Proof.
Assume R is Nagata and let R \to S and \mathfrak m' \subset S be as in (2). Then S is Nagata by Lemma 10.162.5. Hence the local ring S_{\mathfrak m'} is Nagata (Lemma 10.162.6). Thus it is analytically unramified by Lemma 10.162.13. It is clear that (2) implies (3).
Assume (3) holds. Let \mathfrak p \subset R be a prime ideal and let L/\kappa (\mathfrak p) 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 which generate L over \kappa (\mathfrak p). For each i let P_ i(T) = T^{d_ i} + a_{i, 1} T^{d_ i - 1} + \ldots + a_{i, d_ i} be the minimal polynomial for x_ i over \kappa (\mathfrak p). After replacing x_ i by f_ i x_ i for a suitable f_ i \in R, f_ i \not\in \mathfrak p we may assume a_{i, j} \in R/\mathfrak p. In fact, after further multiplying by elements of \mathfrak m, we may assume a_{i, j} \in \mathfrak m/\mathfrak p \subset R/\mathfrak p for all i, j. 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 (3) S is analytically unramified and by Lemma 10.162.10 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 L we win.
\square
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