Lemma 10.156.2. Let $R$ be a Nagata ring. Let $R \to S$ be essentially of finite type with $S$ reduced. Then the integral closure of $R$ in $S$ is finite over $R$.

**Proof.**
As $S$ is essentially of finite type over $R$ it is Noetherian and has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ m$, see Lemma 10.30.6. Since $S$ is reduced we have $S \subset \prod S_{\mathfrak q_ i}$ and each $S_{\mathfrak q_ i} = K_ i$ is a field, see Lemmas 10.24.4 and 10.24.1. It suffices to show that the integral closure $A_ i'$ of $R$ in each $K_ i$ is finite over $R$. This is true because $R$ is Noetherian and $A \subset \prod A_ i'$. Let $\mathfrak p_ i \subset R$ be the prime of $R$ corresponding to $\mathfrak q_ i$. As $S$ is essentially of finite type over $R$ we see that $K_ i = S_{\mathfrak q_ i} = \kappa (\mathfrak q_ i)$ is a finitely generated field extension of $\kappa (\mathfrak p_ i)$. Hence the algebraic closure $L_ i$ of $\kappa (\mathfrak p_ i)$ in $\subset K_ i$ is finite over $\kappa (\mathfrak p_ i)$, see Fields, Lemma 9.26.10. It is clear that $A_ i'$ is the integral closure of $R/\mathfrak p_ i$ in $L_ i$, and hence we win by definition of a Nagata ring.
$\square$

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