28.2.1. In this case the complement $S_ i \setminus E$ is constructible too. Hence there exists an affine scheme $T$ and a morphism $T \to S_ i$ whose image is $S_ i \setminus E$, see Algebra, Lemma 10.29.4. By Lemma 32.4.9 we see that $T \times _{S_ i} S_{i'}$ is empty for all sufficiently large $i'$, and hence $f_{i'i}(S_{i'}) \subset E$ for all sufficiently large $i'$. $\square$

Proof. Choose $i_0 \in I$. Note that $I$ is nonempty as the limit is directed. For convenience we write $S_0 = S_{i_0}$ and $i_0 = 0$. Choose an affine open covering $S_0 = U_{1, 0} \cup \ldots \cup U_{m, 0}$. Denote $U_{j, i} \subset S_ i$ the inverse image of $U_{j, 0}$ under the transition morphism for $i \geq 0$. Denote $U_ j$ the inverse image of $U_{j, 0}$ in $S$. Note that $U_ j = \mathop{\mathrm{lim}}\nolimits _ i U_{j, i}$ is a limit of affine schemes.

We first prove the uniqueness statement: Let $V_ i \subset S_ i$ and $V_{i'} \subset S_{i'}$ quasi-compact opens such that $f_ i^{-1}(V_ i) = f_{i'}^{-1}(V_{i'})$. It suffices to show that $f_{i''i}^{-1}(V_ i \cap U_{j, i''})$ and $f_{i''i'}^{-1}(V_{i'} \cap U_{j, i''})$ become equal for $i''$ large enough. Hence we reduce to the case of a limit of affine schemes. In this case write $S = \mathop{\mathrm{Spec}}(R)$ and $S_ i = \mathop{\mathrm{Spec}}(R_ i)$ for all $i \in I$. We may write $V_ i = S_ i \setminus V(h_1, \ldots , h_ m)$ and $V_{i'} = S_{i'} \setminus V(g_1, \ldots , g_ n)$. The assumption means that the ideals $\sum g_ jR$ and $\sum h_ jR$ have the same radical in $R$. This means that $g_ j^ N = \sum a_{jj'}h_{j'}$ and $h_ j^ N = \sum b_{jj'} g_{j'}$ for some $N \gg 0$ and $a_{jj'}$ and $b_{jj'}$ in $R$. Since $R = \mathop{\mathrm{colim}}\nolimits _ i R_ i$ we can chose an index $i'' \geq i$ such that the equations $g_ j^ N = \sum a_{jj'}h_{j'}$ and $h_ j^ N = \sum b_{jj'} g_{j'}$ hold in $R_{i''}$ for some $a_{jj'}$ and $b_{jj'}$ in $R_{i''}$. This implies that the ideals $\sum g_ jR_{i''}$ and $\sum h_ jR_{i''}$ have the same radical in $R_{i''}$ as desired.

We prove existence: If $S_0$ is affine, then $S_ i = \mathop{\mathrm{Spec}}(R_ i)$ for all $i \geq 0$ and $S = \mathop{\mathrm{Spec}}(R)$ with $R = \mathop{\mathrm{colim}}\nolimits R_ i$. Then $V = S \setminus V(g_1, \ldots , g_ n)$ for some $g_1, \ldots , g_ n \in R$. Choose any $i$ large enough so that each of the $g_ j$ comes from an element $g_{j, i} \in R_ i$ and take $V_ i = S_ i \setminus V(g_{1, i}, \ldots , g_{n, i})$. If $S_0$ is general, then the opens $V \cap U_ j$ are quasi-compact because $S$ is quasi-separated. Hence by the affine case we see that for each $j = 1, \ldots , m$ there exists an $i_ j \in I$ and a quasi-compact open $V_{i_ j} \subset U_{j, i_ j}$ whose inverse image in $U_ j$ is $V \cap U_ j$. Set $i = \max (i_1, \ldots , i_ m)$ and let $V_ i = \bigcup f_{ii_ j}^{-1}(V_{i_ j})$.

