# Torsion-free modules

1. Jan 25, 2009

### hypermonkey2

Hi all, I came across this problem in a book and I can`t seem to crack it.
It says that if we have an integral domain R and M is any non-principal ideal of R,
then
M is torsion-free of rank 1 and is NOT a free R-module.

Why is this true?

cheers

2. Jan 25, 2009

### hypermonkey2

ah so i see why it needs to be torsion free. (M lies in R, so if there is r such that rm=0, R cant be an integral domain...)

but what about the rank? Why must it be 1? I am also surprised that it is not free..