The Stacks project

Proof. Let $Y_0 = \bigcup _{j = 1, \ldots , m} V_{j, 0}$ be a finite affine open covering. Set $U_{j, 0} = f_0^{-1}(V_{j, 0})$. For $i \geq 0$ we denote $V_{j, i}$ the inverse image of $V_{j, 0}$ in $Y_ i$ and $U_{j, i} = f_ i^{-1}(V_{j, i})$. Similarly we have $U_ j = f^{-1}(V_ j)$. Then $U_ j = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_{j, i}$ (see Lemma 32.2.2). Since $U_ j$ is affine by assumption we see that each $U_{j, i}$ is affine for $i$ large enough, see Lemma 32.4.13. As there are finitely many $j$ we can pick an $i$ which works for all $j$. Thus $f_ i$ is affine for $i$ large enough, see Morphisms, Lemma 29.11.3. $\square$

Proof. A finite morphism is affine, see Morphisms, Definition 29.44.1. Hence by Lemma 32.8.2 above after increasing $0$ we may assume that $f_0$ is affine. By writing $Y_0$ as a finite union of affines we reduce to proving the result when $X_0$ and $Y_0$ are affine and map into a common affine $W \subset S_0$. The corresponding algebra statement follows from Algebra, Lemma 10.168.3. $\square$

Proof. Choose a finite affine open covering $Y_0 = \bigcup _{j = 1, \ldots , m} Y_{j, 0}$ such that each $Y_{j, 0}$ maps into an affine open $S_{j, 0} \subset S_0$. For each $j$ let $f_0^{-1}Y_{j, 0} = \bigcup _{k = 1, \ldots , n_ j} X_{k, 0}$ be a finite affine open covering. Since the property of being unramified is local we see that it suffices to prove the lemma for the morphisms of affines $X_{k, i} \to Y_{j, i} \to S_{j, i}$ which are the base changes of $X_{k, 0} \to Y_{j, 0} \to S_{j, 0}$ to $S_ i$. Thus we reduce to the case that $X_0, Y_0, S_0$ are affine

In the affine case we reduce to the following algebra result. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$. For some $0 \in I$ suppose given an $R_0$-algebra map $A_ i \to B_ i$ of finite type. If $R \otimes _{R_0} A_0 \to R \otimes _{R_0} B_0$ is unramified, then for some $i \geq 0$ the map $R_ i \otimes _{R_0} A_0 \to R_ i \otimes _{R_0} B_0$ is unramified. This follows from Algebra, Lemma 10.168.5. $\square$

Proof. A closed immersion is affine, see Morphisms, Lemma 29.11.9. Hence by Lemma 32.8.2 above after increasing $0$ we may assume that $f_0$ is affine. By writing $Y_0$ as a finite union of affines we reduce to proving the result when $X_0$ and $Y_0$ are affine and map into a common affine $W \subset S_0$. The corresponding algebra statement is a consequence of Algebra, Lemma 10.168.4. $\square$

Proof. Apply Lemma 32.8.5 to the diagonal morphism $\Delta _{X_0/S_0} : X_0 \to X_0 \times _{S_0} X_0$. (This is permissible as diagonal morphisms are locally of finite type and the fibre product $X_0 \times _{S_0} X_0$ is quasi-compact and quasi-separated, see Schemes, Lemma 26.21.2, Morphisms, Lemma 29.15.5, and Schemes, Remark 26.21.18. $\square$

Proof. Choose a finite affine open covering $Y_0 = \bigcup _{j = 1, \ldots , m} Y_{j, 0}$ such that each $Y_{j, 0}$ maps into an affine open $S_{j, 0} \subset S_0$. For each $j$ let $f_0^{-1}Y_{j, 0} = \bigcup _{k = 1, \ldots , n_ j} X_{k, 0}$ be a finite affine open covering. Since the property of being flat is local we see that it suffices to prove the lemma for the morphisms of affines $X_{k, i} \to Y_{j, i} \to S_{j, i}$ which are the base changes of $X_{k, 0} \to Y_{j, 0} \to S_{j, 0}$ to $S_ i$. Thus we reduce to the case that $X_0, Y_0, S_0$ are affine

In the affine case we reduce to the following algebra result. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$. For some $0 \in I$ suppose given an $R_0$-algebra map $A_ i \to B_ i$ of finite presentation. If $R \otimes _{R_0} A_0 \to R \otimes _{R_0} B_0$ is flat, then for some $i \geq 0$ the map $R_ i \otimes _{R_0} A_0 \to R_ i \otimes _{R_0} B_0$ is flat. This follows from Algebra, Lemma 10.168.1 part (3). $\square$

Proof. By Lemmas 32.8.7 and 32.8.3 we find an $i$ such that $f_ i$ is flat and finite. On the other hand, $f_ i$ is locally of finite presentation. Hence $f_ i$ is finite locally free by Morphisms, Lemma 29.48.2. If moreover $f$ is finite locally free of degree $d$, then the image of $Y \to Y_ i$ is contained in the open and closed locus $W_ d \subset Y_ i$ over which $f_ i$ has degree $d$. By Lemma 32.4.10 we see that for some $i' \geq i$ the image of $Y_{i'} \to Y_ i$ is contained in $W_ d$. Then $f_{i'}$ will be finite locally free of degree $d$. $\square$

