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 10.158.3. Let $R \to S$ be a ring map. Assume that

  1. $R \to S$ is faithfully flat, and

  2. $S$ is a normal ring.

Then $R$ is a normal ring.

Proof. Since $S$ is reduced it follows that $R$ is reduced. Let $\mathfrak p$ be a prime of $R$. We have to show that $R_{\mathfrak p}$ is a normal domain. Since $S_{\mathfrak p}$ is faithfully over $R_{\mathfrak p}$ too we may assume that $R$ is local with maximal ideal $\mathfrak m$. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak m$. Then we see that $R \to S_{\mathfrak q}$ is faithfully flat (Lemma 10.38.17). Hence we may assume $S$ is local as well. In particular $S$ is a normal domain. Since $R \to S$ is faithfully flat and $S$ is a normal domain we see that $R$ is a domain. Next, suppose that $a/b$ is integral over $R$ with $a, b \in R$. Then $a/b \in S$ as $S$ is normal. Hence $a \in bS$. This means that $a : R \to R/bR$ becomes the zero map after base change to $S$. By faithful flatness we see that $a \in bR$, so $a/b \in R$. Hence $R$ is normal. $\square$


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