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The background to my question, in a nutshell, is: If $k$ is a field and $X$ a $k$-variety, i.e. an integral, separated, finite type $k$-scheme, which discrete rank $1$ valuations on $k(X)$ come from codimension $1$ points on some model of $k(X)$?

Let $A$ be a noetherian local domain with residue field $k_A$, field of fractions $K$ and $v$ a discrete rank $1$ valuation on $K$ dominating $A$, with residue field $k(v)$. The Abhyankar Inequality then reads $$1+trdeg_{k_A} k(v)\leq \dim A,$$ If this inequality is an equality, then the extension $k(v)/k_A$ is finitely generated. There are of course examples where this inequality is a strict inequality, e.g. Example 9 in [1] (which is originally due to Zariski). In this example, $\dim A = 3$, the transcendence degree of $k(v)/k_A$ is $1$, but this extension is not finitely generated.

My question is: Does finite generation of the residue extension imply that Abhyankar's inequality is an equality, in "geometric situations"?

To be precise: Let $k$ be a field, $X$ an integral, separated, finite type $k$-scheme and $K=k(X)$ its function field. Let $v$ be a discrete valuation on $K$ and assume that it has a center $x$ on $X$. Then the above discussion applies with $A=\mathcal{O}_{X,x}$, $k_A=k(x)$, and I ask whether it is known that the Abhyankar inequality is an equality if $k_v/k(x)$ (or equivalently $k_v/k$) is finitely generated.

To tie this formulation to the motivation from the beginning: $v$ comes from a divisor on some model if and only if the Abhyankar inequality is an equality. My question is whether $v$ comes from a divisor if and only if $k(v)/k$ is finitely generated.

[1] http://www.math.univ-toulouse.fr/~vaquie/textes/uniformization.pdf

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  • $\begingroup$ This is not an answer to your precise question but on the comment "$v$ comes from a divisor" by which I suppose you mean "the center of $v$ is a divisor". Let $A=k[X,Y]_{(X,Y)}$ and $v(f(X,Y))=ord_t(f(t,t))$. Then $v$ is a discrete valuation dominating $A$. Clearly, $k_A=k$ and check that $k(v)$ contains $k(X/Y)$. I think it is also true that $k(X/Y)$ is algebraically closed in $k(X,Y)$. So by Abhyankar's inequality $k(v)=k(X/Y)$. But of course the center of $v$ is of co-dimension 2 but equality holds in Abhyankar's inequality. $\endgroup$
    – Manish Kumar
    Commented Jul 3, 2013 at 12:59
  • $\begingroup$ Dear Manish, thanks for your comment! It is true that a discrete valuation for which equality holds in Abhyankar's inequality does not necessarily have a divisor as center on a given model ($\mathbb{A}^2$ in your example). But as soon as equality holds, there exists some model on which the valuation has a divisor as center; if I am not mistaken, then in your example one such model would be the spectrum of $k[X,Y,X/Y]$, which is an open subset of the blow-up of $\mathbb{A}^2$ in the origin. $\endgroup$
    – Lars
    Commented Jul 3, 2013 at 13:36
  • $\begingroup$ Dear Lars, Thanks for the clarification. I think for every valuation $v$ of $k(X)/k$ one should be able find a model of $X$ on which the center of $v$ is a divisor. If the center of $v$ on $X$ is regular then if one takes a sequence of monoidal transformation of $X$ along $v$ then eventually the center will become a divisor. Basically look at the valuations of a system of parameters of the ideal defining the center of $X$. Their gcd is 1. After the monidal transformation these numbers will go down for the center of $v$ on this new model. Eventually one would get that the center is a divisor. $\endgroup$
    – Manish Kumar
    Commented Jul 3, 2013 at 17:25
  • $\begingroup$ Note that resolution of singularities (in fact uniformization itself) implies that $X$ has a model so that the center of $v$ is regular. $\endgroup$
    – Manish Kumar
    Commented Jul 3, 2013 at 17:28
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    $\begingroup$ Hi Manish, it is not true that every discrete valuation on k(X)/k has a divisor as a center on some model. This property is equivalent to the Abhyankar inequality being an equality. In particular, a discrete valuation $v$ has a divisor as center on some model if and only if $trdeg_k k(v)=trdeg_k k(X)-1$. There is an example for a discrete valuation which does not have this property in the paper by Vaquié which I cite. $\endgroup$
    – Lars
    Commented Jul 3, 2013 at 19:58

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