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Let $(R, \mathfrak m)$ be a non-field local domain with fraction field $Q(R)$ . Let $k_{R}:=R/m$.

We know that there is a Valuation ring $(V,\mathfrak m_V)$ such that $R \subseteq V \subsetneq Q(R)$ such that $\mathfrak m_V \cap R=\mathfrak m$ (in fact any local ring between $R$ and $K$ maximal w.r.t. "dominance" is a valuation ring). Then $k_V:=V/\mathfrak m_V$ is an extension field of $k_R$.

My questions is: Let $(R, \mathfrak m)$ be a non-field local domain with fraction field $Q(R)$ and let $k_{R}:=R/m$ ;

Does there exist a Valuation ring $(V,\mathfrak m_V)$ such that $R \subseteq V \subsetneq Q(R)$ and $\mathfrak m_V \cap R=\mathfrak m$ and $V/\mathfrak m_V$ is a finite extension field of $R/m$ ? If this is not always possible, then can we at least have a valuation ring as above with the extension $V/\mathfrak m_V$ over $R/\mathfrak m$ being algebraic ?

UPDATE: As an answer by Johannes Hahn shows ... the algebraic extension claim is true .

The claim of existence of the dominating valuation ring with finite extension of residue field remains unsettled.

One might ask whether this finite extension claim can be answered when $R$ is integrally closed in $Q(R)$ or say if $R$ has small Krull-dimension.

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    $\begingroup$ Trivial remark: if $R$ is regular (in particular Noetherian), then we can actually make $V/\mathfrak m_V$ equal to $R/\mathfrak m$. Indeed, if $x_1,\ldots,x_n$ is a regular system of parameters, then we can make a valuation $v \colon Q(R) \to \mathbb Z^n$ sending $f \in R$ to $(v_{x_1}(f), v_{x_2}(f \mod{x_1}), \ldots,v_{x_n}(f \mod{(x_1,\ldots,x_{n-1})}))$. It seems that in general the Noetherian case should be fine if we allow a finite extension. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 6, 2018 at 17:08
  • $\begingroup$ @R.vanDobbendeBruyn: What do you mean by $v_{x_1}(f)$ ... ? $\endgroup$
    – user111524
    Commented May 6, 2018 at 17:23
  • $\begingroup$ @R.vanDobbendeBruyn: Sorry, I haven't heard of the term "divisorial valuation" ... could you please elaborate or provide a reference ? $\endgroup$
    – user111524
    Commented May 6, 2018 at 17:43
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    $\begingroup$ Because $R$ is regular, so is the localisation $R_{(x_1)}$. But $(x_1)$ is a prime of height one by Krull's Hauptidealsatz, so $R_{(x_1)}$ is a DVR. This defines a valuation $v_{x_1}$ on $Q(R)$, called the divisorial valuation associated with the prime divisor $V(x_1) \subseteq \operatorname{Spec} R$. It's just the classical order of vanishing of a function along a codimension $1$ subvariety. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 6, 2018 at 17:46
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    $\begingroup$ I think that $x_1$ is an irreducible element because it maps to a nonzero element of $\mathfrak m/\mathfrak m^2$. Because regular local rings are UFD's, this implies that $(x_1)$ is prime. The same argument shows that $x_i$ is prime in $R/(x_1,\ldots,x_{i-1})$ for all $i$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 6, 2018 at 18:01

1 Answer 1

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Yes, an algebraic extension is always possible. If one applies Zorn's lemma not to $$\{R\subseteq V\subseteq Q(R) \mid V\,\text{local}\}$$ but to $$\{R\subseteq V\subseteq Q(R) \mid V\,\text{local} \wedge R/\mathfrak{m}_R \subseteq V/\mathfrak{m}_V \,\text{algebraic}\}$$ instead, one can prove the existence of maximal elements and show that those are valuation rings as well using the usual proof as a blue print.

If it is possible to obtain finite extensions, I don't know. In fact I asked this a while back on MSE but got no satisfactory answer.

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  • $\begingroup$ Ah ok ... thanks for that , I will verify the details ... in the question you have linked you have additionally put the assumption that $R$ is normal ... do you not know the answer to a finite extension even then ? $\endgroup$
    – user111524
    Commented May 6, 2018 at 17:55

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