# Noetherian modules

Question: Let $R$ be a Noetherian ring, and $M$ be an $R-$module, show that $M$ is Noetherian if and only if $M$ is finitely generated.

This is a question on my homework, I'm really confused about one thing we proved in class the following theorem.

Theorem: Let $R$ be a ring and $M$ be any $R-$module. Then $M$ is Noetherian if and only if every submodule of $M$ is finitely generated.

Can't I just consider $M$ to be a submodule of it self to prove the question? Thank you.

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You need to be careful. It is not always the case that a submodule of a finitely generated module is itself finitely generated (you will need to use that the ring is Noetherian at some point). – Tobias Kildetoft Feb 4 '13 at 15:31
Just to clarify: are you saying that what you've labelled "question" is the homework and what you've labelled "theorem" is what you did in class? The theorem provides a proof of the question, but not the other way round. I doubt this is the intention though - maybe try to understand the proof of the theorem and turn it into a proof of the statement in the question. – Matthew Pressland Feb 4 '13 at 15:33
@Tobias but if every submodule of $M$ is finitely generated can't I consider $M$ as a submodule of it self?! – i.a.m Feb 4 '13 at 17:00
Yes, but you need to show that all submodules of $M$ are finitely generated. – Tobias Kildetoft Feb 4 '13 at 17:03
No, that is precisely where you need to be careful. Any ring is finitely generated as a module over itself. But clearly not all rings are Noetherian. – Tobias Kildetoft Feb 4 '13 at 17:08

Yes you can, but it gives you only one side of your homework. The real thing you have to prove is that given $R$ noetherian and $M$ finitely generated, all the submodules of $M$ are finitely generated.
@damianl but the theorem works for any $R$ is this not true? – i.a.m Feb 4 '13 at 16:54
@damianl but if $M$ is finitely generated then every submodule of it must be finitely generated right? – i.a.m Feb 4 '13 at 17:03
No ! This is precisely with the noetherian hypothesis on $R$ is so important in maths. – Damien L Feb 4 '13 at 17:04