The Stacks project

Lemma 10.165.5. Let $k$ be a field. Let $A, B$ be $k$-algebras. Assume $A$ is geometrically normal over $k$ and $B$ is a normal ring. Then $A \otimes _ k B$ is a normal ring.

Proof. Let $\mathfrak r$ be a prime ideal of $A \otimes _ k B$. Denote $\mathfrak p$, resp. $\mathfrak q$ the corresponding prime of $A$, resp. $B$. Then $(A \otimes _ k B)_{\mathfrak r}$ is a localization of $A_{\mathfrak p} \otimes _ k B_{\mathfrak q}$. Hence it suffices to prove the result for the ring $A_{\mathfrak p} \otimes _ k B_{\mathfrak q}$, see Lemma 10.37.13 and Lemma 10.165.3. Thus we may assume $A$ and $B$ are domains.

Assume that $A$ and $B$ are domains with fractions fields $K$ and $L$. Note that $B$ is the filtered colimit of its finite type normal $k$-sub algebras (as $k$ is a Nagata ring, see Proposition 10.162.16, and hence the integral closure of a finite type $k$-sub algebra is still a finite type $k$-sub algebra by Proposition 10.162.15). By Lemma 10.37.17 we reduce to the case that $B$ is of finite type over $k$.

Assume that $A$ and $B$ are domains with fractions fields $K$ and $L$ and $B$ of finite type over $k$. In this case the ring $K \otimes _ k B$ is of finite type over $K$, hence Noetherian (Lemma 10.31.1). In particular $K \otimes _ k B$ has finitely many minimal primes (Lemma 10.31.6). Since $A \to A \otimes _ k B$ is flat, this implies that $A \otimes _ k B$ has finitely many minimal primes (by going down for flat ring maps – Lemma 10.39.19 – these primes all lie over $(0) \subset A$). Thus it suffices to prove that $A \otimes _ k B$ is integrally closed in its total ring of fractions (Lemma 10.37.16).

We claim that $K \otimes _ k B$ and $A \otimes _ k L$ are both normal rings. If this is true then any element $x$ of $Q(A \otimes _ k B)$ which is integral over $A \otimes _ k B$ is (by Lemma 10.37.12) contained in $K \otimes _ k B \cap A \otimes _ k L = A \otimes _ k B$ and we're done. Since $A \otimes _ K L$ is a normal ring by assumption, it suffices to prove that $K \otimes _ k B$ is normal.

As $A$ is geometrically normal over $k$ we see $K$ is geometrically normal over $k$ (Lemma 10.165.3) hence $K$ is geometrically reduced over $k$. Hence $K = \bigcup K_ i$ is the union of finitely generated field extensions of $k$ which are geometrically reduced (Lemma 10.43.2). Each $K_ i$ is the localization of a smooth $k$-algebra (Lemma 10.158.10). So $K_ i \otimes _ k B$ is the localization of a smooth $B$-algebra hence normal (Lemma 10.163.9). Thus $K \otimes _ k B$ is a normal ring (Lemma 10.37.17) and we win. $\square$

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 06DF. Beware of the difference between the letter 'O' and the digit '0'.