Lemma 15.50.4. Let $R$ be a Noetherian ring. Then $R$ is a G-ring if and only if for every finite free ring map $R \to S$ the formal fibres of $S$ are regular rings.
Proof. Assume that for any finite free ring map $R \to S$ the ring $S$ has regular formal fibres. Let $\mathfrak q \subset \mathfrak p \subset R$ be primes and let $\kappa (\mathfrak q) \subset L$ be a finite purely inseparable extension. To show that $R$ is a G-ring it suffices to show that
\[ R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak q)} L \]
is a regular ring. Choose a finite free extension $R \to R'$ such that $\mathfrak q' = \mathfrak qR'$ is a prime and such that $\kappa (\mathfrak q')$ is isomorphic to $L$ over $\kappa (\mathfrak q)$, see Algebra, Lemma 10.159.3. By Algebra, Lemma 10.97.8 we have
\[ R_\mathfrak p^\wedge \otimes _ R R' = \prod (R'_{\mathfrak p_ i'})^\wedge \]
where $\mathfrak p_ i'$ are the primes of $R'$ lying over $\mathfrak p$. Thus we have
\[ R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak q)} L = R_\mathfrak p^\wedge \otimes _ R R' \otimes _{R'} \kappa (\mathfrak q') = \prod (R'_{\mathfrak p_ i'})^\wedge \otimes _{R'_{\mathfrak p'_ i}} \kappa (\mathfrak q') \]
Our assumption is that the rings on the right are regular, hence the ring on the left is regular too. Thus $R$ is a G-ring. The converse follows from Lemma 15.50.3. $\square$
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