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*}

Proposition 10.88.5. Let $M$ be an $R$-module. The following are equivalent:

  1. $M$ is Mittag-Leffler.

  2. For every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, the canonical map $M \otimes _ R \left( \prod _{\alpha } Q_{\alpha } \right) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$ is injective.

Proof. First we prove (1) implies (2). Suppose $M$ is Mittag-Leffler and let $x$ be in the kernel of $M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$. Write $M$ as a colimit $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ of a directed system of finitely presented modules $M_ i$. Then $M \otimes _ R (\prod _{\alpha } Q_{\alpha })$ is the colimit of $M_ i \otimes _ R (\prod _{\alpha } Q_{\alpha })$. So $x$ is the image of an element $x_ i \in M_ i \otimes _ R (\prod _{\alpha } Q_{\alpha })$. We must show that $x_ i$ maps to $0$ in $M_ j \otimes _ R (\prod _{\alpha } Q_{\alpha })$ for some $j \geq i$. Since $M$ is Mittag-Leffler, we may choose $j \geq i$ such that $M_ i \to M_ j$ and $M_ i \to M$ dominate each other. Then consider the commutative diagram

\[ \xymatrix{ M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] & \prod _{\alpha } (M \otimes _ R Q_{\alpha }) \\ M_ i \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } \ar[d] \ar[u] & \prod _{\alpha } (M_ i \otimes _ R Q_{\alpha }) \ar[d] \ar[u] \\ M_ j \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } & \prod _{\alpha } (M_ j \otimes _ R Q_{\alpha }) } \]

whose bottom two horizontal maps are isomorphisms, according to Proposition 10.88.3. Since $x_ i$ maps to $0$ in $\prod _{\alpha } (M \otimes _ R Q_{\alpha })$, its image in $\prod _{\alpha } (M_ i \otimes _ R Q_{\alpha })$ is in the kernel of the map $\prod _{\alpha } (M_ i \otimes _ R Q_{\alpha }) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$. But this kernel equals the kernel of $\prod _{\alpha } (M_ i \otimes _ R Q_{\alpha }) \to \prod _{\alpha } (M_ j \otimes _ R Q_{\alpha })$ according to the choice of $j$. Thus $x_ i$ maps to $0$ in $\prod _{\alpha } (M_ j \otimes _ R Q_{\alpha })$ and hence to $0$ in $M_ j \otimes _ R (\prod _{\alpha } Q_{\alpha })$.

Now suppose (2) holds. We prove $M$ satisfies formulation (1) of being Mittag-Leffler from Proposition 10.87.6. Let $f: P \to M$ be a map from a finitely presented module $P$ to $M$. Choose a set $B$ of representatives of the isomorphism classes of finitely presented $R$-modules. Let $A$ be the set of pairs $(Q, x)$ where $Q \in B$ and $x \in \mathop{\mathrm{Ker}}(P \otimes Q \to M \otimes Q)$. For $\alpha = (Q, x) \in A$, we write $Q_{\alpha }$ for $Q$ and $x_{\alpha }$ for $x$. Consider the commutative diagram

\[ \xymatrix{ M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] & \prod _{\alpha } (M \otimes _ R Q_{\alpha }) \\ P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } \ar[u] & \prod _{\alpha } (P \otimes _ R Q_{\alpha }) \ar[u] . } \]

The top arrow is an injection by assumption, and the bottom arrow is an isomorphism by Proposition 10.88.3. Let $x \in P \otimes _ R (\prod _{\alpha } Q_{\alpha })$ be the element corresponding to $(x_{\alpha }) \in \prod _{\alpha } (P \otimes _ R Q_{\alpha })$ under this isomorphism. Then $x \in \mathop{\mathrm{Ker}}( P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to M \otimes _ R (\prod _{\alpha } Q_{\alpha }))$ since the top arrow in the diagram is injective. By Lemma 10.88.4, we get a finitely presented module $P'$ and a map $f': P \to P'$ such that $f: P \to M$ factors through $f'$ and $x \in \mathop{\mathrm{Ker}}(P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to P' \otimes _ R (\prod _{\alpha } Q_{\alpha }))$. We have a commutative diagram

\[ \xymatrix{ P' \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } & \prod _{\alpha } (P' \otimes _ R Q_{\alpha }) \\ P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } \ar[u] & \prod _{\alpha } (P \otimes _ R Q_{\alpha }) \ar[u] . } \]

where both the top and bottom arrows are isomorphisms by Proposition 10.88.3. Thus since $x$ is in the kernel of the left vertical map, $(x_{\alpha })$ is in the kernel of the right vertical map. This means $x_{\alpha } \in \mathop{\mathrm{Ker}}(P \otimes _ R Q_{\alpha } \to P' \otimes _ R Q_{\alpha })$ for every $\alpha \in A$. By the definition of $A$ this means $\mathop{\mathrm{Ker}}(P \otimes _ R Q \to P' \otimes _ R Q) \supset \mathop{\mathrm{Ker}}(P \otimes _ R Q \to M \otimes _ R Q)$ for all finitely presented $Q$ and, since $f: P \to M$ factors through $f': P \to P'$, actually equality holds. By Lemma 10.87.3, $f$ and $f'$ dominate each other. $\square$


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