Question on a torsion-free module, not finitely generated, not free over PID

Question

A classic example is $\mathbb{Q}$ as an abelian group. In order to be free, it would require a proper subgroup to be the basis. It's clear to me, on an intuitive level, that no such subgroup exists. But I'm having trouble making this intuition rigorous. My hunch is that such a subgroup would have to be of finite index, and there can be none such subgroup.

But why exactly can there be none in this case? Also, why in general must a basis be of finite index when considered as a subgroup? I'm looking for a formal way to express these properties.

Answer 1

Theorem. Let $R$ be an integral domain. Then $Q$, the field of fractions of $R$, is a free $R$-module if and only if $R$ is a field.

One direction is obvious.

Suppose $Q$ is free and that $x,y\in Q$ are nonzero. We can write $x=a/c$, $y=b/c$, so $$ b\frac{a}{c}+(-a)\frac{b}{c}=0 $$ Therefore $x$ and $y$ are not linearly independent over $R$. Hence $Q$ must have rank $1$