Lemma 10.89.11. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. If $S$ is Mittag-Leffler as an $R$-module, and $M$ is flat and Mittag-Leffler as an $S$-module, then $M$ is Mittag-Leffler as an $R$-module.
Proof. We deduce this from the characterization of Proposition 10.89.5. Namely, suppose that $Q_\alpha $ is a family of $R$-modules. Consider the composition
\[ \xymatrix{ M \otimes _ R \prod _\alpha Q_\alpha = M \otimes _ S S \otimes _ R \prod _\alpha Q_\alpha \ar[d] \\ M \otimes _ S \prod _\alpha (S \otimes _ R Q_\alpha ) \ar[d] \\ \prod _\alpha (M \otimes _ S S \otimes _ R Q_\alpha ) = \prod _\alpha (M \otimes _ R Q_\alpha ) } \]
The first arrow is injective as $M$ is flat over $S$ and $S$ is Mittag-Leffler over $R$ and the second arrow is injective as $M$ is Mittag-Leffler over $S$. Hence $M$ is Mittag-Leffler over $R$. $\square$
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Comment #9893 by Quentin on
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