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 10.87.10. Let $R$ be a ring and $M$ an $R$-module. Then $M$ is Mittag-Leffler if and only if for every finite free $R$-module $F$ and module map $f: F \to M$, there exists a finitely presented $R$-module $Q$ and a module map $g : F \to Q$ such that $g$ and $f$ dominate each other, i.e., $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ N) = \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ N)$ for every $R$-module $N$.

Proof. Since the condition is clear weaker than condition (1) of Proposition 10.87.6 we see that a Mittag-Leffler module satisfies the condition. Conversely, suppose that $M$ satisfies the condition and that $f : P \to M$ is an $R$-module map from a finitely presented $R$-module $P$ into $M$. Choose a surjection $F \to P$ where $F$ is a finite free $R$-module. By assumption we can find a map $F \to Q$ where $Q$ is a finitely presented $R$-module such that $F \to Q$ and $F \to M$ dominate each other. In particular, the kernel of $F \to Q$ contains the kernel of $F \to P$, hence we obtain an $R$-module map $g : P \to Q$ such that $F \to Q$ is equal to the composition $F \to P \to Q$. Let $N$ be any $R$-module and consider the commutative diagram

\[ \xymatrix{ F \otimes _ R N \ar[d] \ar[r] & Q \otimes _ R N \\ P \otimes _ R N \ar[ru] \ar[r] & M \otimes _ R N } \]

By assumption the kernels of $F \otimes _ R N \to Q \otimes _ R N$ and $F \otimes _ R N \to M \otimes _ R N$ are equal. Hence, as $F \otimes _ R N \to P \otimes _ R N$ is surjective, also the kernels of $P \otimes _ R N \to Q \otimes _ R N$ and $P \otimes _ R N \to M \otimes _ R N$ are equal. $\square$


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