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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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  • $\begingroup$ can you see what happens if you take M' maximal? $\endgroup$ – Mariano Suárez-Álvarez Oct 12 '12 at 5:29
  • $\begingroup$ @MarianoSuárez-Alvarez: any maximal? any "limit" of ascending chain? $\endgroup$ – Aspirin Oct 12 '12 at 5:34
  • $\begingroup$ 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... $\endgroup$ – user26857 Oct 12 '12 at 7:40
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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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  • $\begingroup$ @navigetor23 Anyone who defines "maximal submodule" and allows $M$ to be considered is wasting their time. That is why the overwhelming majority of sources require by definition maximal submodules to be proper submodules. $\endgroup$ – rschwieb Oct 12 '12 at 14:20
  • $\begingroup$ A maximal submodule may not exist :). @navigator23: you are wasting your time deletting comments and answers :). $\endgroup$ – user18119 Oct 12 '12 at 20:17
  • $\begingroup$ @qiL Of course a maximal submodule must exist: the module is Noetherian. $\endgroup$ – rschwieb Oct 13 '12 at 11:08
  • $\begingroup$ not always, even for Noetherian module. :) OK I will say why in a few hours if nobody notices the problem in the statement of the OP. $\endgroup$ – user18119 Oct 13 '12 at 12:36
  • $\begingroup$ @QiL Well under reasonable assumptions like the axiom of choice, "Noetherian" is equivalent to having the maximal condition on any subset of its submodules, so I choose the set to be the proper submodules and I have maximal submodules. $\endgroup$ – rschwieb Oct 14 '12 at 0:53

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