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 $x$.
$\square$

## Comments (0)