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.

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 $x$?

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.

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    $\begingroup$ Try to do this for $X = \mathbb{A}^2$ first. Can you find a huge field $F$ which contains all the function fields $k(c)$ as subfields, where $c$ can be any curve in $X$? If you can do this and you're in characteristic zero, then you can try $V = F[[t]]$; the local ring $R := \mathcal{O}_{X,c}$ at $c$ is a DVR since $X$ is smooth; it embeds into its completion $\widehat{R}$ which by Cohen's structure theorem is just $k(c)[[t_c]]$ for some uniformizer $t_c$, and this can then be embedded into $V = F[[t]]$ via $t_c \mapsto t$. Then the preimage of $tV$ in $R$ is its maximal ideal. $\endgroup$
    – user91132
    Dec 31, 2010 at 13:06

1 Answer 1


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.

  • $\begingroup$ Thanks! For a general base field $k$, I guess you'd use any field which contains the algebraic closure of $k(x_1, x_2, \dots)$ in place of $\mathbb C$. $\endgroup$ Dec 30, 2013 at 14:48

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