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Nonzero R-Module over commuttaive ring, all submodules free => R PID?

  1. Aug 9, 2011 #1
    Let R be a commutative ring with 1. If there exists a non-zero R-module M such that every submodule of M is free, then R is a PID.

    I remember proving something similar to this, assuming submodules of all R-modules are free, but I'm not too sure about this question. The direction I am headed in is to consider M as an I-module. As IM->N, N has a basis {n_i}. After playing around a bit I get lost.
     
  2. jcsd
  3. Aug 9, 2011 #2
    This means in particular that there exists a set I such that the submodules of [itex]R^I[/itex] are all free. Can you show that this implies that all ideals are free?
     
  4. Aug 9, 2011 #3
    I am not sure what you mean by the notation [itex]R^I[/itex]. As M is a specific R-module who's submodules are free, wouldn't I have to consider only the sub-modules of M?
     
  5. Aug 9, 2011 #4
    Yes, but in particular, M is free. Thus M is isomorphic to [itex]R^I[/itex] (or [itex]\oplus_{i\in I} R[/itex] if you prefer that)
     
  6. Aug 9, 2011 #5
    Ah, gotcha. Then R (and hence any ideal) itself can be regarded as a submodule of [itex]\oplus R[/itex] by the isomorphism. Then this reduces to the other problem?

    So it seems any free module on a commutative ring R contains R and its ideals as a submodule (at least isomorphically).
     
  7. Aug 9, 2011 #6
    Indeed; so the problem reduces to the ideals of R. You know that they are free (as R-module), and you need to show that they are generated by 1 element.
     
  8. Aug 9, 2011 #7
    Yes, I did that as a homework problem at some point in my last algebra class. A contradiction arises assuming there were two or more basis elements and choosing the coefficients appropriately.

    Thanks for your help.
     
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