Lemma 28.7.9. Let $X$ be an integral normal scheme. Then $\Gamma (X, \mathcal{O}_ X)$ is a normal domain.

** The ring of functions on a normal scheme is normal. **

**Proof.**
Set $R = \Gamma (X, \mathcal{O}_ X)$. It is clear that $R$ is a domain. Suppose $f = a/b$ is an element of its fraction field which is integral over $R$. Say we have $f^ d + \sum _{i = 0, \ldots , d - 1} a_ i f^ i = 0$ with $a_ i \in R$. Let $U \subset X$ be a nonempty affine open. Since $b \in R$ is not zero and since $X$ is integral we see that also $b|_ U \in \mathcal{O}_ X(U)$ is not zero. Hence $a/b$ is an element of the fraction field of $\mathcal{O}_ X(U)$ which is integral over $\mathcal{O}_ X(U)$ (because we can use the same polynomial $f^ d + \sum _{i = 0, \ldots , d - 1} a_ i|_ U f^ i = 0$ on $U$). Since $\mathcal{O}_ X(U)$ is a normal domain (Lemma 28.7.4), we see that $f_ U = (a|_ U)/(b|_ U) \in \mathcal{O}_ X(U)$. It is clear that $f_ U|_ V = f_ V$ whenever $V \subset U \subset X$ are nonempty affine open. Hence the local sections $f_ U$ glue to an element $g \in R = \Gamma (X, \mathcal{O}_ X)$. Then $bg$ and $a$ restrict to the same element of $\mathcal{O}_ X(U)$ for all $U$ as above, hence $bg = a$, in other words, $g$ maps to $f$ in the fraction field of $R$.
$\square$

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