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. $\square$
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