TITLE: Is this module finitely generated?
QUESTION [5 upvotes]: Suppose $M$ is a $A$-module, $A$ is a commutative ring with 1, such that for every countably generated submodule $N$ of $M$, there exists a finitely generated submodule $L$ which contains $N$. 

Must $M$ be finitely generated?

(Maybe it should be tagged by set-theory? )
Thanks.

REPLY [4 votes]: Let 

$X$ be an uncountable set,
$F$ be a field,
$A$ be the ring of functions $X \to F$ which are constant except possibly on a countable subset of $X$,
$M$ be the left $A$-module of functions $X \to F$ which are zero except possibly on a countable subset of $X$. 

Then every countably generated submodule of $M$ is in fact contained in a submodule generated by one element (given a sequence $m_1, m_2, ... \in M$, the submodule they generate is contained in the submodule generated by an $m$ which is nonzero whenever any of the $m_i$ is nonzero), but $M$ itself is uncountably generated. Both properties follow from the fact that a countable union of countable subsets of $X$ is countable.