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 10.157.9. Let $\varphi : R \to S$ be a ring map. Assume

  1. $\varphi $ is smooth,

  2. $R$ is normal.

Then $S$ is normal.

Proof. Observe that $R \to S$ is flat with regular fibres (see the list of results on smooth ring maps in Section 10.140). In particular, the fibres are normal. Thus if $R$ is Noetherian, then $S$ is Noetherian and we get the result from Lemma 10.157.8.

The general case. First note that $R$ is reduced and hence $S$ is reduced by Lemma 10.157.7. Let $\mathfrak q$ be a prime of $S$ and let $\mathfrak p$ be the corresponding prime of $R$. Note that $R_{\mathfrak p}$ is a normal domain. We have to show that $S_{\mathfrak q}$ is a normal domain. To do this we may replace $R$ by $R_{\mathfrak p}$ and $S$ by $S_{\mathfrak p}$. Hence we may assume that $R$ is a normal domain.

Assume $R \to S$ smooth, and $R$ a normal domain. We may find a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes _{R_0} S_0$, see remark (10) in Section 10.140. As $R_0$ is a Nagata domain (see Proposition 10.156.16) we see that its integral closure $R_0'$ is finite over $R_0$. Moreover, as $R$ is a normal domain it is clear that $R_0' \subset R$. Hence we may replace $R_0$ by $R_0'$ and $S_0$ by $R_0' \otimes _{R_0} S_0$ and assume that $R_0$ is a normal Noetherian domain. By the first paragraph of the proof we conclude that $S_0$ is a normal ring (it need not be a domain of course). In this way we see that $R = \bigcup R_\lambda $ is the union of normal Noetherian domains and correspondingly $S = \mathop{\mathrm{colim}}\nolimits R_\lambda \otimes _{R_0} S_0$ is the colimit of normal rings. This implies that $S$ is a normal ring. Some details omitted. $\square$


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