The Stacks project

Proposition 15.50.5. Let $R$ be a $P$-ring where $P$ satisfies (A), (B), (C), and (D). If $R \to S$ is essentially of finite type then $S$ is a $P$-ring.

Proof. Since being a $P$-ring is a property of the local rings it is clear that a localization of a $P$-ring is a $P$-ring. Conversely, if every localization at a prime is a $P$-ring, then the ring is a $P$-ring. Thus it suffices to show that $S_\mathfrak q$ is a $P$-ring for every finite type $R$-algebra $S$ and every prime $\mathfrak q$ of $S$. Writing $S$ as a quotient of $R[x_1, \ldots , x_ n]$ we see from Lemma 15.50.3 that it suffices to prove that $R[x_1, \ldots , x_ n]$ is a $P$-ring. By induction on $n$ it suffices to prove that $R[x]$ is a $P$-ring. Let $\mathfrak q \subset R[x]$ be a maximal ideal. By Lemma 15.50.4 it suffices to show that the fibres of

\[ R[x]_\mathfrak q \longrightarrow R[x]_\mathfrak q^\wedge \]

have $P$. If $\mathfrak q$ lies over $\mathfrak p \subset R$, then we may replace $R$ by $R_\mathfrak p$. Hence we may assume that $R$ is a Noetherian local $P$-ring with maximal ideal $\mathfrak m$ and that $\mathfrak q \subset R[x]$ lies over $\mathfrak m$. Note that there is a unique prime $\mathfrak q' \subset R^\wedge [x]$ lying over $\mathfrak q$. Consider the diagram

\[ \xymatrix{ R[x]_\mathfrak q^\wedge \ar[r] & (R^\wedge [x]_{\mathfrak q'})^\wedge \\ R[x]_\mathfrak q \ar[r] \ar[u] & R^\wedge [x]_{\mathfrak q'} \ar[u] } \]

Since $R$ is a $P$-ring the fibres of $R[x] \to R^\wedge [x]$ have $P$ because they are base changes of the fibres of $R \to R^\wedge $ by a finitely generated field extension so (A) applies. Hence the fibres of the lower horizontal arrow have $P$ for example by Lemma 15.50.2. The right vertical arrow is regular because $R^\wedge $ is a G-ring (Propositions 15.49.6 and 15.49.10). It follows that the fibres of the composition $R[x]_\mathfrak q \to (R^\wedge [x]_{\mathfrak q'})^\wedge $ have $P$ by (C). Hence the fibres of the left vertical arrow have $P$ by (D) and the proof is complete. $\square$


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