Lemma 10.166.6. Let $k'/k$ be a separable algebraic field extension. Let $A$ be an algebra over $k'$. Then $A$ is geometrically regular over $k$ if and only if it is geometrically regular over $k'$.

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 regular over $k'$, then $A$ is geometrically regular over $k$.

Assume $A$ is geometrically regular over $k$. Since $k'$ is the filtered colimit of finite extensions of $k$ we may assume by Lemma 10.166.5 that $k'/k$ is finite separable. Consider the ring maps

$k' \to A \otimes _ k k' \to A.$

Note that $A \otimes _ k k'$ is geometrically regular over $k'$ as a base change of $A$ to $k'$. Note that $A \otimes _ k k' \to A$ is the base change of $k' \otimes _ k k' \to k'$ by the map $k' \to A$. Since $k'/k$ is an étale extension of rings, we see that $k' \otimes _ k k' \to k'$ is étale (Lemma 10.143.3). Hence $A$ is geometrically regular over $k'$ by Lemma 10.166.4. $\square$

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