Characterize local Nagata rings

Thanks to J\'anos Koll\'ar
http://stacks.math.columbia.edu/tag/0331#comment-1524

Comment: Clearly this proof is not optimal. On the other hand it is
sometimes a good idea to use a little bit more theory in order for
the reader to see that it exists and the lemma is not just some
completely random fact...

algebra.tex

@@ -41686,7 +41686,10 @@ \section{Nagata rings}
 \begin{enumerate}
 \item $R$ is Nagata,
 \item for $R \to S$ finite with $S$ a domain and $\mathfrak m' \subset S$
-maximal the local ring $S_{\mathfrak m'}$ is analytically unramified.
+maximal the local ring $S_{\mathfrak m'}$ is analytically unramified,
+\item for $(R, \mathfrak m) \to (S, \mathfrak m')$ finite
+local homomorphism with $S$ a domain, then $S$ is analytically
+unramified.
 \end{enumerate}
 \end{lemma}
 
@@ -41696,9 +41699,10 @@ \section{Nagata rings}
 Hence the local ring $S_{\mathfrak m'}$ is Nagata
 (Lemma \ref{lemma-nagata-localize}). Thus it is analytically
 unramified by Lemma \ref{lemma-local-nagata-domain-analytically-unramified}.
+It is clear that (2) implies (3).
 
 \medskip\noindent
-Assume (2) holds. Let $\mathfrak p \subset R$ be a prime ideal and
+Assume (3) holds. Let $\mathfrak p \subset R$ be a prime ideal and
 let $f.f.(R/\mathfrak p) \subset L$ be a finite extension of fields.
 To prove (1) we have to show that the integral closure of $R/\mathfrak p$
 is finite over $R/\mathfrak p$. Choose $x_1, \ldots, x_n \in L$
@@ -41713,7 +41717,7 @@ \section{Nagata rings}
 Having done this let $S = R/\mathfrak p[x_1, \ldots, x_n] \subset L$.
 Then $S$ is finite over $R$, a domain, and $S/\mathfrak m S$ is a quotient
 of $R/\mathfrak m[T_1, \ldots, T_n]/(T_1^{d_1}, \ldots, T_n^{d_n})$.
-Hence $S$ is local. By (2) $S$ is analytically unramified and by
+Hence $S$ is local. By (3) $S$ is analytically unramified and by
 Lemma \ref{lemma-analytically-unramified-easy}
 we find that its integral closure $S'$ in $L$ is finite over $S$.
 Since $S'$ is also the integral closure of $R/\mathfrak p$ in

more-algebra.tex

@@ -10232,27 +10232,6 @@ \section{Excellent rings}
 the proof that localization preserves (quasi-)excellency.
 \end{proof}
 
-\begin{lemma}
-\label{lemma-quasi-excellent-nagata}
-A quasi-excellent ring is Nagata.
-\end{lemma}
-
-\begin{proof}
-Let $R$ be quasi-excellent.
-Using that a finite type algebra over $R$ is quasi-excellent
-(Lemma \ref{lemma-finite-type-over-excellent}) we see that
-it suffices to show that any quasi-excellent domain is N-1, see
-Algebra, Lemma \ref{algebra-lemma-check-universally-japanese}.
-Applying Algebra, Lemma \ref{algebra-lemma-characterize-N-1}
-(and using that a quasi-excellent ring is J-2) we reduce
-to showing that a quasi-excellent local domain $R$ is N-1.
-As $R \to R^\wedge$ is regular we see that $R^\wedge$
-is reduced by Lemma \ref{lemma-reduced-goes-up}.
-In other words, $R$ is analytically unramified.
-Hence $R$ is N-1 by
-Algebra, Lemma \ref{algebra-lemma-analytically-unramified-easy}.
-\end{proof}
-
 \begin{proposition}
 \label{proposition-ubiquity-excellent}
 The following types of rings are excellent:
@@ -10277,6 +10256,79 @@ \section{Excellent rings}
 \ref{algebra-remark-Noetherian-complete-local-ring-universally-catenary}.
 \end{proof}
 
+\noindent
+The material developed above has some consequences for Nagata rings.
+
+\begin{lemma}
+\label{lemma-Nagata-local-ring}
+Let $(A, \mathfrak m)$ be a Noetherian local ring.
+The following are equivalent
+\begin{enumerate}
+\item $A$ is Nagata, and
+\item the formal fibres of $A$ are geometrically reduced.
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+Assume (2). By
+Algebra, Lemma \ref{algebra-lemma-local-nagata-and-analytically-unramified}
+we have to show that if $A \to B$ is finite, $B$ is a domain,
+and $\mathfrak m' \subset B$ is a maximal ideal, then $B_{\mathfrak m'}$
+is analytically unramified.
+Combining Lemma \ref{lemma-formal-fibres-reduced} and
+\ref{lemma-check-P-ring-maximal-ideals} and
+Proposition \ref{proposition-finite-type-over-P-ring}
+we see that the formal fibres of $B_{\mathfrak m'}$ are
+geometrically reduced. In particular
+$B_{\mathfrak m'}^\wedge \otimes_B f.f.(B)$ is reduced
+and it follows that $B_{\mathfrak m'}^\wedge$ is reduced, i.e.,
+$B_{\mathfrak m'}$ is analytically unramified.
+
+\medskip\noindent
+Assume (1). Let $\mathfrak q \subset A$ be a prime ideal
+and let $\kappa(\mathfrak q) \subset K$ be a finite extension.
+We have to show that $A^\wedge \otimes_A K$ is reduced.
+Let $A/\mathfrak q \subset B \subset K$ be a local subring
+finite over $A$ whose fraction field is $K$.
+To construct $B$ choose $x_1, \ldots, x_n \in K$
+which generate $K$ over $\kappa(\mathfrak q)$
+and which satisfy monic polynomials
+$P_i(T) = T^{d_i} + a_{i, 1} T^{d_i - 1} + \ldots + a_{i, d_i} = 0$
+with $a_{i, j} \in \mathfrak m$. Then let $B$ be the $A$-subalgebra
+of $K$ generated by $x_1, \ldots, x_n$. (For more details see
+the proof of Algebra, Lemma
+\ref{algebra-lemma-local-nagata-and-analytically-unramified}.)
+Then
+$$
+A^\wedge \otimes_A K =
+(A^\wedge \otimes_A B)_\mathfrak q =
+B^\wedge_\mathfrak q
+$$
+Since $B^\wedge$ is reduced by  Algebra, Lemma
+\ref{algebra-lemma-local-nagata-and-analytically-unramified}
+the proof is complete.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-quasi-excellent-nagata}
+A quasi-excellent ring is Nagata.
+\end{lemma}
+
+\begin{proof}
+Let $R$ be quasi-excellent.
+Using that a finite type algebra over $R$ is quasi-excellent
+(Lemma \ref{lemma-finite-type-over-excellent}) we see that
+it suffices to show that any quasi-excellent domain is N-1, see
+Algebra, Lemma \ref{algebra-lemma-check-universally-japanese}.
+Applying Algebra, Lemma \ref{algebra-lemma-characterize-N-1}
+(and using that a quasi-excellent ring is J-2) we reduce
+to showing that a quasi-excellent local domain $R$ is N-1.
+As $R \to R^\wedge$ is regular we see that $R^\wedge$
+is reduced by Lemma \ref{lemma-reduced-goes-up}.
+In other words, $R$ is analytically unramified.
+Hence $R$ is N-1 by
+Algebra, Lemma \ref{algebra-lemma-analytically-unramified-easy}.
+\end{proof}