Lemma 10.89.8. Let $M_1 \to M_2 \to M_3 \to 0$ be an exact sequence of $R$-modules. If $M_1$ is finitely generated and $M_2$ is Mittag-Leffler, then $M_3$ is Mittag-Leffler.

Proof. For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, since tensor product is right exact, we have a commutative diagram

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

with exact rows. By Proposition 10.89.2 the left vertical arrow is surjective. By Proposition 10.89.5 the middle vertical arrow is injective. A diagram chase shows the right vertical arrow is injective. Hence $M_3$ is Mittag-Leffler by Proposition 10.89.5. $\square$

There are also:

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).