Lemma 33.10.3. Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent

1. $X$ is geometrically normal,

2. $X_{k'}$ is a normal scheme for every field extension $k'/k$,

3. $X_{k'}$ is a normal scheme for every finitely generated field extension $k'/k$,

4. $X_{k'}$ is a normal scheme for every finite purely inseparable field extension $k'/k$,

5. for every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is geometrically normal (see Algebra, Definition 10.165.2), and

6. $X_{k^{perf}}$ is a normal scheme.

Proof. Assume (1). Then for every field extension $k \subset k'$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is normal. By definition this means that $X_{k'}$ is normal. Hence (2).

It is clear that (2) implies (3) implies (4).

Assume (4) and let $U \subset X$ be an affine open subscheme. Then $U_{k'}$ is a normal scheme for any finite purely inseparable extension $k \subset k'$ (including $k = k'$). This means that $k' \otimes _ k \mathcal{O}(U)$ is a normal ring for all finite purely inseparable extensions $k \subset k'$. Hence $\mathcal{O}(U)$ is a geometrically normal $k$-algebra by definition. Hence (4) implies (5).

Assume (5). For any field extension $k \subset k'$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_ X(U) \otimes _ k k'$ where $U$ is affine open in $X$ (see Schemes, Section 26.17). Hence $X_{k'}$ is normal. So (1) holds.

The equivalence of (5) and (6) follows from the definition of geometrically normal algebras and the equivalence (just proved) of (3) and (4). $\square$

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