Lemma 15.50.3. Let $R \to R'$ be a finite type map of Noetherian rings and let
\[ \xymatrix{ \mathfrak q' \ar[r] & \mathfrak p' \ar[r] & R' \\ \mathfrak q \ar[r] \ar@{-}[u] & \mathfrak p \ar[r] \ar@{-}[u] & R \ar[u] } \]
be primes. Assume $R \to R'$ is quasi-finite at $\mathfrak p'$.
If the formal fibre $R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q)$ is geometrically regular over $\kappa (\mathfrak q)$, then the formal fibre $R'_{\mathfrak p'} \otimes _{R'} \kappa (\mathfrak q')$ is geometrically regular over $\kappa (\mathfrak q')$.
If the formal fibres of $R_\mathfrak p$ are geometrically regular, then the formal fibres of $R'_{\mathfrak p'}$ are geometrically regular.
If $R \to R'$ is quasi-finite and $R$ is a G-ring, then $R'$ is a G-ring.
Proof.
It is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3). Assume $R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q)$ is geometrically regular over $\kappa (\mathfrak q)$. By Algebra, Lemma 10.124.3 we see that
\[ R_\mathfrak p^\wedge \otimes _ R R' = (R'_{\mathfrak p'})^\wedge \times B \]
for some $R_\mathfrak p^\wedge $-algebra $B$. Hence $R'_{\mathfrak p'} \to (R'_{\mathfrak p'})^\wedge $ is a factor of a base change of the map $R_\mathfrak p \to R_\mathfrak p^\wedge $. It follows that $(R'_{\mathfrak p'})^\wedge \otimes _{R'} \kappa (\mathfrak q')$ is a factor of
\[ R_\mathfrak p^\wedge \otimes _ R R' \otimes _{R'} \kappa (\mathfrak q') = R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak q)} \kappa (\mathfrak q'). \]
Thus the result follows as extension of base field preserves geometric regularity, see Algebra, Lemma 10.166.1.
$\square$
Comments (1)
Comment #9785 by Shizhang on