Lemma 10.42.9. Let $k \subset k'$ be a separable algebraic field extension. Let $A$ be an algebra over $k'$. Then $A$ is geometrically reduced over $k$ if and only if it is geometrically reduced over $k'$.

**Proof.**
Assume $A$ is geometrically reduced over $k'$. Let $K/k$ be a field extension. Then $K \otimes _ k k'$ is a reduced ring by Lemma 10.42.6. Hence by Lemma 10.42.5 we find that $K \otimes _ k A = (K \otimes _ k k') \otimes _{k'} A$ is reduced.

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

Since $k' \otimes _ k k' \to k'$ is a localization by Lemma 10.42.8, we see that $K \otimes _{k'} A$ is a localization of a reduced algebra, hence reduced. $\square$

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