Lemma 10.162.12. Let $(R, \mathfrak m)$ be a Noetherian local domain. Let $x \in \mathfrak m$. Assume
$x \not= 0$,
$R/xR$ has no embedded primes, and
for each associated prime $\mathfrak p \subset R$ of $R/xR$ we have
the local ring $R_{\mathfrak p}$ is regular, and
$\mathfrak p$ is analytically unramified.
Then $R$ is analytically unramified.
Proof. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the associated primes of the $R$-module $R/xR$. Since $R/xR$ has no embedded primes we see that each $\mathfrak p_ i$ has height $1$, and is a minimal prime over $(x)$. For each $i$, let $\mathfrak q_{i1}, \ldots , \mathfrak q_{is_ i}$ be the associated primes of the $R^\wedge $-module $R^\wedge /\mathfrak p_ iR^\wedge $. By Lemma 10.162.11 we see that $(R^\wedge )_{\mathfrak q_{ij}}$ is regular. By Lemma 10.65.3 we see that
Let $y \in R^\wedge $ with $y^2 = 0$. As $(R^\wedge )_{\mathfrak q_{ij}}$ is regular, and hence a domain (Lemma 10.106.2) we see that $y$ maps to zero in $(R^\wedge )_{\mathfrak q_{ij}}$. Hence $y$ maps to zero in $R^\wedge /xR^\wedge $ by Lemma 10.63.19. Hence $y = xy'$. Since $x$ is a nonzerodivisor (as $R \to R^\wedge $ is flat) we see that $(y')^2 = 0$. Hence we conclude that $y \in \bigcap x^ nR^\wedge = (0)$ (Lemma 10.51.4). $\square$
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