Lemma 10.159.1. Let $k$ be a field. Let $A$ be a $k$-algebra. The following properties of $A$ are equivalent:

$k' \otimes _ k A$ is a normal ring for every field extension $k'/k$,

$k' \otimes _ k A$ is a normal ring for every finitely generated field extension $k'/k$,

$k' \otimes _ k A$ is a normal ring for every finite purely inseparable extension $k'/k$,

$k^{perf} \otimes _ k A$ is a normal ring.

Here normal ring is defined in Definition 10.36.11.

**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 10.158.3. In this way we see that (4) $\Rightarrow $ (3).

Assume (2) and let $k \subset k'$ 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 10.36.17. Hence (1) holds.

Assume (3) and let $k \subset K$ be a finitely generated field extension. By Lemma 10.44.3 we can find a diagram

\[ \xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] } \]

where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ is separable. By Lemma 10.152.10 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 10.135.4) and ascent of normality along smooth maps (Lemma 10.157.9). 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 10.36.13). 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 10.158.3). This proves the lemma.
$\square$

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