The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.158.7. Let $R \to S$ be a ring map. Assume that

  1. $R \to S$ is smooth and surjective on spectra, and

  2. $S$ is a Nagata ring.

Then $R$ is a Nagata ring.

Proof. Recall that a Nagata ring is the same thing as a Noetherian universally Japanese ring (Proposition 10.156.15). We have already seen that $R$ is Noetherian in Lemma 10.158.1. Let $R \to A$ be a finite type ring map into a domain. According to Lemma 10.156.3 it suffices to check that $A$ is N-1. It is clear that $B = A \otimes _ R S$ is a finite type $S$-algebra and hence Nagata (Proposition 10.156.15). Since $A \to B$ is smooth (Lemma 10.135.4) we see that $B$ is reduced (Lemma 10.157.7). Since $B$ is Noetherian it has only a finite number of minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$ (see Lemma 10.30.6). As $A \to B$ is flat each of these lies over $(0) \subset A$ (by going down, see Lemma 10.38.18) The total ring of fractions $Q(B)$ is the product of the $L_ i = \kappa (\mathfrak q_ i)$ (Lemmas 10.24.4 and 10.24.1). Moreover, the integral closure $B'$ of $B$ in $Q(B)$ is the product of the integral closures $B_ i'$ of the $B/\mathfrak q_ i$ in the factors $L_ i$ (compare with Lemma 10.36.16). Since $B$ is universally Japanese the ring extensions $B/\mathfrak q_ i \subset B_ i'$ are finite and we conclude that $B' = \prod B_ i'$ is finite over $B$. Since $A \to B$ is flat we see that any nonzerodivisor on $A$ maps to a nonzerodivisor on $B$. The corresponding map

\[ Q(A) \otimes _ A B = (A \setminus \{ 0\} )^{-1}A \otimes _ A B = (A \setminus \{ 0\} )^{-1}B \to Q(B) \]

is injective (we used Lemma 10.11.15). Via this map $A'$ maps into $B'$. This induces a map

\[ A' \otimes _ A B \longrightarrow B' \]

which is injective (by the above and the flatness of $A \to B$). Since $B'$ is a finite $B$-module and $B$ is Noetherian we see that $A' \otimes _ A B$ is a finite $B$-module. Hence there exist finitely many elements $x_ i \in A'$ such that the elements $x_ i \otimes 1$ generate $A' \otimes _ A B$ as a $B$-module. Finally, by faithful flatness of $A \to B$ we conclude that the $x_ i$ also generated $A'$ as an $A$-module, and we win. $\square$


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