The Stacks project

Proof. Assume the formal fibres of $R_\mathfrak m$ have $P$ for all maximal ideals $\mathfrak m$ of $R$. Let $\mathfrak p$ be a prime of $R$ and choose a maximal ideal $\mathfrak p \subset \mathfrak m$. Since $R_\mathfrak m \to R_\mathfrak m^\wedge $ is faithfully flat we can choose a prime $\mathfrak p'$ if $R_\mathfrak m^\wedge $ lying over $\mathfrak pR_\mathfrak m$. Consider the commutative diagram

By assumption the fibres of the ring map $R_\mathfrak m \to R_\mathfrak m^\wedge $ have $P$. By Proposition 15.50.6 $(R_\mathfrak m^\wedge )_{\mathfrak p'} \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ is regular. The localization $R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge )_{\mathfrak p'}$ is regular. Hence $R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ is regular by Lemma 15.41.4. Hence the fibres of $R_\mathfrak m \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ have $P$ by (C). Since $R_\mathfrak m \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ factors through the localization $R_\mathfrak p$, also the fibres of $R_\mathfrak p \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ have $P$. Thus we may apply (D) to see that the fibres of $R_\mathfrak p \to R_\mathfrak p^\wedge $ have $P$. $\square$