Lemma 15.50.7. Let $R$ be a Noetherian ring. Then $R$ is a G-ring if and only if $R_\mathfrak m$ has geometrically regular formal fibres for every maximal ideal $\mathfrak m$ of $R$.

**Proof.**
Assume $R_\mathfrak m \to R_\mathfrak m^\wedge $ is regular for every maximal ideal $\mathfrak m$ 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 ring map $R_\mathfrak m \to R_\mathfrak m^\wedge $ is regular. 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 \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ is regular by Lemma 15.41.4. Since it factors through the localization $R_\mathfrak p$, also the ring map $R_\mathfrak p \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ is regular. Thus we may apply Lemma 15.41.7 to see that $R_\mathfrak p \to R_\mathfrak p^\wedge $ is regular. $\square$

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## Comments (1)

Comment #9786 by Shizhang on