Lemma 33.12.4. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k'/k$ be a finitely generated 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 regular at $x$,

$X_{k'}$ is geometrically regular at $x'$.

In particular, $X$ is geometrically regular over $k$ if and only if $X_{k'}$ is geometrically regular over $k'$.

**Proof.**
It is clear that (1) implies (2). Assume (2). Let $k \subset k''$ 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, finitely generated, field extension $k \subset k'''$ (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 of Noetherian local rings and hence faithfully flat. By (2) we see that the local ring on the right is regular. Thus by Algebra, Lemma 10.110.9 we conclude that $\mathcal{O}_{X_{k''}, x''}$ is regular. By Lemma 33.12.2 we see that $X$ is geometrically regular at $x$.
$\square$

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