32.8 Descending properties of morphisms

This section is the analogue of Section 32.4 for properties of morphisms over $S$. We will work in the following situation.

Situation 32.8.1. Let $S = \mathop{\mathrm{lim}}\nolimits S_ i$ be a limit of a directed system of schemes with affine transition morphisms (Lemma 32.2.2). Let $0 \in I$ and let $f_0 : X_0 \to Y_0$ be a morphism of schemes over $S_0$. Assume $S_0$, $X_0$, $Y_0$ are quasi-compact and quasi-separated. Let $f_ i : X_ i \to Y_ i$ be the base change of $f_0$ to $S_ i$ and let $f : X \to Y$ be the base change of $f_0$ to $S$.

Lemma 32.8.2. Notation and assumptions as in Situation 32.8.1. If $f$ is affine, then there exists an index $i \geq 0$ such that $f_ i$ is affine.

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. $\square$

Lemma 32.8.3. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is a finite morphism, and

2. $f_0$ is locally of finite type,

then there exists an $i \geq 0$ such that $f_ i$ is finite.

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. $\square$

Lemma 32.8.4. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is unramified, and

2. $f_0$ is locally of finite type,

then there exists an $i \geq 0$ such that $f_ i$ is unramified.

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. $\square$

Lemma 32.8.5. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is a closed immersion, and

2. $f_0$ is locally of finite type,

then there exists an $i \geq 0$ such that $f_ i$ is a closed immersion.

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. $\square$

Lemma 32.8.6. Notation and assumptions as in Situation 32.8.1. If $f$ is separated, then $f_ i$ is separated for some $i \geq 0$.

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. $\square$

Lemma 32.8.7. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is flat,

2. $f_0$ is locally of finite presentation,

then $f_ i$ is flat for some $i \geq 0$.

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). $\square$

Lemma 32.8.8. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is finite locally free (of degree $d$),

2. $f_0$ is locally of finite presentation,

then $f_ i$ is finite locally free (of degree $d$) for some $i \geq 0$.

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 $d$. $\square$

Lemma 32.8.9. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is smooth,

2. $f_0$ is locally of finite presentation,

then $f_ i$ is smooth for some $i \geq 0$.

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. $\square$

Lemma 32.8.10. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is étale,

2. $f_0$ is locally of finite presentation,

then $f_ i$ is étale for some $i \geq 0$.

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. $\square$

Lemma 32.8.11. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is an isomorphism, and

2. $f_0$ is locally of finite presentation,

then $f_ i$ is an isomorphism for some $i \geq 0$.

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. $\square$

Lemma 32.8.12. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is an open immersion, and

2. $f_0$ is locally of finite presentation,

then $f_ i$ is an open immersion for some $i \geq 0$.

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. $\square$

Lemma 32.8.13. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is an immersion, and

2. $f_0$ is locally of finite type,

then $f_ i$ is an immersion for some $i \geq 0$.

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. $\square$

Lemma 32.8.14. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is a monomorphism, and

2. $f_0$ is locally of finite type,

then $f_ i$ is a monomorphism for some $i \geq 0$.

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. $\square$

Lemma 32.8.15. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is surjective, and

2. $f_0$ is locally of finite presentation,

then there exists an $i \geq 0$ such that $f_ i$ is surjective.

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. $\square$

Lemma 32.8.16. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is syntomic, and

2. $f_0$ is locally of finite presentation,

then there exists an $i \geq 0$ such that $f_ i$ is syntomic.

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. $\square$

Comment #5469 by Harry Gindi on

Hey, I once called this argument called 'smearing out from the generic fibre', and I was looking for it and couldn't find it. Maybe you could add a remark in the introduction noting that they are the same thing?

Comment #5470 by Harry Gindi on

*heard this argument called

Comment #5682 by on

Maybe now google will find it? I'm not sure I agree with the name though.

Comment #6653 by Akhil Mathew on

Does the reference at the end of the proof of Lemma 32.8.15 (Lemma 0C3L) mean to point somewhere else (currently to 07BV)?

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).