# The Stacks Project

## Tag 0358

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 $f^d + \sum_{i = 1, \ldots, d} a_i f^i = 0$ 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 $f^d + \sum_{i = 1, \ldots, d} a_i|_U f^i = 0$ 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. $\square$

The code snippet corresponding to this tag is a part of the file properties.tex and is located in lines 887–891 (see updates for more information).

\begin{lemma}
\label{lemma-normal-integral-sections}
Let $X$ be an integral normal scheme.
Then $\Gamma(X, \mathcal{O}_X)$ is a normal domain.
\end{lemma}

\begin{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
$f^d + \sum_{i = 1, \ldots, d} a_i f^i = 0$ 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
$f^d + \sum_{i = 1, \ldots, d} a_i|_U f^i = 0$ on $U$).
Since $\mathcal{O}_X(U)$ is a normal domain
(Lemma \ref{lemma-locally-normal}), 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.
\end{proof}

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