When can closedness of a constructible set be checked using DVRs?

Question

Let $S$ a Noetherian scheme and $X \subseteq |S|$ be a constructible subset of its topological space defined by some point-wise property $P$. To check whether $X$ is closed, it suffices to show that the property $P$ is stable under specialization, which reduces to checking the same condition on specializations of height 1. That is, we may reduce to the case that $S = \operatorname{Spec} A$ is a 1-dimensional, local Noetherian ring.

Often, in the literature, I see arguments now assume that $S$ is a DVR. This reduction corresponds to assuming that $S$ is regular. Why can one make this reduction? There are many examples inside of affine space of height one specializations that do not give rise to DVRs, e.g. the stalk at a singularity on a curve.

An example can be found in Conrad's reductive group schemes, the proof of Proposition 3.1.12. Brian is generally a careful guy, so I assume the reduction is valid, but I haven't found a justification for this in e.g. EGA IV3.9.

Answer 1

Sergey Guminov in the comments is correct - it won't literally be the case that if $s_0\rightsquigarrow s_1$ is a specialization in $S$, then $\mathcal{O}_{\overline{s_0},s_1}$ is a DVR. But what is true is that there exists a morphism $f:\operatorname{Spec} A\to S$ where $A$ is a DVR and the images of the generic and special point are $s_0,s_1$ respectively. For reference, Hartshorne exercise II.4.11(a) lays out how to prove it:

Exercise II.4.11. If you are willing to do some harder commutative algebra, and stick to noetherian schemes, then we can express the valuative criteria of separatedness and properness using only discrete valuation rings.

(a) If $\mathcal{O},\mathfrak{m}$ is a noetherian local domain with quotient field $K$, and if $L$ is a finitely generated field extension of $K$, then there exists a discrete valuation ring $R$ of $L$ dominating $\mathcal{O}$. Prove this in the following steps. By taking a polynomial ring over $\mathcal{O}$, reduce to the case where $L$ is a finite extension field of $K$. Then show that for a suitable choice of generators $x_1,\cdots,x_n$ of $\mathfrak{m}$, the ideal $\mathfrak{a}=(x_1)$ in $\mathcal{O}'=\mathcal{O}[x_2/x_1,\cdots,x_n/x_1]$ is not equal to the unit ideal. Then let $\mathfrak{p}$ be a minimal prime ideal of $\mathfrak{a}$, and let $\mathcal{O}'_{\mathfrak{p}}$ be the localization of $\mathcal{O}'$ at $\mathfrak{p}$. This is a noetherian local domain of dimension $1$ dominating $\mathcal{O}$. Let $\widetilde{\mathcal{O}'}_{\mathfrak{P}}$ be the integral closure of $\mathcal{O}'_{\mathfrak{p}}$ in $L$. Use the theorem of Krull-Akizuki (see Nagata [7, p. 115]) to show that $\widetilde{\mathcal{O}'}_{\mathfrak{P}}$ is noetherian of dimension $1$. Finally, take $R$ to be a localization of $\widetilde{\mathcal{O}'}_{\mathfrak{P}}$ at one of its maximal ideals.

Here's the Krull-Akizuki theorem referenced above:

Theorem.: Let $A$ be a reduced noetherian ring of dimension at most one with $K$ its ring of fractions. If $B$ is a subring of a finite extension $L$ of $K$ containing $A$, then $B$ is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal $I$ of $B$, the quotient $B/I$ is finite over $A$.

This gets you the required $f$ and $A$, and now (assuming your condition plays nicely with base change) you can really assume that $S$ is the spectrum of a DVR.