Email from Juan Pablo Acosta Lopez dated 12/20/14.

Lemma 10.95.1. Let $R \to S$ be a faithfully flat ring map. Let $M$ be an $R$-module. If the $S$-module $M \otimes _ R S$ is Mittag-Leffler, then $M$ is Mittag-Leffler.

Proof. Write $M = \mathop{\mathrm{colim}}\nolimits _{i\in I} M_ i$ as a directed colimit of finitely presented $R$-modules $M_ i$. Using Proposition 10.88.6, we see that we have to prove that for each $i \in I$ there exists $i \leq j$, $j\in I$ such that $M_ i\rightarrow M_ j$ dominates $M_ i\rightarrow M$.

Take $N$ the pushout

$\xymatrix{ M_ i \ar[r] \ar[d] & M_ j \ar[d] \\ M \ar[r] & N }$

Then the lemma is equivalent to the existence of $j$ such that $M_ j\rightarrow N$ is universally injective, see Lemma 10.88.4. Observe that the tensorization by $S$

$\xymatrix{ M_ i\otimes _ R S \ar[r] \ar[d] & M_ j\otimes _ R S \ar[d] \\ M\otimes _ R S \ar[r] & N\otimes _ R S }$

Is a pushout diagram. So because $M \otimes _ R S = \mathop{\mathrm{colim}}\nolimits _{i\in I} M_ i \otimes _ R S$ expresses $M\otimes _ R S$ as a colimit of $S$-modules of finite presentation, and $M\otimes _ R S$ is Mittag-Leffler, there exists $j \geq i$ such that $M_ j\otimes _ R S\rightarrow N\otimes _ R S$ is universally injective. So using that $R\rightarrow S$ is faithfully flat we conclude that $M_ j\rightarrow N$ is universally injective too. $\square$

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