  Let $R$ be an associative ring with ${\bf 1}$ which is not commutative.
Assume that any non-zero commutator $v\in R$ satisfies: $v^2$ is in the center
of $R$ and $v$ is not a zero-divisor. (Note that our assumptions do not include
finite dimensionality.) We prove that $R$ has no zero divisors, and that if
${\rm char(R)}\ne 2,$ then the localization of $R$ at its center is a
quaternion division algebra.
