Lemma 15.51.3. 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 $\mathfrak p'$. Assume $P$ satisfies (A) and (B).
If $\kappa (\mathfrak q) \to R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q)$ has $P$, then $\kappa (\mathfrak q') \to R'_{\mathfrak p'} \otimes _{R'} \kappa (\mathfrak q')$ has $P$.
If the formal fibres of $R_\mathfrak p$ have $P$, then the formal fibres of $R'_{\mathfrak p'}$ have $P$.
If $R \to R'$ is quasi-finite and $R$ is a $P$-ring, then $R'$ is a $P$-ring.
Proof.
It is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3). Assume $P$ holds for $\kappa (\mathfrak q) \to R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q)$. 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 from the assumptions on $P$.
$\square$
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