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I'm trying to prove the following:

An $R$-module $M$ is finitely generated if and only if for every sequence of finitely generated submodules $N_0 \subset N_1 \subset \dots \subset N_n \subset \dots$ we have that $\cup_{n \in \mathbb N}N_n$ is finitely generated.

I have proved that that this chain condition implies $M$ is finitely generated by applying Zorn's lemma in the set of the finitely generated submodules and showing that the maximal element must be $M$. I'm having trouble on proving the converse.

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    $\begingroup$ The other direction is false in general. Take a non-noetherian ring (which is clearly finitely generated as a module over itself) and an ideal, which is not finitely generated, but generated by a countable set. Then you can easily construct such a chain of finitely generated ideals, such that the union is your ideal you started with. Maybe $R$ is assumed to be noetherian? Then the other direction is quite trivial. $\endgroup$ – MooS Sep 24 '15 at 6:40

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