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

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
$f^d + \sum_{i = 0, \ldots, d - 1} a_i f^i = 0$ 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 \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}

## Comments (0)

There are no comments yet for this tag.

There are also 2 comments on Section 27.7: Properties of Schemes.

## Add a comment on tag 0358

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

This captcha seems more appropriate than the usual illegible gibberish, right?