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*}

Remark 10.87.13. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module which is Mittag-Leffler as an $R$-module. Then it is in general not the case that $M$ is Mittag-Leffler as an $S$-module. For example suppose that $S$ is the ring of dual numbers over $R$, i.e., $S = R \oplus R\epsilon $ with $\epsilon ^2 = 0$. Then an $S$-module consists of an $R$-module $M$ endowed with a square zero $R$-linear endomorphism $\epsilon : M \to M$. Now suppose that $M_0$ is an $R$-module which is not Mittag-Leffler. Choose a presentation $F_1 \xrightarrow {u} F_0 \to M_0 \to 0$ with $F_1$ and $F_0$ free $R$-modules. Set $M = F_1 \oplus F_0$ with

\[ \epsilon = \left( \begin{matrix} 0 & 0 \\ u & 0 \end{matrix} \right) : M \longrightarrow M. \]

Then $M/\epsilon M \cong F_1 \oplus M_0$ is not Mittag-Leffler over $R = S/\epsilon S$, hence not Mittag-Leffler over $S$ (see Lemma 10.87.12). On the other hand, $M/\epsilon M = M \otimes _ S S/\epsilon S$ which would be Mittag-Leffler over $S$ if $M$ was, see Lemma 10.87.9.

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