= Right denominator set
{title2=$sr=0\Rightarrow rt=0\text{ for some }t\in S$}
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 $rs^{-1}$ 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.
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