Lemma 10.89.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules. Then:

1. If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.

2. If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.

Proof. For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules we have a commutative diagram

$\xymatrix{ 0 \ar[r] & M_1 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_2 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_3 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \prod _{\alpha }(M_1 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_2 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_3 \otimes Q_{\alpha })\ar[r] & 0 }$

with exact rows. Thus (1) and (2) follow from Proposition 10.89.5. $\square$

## Comments (4)

Comment #2973 by on

Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J.

Comment #2974 by on

Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)

Comment #2975 by on

Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)

Comment #3098 by on

OK, yes. I've added this as a separate lemma. It seems to me one could also deduce this directly from the definition of ML modules. If you have a minute, please check the changes to see if you agree. Thanks!

There are also:

• 4 comment(s) on Section 10.89: Interchanging direct products with tensor

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