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 10.138.7. Let $k$ be a field of characteristic $0$. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset S$ be a prime. The following are equivalent:

  1. The algebra $S$ is smooth at $\mathfrak q$ over $k$.

  2. The $S_{\mathfrak q}$-module $\Omega _{S/k, \mathfrak q}$ is (finite) free.

  3. The ring $S_{\mathfrak q}$ is regular.

Proof. In characteristic zero any field extension is separable and hence the equivalence of (1) and (3) follows from Lemma 10.138.5. Also (1) implies (2) by definition of smooth algebras. Assume that $\Omega _{S/k, \mathfrak q}$ is free over $S_{\mathfrak q}$. We are going to use the notation and observations made in the proof of Lemma 10.138.5. So $R = S_{\mathfrak q}$ with maximal ideal $\mathfrak m$ and residue field $\kappa $. Our goal is to prove $R$ is regular.

If $\mathfrak m/\mathfrak m^2 = 0$, then $\mathfrak m = 0$ and $R \cong \kappa $. Hence $R$ is regular and we win.

If $\mathfrak m/ \mathfrak m^2 \not= 0$, then choose any $f \in \mathfrak m$ whose image in $\mathfrak m/ \mathfrak m^2$ is not zero. By Lemma 10.138.4 we see that $\text{d}f$ has nonzero image in $\Omega _{R/k}/\mathfrak m\Omega _{R/k}$. By assumption $\Omega _{R/k} = \Omega _{S/k, \mathfrak q}$ is finite free and hence by Nakayama's Lemma 10.19.1 we see that $\text{d}f$ generates a direct summand. We apply Lemma 10.138.6 to deduce that $f$ is a nonzerodivisor in $R$. Furthermore, by Lemma 10.130.9 we get an exact sequence

\[ (f)/(f^2) \to \Omega _{R/k} \otimes _ R R/fR \to \Omega _{(R/fR)/k} \to 0 \]

This implies that $\Omega _{(R/fR)/k}$ is finite free as well. Hence by induction we see that $R/fR$ is a regular local ring. Since $f \in \mathfrak m$ was a nonzerodivisor we conclude that $R$ is regular, see Lemma 10.105.7. $\square$


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