The statement on coverings follows from the uniqueness statement for the opens $V_{1, i} \cup \ldots \cup V_{n, i}$ and $S_ i$ of $S_ i$. $\square$

Proof. Choose $i_0 \in I$. Note that $I$ is nonempty as the limit is directed. For convenience we write $S_0 = S_{i_0}$ and $i_0 = 0$. Let $s \in S$. We may choose an affine open $U_0 \subset S_0$ containing $f_0(s)$. Since $S$ is quasi-affine we may choose an element $a \in \Gamma (S, \mathcal{O}_ S)$ such that $s \in D(a) \subset f_0^{-1}(U_0)$, and such that $D(a)$ is affine. By Lemma 32.4.7 there exists an $i \geq 0$ such that $a$ comes from an element $a_ i \in \Gamma (S_ i, \mathcal{O}_{S_ i})$. For any index $j \geq i$ we denote $a_ j$ the image of $a_ i$ in the global sections of the structure sheaf of $S_ j$. Consider the opens $D(a_ j) \subset S_ j$ and $U_ j = f_{j0}^{-1}(U_0)$. Note that $U_ j$ is affine and $D(a_ j)$ is a quasi-compact open of $S_ j$, see Properties, Lemma 28.26.4 for example. Hence we may apply Lemma 32.4.11 to the opens $U_ j$ and $U_ j \cup D(a_ j)$ to conclude that $D(a_ j) \subset U_ j$ for some $j \geq i$. For such an index $j$ we see that $D(a_ j) \subset S_ j$ is an affine open (because $D(a_ j)$ is a standard affine open of the affine open $U_ j$) containing the image $f_ j(s)$.

We conclude that for every $s \in S$ there exist an index $i \in I$, and a global section $a \in \Gamma (S_ i, \mathcal{O}_{S_ i})$ such that $D(a) \subset S_ i$ is an affine open containing $f_ i(s)$. Because $S$ is quasi-compact we may choose a single index $i \in I$ and global sections $a_1, \ldots , a_ m \in \Gamma (S_ i, \mathcal{O}_{S_ i})$ such that each $D(a_ j) \subset S_ i$ is affine open and such that $f_ i : S \to S_ i$ has image contained in the union $W_ i = \bigcup _{j = 1, \ldots , m} D(a_ j)$. For $i' \geq i$ set $W_{i'} = f_{i'i}^{-1}(W_ i)$. Since $f_ i^{-1}(W_ i)$ is all of $S$ we see (by Lemma 32.4.11 again) that for a suitable $i' \geq i$ we have $S_{i'} = W_{i'}$. Thus we may replace $i$ by $i'$ and assume that $S_ i = \bigcup _{j = 1, \ldots , m} D(a_ j)$. This implies that $\mathcal{O}_{S_ i}$ is an ample invertible sheaf on $S_ i$ (see Properties, Definition 28.26.1) and hence that $S_ i$ is quasi-affine, see Properties, Lemma 28.27.1. Hence we win. $\square$

Proof. By Lemma 32.4.12 we may assume that $S_0$ is quasi-affine for some $0 \in I$. Set $R_0 = \Gamma (S_0, \mathcal{O}_{S_0})$. Then $S_0$ is a quasi-compact open of $T_0 = \mathop{\mathrm{Spec}}(R_0)$. Denote $j_0 : S_0 \to T_0$ the corresponding quasi-compact open immersion. For $i \geq 0$ set $\mathcal{A}_ i = f_{i0, *}\mathcal{O}_{S_ i}$. Since $f_{i0}$ is affine we see that $S_ i = \underline{\mathop{\mathrm{Spec}}}_{S_0}(\mathcal{A}_ i)$. Set $T_ i = \underline{\mathop{\mathrm{Spec}}}_{T_0}(j_{0, *}\mathcal{A}_ i)$. Then $T_ i \to T_0$ is affine, hence $T_ i$ is affine. Thus $T_ i$ is the spectrum of

