The ring of functions on a normal scheme is normal.

Lemma 27.7.9. Let $X$ be an integral normal scheme. Then $\Gamma (X, \mathcal{O}_ X)$ is a normal domain.

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 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 27.7.2), we see that $f_ U = (a|_ U)/(b|_ U) \in \mathcal{O}_ X(U)$. It is easy to see that $f_ U|_ V = f_ V$ whenever $V \subset U \subset X$ are affine open. Hence the local sections $f_ U$ glue to a global section $f$ as desired. $\square$

Comment #3033 by Brian Lawrence on

Suggested slogan: The ring of functions on an integral normal scheme is a normal domain.

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