If $X$ is a noetherian scheme with points $x$ and $\xi$ so that $x$ is in the closure of $\{\xi\}$, then there exists a discrete valuation ring $V$ and a map $Spec(V)\to X$ sending the generic point to $\xi$ and the closed point to $x$. That is, any closure relation among point of $X$ can be witnessed by a map from a DVR. This follows from the commutative algebra fact that any noetherian local ring is dominated by a DVR. (Applied to the local ring of $x$ in $\overline{\{\xi\}}$)


I'd like to know if I can rearrange the quantifiers: can I choose a single $V$ that will witness any closure relation among points of $X$? The answer is NO, because the characteristics of residue fields of points of $X$ can jump around. For example, if $X=Spec(\mathbb Z)$ and the residue field of $V$ has characteristic $p$, there's no way $V$ can witness the specialization $(q)\in\overline{\{(0)\}}$ for a prime $q$ different from $p$. If I remove this problem by requiring $X$ to be of finite type over a field, I feel like the answer might be yes.



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> If $X$ is a scheme of finite type (may as well be a variety) over a field $k$, does there exist a DVR $V$ so that for any closure relation $x\in\overline{\{\xi\}}$ among points of $X$, there is a map $Spec(V)\to X$ sending the generic point to $\xi$ and the closed point to $x$?
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**Remark 1:** I think it's worth pointing out one way to construct these witnessing maps from DVRs. The local ring at the generic point of a divisor on a smooth variety *is* a DVR, so if $\overline{\{\xi\}}$ is smooth and $\overline{\{x\}}$ is a divisor, it's easy. Otherwise, you can (in characteristic zero) resolve the singularities of $\overline{\{\xi\}}$ and blow up a point that maps to $x$. Now you have a smooth variety whose generic point maps to $\xi$ with a divisor whose generic point maps to $x$, so the local ring at the generic point of the divisor is a DVR that witnesses the specialization. Perhaps reasoning like this can somehow be used to construct $V$.


**Remark 2:** If $V$ exists, it's huge. For example, if $X=\mathbb A^2$, then the generic point of $X$ specializes to the generic point of any curve in $\mathbb A^2$, so the residue field of $V$ will have to contain the function fields of curves of arbitrarily large genus. Maybe there's a small way to accomplish this, but I don't mind if $V$ ends up being a ginormous ultraproduct or something.



Can a single DVR witness all specializations on a variety?
I think you already answered the question yourself and so did Konstantin Ardakov. But this question was still marked as open...


Suppose that the ground field is the field Q of rational numbers. Then let V be the power series ring C[[t]] where C is the field of complex numbers. This is an example.


To prove it, as you say, it suffices to prove that any local domain A which is a localization of a smooth algebra over Q, can be dominated by C[[t]]. Choose a regular system of parameters x\_1, ..., x\_d in the maximal ideal of A. Consider the quotient B of A(t\_1, ..., t\_d) by x\_i - t\_i x\_1 where t\_1, ..., t\_d are variables and where A(t\_1, ..., t\_d) is the localization of A[t\_1, ..., t\_d] at the prime ideal m\_A(A[t\_1, ..., t\_d]). Then B is a regular local ring of dimension 1 and A injects into B. The completion of B is K[[t]] where K is a finitely generated extension of Q. Hence B, and therefore A, is dominated by C[[t]].


An Italian geometer might replace the use of C by a universal overfield.


