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 10.138.6. Let $R \to S$ be a $\mathbf{Q}$-algebra map. Let $f \in S$ be such that $\Omega _{S/R} = S \text{d}f \oplus C$ for some $S$-submodule $C$. Then

  1. $f$ is not nilpotent, and

  2. if $S$ is a Noetherian local ring, then $f$ is a nonzerodivisor in $S$.

Proof. For $a \in S$ write $\text{d}(a) = \theta (a)\text{d}f + c(a)$ for some $\theta (a) \in S$ and $c(a) \in C$. Consider the $R$-derivation $S \to S$, $a \mapsto \theta (a)$. Note that $\theta (f) = 1$.

If $f^ n = 0$ with $n > 1$ minimal, then $0 = \theta (f^ n) = n f^{n - 1}$ contradicting the minimality of $n$. We conclude that $f$ is not nilpotent.

Suppose $fa = 0$. If $f$ is a unit then $a = 0$ and we win. Assume $f$ is not a unit. Then $0 = \theta (fa) = f\theta (a) + a$ by the Leibniz rule and hence $a \in (f)$. By induction suppose we have shown $fa = 0 \Rightarrow a \in (f^ n)$. Then writing $a = f^ nb$ we get $0 = \theta (f^{n + 1}b) = (n + 1)f^ nb + f^{n + 1}\theta (b)$. Hence $a = f^ n b = -f^{n + 1}\theta (b)/(n + 1) \in (f^{n + 1})$. Since in the Noetherian local ring $S$ we have $\bigcap (f^ n) = 0$, see Lemma 10.50.4 we win. $\square$


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