Consider the group ring $\mathbb{Q} F_2$ with rational coefficients over the free group in two generators ($a$ and $b$).
It is not an Ore-Domain: While it does not contain any zero-divisors, for example the elements $1+a$ and $1+b$ do not have a common (left or right) multiple.
My Question is: Given $x, y \in \mathbb{Q} F_2$, what are criteria to decide if $x$ and $y$ have a common (say right) multiple?
Of course, if $x$ and $y$ commute, they have $xy = yx$ as a common multiple. Also, if $y$ is a multiple of $x$, then $y$ is a common multiple of $x$ and $y$. Are there cases where $x$ and $y$ fulfill neither of these conditions and still have a common multiple?
I am not sure if things become easier if we take coefficients in $\mathbb{C}$ instead, or if we take another group (without trivializing the problem), but an answer in these cases would also be appreciated.