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 10.87.11. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. If $M$ is a Mittag-Leffler module over $S$ then $M$ is a Mittag-Leffler module over $R$.

Proof. Assume $M$ is a Mittag-Leffler module over $S$. Write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ as a directed colimit of finitely presented $S$-modules $M_ i$. As $M$ is Mittag-Leffler over $S$ there exists for each $i$ an index $j \geq i$ such that for all $k \geq j$ there is a factorization $f_{ij} = h \circ f_{ik}$ (where $h$ depends on $i$, the choice of $j$ and $k$). Note that by Lemma 10.35.23 the modules $M_ i$ are also finitely presented as $R$-modules. Moreover, all the maps $f_{ij}, f_{ik}, h$ are maps of $R$-modules. Thus we see that the system $(M_ i, f_{ij})$ satisfies the same condition when viewed as a system of $R$-modules. Thus $M$ is Mittag-Leffler as an $R$-module. $\square$

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