The Stacks project

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

whose bottom two horizontal maps are isomorphisms, according to Proposition 10.89.3. 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 10.88.6. 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

The top arrow is an injection by assumption, and the bottom arrow is an isomorphism by Proposition 10.89.3. 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 10.89.4, 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

where both the top and bottom arrows are isomorphisms by Proposition 10.89.3. 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 10.88.3, $f$ and $f'$ dominate each other. $\square$