Two questions about discrete valuation rings of varieties

Question

Let $X$ be a proper, normal variety over $\mathbb{C}$, and $k(X)$ be its field of rational functions. I think the following two statements are true, but I was unable to give a proof or find the references:

(1) For any $f \in k(X), f \neq 0$, there are only finite discrete valuations $v$ of $k(X)$ such that $v(f) \neq 0$.

(2) For any discrete valuation $v$ of $k(X)$, there exists a variety $Y$, birational to $X$, and a Weil divisor $E$ such that the valuation given by this divisor is the same as $v$.

Any suggestion for either problem is welcome!

Answer 1

$\def\OO{{\mathcal O}}\def\CC{{\mathbb C}}$So it turns out that the situation is actually much more complicated when your variety is of dimension greater than 1.

First some background on Zariski-Riemann spaces. Let $K/k$ be a field extension (in our case $k(X)/\CC$). Then the Zariski-Riemann space $Z(K/k)$ is the space of all valuation rings $k \subset \OO_v \subset K$. I will sometimes denote the point in $Z$ by just the valuation $v$ where convenient though this is technically abuse of notation.

We topologize this space by defining open sets to be of the form

$$ \mathscr{V}(A) := \{\OO_v \in Z : A \subset \OO_v\} $$

where $k \subset A \subset K$ is a subalgebra finitely generated over $k$. Then this space inherits a sheaf of rings by

$$ \mathcal{O}(U) := \bigcap_{v \in U} \mathcal{O}_v $$

which makes it a locally ringed space with local ring at each point the valuation that that point represents.

In the case that $k$ is algebraically closed of characteristic $0$ (possibly unnecessary assumptions?) and the transcendence degree of $K/k$ is $1$, the Zariski-Riemann space is exactly the unique smooth projective curve over $k$ with function field $K$. For higher dimensions, the Zariski Riemann space is not a scheme, just a locally ringed space.

The point of all this is that in higher dimensions, there are many more points of the Zariski Riemann space than valuations that come from divisors on some birational model of the field which I will call divisorial valuations.

Let's take the example of a surface $S$ over $\CC$ and look at the Zariski-Riemann space of $k(S)/\CC$. Then on top of the trivial valuation and the divisorial valuation, we have a valuation corresponding to the germ of a curve $C$ going through a fixed point $p$ on some birational variety $\tilde{S}$. This is a rank two discrete valuation with value group isomorphic to $\mathbb{Z} \times \mathbb{Z}$ lexicographically ordered that gives the order of vanishing of a function $f$ along the curve, and then at the point $p$ as a function on the curve.

Even if we restrict to rank one discrete valuations (i.e. value group $\mathbb{Z}$) there are non algebraic valuations given by the order of vanishing along some analytic germ like $y = e^x$ in the plane. This gives a rank one discrete valuation that can't come from a divisor on some birational model.

In higher dimensions the situation gets even more complicated and in fact as far as I can tell, a complete classification of the points in the Zariski-Riemann space is not well understand. In general you will at least get a rank $k$ discrete valuation with value group the lexicographically ordered $\mathbb{Z}^k$ given by the order of vanishing along the germ of an inclusion of subvarieties $V_1 \supset V_2 \supset \ldots \supset V_k$ where $\operatorname{codim} V_i = i$.

However, according to someone I asked, the divisorial discrete valuations are dense in the Zariski-Riemann space. So "most" valuations come from a divisor on some birational model. I have not been able to find a reference for this though as the literature on Zariski-Riemann spaces is sparse.

Answer 2

$\textbf{Wrong answer. See Comments!}$

$\def\OO{{\mathcal O}}\def\CC{{\mathbb C}}$I think the following argument shows that in fact $v$ is the valuation associated to a prime Weil divisor on $X$ from which part $(1)$ follows as well.

Let $v$ be such a discrete valuation. Consider the valuation ring $\mathcal{O}_v \subset k(X)$ corresponding to $v$. Now by the valuative criterion of properness, we have a lift of the map $\operatorname{Spec} \OO_v \to \operatorname{Spec} \CC$ to a morphism $\operatorname{Spec} \OO_v \to X$ that commutes with the morphism from $\operatorname{Spec} k(X)$. So the generic point of $\OO_v$ is sent to the generic point of $X$ and the closed point of $\OO_v$ is sent to a point $p \in X$.

This induces a morphism of local rings $\OO_p \to \OO_v$ commuting with the inclusions to $k(X)$. Thus $\OO_p$ includes into $\OO_v$ as subrings of $k(X)$ but $X$ is normal so $\OO_p$ is integrally closed and so is $\OO_v$ since it is a DVR thus $\OO_p = \OO_v$. The dimension of $\OO_v$ is $1$ and so $p$ is a height one prime. That is, $p$ is the generic point of a prime Weil divisor $D$ on $X$. But the valuation associated to a divisor is exactly the valuation of the local ring at the generic point of that divisor.