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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!

rschwieb
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Aspirin
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1 Answers1

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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!

rschwieb
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  • @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. – rschwieb Oct 12 '12 at 14:20
  • A maximal submodule may not exist :). @navigator23: you are wasting your time deletting comments and answers :). –  Oct 12 '12 at 20:17
  • @qiL Of course a maximal submodule must exist: the module is Noetherian. – rschwieb Oct 13 '12 at 11:08
  • 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. –  Oct 13 '12 at 12:36
  • @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. – rschwieb Oct 14 '12 at 00:53
  • Dear rschwieb, sorry for wasting your time. The point is $M$ must be non-zero. Assuming the axiome of choice, the existence of maximal submodules is valid over any ring, provided that $M\ne 0$ (not necessarily finitely generated) ! This is like the existence of a maximal ideal in a non-zero ring. –  Oct 14 '12 at 07:19
  • @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. – rschwieb Oct 14 '12 at 13:08