Lemma 15.40.6. Let $k \subset K$ be a field extension. Then $k \to K$ is a regular ring map if and only if $K$ is a separable field extension of $k$.
Proof. If $k \to K$ is regular, then $K$ is geometrically reduced over $k$, hence $K$ is separable over $k$ by Algebra, Proposition 10.152.9. Conversely, if $K/k$ is separable, then $K$ is a colimit of smooth $k$-algebras, see Algebra, Lemma 10.152.11 hence is regular by Lemma 15.40.5. $\square$
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Your email address will not be published.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.