Two questions about discrete valuation rings of varieties 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!

$\textbf{Wrong answer. See Comments!}$
$\def\OO{{\mathcal O}}\def\CC{{\mathbb C}}$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.
