Lemma 15.42.4. Let $A$ be a Noetherian local ring. Then $A$ is regular if and only if $A^\wedge $ is so.

**Proof.**
If $A^\wedge $ is regular, then $A$ is regular by Algebra, Lemma 10.109.9. Assume $A$ is regular. Let $\mathfrak m$ be the maximal ideal of $A$. Then $\dim _{\kappa (\mathfrak m)} \mathfrak m/\mathfrak m^2 = \dim (A) = \dim (A^\wedge )$ (Lemma 15.42.1). On the other hand, $\mathfrak mA^\wedge $ is the maximal ideal of $A^\wedge $ and hence $\mathfrak m_{A^\wedge }$ is generated by at most $\dim (A^\wedge )$ elements. Thus $A^\wedge $ is regular. (You can also use Algebra, Lemma 10.111.8.)
$\square$

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