Lemma 33.10.2. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent
$X$ is geometrically normal at $x$,
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
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 10.18.3). 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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)