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 10.100.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Let $A$ be a set and let $x_\alpha \in M$, $\alpha \in A$ be a collection of elements of $M$. Assume

  1. $I$ is nilpotent,

  2. $\{ \overline{x}_\alpha \} _{\alpha \in A}$ forms a basis for $M/IM$ over $R/I$, and

  3. $\text{Tor}_1^ R(R/I, M) = 0$.

Then $M$ is free on $\{ x_\alpha \} _{\alpha \in A}$ over $R$.

Proof. Let $R$, $I$, $M$, $\{ x_\alpha \} _{\alpha \in A}$ be as in the lemma and satisfy assumptions (1), (2), and (3). By Nakayama's Lemma 10.19.1 the elements $x_\alpha $ generate $M$ over $R$. The assumption $\text{Tor}_1^ R(R/I, M) = 0$ implies that we have a short exact sequence

\[ 0 \to I \otimes _ R M \to M \to M/IM \to 0. \]

Let $\sum f_\alpha x_\alpha = 0$ be a relation in $M$. By choice of $x_\alpha $ we see that $f_\alpha \in I$. Hence we conclude that $\sum f_\alpha \otimes x_\alpha = 0$ in $I \otimes _ R M$. The map $I \otimes _ R M \to I/I^2 \otimes _{R/I} M/IM$ and the fact that $\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M/IM$ implies that $f_\alpha \in I^2$! Hence we conclude that there are no relations among the images of the $x_\alpha $ in $M/I^2M$. In other words, we see that $M/I^2M$ is free with basis the images of the $x_\alpha $. Using the map $I \otimes _ R M \to I/I^3 \otimes _{R/I^2} M/I^2M$ we then conclude that $f_\alpha \in I^3$! And so on. Since $I^ n = 0$ for some $n$ by assumption (1) we win. $\square$


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