Lemma 10.156.5. Let $R$ be a Nagata ring. If $R \to S$ is a quasi-finite ring map (for example finite) then $S$ is a Nagata ring also.
Proof. First note that $S$ is Noetherian as $R$ is Noetherian and a quasi-finite ring map is of finite type. Let $\mathfrak q \subset S$ be a prime ideal, and set $\mathfrak p = R \cap \mathfrak q$. Then $R/\mathfrak p \subset S/\mathfrak q$ is quasi-finite and hence we conclude that $S/\mathfrak q$ is N-2 by Lemma 10.155.5 as desired. $\square$
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