Lemma 33.10.2. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent

1. $X$ is geometrically normal at $x$,

2. for every finite purely inseparable field extension $k'$ of $k$ and $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is normal, and

3. the ring $\mathcal{O}_{X, x}$ is geometrically normal over $k$ (see Algebra, Definition 10.165.2).

Proof. It is clear that (1) implies (2). Assume (2). Let $k'/k$ be a finite purely inseparable field extension (for example $k = k'$). Consider the ring $\mathcal{O}_{X, x} \otimes _ k k'$. By Algebra, Lemma 10.46.7 its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma 10.18.2). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes _ k k'$. By assumption this is a normal ring. Hence we deduce (3) by Algebra, Lemma 10.165.1.

Assume (3). Let $k'/k$ be a field extension. Since $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma 29.9.4). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes _ k k'$. Hence it is normal by assumption and (1) is proved. $\square$

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