Why is M Torsion-free & Rank 1 but Not a Free R-Module?

hypermonkey2 · Jan 25, 2009

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

hypermonkey2 · Jan 25, 2009

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 can't be an integral domain...)

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

Why is M Torsion-free & Rank 1 but Not a Free R-Module?

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