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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 10.156.10. Let $(R, \mathfrak m)$ be a Noetherian local ring.

  1. If $R$ is analytically unramified, then $R$ is reduced.

  2. If $R$ is analytically unramified, then each minimal prime of $R$ is analytically unramified.

  3. If $R$ is reduced with minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and each $\mathfrak q_ i$ is analytically unramified, then $R$ is analytically unramified.

  4. If $R$ is analytically unramified, then the integral closure of $R$ in its total ring of fractions $Q(R)$ is finite over $R$.

  5. If $R$ is a domain and analytically unramified, then $R$ is N-1.

Proof. In this proof we will use the remarks immediately following Definition 10.156.9. As $R \to R^\wedge $ is a faithfully flat local ring homomorphism it is injective and (1) follows.

Let $\mathfrak q$ be a minimal prime of $R$, and assume $R$ is analytically unramified. Then $\mathfrak q$ is an associated prime of $R$ (see Proposition 10.62.6). Hence there exists an $f \in R$ such that $\{ x \in R \mid fx = 0\} = \mathfrak q$. Note that $(R/\mathfrak q)^\wedge = R^\wedge /\mathfrak q^\wedge $, and that $\{ x \in R^\wedge \mid fx = 0\} = \mathfrak q^\wedge $, because completion is exact (Lemma 10.96.2). If $x \in R^\wedge $ is such that $x^2 \in \mathfrak q^\wedge $, then $fx^2 = 0$ hence $(fx)^2 = 0$ hence $fx = 0$ hence $x \in \mathfrak q^\wedge $. Thus $\mathfrak q$ is analytically unramified and (2) holds.

Assume $R$ is reduced with minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and each $\mathfrak q_ i$ is analytically unramified. Then $R \to R/\mathfrak q_1 \times \ldots \times R/\mathfrak q_ t$ is injective. Since completion is exact (see Lemma 10.96.2) we see that $R^\wedge \subset (R/\mathfrak q_1)^\wedge \times \ldots \times (R/\mathfrak q_ t)^\wedge $. Hence (3) is clear.

Assume $R$ is analytically unramified. Let $\mathfrak p_1, \ldots , \mathfrak p_ s$ be the minimal primes of $R^\wedge $. Then we see that

\[ Q(R^\wedge ) = R^\wedge _{\mathfrak p_1} \times \ldots \times R^\wedge _{\mathfrak p_ s} \]

with each $R^\wedge _{\mathfrak p_ i}$ a field as $R^\wedge $ is reduced (see Lemma 10.24.4). Hence the integral closure $S$ of $R^\wedge $ in $Q(R^\wedge )$ is equal to $S = S_1 \times \ldots \times S_ s$ with $S_ i$ the integral closure of $R^\wedge /\mathfrak p_ i$ in its fraction field. In particular $S$ is finite over $R^\wedge $. Denote $R'$ the integral closure of $R$ in $Q(R)$. As $R \to R^\wedge $ is flat we see that $R' \otimes _ R R^\wedge \subset Q(R) \otimes _ R R^\wedge \subset Q(R^\wedge )$. Moreover $R' \otimes _ R R^\wedge $ is integral over $R^\wedge $ (Lemma 10.35.13). Hence $R' \otimes _ R R^\wedge \subset S$ is a $R^\wedge $-submodule. As $R^\wedge $ is Noetherian it is a finite $R^\wedge $-module. Thus we may find $f_1, \ldots , f_ n \in R'$ such that $R' \otimes _ R R^\wedge $ is generated by the elements $f_ i \otimes 1$ as a $R^\wedge $-module. By faithful flatness we see that $R'$ is generated by $f_1, \ldots , f_ n$ as an $R$-module. This proves (4).

Part (5) is a special case of part (4). $\square$


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