For some reason, I'm stuck in the notion that taking $R$ as a free $R$-module, with basis $B$, induces a free abelian group on the additive group in $R$ with basis $B$ (i.e. the abelian group in the ring $R$ would have the same basis $B$).
To convince myself otherwise, I'm looking for a counter-example. I can sort of see why it might not be true - intuitively, the coefficients used to describe elements as combinations of basis elements are not the same (one has coefficients in $R$ while the other has integer coefficients), but, being stuck with the usual example of $R=\mathbb{Z}$ (for which it does hold true), I'm unable to find a counter example. Any ideas?
Since any vector space is a free module over itself, in particular so is $\mathbb{R}$ with basis $B=\{1\}$.
$\mathbb{R}$ is not free abelian with basis $\{1\}$ since, e.g. there is no integer $n$ such that $\pi = n$.