The Stacks project

Proof. It is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3) and (1) $\Rightarrow $ (4).

If $k'/k$ is a finite purely inseparable extension, then there is an embedding $k' \to k^{perf}$ of $k$-extensions. The ring map $k' \otimes _ k A \to k^{perf} \otimes _ k A$ is faithfully flat, hence $k' \otimes _ k A$ is normal if $k^{perf} \otimes _ k A$ is normal by Lemma 10.164.3. In this way we see that (4) $\Rightarrow $ (3).

Assume (2) and let $k'/k$ be any field extension. Then we can write $k' = \mathop{\mathrm{colim}}\nolimits _ i k_ i$ as a directed colimit of finitely generated field extensions. Hence we see that $k' \otimes _ k A = \mathop{\mathrm{colim}}\nolimits _ i k_ i \otimes _ k A$ is a directed colimit of normal rings. Thus we see that $k' \otimes _ k A$ is a normal ring by Lemma 10.37.17. Hence (1) holds.

Assume (3) and let $K/k$ be a finitely generated field extension. By Lemma 10.45.3 we can find a diagram

where $k'/k$, $K'/K$ are finite purely inseparable field extensions such that $K'/k'$ is separable. By Lemma 10.158.10 there exists a smooth $k'$-algebra $B$ such that $K'$ is the fraction field of $B$. Now we can argue as follows: Step 1: $k' \otimes _ k A$ is a normal ring because we assumed (3). Step 2: $B \otimes _{k'} k' \otimes _ k A$ is a normal ring as $k' \otimes _ k A \to B \otimes _{k'} k' \otimes _ k A$ is smooth (Lemma 10.137.4) and ascent of normality along smooth maps (Lemma 10.163.9). Step 3. $K' \otimes _{k'} k' \otimes _ k A = K' \otimes _ k A$ is a normal ring as it is a localization of a normal ring (Lemma 10.37.13). Step 4. Finally $K \otimes _ k A$ is a normal ring by descent of normality along the faithfully flat ring map $K \otimes _ k A \to K' \otimes _ k A$ (Lemma 10.164.3). This proves the lemma. $\square$

Proof. This is clear as being a normal ring is checked at the localizations at prime ideals. $\square$

Proof. This is true because $k^{perf} \otimes _ k K$ is a field. Namely, it is reduced by Lemma 10.43.6. By Lemma 10.45.4 (or by Definition 10.45.5) the field extension $k^{perf}/k$ is purely inseparable. Hence by Lemma 10.46.10 the ring $k^{perf} \otimes _ k K$ has a unique prime ideal. A reduced ring with a unique prime ideal is a field. $\square$

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. $\square$

Proof. Let $L/k$ be a finite purely inseparable field extension. Then $L' = k' \otimes _ k L$ is a field (see material in Fields, Section 9.28) and $A \otimes _ k L = A \otimes _{k'} L'$. Hence if $A$ is geometrically normal over $k'$, then $A$ is geometrically normal over $k$.

Assume $A$ is geometrically normal over $k$. Let $K/k'$ be a field extension. Then

Since $k' \otimes _ k k' \to k'$ is a localization by Lemma 10.43.8, we see that $K \otimes _{k'} A$ is a localization of a normal ring, hence normal. $\square$