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?



Relative valuative criteria of properness for flat morphisms
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.