Proof. Being smooth is local on the source and the target (Morphisms, Lemma 29.34.2) hence we may assume $S_0, X_0, Y_0$ affine (details omitted). The corresponding algebra fact is Algebra, Lemma 10.168.8. $\square$

Proof. Being étale is local on the source and the target (Morphisms, Lemma 29.36.2) hence we may assume $S_0, X_0, Y_0$ affine (details omitted). The corresponding algebra fact is Algebra, Lemma 10.168.7. $\square$

Proof. By Lemmas 32.8.10 and 32.8.5 we can find an $i$ such that $f_ i$ is flat and a closed immersion. Then $f_ i$ identifies $X_ i$ with an open and closed subscheme of $Y_ i$, see Morphisms, Lemma 29.26.2. By assumption the image of $Y \to Y_ i$ maps into $f_ i(X_ i)$. Thus by Lemma 32.4.10 we find that $Y_{i'}$ maps into $f_ i(X_ i)$ for some $i' \geq i$. It follows that $X_{i'} \to Y_{i'}$ is surjective and we win. $\square$

Proof. By Lemma 32.8.10 we can find an $i$ such that $f_ i$ is étale. Then $V_ i = f_ i(X_ i)$ is a quasi-compact open subscheme of $Y_ i$ (Morphisms, Lemma 29.36.13). let $V$ and $V_{i'}$ for $i' \geq i$ be the inverse image of $V_ i$ in $Y$ and $Y_{i'}$. Then $f : X \to V$ is an isomorphism (namely it is a surjective open immersion). Hence by Lemma 32.8.11 we see that $X_{i'} \to V_{i'}$ is an isomorphism for some $i' \geq i$ as desired. $\square$

Proof. There exists an open $V \subset Y$ such that the morphism $f$ factors as $X \to V \to Y$ and such that $X \to V$ is a closed immersion, see discussion in Schemes, Section 26.10. Since $X$ is quasi-compact, we may and do assume $V$ is a quasi-compact open of $Y$. By Lemma 32.4.11 after increasing $0$ we can find a quasi-compact open $V_0 \subset Y_0$ such that $V$ is the inverse image of $V_0$. Then the inverse image of $V_0$ in $X_0$ is a quasi-compact open whose inverse image in $X$ is $X$. Hence by the same lemma applied to $X = \mathop{\mathrm{lim}}\nolimits X_ i$ we may assume after increasing $0$ that we have the factorization $X_0 \to V_0 \to Y_0$. Then for large enough $i \geq 0$ the morphism $X_ i \to V_ i$ where $V_ i = Y_ i \times _{Y_0} V_0$ is a closed immersion by Lemma 32.8.5 and the proof is complete. $\square$

Proof. Recall that a morphism of schemes $V \to W$ is a monomorphism if and only if the diagonal $V \to V \times _ W V$ is an isomorphism (Schemes, Lemma 26.23.2). The morphism $X_0 \to X_0 \times _{Y_0} X_0$ is locally of finite presentation by Morphisms, Lemma 29.21.12. Since $X_0 \times _{Y_0} X_0$ is quasi-compact and quasi-separated (Schemes, Remark 26.21.18) we conclude from Lemma 32.8.11 that $\Delta _ i : X_ i \to X_ i \times _{Y_ i} X_ i$ is an isomorphism for some $i \geq 0$. For this $i$ the morphism $f_ i$ is a monomorphism. $\square$

Proof. The morphism $f_0$ is of finite presentation. Hence $E = f_0(X_0)$ is a constructible subset of $Y_0$, see Morphisms, Lemma 29.22.2. Since $f_ i$ is the base change of $f_0$ by $Y_ i \to Y_0$ we see that the image of $f_ i$ is the inverse image of $E$ in $Y_ i$. Moreover, we know that $Y \to Y_0$ maps into $E$. Hence we win by Lemma 32.4.10. $\square$

Proof. Choose a finite affine open covering $Y_0 = \bigcup _{j = 1, \ldots , m} Y_{j, 0}$ such that each $Y_{j, 0}$ maps into an affine open $S_{j, 0} \subset S_0$. For each $j$ let $f_0^{-1}Y_{j, 0} = \bigcup _{k = 1, \ldots , n_ j} X_{k, 0}$ be a finite affine open covering. Since the property of being syntomic is local we see that it suffices to prove the lemma for the morphisms of affines $X_{k, i} \to Y_{j, i} \to S_{j, i}$ which are the base changes of $X_{k, 0} \to Y_{j, 0} \to S_{j, 0}$ to $S_ i$. Thus we reduce to the case that $X_0, Y_0, S_0$ are affine

In the affine case we reduce to the following algebra result. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$. For some $0 \in I$ suppose given an $R_0$-algebra map $A_ i \to B_ i$ of finite presentation. If $R \otimes _{R_0} A_0 \to R \otimes _{R_0} B_0$ is syntomic, then for some $i \geq 0$ the map $R_ i \otimes _{R_0} A_0 \to R_ i \otimes _{R_0} B_0$ is syntomic. This follows from Algebra, Lemma 10.168.9. $\square$