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

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In summary, if R is a commutative ring with 1 and there exists a non-zero R-module M such that every submodule of M is free, then R is a PID. This can be proven by considering M as an I-module and showing that all ideals of R are generated by 1 element.
  • #1
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?
 
  • #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?
 
  • #4
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)
 
  • #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).
 
  • #6
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.
 
  • #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.
 

1. What is a nonzero R-module over a commutative ring?

A nonzero R-module is a mathematical structure that consists of a commutative ring R and a set of elements that can be added and multiplied by elements of R. This set of elements is known as a module and is analogous to a vector space in linear algebra. Nonzero R-modules are those that contain at least one nonzero element.

2. What does it mean for all submodules to be free in an R-module?

If all the submodules of an R-module are free, it means that they are all isomorphic to direct sums of copies of the ring R. This essentially means that the submodules are independent and do not have any nontrivial relations between them.

3. What is a PID in the context of R-modules?

A PID (Principal Ideal Domain) is a commutative ring in which every ideal can be generated by a single element. In the context of R-modules, this means that every submodule can be generated by a single element.

4. How are nonzero R-modules over commutative rings related to PIDs?

If a nonzero R-module over a commutative ring has all submodules that are free, then the ring is necessarily a PID. This is known as the structure theorem for PIDs.

5. Can an R-module be a PID if it has non-free submodules?

No, an R-module cannot be a PID if it has non-free submodules. The structure theorem for PIDs states that in order for an R-module to be a PID, all submodules must be free. If even one submodule is non-free, then the R-module cannot be a PID.

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