
Lemma 10.88.10. If $M = \bigoplus _{i \in I} M_ i$ is a direct sum of $R$-modules, then $M$ is Mittag-Leffler if and only if each $M_ i$ is Mittag-Leffler.

Proof. The “only if” direction follows from Lemma 10.88.7 (1) and the fact that a split short exact sequence is universally exact. The converse follows from Lemma 10.88.9 but we can also argue it directly as follows. First note that if $I$ is finite then this follows from Lemma 10.88.7 (2). For general $I$, if all $M_ i$ are Mittag-Leffler then we prove the same of $M$ by verifying condition (1) of Proposition 10.87.6. Let $f: P \to M$ be a map from a finitely presented module $P$. Then $f$ factors as $P \xrightarrow {f'} \bigoplus _{i' \in I'} M_{i'} \hookrightarrow \bigoplus _{i \in I} M_ i$ for some finite subset $I'$ of $I$. By the finite case $\bigoplus _{i' \in I'} M_{i'}$ is Mittag-Leffler and hence there exists a finitely presented module $Q$ and a map $g: P \to Q$ such that $g$ and $f'$ dominate each other. Then also $g$ and $f$ dominate each other. $\square$

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).