Lemma 10.50.11. Let $A$ be a normal domain with fraction field $K$.

For every $x \in K$, $x \not\in A$ there exists a valuation ring $A \subset V \subset K$ with fraction field $K$ such that $x \not\in V$.

If $A$ is local, we can moreover choose $V$ which dominates $A$.

In other words, $A$ is the intersection of all valuation rings in $K$ containing $A$ and if $A$ is local, then $A$ is the intersection of all valuation rings in $K$ dominating $A$.

**Proof.**
Suppose $x \in K$, $x \not\in A$. Consider $B = A[x^{-1}]$. Then $x \not\in B$. Namely, if $x = a_0 + a_1x^{-1} + \ldots + a_ d x^{-d}$ then $x^{d + 1} - a_0x^ d - \ldots - a_ d = 0$ and $x$ is integral over $A$ in contradiction with the fact that $A$ is normal. Thus $x^{-1}$ is not a unit in $B$. Thus $V(x^{-1}) \subset \mathop{\mathrm{Spec}}(B)$ is not empty (Lemma 10.17.2), and we can choose a prime $\mathfrak p \subset B$ with $x^{-1} \in \mathfrak p$. Choose a valuation ring $V \subset K$ dominating $B_\mathfrak p$ (Lemma 10.50.2). Then $x \not\in V$ as $x^{-1} \in \mathfrak m_ V$.

If $A$ is local, then we claim that $x^{-1} B + \mathfrak m_ A B \not= B$. Namely, if $1 = (a_0 + a_1x^{-1} + \ldots + a_ d x^{-d})x^{-1} + a'_0 + \ldots + a'_ d x^{-d}$ with $a_ i \in A$ and $a'_ i \in \mathfrak m_ A$, then we'd get

\[ (1 - a'_0) x^{d + 1} - (a_0 + a'_1) x^ d - \ldots - a_ d = 0 \]

Since $a'_0 \in \mathfrak m_ A$ we see that $1 - a'_0$ is a unit in $A$ and we conclude that $x$ would be integral over $A$, a contradiction as before. Then choose the prime $\mathfrak p \supset x^{-1} B + \mathfrak m_ A B$ we find $V$ dominating $A$.
$\square$

## Comments (1)

Comment #3227 by Giulio on

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