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

• hypermonkey2
In summary, M is torsion-free because it does not contain any elements that can be multiplied by a non-zero element in the ring R to equal zero. It is also rank 1 because it can be generated by a single element. However, M is not a free R-module because it does not have a basis. It may be a free module over a different ring, as the definition of a free module depends on the ring. Some examples of rank 1 but not free R-modules include the module of all rational numbers over the ring of integers and the module of all real numbers over the ring of continuous functions on the interval [0,1].
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

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..

## 1. Why is M torsion-free?

M is torsion-free because it does not contain any elements that can be multiplied by a non-zero element in the ring R to equal zero. In other words, there are no elements in M that have a non-trivial annihilator in R.

## 2. Why is M rank 1?

M is rank 1 because it can be generated by a single element. This means that there is a single element in M that can generate all other elements through multiplication by elements in the ring R.

## 3. Why is M not a free R-module?

M is not a free R-module because it does not have a basis. In other words, there is no set of elements in M that can generate all other elements through linear combinations with coefficients in R.

## 4. Can M be a free module over a different ring?

Yes, M may be a free module over a different ring. The definition of a free module depends on the ring, so M may satisfy the conditions for being a free module over one ring but not over another.

## 5. What are some examples of rank 1 but not free R-modules?

One example is the module of all rational numbers over the ring of integers. Another example is the module of all real numbers over the ring of continuous functions on the interval [0,1]. In both cases, the modules are rank 1 because they can be generated by a single element, but they are not free because they do not have a basis.

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