Lemma 10.88.12. Let $R$ be a ring. Let $S = R/I$ for some finitely generated ideal $I$. Let $M$ be an $S$-module. Then $M$ is a Mittag-Leffler module over $R$ if and only if $M$ is a Mittag-Leffler module over $S$.

**Proof.**
One implication follows from Lemma 10.88.11. To prove the other, assume $M$ is Mittag-Leffler as an $R$-module. Write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ as a directed colimit of finitely presented $S$-modules. As $I$ is finitely generated, the ring $S$ is finite and finitely presented as an $R$-algebra, hence the modules $M_ i$ are finitely presented as $R$-modules, see Lemma 10.36.23. Next, let $N$ be any $S$-module. Note that for each $i$ we have $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N) = \mathop{\mathrm{Hom}}\nolimits _ S(M_ i, N)$ as $R \to S$ is surjective. Hence the condition that the inverse system $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N))_ i$ satisfies Mittag-Leffler, implies that the system $(\mathop{\mathrm{Hom}}\nolimits _ S(M_ i, N))_ i$ satisfies Mittag-Leffler. Thus $M$ is Mittag-Leffler over $S$ by definition.
$\square$

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