Lemma 10.89.11 (05CT)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.89: Interchanging direct products with tensor
Lemma 10.89.11 (cite)

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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 $R$ . $\square$

Comments (1)

Comment #9893 by Quentin on December 11, 2024 at 07:37

The hypothesis that $M$ be a Mittag-Leffler $S$ -module is redundant: by 10.88.8, this follows from from $M$ being a flat $S$ -module.

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