
Lemma 15.50.4. Let $R$ be a Noetherian ring. Assume $P$ satisfies (C) and (D). Then $R$ is a $P$-ring if and only if the formal fibres of $R_\mathfrak m$ have $P$ for every maximal ideal $\mathfrak m$ of $R$.

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

$\xymatrix{ R_\mathfrak m^\wedge \ar[r] & (R_\mathfrak m^\wedge )_{\mathfrak p'} \ar[r] & (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge \\ R_\mathfrak m \ar[u] \ar[r] & R_\mathfrak p \ar[u] \ar[r] & R_\mathfrak p^\wedge \ar[u] }$

By assumption the fibres of the ring map $R_\mathfrak m \to R_\mathfrak m^\wedge$ have $P$. By Proposition 15.49.6 $(R_\mathfrak m^\wedge )_{\mathfrak p'} \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge$ 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 15.40.4. 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 $P$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).