Lemma 15.50.14. Let $A$ be a G-ring. Let $I \subset A$ be an ideal and let $A^\wedge $ be the completion of $A$ with respect to $I$. Then $A \to A^\wedge $ is regular.

[Theorem 79, MatCA]

**Proof.**
The ring map $A \to A^\wedge $ is flat by Algebra, Lemma 10.97.2. The ring $A^\wedge $ is Noetherian by Algebra, Lemma 10.97.6. Thus it suffices to check the third condition of Lemma 15.41.2. Let $\mathfrak m' \subset A^\wedge $ be a maximal ideal lying over $\mathfrak m \subset A$. By Algebra, Lemma 10.96.6 we have $IA^\wedge \subset \mathfrak m'$. Since $A^\wedge /IA^\wedge = A/I$ we see that $I \subset \mathfrak m$, $\mathfrak m/I = \mathfrak m'/IA^\wedge $, and $A/\mathfrak m = A^\wedge /\mathfrak m'$. Since $A^\wedge /\mathfrak m'$ is a field, we conclude that $\mathfrak m$ is a maximal ideal as well. Then $A_\mathfrak m \to A^\wedge _{\mathfrak m'}$ is a flat local ring homomorphism of Noetherian local rings which identifies residue fields and such that $\mathfrak m A^\wedge _{\mathfrak m'} = \mathfrak m'A^\wedge _{\mathfrak m'}$. Thus it induces an isomorphism on complete local rings, see Lemma 15.43.9. Let $(A_\mathfrak m)^\wedge $ be the completion of $A_\mathfrak m$ with respect to its maximal ideal. The ring map

is faithfully flat (Algebra, Lemma 10.97.3). Thus we can apply Lemma 15.41.7 to the ring maps

to conclude because $A_\mathfrak m \to (A_\mathfrak m)^\wedge $ is regular as $A$ is a G-ring. $\square$

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