Write $S = \mathop{\mathrm{Spec}}(R)$. We have $R = \mathop{\mathrm{colim}}\nolimits _ i R_ i$ by Lemma 32.4.7. Hence also $S = \mathop{\mathrm{lim}}\nolimits _ i T_ i$. As formation of the relative spectrum commutes with base change, the inverse image of the open $S_0 \subset T_0$ in $T_ i$ is $S_ i$. Let $Z_0 = T_0 \setminus S_0$ and let $Z_ i \subset T_ i$ be the inverse image of $Z_0$. As $S_ i = T_ i \setminus Z_ i$, it suffices to show that $Z_ i$ is empty for some $i$. Assume $Z_ i$ is nonempty for all $i$ to get a contradiction. By Lemma 32.4.8 there exists a point $s$ of $S = \mathop{\mathrm{lim}}\nolimits T_ i$ which maps to a point of $Z_ i$ for every $i$. But $S = \mathop{\mathrm{lim}}\nolimits _ i S_ i$, and hence we arrive at a contradiction by Lemma 32.4.6. $\square$

Proof. Choose a finite affine open covering $S_0 = U_{0, 1} \cup \ldots \cup U_{0, m}$. Set $U_{i, j} \subset S_ i$ and $U_ j \subset S$ equal to the inverse image of $U_{0, j}$. Note that $U_{i, j}$ and $U_ j$ are affine. As $S$ is separated the intersections $U_{j_1} \cap U_{j_2}$ are affine. Since $U_{j_1} \cap U_{j_2} = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_{i, j_1} \cap U_{i, j_2}$ we see that $U_{i, j_1} \cap U_{i, j_2}$ is affine for large $i$ by Lemma 32.4.13. To show that $S_ i$ is separated for large $i$ it now suffices to show that

is surjective for large $i$ (Schemes, Lemma 26.21.7).

To get rid of the annoying indices, assume we have affine opens $U, V \subset S_0$ such that $U \cap V$ is affine too. Let $U_ i, V_ i \subset S_ i$, resp. $U, V \subset S$ be the inverse images. We have to show that $\mathcal{O}(U_ i) \otimes \mathcal{O}(V_ i) \to \mathcal{O}(U_ i \cap V_ i)$ is surjective for $i$ large enough and we know that $\mathcal{O}(U) \otimes \mathcal{O}(V) \to \mathcal{O}(U \cap V)$ is surjective. Note that $\mathcal{O}(U_0) \otimes \mathcal{O}(V_0) \to \mathcal{O}(U_0 \cap V_0)$ is of finite type, as the diagonal morphism $S_ i \to S_ i \times S_ i$ is an immersion (Schemes, Lemma 26.21.2) hence locally of finite type (Morphisms, Lemmas 29.15.2 and 29.15.5). Thus we can choose elements $f_{0, 1}, \ldots , f_{0, n} \in \mathcal{O}(U_0 \cap V_0)$ which generate $\mathcal{O}(U_0 \cap V_0)$ over $\mathcal{O}(U_0) \otimes \mathcal{O}(V_0)$. Observe that for $i \geq 0$ the diagram of schemes

is cartesian. Thus we see that the images $f_{i, 1}, \ldots , f_{i, n} \in \mathcal{O}(U_ i \cap V_ i)$ generate $\mathcal{O}(U_ i \cap V_ i)$ over $\mathcal{O}(U_ i) \otimes \mathcal{O}(V_0)$ and a fortiori over $\mathcal{O}(U_ i) \otimes \mathcal{O}(V_ i)$. By assumption the images $f_1, \ldots , f_ n \in \mathcal{O}(U \otimes V)$ are in the image of the map $\mathcal{O}(U) \otimes \mathcal{O}(V) \to \mathcal{O}(U \cap V)$. Since $\mathcal{O}(U) \otimes \mathcal{O}(V) = \mathop{\mathrm{colim}}\nolimits \mathcal{O}(U_ i) \otimes \mathcal{O}(V_ i)$ we see that they are in the image of the map at some finite level and the lemma is proved. $\square$

