Ore condition

Remarks

Example: if $R$ is commutative, every multiplicative set $S$ satisfies both Ore conditions: take $t = s$ and $b = a$. Every right Noetherian domain is a right Ore domain.

Non-example: let $k$ be a field and $R = k\langle x, y \rangle$ the free algebra, with $S = R \setminus \set{0}$. Each nonzero element of $xS$ is a combination of words beginning with $x$, and each nonzero element of $yR$ is a combination of words beginning with $y$. So $xS \cap yR = \set{0}$, while $0 \notin xS$ since $R$ is a domain. Hence $xS \cap yR = \emptyset$ and the right Ore condition fails for $a = x$ and $s = y$.

The right Ore condition says that a left fraction $s^{-1} a$ can be rewritten as a right fraction $b t^{-1}$, since $a t = s b$ gives $s^{-1} a = b t^{-1}$ once $s$ and $t$ are invertible. A multiplicative set that satisfies the right Ore condition and is right reversible (if $s a = 0$ with $s \in S$, then $a t = 0$ for some $t \in S$) is a right denominator set. If $S$ contains no zero divisors, reversibility is automatic.