Lemma 10.50.3. Let $A$ be a valuation ring. Then $A$ is a normal domain.

Proof. Suppose $x$ is in the field of fractions of $A$ and integral over $A$. Let $A'$ denote the subring of $K$ generated by $A$ and $x$. Since $A\subset A'$ is an integral extension, we see by Lemma 10.36.17 that there is a prime ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates $A$. Since $A$ is a valuation ring we conclude that $A=A'_{\mathfrak m'}$ and therefore that $x\in A$. $\square$

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