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 10.87.12. Let $R$ be a ring. Let $S = R/I$ for some finitely generated ideal $I$. Let $M$ be an $S$-module. Then $M$ is a Mittag-Leffler module over $R$ if and only if $M$ is a Mittag-Leffler module over $S$.

Proof. One implication follows from Lemma 10.87.11. To prove the other, assume $M$ is Mittag-Leffler as an $R$-module. Write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ as a directed colimit of finitely presented $S$-modules. As $I$ is finitely generated, the ring $S$ is finite and finitely presented as an $R$-algebra, hence the modules $M_ i$ are finitely presented as $R$-modules, see Lemma 10.35.23. Next, let $N$ be any $S$-module. Note that for each $i$ we have $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N) = \mathop{\mathrm{Hom}}\nolimits _ S(M_ i, N)$ as $R \to S$ is surjective. Hence the condition that the inverse system $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N))_ i$ satisfies Mittag-Leffler, implies that the system $(\mathop{\mathrm{Hom}}\nolimits _ S(M_ i, N))_ i$ satisfies Mittag-Leffler. Thus $M$ is Mittag-Leffler over $S$ by definition. $\square$


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