The Stacks project

The ring of functions on a normal scheme is normal.

Lemma 28.7.9. 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 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 28.7.2), 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 $R$. $\square$


Comments (3)

Comment #3033 by Brian Lawrence on

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

Comment #8517 by on

At the end of the proof, instead of "hence the local sections glue to a global section as desired," one could be a little bit more explanatory by saying "hence the local sections glue to a global section . In particular, for all open affine ; whence , so ."

There are also:

  • 3 comment(s) on Section 28.7: Normal schemes

Post a comment

Your email address will not be published.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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

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 input field. As a reminder, this is tag 0358. Beware of the difference between the letter 'O' and the digit '0'.