The Stacks project

Lemma 10.164.3. Let $R \to S$ be a ring map. Assume that

  1. $R \to S$ is faithfully flat, and

  2. $S$ is a normal ring.

Then $R$ is a normal ring.

Proof. Since $S$ is reduced it follows that $R$ is reduced. Let $\mathfrak p$ be a prime of $R$. We have to show that $R_{\mathfrak p}$ is a normal domain. Since $S_{\mathfrak p}$ is faithfully over $R_{\mathfrak p}$ too we may assume that $R$ is local with maximal ideal $\mathfrak m$. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak m$. Then we see that $R \to S_{\mathfrak q}$ is faithfully flat (Lemma 10.39.17). Hence we may assume $S$ is local as well. In particular $S$ is a normal domain. Since $R \to S$ is faithfully flat and $S$ is a normal domain we see that $R$ is a domain. Next, suppose that $a/b$ is integral over $R$ with $a, b \in R$. Then $a/b \in S$ as $S$ is normal. Hence $a \in bS$. This means that $a : R \to R/bR$ becomes the zero map after base change to $S$. By faithful flatness we see that $a \in bR$, so $a/b \in R$. Hence $R$ is normal. $\square$

Comments (0)

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Your email address will not be published.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 033G. Beware of the difference between the letter 'O' and the digit '0'.