Proof. The assumption means there are finitely many sections $s_1, \ldots , s_ m \in \Gamma (S, \mathcal{L})$ such that $S_{s_ j}$ is affine and such that $S = \bigcup S_{s_ j}$, see Properties, Definition 28.26.1. By Lemma 32.4.7 we can find an $i \in I$ and sections $s_{i, j} \in \Gamma (S_ i, \mathcal{L}_ i)$ mapping to $s_ j$. By Lemma 32.4.13 we may, after increasing $i$, assume that $(S_ i)_{s_{i, j}}$ is affine for $j = 1, \ldots , m$. By Lemma 32.4.11 we may, after increasing $i$ a last time, assume that $S_ i = \bigcup (S_ i)_{s_{i, j}}$. Then $\mathcal{L}_ i$ is ample by definition. $\square$

Proof. Proof of (1). Choose $0 \in I$ and a finite affine open covering $X_0 = U_{0, 1} \cup \ldots \cup U_{0, m}$ with the property that $U_{0, j}$ maps into an affine open $W_ j \subset S$. Let $V_ j \subset Y$, resp. $U_{i, j} \subset X_ i$, $i \geq 0$, resp. $U_ j \subset X$ be the inverse image of $U_{0, j}$. It suffices to prove that $V_ j \to U_{i, j}$ is a closed immersion for $i$ sufficiently large and we know that $V_ j \to U_ j$ is a closed immersion. Thus we reduce to the following algebra fact: If $A = \mathop{\mathrm{colim}}\nolimits A_ i$ is a directed colimit of $R$-algebras, $A \to B$ is a surjection of $R$-algebras, and $B$ is a finitely generated $R$-algebra, then $A_ i \to B$ is surjective for $i$ sufficiently large.

Proof of (2). Choose $0 \in I$. Choose a quasi-compact open $X'_0 \subset X_0$ such that $Y \to X_0$ factors through $X'_0$. After replacing $X_ i$ by the inverse image of $X'_0$ for $i \geq 0$ we may assume all $X_ i'$ are quasi-compact and quasi-separated. Let $U \subset X$ be a quasi-compact open such that $Y \to X$ factors through a closed immersion $Y \to U$ ($U$ exists as $Y$ is quasi-compact). By Lemma 32.4.11 we may assume that $U = \mathop{\mathrm{lim}}\nolimits U_ i$ with $U_ i \subset X_ i$ quasi-compact open. By part (1) we see that $Y \to U_ i$ is a closed immersion for some $i$. Thus (2) holds.

Proof of (3). Working affine locally on $X_0$ for some $0 \in I$ as in the proof of (1) we reduce to the following algebra fact: If $A = \mathop{\mathrm{lim}}\nolimits A_ i$ is a directed colimit of $R$-algebras with surjective transition maps and $A$ of finite presentation over $A_0$, then $A = A_ i$ for some $i$. Namely, write $A = A_0/(f_1, \ldots , f_ n)$. Pick $i$ such that $f_1, \ldots , f_ n$ map to zero under the surjective map $A_0 \to A_ i$. $\square$

Proof. Let $0 \in I$. Note that $I$ is nonempty as the limit is directed. As $X_0$ is quasi-compact we can find finitely many affine opens $U_1, \ldots , U_ n \subset S$ such that $X_0 \to S$ maps into $U_1 \cup \ldots \cup U_ n$. Denote $h_ i : X_ i \to S$ the structure morphism. It suffices to check that for some $i \geq 0$ the morphisms $h_ i^{-1}(U_ j) \to U_ j$ are separated for $j = 1, \ldots , n$. Since $S$ is quasi-separated the morphisms $U_ j \to S$ are quasi-compact. Hence $h_ i^{-1}(U_ j)$ is quasi-compact and quasi-separated. In this way we reduce to the case $S$ affine. In this case we have to show that $X_ i$ is separated and we know that $X$ is separated. Thus the lemma follows from Lemma 32.4.14. $\square$

Proof. Choose a finite affine open covering $S = \bigcup _{j = 1, \ldots , n} V_ j$. Denote $f : X \to S$ and $f_ i : X_ i \to S$ the structure morphisms. For each $j$ the scheme $f^{-1}(V_ j) = \mathop{\mathrm{lim}}\nolimits _ i f_ i^{-1}(V_ j)$ is affine (as a finite morphism is affine by definition). Hence by Lemma 32.4.13 there exists an $i \in I$ such that each $f_ i^{-1}(V_ j)$ is affine. In other words, $f_ i : X_ i \to S$ is affine for $i$ large enough, see Morphisms, Lemma 29.11.3. $\square$

