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 10.88.11. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. If $S$ is Mittag-Leffler as an $R$-module, and $M$ is flat and Mittag-Leffler as an $S$-module, then $M$ is Mittag-Leffler as an $R$-module.

Proof. We deduce this from the characterization of Proposition 10.88.5. Namely, suppose that $Q_\alpha $ is a family of $R$-modules. Consider the composition

\[ \xymatrix{ M \otimes _ R \prod _\alpha Q_\alpha = M \otimes _ S S \otimes _ R \prod _\alpha Q_\alpha \ar[d] \\ M \otimes _ S \prod _\alpha (S \otimes _ R Q_\alpha ) \ar[d] \\ \prod _\alpha (M \otimes _ S \otimes _ R Q_\alpha ) = \prod _\alpha (M \otimes _ R Q_\alpha ) } \]

The first arrows is injective as $M$ is flat over $S$ and $S$ is Mittag-Leffler over $R$ and the second arrow is injective as $M$ is Mittag-Leffler over $S$. Hence $M$ is Mittag-Leffler over $R$. $\square$


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