Lemma 10.43.9. Let k'/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 10.43.6. Hence by Lemma 10.43.5 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
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 reduced algebra, hence reduced. \square
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