Proof. By Lemma 32.4.18 we may assume $X_ i \to S$ is affine for all $i$. Choose a finite affine open covering $S = \bigcup _{j = 1, \ldots , n} V_ j$. Denote $f : X \to S$ and $f_ i : X_ i \to S$ the structure morphisms. It suffices to show that there exists an $i$ such that $f_ i^{-1}(V_ j)$ is finite over $V_ j$ for $j = 1, \ldots , m$ (Morphisms, Lemma 29.44.3). Namely, for $i' \geq i$ the composition $X_{i'} \to X_ i \to S$ will be finite as a composition of finite morphisms (Morphisms, Lemma 29.44.5). This reduces us to the affine case: Let $R$ be a ring and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ with $R \to A$ integral and $A_ i \to A_{i'}$ finite for all $i \leq i'$. Moreover $R \to A_ i$ is of finite type for all $i$. Goal: Show that $A_ i$ is finite over $R$ for some $i$. To prove this choose an $i \in I$ and pick generators $x_1, \ldots , x_ m \in A_ i$ of $A_ i$ as an $R$-algebra. Since $A$ is integral over $R$ we can find monic polynomials $P_ j \in R[T]$ such that $P_ j(x_ j) = 0$ in $A$. Thus there exists an $i' \geq i$ such that $P_ j(x_ j) = 0$ in $A_{i'}$ for $j = 1, \ldots , m$. Then the image $A'_ i$ of $A_ i$ in $A_{i'}$ is finite over $R$ by Algebra, Lemma 10.36.5. Since $A'_ i \subset A_{i'}$ is finite too we conclude that $A_{i'}$ is finite over $R$ by Algebra, Lemma 10.7.3. $\square$

Proof. By Lemma 32.4.18 we may assume $X_ i \to S$ is affine for all $i$. Choose a finite affine open covering $S = \bigcup _{j = 1, \ldots , n} V_ j$. Denote $f : X \to S$ and $f_ i : X_ i \to S$ the structure morphisms. It suffices to show that there exists an $i$ such that $f_ i^{-1}(V_ j)$ is a closed subscheme of $V_ j$ for $j = 1, \ldots , m$ (Morphisms, Lemma 29.2.1). This reduces us to the affine case: Let $R$ be a ring and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ with $R \to A$ surjective and $A_ i \to A_{i'}$ surjective for all $i \leq i'$. Moreover $R \to A_ i$ is of finite type for all $i$. Goal: Show that $R \to A_ i$ is surjective for some $i$. To prove this choose an $i \in I$ and pick generators $x_1, \ldots , x_ m \in A_ i$ of $A_ i$ as an $R$-algebra. Since $R \to A$ is surjective we can find $r_ j \in R$ such that $r_ j$ maps to $x_ j$ in $A$. Thus there exists an $i' \geq i$ such that $r_ j$ maps to the image of $x_ j$ in $A_{i'}$ for $j = 1, \ldots , m$. Since $A_ i \to A_{i'}$ is surjective this implies that $R \to A_{i'}$ is surjective. $\square$

Proof. Choose an open subscheme $U \subset S$ such that $X \to S$ factors as a closed immersion $X \to U$ composed with the inclusion morphism $U \to S$. Since $X$ is quasi-compact, we may shrink $U$ and assume $U$ is quasi-compact. Denote $V_ i \subset X_ i$ the inverse image of $U$. Since $V_ i$ pulls back to $X$ we see that $V_ i = X_ i$ for all $i$ large enough by Lemma 32.4.11. Thus we may assume $X = \mathop{\mathrm{lim}}\nolimits X_ i$ in the category of schemes over $U$. Then we see that $X_ i \to U$ is a closed immersion for $i$ large enough by Lemma 32.4.20. This proves the lemma. $\square$