Lemma 10.159.5. 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 10.36.13 and Lemma 10.159.3. 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 10.156.16, and hence the integral closure of a finite type $k$-sub algebra is still a finite type $k$-sub algebra by Proposition 10.156.15). By Lemma 10.36.17 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 10.30.1). In particular $K \otimes _ k B$ has finitely many minimal primes (Lemma 10.30.6). 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 10.38.18 – 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 10.36.16).

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 10.36.12) 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 10.159.3) 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 10.42.2). Each $K_ i$ is the localization of a smooth $k$-algebra (Lemma 10.152.10). So $K_ i \otimes _ k B$ is the localization of a smooth $B$-algebra hence normal (Lemma 10.157.9). Thus $K \otimes _ k B$ is a normal ring (Lemma 10.36.17) and we win. $\square$

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