Remark 10.93.2. Lemma 10.93.1 does not hold without the countable generation assumption. For example, the $\mathbf Z$-module $M = \mathbf{Z}[[x]]$ is flat and Mittag-Leffler but not projective. It is Mittag-Leffler by Lemma 10.91.4. Subgroups of free abelian groups are free, hence a projective $\mathbf Z$-module is in fact free and so are its submodules. Thus to show $M$ is not projective it suffices to produce a non-free submodule. Fix a prime $p$ and consider the submodule $N$ consisting of power series $f(x) = \sum a_ i x^ i$ such that for every integer $m \geq 1$, $p^ m$ divides $a_ i$ for all but finitely many $i$. Then $\sum a_ i p^ i x^ i$ is in $N$ for all $a_ i \in \mathbf{Z}$, so $N$ is uncountable. Thus if $N$ were free it would have uncountable rank and the dimension of $N/pN$ over $\mathbf{Z}/p$ would be uncountable. This is not true as the elements $x^ i \in N/pN$ for $i \geq 0$ span $N/pN$.

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