Lemma 10.156.13. Let $(R, \mathfrak m)$ be a local ring. If $R$ is Noetherian, a domain, and Nagata, then $R$ is analytically unramified.

**Proof.**
By induction on $\dim (R)$. The case $\dim (R) = 0$ is trivial. Hence we assume $\dim (R) = d$ and that the lemma holds for all Noetherian Nagata domains of dimension $< d$.

Let $R \subset S$ be the integral closure of $R$ in the field of fractions of $R$. By assumption $S$ is a finite $R$-module. By Lemma 10.156.5 we see that $S$ is Nagata. By Lemma 10.111.4 we see $\dim (R) = \dim (S)$. Let $\mathfrak m_1, \ldots , \mathfrak m_ t$ be the maximal ideals of $S$. Each of these lies over the maximal ideal $\mathfrak m$ of $R$. Moreover

for sufficiently large $n$ as $S/\mathfrak mS$ is Artinian. By Lemma 10.96.2 $R^\wedge \to S^\wedge $ is an injective map, and by the Chinese Remainder Lemma 10.14.4 combined with Lemma 10.95.9 we have $S^\wedge = \prod S^\wedge _ i$ where $S^\wedge _ i$ is the completion of $S$ with respect to the maximal ideal $\mathfrak m_ i$. Hence it suffices to show that $S_{\mathfrak m_ i}$ is analytically unramified. In other words, we have reduced to the case where $R$ is a Noetherian normal Nagata domain.

Assume $R$ is a Noetherian, normal, local Nagata domain. Pick a nonzero $x \in \mathfrak m$ in the maximal ideal. We are going to apply Lemma 10.156.12. We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is clear. We have that $R/xR$ has no embedded primes by Lemma 10.151.6. Thus property (2) holds. The same lemma also tells us each associated prime $\mathfrak p$ of $R/xR$ has height $1$. Hence $R_{\mathfrak p}$ is a $1$-dimensional normal domain hence regular (Lemma 10.118.7). Thus (3)(a) holds. Finally (3)(b) holds by induction hypothesis, since $R/\mathfrak p$ is Nagata (by Lemma 10.156.5 or directly from the definition). Thus we conclude $R$ is analytically unramified. $\square$

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