Lemma 33.6.6. Let k be a field. Let X be a scheme over k. Let k'/k be a field extension. Let x \in X be a point, and let x' \in X_{k'} be a point lying over x. The following are equivalent
X is geometrically reduced at x,
X_{k'} is geometrically reduced at x'.
In particular, X is geometrically reduced over k if and only if X_{k'} is geometrically reduced over k'.
Proof.
It is clear that (1) implies (2). Assume (2). Let k''/k be a finite purely inseparable field extension and let x'' \in X_{k''} be a point lying over x (actually it is unique). We can find a common field extension k'''/k (i.e. with both k' \subset k''' and k'' \subset k''') and a point x''' \in X_{k'''} lying over both x' and x''. Consider the map of local rings
\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.
This is a flat local ring homomorphism and hence faithfully flat. By (2) we see that the local ring on the right is reduced. Thus by Algebra, Lemma 10.164.2 we conclude that \mathcal{O}_{X_{k''}, x''} is reduced. Thus by Lemma 33.6.5 we conclude that X is geometrically reduced at x.
\square
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