Lemma 10.165.6. Let $k'/k$ be a separable algebraic field extension. Let $A$ be an algebra over $k'$. Then $A$ is geometrically normal over $k$ if and only if it is geometrically normal 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 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

$K \otimes _{k'} A = (K \otimes _ k A) \otimes _{(k' \otimes _ k k')} k'$

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. $\square$

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