For an integral domain $R$ , let $Frac(R)$ denote its field of fractions . Then $R$ is embedded in $Frac(R)$ and we can consider $Frac(R)$ as an $R$-module. Can we characterize those integral domains $R$ such that every proper non-zero submodule of the $R$-module $Frac(R)$ is free ?
Obviously since every ideal of $R$ is a $R$-submodule of $Frac(R)$ , so every non-zero ideal $I$ of $R$ is free as an $R$-module , hence $I$ must be principal ideal i.e. $R$ must be a PID . But surely $R=\mathbb Z$ doesn't satisfy the conditions of my question , hence the class of integral domains satisfying my conditions is a strict subclass of PID s .
If $M\subset Frac(R)$ is a nonzero submodule which is free, then it must be cyclic, since $M\otimes_R Frac(R)=Frac(R)$ is free of rank $1$ over $Frac(R)$. Or by a more elementary argument, if $x=\frac{a}{b}$ and $y=\frac{c}{d}$ are two distinct nonzero elements of $M$ with $a,b,c,d\in R$, then $bcx=ady$, which is a nontrivial relation between $x$ and $y$. So $x$ and $y$ cannot both be part of a free generating set for $M$, and $M$ cannot be generated freely by more than one element. So this is equivalent to your previous question: the answer is that $R$ satisfies your condition iff $R$ is a field or a DVR.