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Let $f: X\rightarrow S$ be a flat quasi-projective morphism, where $X$ is a smooth variety, and $S$ is a discrete valuation ring. Then we know that $f$ is proper morphism if and only if it satisfies the well-known valuative criterion of properness. The evaluative criterion of properness says the following.

Given a DVR $T$ (with the closed point $0$) and maps $T\rightarrow S$ and $T\setminus 0\rightarrow X$ such that the obvious square commutes, there exists a unique lift $T\rightarrow X$ such that all the triangles commute.

My questions are the following.

  1. If we know that the map $f$ is flat, then is it enough to check the criterion for DVR T (over $S$) such that the composite map $T\setminus 0\rightarrow X\rightarrow S$ is flat?

  2. Suppose $f: X\rightarrow S$ is any flat morphism (may not be quasi-projective) then is it enough to check the criterion for DVR T (over $S$) such that the composite map $T\setminus 0\rightarrow X\rightarrow S$ is flat?

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  • $\begingroup$ The composite map corresponds to a homomorphism of rings $R\rightarrow K$, where $R$ is a DVR and $K$ a field. Such a map is always flat. $\endgroup$
    – abx
    Aug 27 at 16:24
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    $\begingroup$ @abx The following is a counterexample to what you said. $\mathbb C[|t|]\rightarrow \mathbb C(t)$ mapping $t\mapsto 0$. $\endgroup$
    – Bappa
    Aug 27 at 17:17
  • $\begingroup$ OK, but that means that $T$ maps to the closed point of $S$ — not a very interesting case for the valuative criterion. $\endgroup$
    – abx
    Aug 27 at 18:22
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    $\begingroup$ @abx this is exactly the interesting part of the question and why the flatness assumption on $f$ is necessary. For example, without flatness you could let $X$ be the union of a proper $S$-scheme with a non-proper scheme lying over the closed point of $S$. Then the flat valuative criterion would be satisfied but the morphism would not be proper. Or even let $X$ be a scheme over the closed point of $S$ viewed as an $S$-scheme. Then the flat valuative criterion is trivially satisfied but testing against $T$ which are flat over $S$ tells you nothing about $X$. $\endgroup$ Aug 27 at 22:10
  • $\begingroup$ @Dori Bejleri: Thank you for this detailed comment. $\endgroup$
    – abx
    Aug 28 at 6:42

1 Answer 1

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Let $S = \operatorname{Spec} R$ where $R$ is a DVR with generic point $\eta$ and closed point $0$. A homomorphism $R \to T$ is flat if and only if it is injective and more generally, $f : X \to S$ is flat if and only if $X$ is equal to the scheme theoretic closure of the generic fiber $X_{\eta}$.

In particular, $T$ is a DVR with fraction field $K$, then $R \to T$ is flat if and only if $R \to K$ is injective, so that $K$ is an extension of the fraction field of $R$. Thus, a test diagram $\require{AMScd}$ \begin{CD} \operatorname{Spec}K @>>> X\\ @VVV @VV f V\\ \operatorname{Spec} T @>>> S \end{CD} in the valuative criterion satisfies the flatness condition if and only if $\operatorname{Spec} K$ maps to the generic fiber of $f$. It follows that the generic fiber of $f$ is proper over $\eta$ so the only thing that can go wrong is that is the central fiber of $f$ is "missing points".

Since $f$ is quasi-projective we can take the closure of $X$ inside $\mathbb{P}^n_S$ for some $n$ and obtain a flat and proper map $g : \bar{X} \to S$. It suffices to show that $X = \bar{X}$. Let $x \in \bar{X}_0$ be a point of the special fiber. By flatness of $g$, there exists a point $y \in \bar{X}_\eta = X_\eta$ specializing to $x$. Since everything is Noetherian, we can witness this specialization via a DVR $T$ and a map $\operatorname{Spec} T \to \bar{X}$ with generic point $\operatorname{Spec} K$ mapping to $y$ and closed point mapping to $x$. Then $T$ is flat over $R$ so by assumption we have an extension $\operatorname{Spec} T \to X \subset \bar{X}$ with $\operatorname{Spec}K \mapsto y$. Since $\bar{X}$ is separated, these two maps $\operatorname{Spec}T \to \bar{X}$ must agree so $x \in X$ and $X = \bar{X}$.

Edit: If we drop quasi-projectivity but assume that $f$ is separated, then by Nagata's Compactification Theorem, there exists an open immersion $X \subset \bar{X}$ and a proper morphism $g : \bar{X} \to S$ extending $f$. Then we can run the argument as before. Since any map between separated schemes ($X$ is a variety and $S$ is affine) is separated, then we are done.

If we ask the same question for general schemes where $f$ is not separated or $X$ is not quasi-compact, then I imagine things can go wrong but I'm not sure.

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  • $\begingroup$ I am very sorry, but I am going to make an annoying edit to my question. I did not mean to assume that map f is quasi-projective. $\endgroup$
    – Bappa
    Aug 27 at 18:17
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    $\begingroup$ @Bappa No worries, I have edited the answer! $\endgroup$ Aug 27 at 22:08
  • $\begingroup$ Thanks a lot for the answers. Can you tell me any reference where I can read about the flat closure of a flat morphism? or in general the flat closure of schemes. It may help to get a proof for general morphisms where f is not separated (In this case the valuative criterion of properness should ask only for the existence of a lift, not the uniqueness). $\endgroup$
    – Bappa
    Aug 27 at 23:00
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    $\begingroup$ @Bappa A reference for flatness and flat closures over a DVR is Hartshorne III.9, Propositions 9.7 and 9.8. $\endgroup$ Aug 28 at 2:40
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    $\begingroup$ It seems that you also need something like $f$ being essentially of finite type or something. And for non-Noetherian schemes, one should test over all valuation rings rather than just DVRs. $\endgroup$
    – Z. M
    Aug 28 at 11:42

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