Lemma 15.51.6. Let A be a P-ring where P satisfies (B) and (D). Let I \subset A be an ideal and let A^\wedge be the completion of A with respect to I. Then the fibres of A \to A^\wedge have P.
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 15.51.2. 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 (D) to the ring maps
to conclude because the fibres of A_\mathfrak m \to (A_\mathfrak m)^\wedge have P as A is a P-ring. \square
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