Lemma 10.164.7. 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 10.162.15). We have already seen that $R$ is Noetherian in Lemma 10.164.1. Let $R \to A$ be a finite type ring map into a domain. According to Lemma 10.162.3 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 10.162.15). Since $A \to B$ is smooth (Lemma 10.137.4) we see that $B$ is reduced (Lemma 10.163.7). Since $B$ is Noetherian it has only a finite number of minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$ (see Lemma 10.31.6). As $A \to B$ is flat each of these lies over $(0) \subset A$ (by going down, see Lemma 10.39.19) The total ring of fractions $Q(B)$ is the product of the $L_ i = \kappa (\mathfrak q_ i)$ (Lemmas 10.25.4 and 10.25.1). 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 10.37.16). 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 10.12.15). 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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