Lemma 10.166.1. Let $k$ be a field. Let $A$ be a $k$-algebra. Assume $A$ is Noetherian. The following properties of $A$ are equivalent:

$k' \otimes _ k A$ is regular for every finitely generated field extension $k'/k$, and

$k' \otimes _ k A$ is regular for every finite purely inseparable extension $k'/k$.

Here regular ring is as in Definition 10.110.7.

**Proof.**
The lemma makes sense by the remarks preceding the lemma. It is clear that (1) $\Rightarrow $ (2).

Assume (2) and let $K/k$ be a finitely generated field extension. By Lemma 10.45.3 we can find a diagram

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

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 regular ring because we assumed (2). Step 2: $B \otimes _{k'} k' \otimes _ k A$ is a regular ring as $k' \otimes _ k A \to B \otimes _{k'} k' \otimes _ k A$ is smooth (Lemma 10.137.4) and ascent of regularity along smooth maps (Lemma 10.163.10). Step 3. $K' \otimes _{k'} k' \otimes _ k A = K' \otimes _ k A$ is a regular ring as it is a localization of a regular ring (immediate from the definition). Step 4. Finally $K \otimes _ k A$ is a regular ring by descent of regularity along the faithfully flat ring map $K \otimes _ k A \to K' \otimes _ k A$ (Lemma 10.164.4). This proves the lemma.
$\square$

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