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\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.156.12. Let $(R, \mathfrak m)$ be a Noetherian local domain. Let $x \in \mathfrak m$. Assume

  1. $x \not= 0$,

  2. $R/xR$ has no embedded primes, and

  3. for each associated prime $\mathfrak p \subset R$ of $R/xR$ we have

    1. the local ring $R_{\mathfrak p}$ is regular, and

    2. $\mathfrak p$ 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 10.156.11 we see that $(R^\wedge )_{\mathfrak q_{ij}}$ is regular. By Lemma 10.64.3 we see that

\[ \text{Ass}_{R^\wedge }(R^\wedge /xR^\wedge ) = \bigcup \nolimits _{\mathfrak p \in \text{Ass}_ R(R/xR)} \text{Ass}_{R^\wedge }(R^\wedge /\mathfrak pR^\wedge ) = \{ \mathfrak q_{ij}\} . \]

Let $y \in R^\wedge $ with $y^2 = 0$. As $(R^\wedge )_{\mathfrak q_{ij}}$ is regular, and hence a domain (Lemma 10.105.2) 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 10.62.19. 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 10.50.4). $\square$


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