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Let $M$ be a noetherian module over noetherian ring $A$.

How to prove that there exists submodule $N\subset M$ such that $M/N\cong A/\mathfrak{p}$ for some prime ideal $\mathfrak{p}\in A$.

Is it true that any submodule of noetherian module over noetherian ring is noetherian? (because it is finitely generated as submodule of noetherian module?)

Thanks a lot!

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    In fact the ideal $\mathfrak{p}$ can be taken maximal: choose $N$ maximal in the set of proper submodules of $M$ (why there is such $N$?); then $M/N$ is a simple module, so...2012-10-12

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Hints for the main question: (under the assumption your ring has unity)

  1. $M/N$ is a simple module when $N$ is a maximal submodule.

  2. A simple module is isomorphic to $R/I$ for some maximal ideal $I\lhd R$.

  3. You probably are aware of some relationship between maximal and prime ideals...

You should be able to reason out why the submodules of a Noetherian module $M$ over any ring are Noetherian. Consider an ascending chain in the submodule $N\subseteq M$ ... and remember it is a chain in $M$ too!

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    @QiL You're right that it is a waste of time. I prefer not to get hung up on things like this unless they're critical. Posters frequently miss hypotheses much more important than this, and it's OK to make little assumptions.2012-10-14