
Lemma 10.156.14. Let $(R, \mathfrak m)$ be a Noetherian local ring. The following are equivalent

1. $R$ is Nagata,

2. for $R \to S$ finite with $S$ a domain and $\mathfrak m' \subset S$ maximal the local ring $S_{\mathfrak m'}$ is analytically unramified,

3. for $(R, \mathfrak m) \to (S, \mathfrak m')$ finite local homomorphism with $S$ a domain, then $S$ is analytically unramified.

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 10.156.5. Hence the local ring $S_{\mathfrak m'}$ is Nagata (Lemma 10.156.6). Thus it is analytically unramified by Lemma 10.156.13. 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 10.156.10 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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