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

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Discussion Overview

The discussion revolves around the properties of a non-zero R-module M over a commutative ring R, specifically exploring the implications of all submodules of M being free. Participants consider whether this condition leads to the conclusion that R is a principal ideal domain (PID).

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that if every submodule of a non-zero R-module M is free, then R must be a PID.
  • Another participant suggests that this implies the existence of a set I such that the submodules of R^I are all free, questioning how this leads to the conclusion that all ideals are free.
  • There is a clarification regarding the notation R^I, with some participants noting that M is a specific R-module and should be considered in that context.
  • It is noted that M being free implies it is isomorphic to R^I, leading to the conclusion that R and its ideals can be regarded as submodules of a free module.
  • One participant recalls a previous homework problem where they showed that a contradiction arises if there are two or more basis elements, suggesting a path to proving that ideals are generated by one element.

Areas of Agreement / Disagreement

Participants appear to agree on the implications of M being free and its relationship to the ideals of R, but the discussion does not reach a consensus on whether the original claim about R being a PID is definitively proven.

Contextual Notes

There are unresolved assumptions regarding the nature of the ideals and their generation, as well as the specific properties of the module M that may affect the conclusions drawn.

daveyp225
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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.
 
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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?
 
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?
 
daveyp225 said:
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?

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)
 
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).
 
daveyp225 said:
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).

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.
 
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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