The Stacks project

\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 10.156.11. Let $R$ be a Noetherian local ring. Let $\mathfrak p \subset R$ be a prime. Assume

  1. $R_{\mathfrak p}$ is a discrete valuation ring, and

  2. $\mathfrak p$ is analytically unramified.

Then for any associated prime $\mathfrak q$ of $R^\wedge /\mathfrak pR^\wedge $ the local ring $(R^\wedge )_{\mathfrak q}$ is a discrete valuation ring.

Proof. Assumption (2) says that $R^\wedge /\mathfrak pR^\wedge $ is a reduced ring. Hence an associated prime $\mathfrak q \subset R^\wedge $ of $R^\wedge /\mathfrak pR^\wedge $ is the same thing as a minimal prime over $\mathfrak pR^\wedge $. In particular we see that the maximal ideal of $(R^\wedge )_{\mathfrak q}$ is $\mathfrak p(R^\wedge )_{\mathfrak q}$. Choose $x \in R$ such that $xR_{\mathfrak p} = \mathfrak pR_{\mathfrak p}$. By the above we see that $x \in (R^\wedge )_{\mathfrak q}$ generates the maximal ideal. As $R \to R^\wedge $ is faithfully flat we see that $x$ is a nonzerodivisor in $(R^\wedge )_{\mathfrak q}$. Hence we win. $\square$

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