2
$\begingroup$

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!

  • 0
    can you see what happens if you take M' maximal?2012-10-12
  • 0
    @MarianoSuárez-Alvarez: any maximal? any "limit" of ascending chain?2012-10-12
  • 0
    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

1 Answers 1

2

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!

  • 0
    @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.2012-10-12
  • 0
    A maximal submodule may not exist :). @navigator23: you are wasting your time deletting comments and answers :).2012-10-12
  • 0
    @qiL Of course a maximal submodule *must* exist: the module is Noetherian.2012-10-13
  • 0
    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.2012-10-13
  • 0
    @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.2012-10-14
  • 0
    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.2012-10-14
  • 0
    @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