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.)
$\square$

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