# Relative valuative criteria of properness for flat morphisms

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?

• 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.
– abx
Aug 27, 2022 at 16:24
• @abx The following is a counterexample to what you said. $\mathbb C[|t|]\rightarrow \mathbb C(t)$ mapping $t\mapsto 0$. Aug 27, 2022 at 17:17
• OK, but that means that $T$ maps to the closed point of $S$ — not a very interesting case for the valuative criterion.
– abx
Aug 27, 2022 at 18:22
• @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 $X$. Aug 27, 2022 at 22:10
• @Dori Bejleri: Thank you for this detailed comment.
– abx
Aug 28, 2022 at 6:42

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.

• 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. Aug 27, 2022 at 18:17
• @Bappa No worries, I have edited the answer! Aug 27, 2022 at 22:08
• 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). Aug 27, 2022 at 23:00
• @Bappa A reference for flatness and flat closures over a DVR is Hartshorne III.9, Propositions 9.7 and 9.8. Aug 28, 2022 at 2:40
• 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.
– Z. M
Aug 28, 2022 at 11:42