Lemma 10.88.11. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. If $M$ is a Mittag-Leffler module over $S$ then $M$ is a Mittag-Leffler module over $R$.

**Proof.**
Assume $M$ is a Mittag-Leffler module over $S$. Write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ as a directed colimit of finitely presented $S$-modules $M_ i$. As $M$ is Mittag-Leffler over $S$ there exists for each $i$ an index $j \geq i$ such that for all $k \geq j$ there is a factorization $f_{ij} = h \circ f_{ik}$ (where $h$ depends on $i$, the choice of $j$ and $k$). Note that by Lemma 10.36.23 the modules $M_ i$ are also finitely presented as $R$-modules. Moreover, all the maps $f_{ij}, f_{ik}, h$ are maps of $R$-modules. Thus we see that the system $(M_ i, f_{ij})$ satisfies the same condition when viewed as a system of $R$-modules. Thus $M$ is Mittag-Leffler as an $R$-module.
$\square$

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