Lemma 10.165.4. Let $k$ be a field. Let $K/k$ be a separable field extension. Then $K$ is geometrically normal over $k$.
Proof. This is true because $k^{perf} \otimes _ k K$ is a field. Namely, it is reduced by Lemma 10.43.6. By Lemma 10.45.4 (or by Definition 10.45.5) the field extension $k^{perf}/k$ is purely inseparable. Hence by Lemma 10.46.10 the ring $k^{perf} \otimes _ k K$ has a unique prime ideal. A reduced ring with a unique prime ideal is a field. $\square$
Post a comment
Your email address will not be published.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 (2)
Comment #8096 by Rubén Muñoz--Bertrand on
Comment #8209 by Stacks Project on