Proposition 41.19.4. Let $A$, $B$ be Noetherian local rings. Let $f : A \to B$ be an étale homomorphism of local rings. Then $A$ is regular if and only if $B$ is so.
Proof. If $B$ is regular, then $A$ is regular by Algebra, Lemma 10.110.9. Assume $A$ is regular. Let $\mathfrak m$ be the maximal ideal of $A$. Then $\dim _{\kappa (\mathfrak m)} \mathfrak m/\mathfrak m^2 = \dim (A) = \dim (B)$ (see Lemma 41.19.1). On the other hand, $\mathfrak mB$ is the maximal ideal of $B$ and hence $\mathfrak m_ B/\mathfrak m_ B = \mathfrak mB/\mathfrak m^2B$ is generated by at most $\dim (B)$ elements. Thus $B$ is regular. (You can also use the slightly more general Algebra, Lemma 10.112.8.) $\square$
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Your email address will not be published.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 (0)