- #1
daveyp225
- 88
- 0
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.
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.