 Right denominator set

A multiplicative set $S$ containing $1$ and not $0$ is a right denominator set if it satisfies the right Ore condition and right reversibility: $sr=0$, with $s\in S$, implies $rt=0$ for some $t\in S$. Its Ore localization consists of right fractions $r{s}^{-1}$ and has kernel $\{r:rt=0\text{ for some }t\in S\}$. This condition permits zero-divisor denominators; it is automatic for a set of regular elements of a ring. When all denominators are regular, the localization map is injective.