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

1. Aug 9, 2011

### daveyp225

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. Aug 9, 2011

### micromass

This means in particular that there exists a set I such that the submodules of $R^I$ are all free. Can you show that this implies that all ideals are free?

3. Aug 9, 2011

### daveyp225

I am not sure what you mean by the notation $R^I$. As M is a specific R-module who's submodules are free, wouldn't I have to consider only the sub-modules of M?

4. Aug 9, 2011

### micromass

Yes, but in particular, M is free. Thus M is isomorphic to $R^I$ (or $\oplus_{i\in I} R$ if you prefer that)

5. Aug 9, 2011

### daveyp225

Ah, gotcha. Then R (and hence any ideal) itself can be regarded as a submodule of $\oplus R$ 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).

6. Aug 9, 2011

### micromass

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.

7. Aug 9, 2011

### daveyp225

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.