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 10.76.4. Let $R$ be a ring. Let $I \subset R$ be a nilpotent ideal. Let $\overline{P}$ be a projective $R/I$-module. Then there exists a projective $R$-module $P$ such that $P/IP \cong \overline{P}$.

Proof. By Lemma 10.76.2 we can choose a set $A$ and a direct sum decomposition $\bigoplus _{\alpha \in A} R/I = \overline{P} \oplus \overline{K}$ for some $R/I$-module $\overline{K}$. Write $F = \bigoplus _{\alpha \in A} R$ for the free $R$-module on $A$. Choose a lift $p : F \to F$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$ of $\bigoplus _{\alpha \in A} R/I$. Note that $p^2 - p \in \text{End}_ R(F)$ is a nilpotent endomorphism of $F$ (as $I$ is nilpotent and the matrix entries of $p^2 - p$ are in $I$; more precisely, if $I^ n = 0$, then $(p^2 - p)^ n = 0$). Hence by Lemma 10.31.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$. $\square$


Comments (2)

Comment #1192 by Mohamed Hashi on

In the statement, should be a projective -module.

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  • 2 comment(s) on Section 10.76: Projective modules

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