Right Ore condition
= Right Ore condition
{title2=$au=sb$}
A multiplicative subset $S\subseteq A$ with $1\in S$ and $0\notin S$ of a <ring> $A$ satisfies the right <Ore condition> if for every $a\in A$ and $s\in S$ there are $u\in S$ and $b\in A$ such that $au=sb$. For a <noncommutative domain>, taking $S=A\setminus\{0\}$ gives a <ring> of right fractions $as^{-1}$; every nonzero fraction is invertible. The usual additional denominator reversibility condition is automatic when the elements of $S$ are nonzero in a <noncommutative domain>.