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

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The discussion centers on the properties of a module M over an integral domain R, specifically when M is a non-principal ideal. It is established that M is torsion-free and has rank 1, but it is not a free R-module. The reasoning provided indicates that if M were torsion, it would contradict the integral domain property of R. The rank being 1 is derived from the structure of non-principal ideals, which can only generate a single element without forming a basis, thus confirming M's non-freeness.

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hypermonkey2
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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
 
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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..
 

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