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  "body": "\n## Definition\n\nLet $R$ be a [[ring|0B1261B5ED]] with identity.\nLet $S \\subseteq R$ be a [[multiplicative set|AN9SKEZ3VY]]: $1 \\in S$ and $st \\in S$ for all $s, t \\in S$.\nThe set $S$ satisfies the *right Ore condition* if for all $a \\in R$ and $s \\in S$ there exist $b \\in R$ and $t \\in S$ with\n$$a t = s b.$$\nEquivalently, $aS \\cap sR \\ne \\emptyset$ for all $a \\in R$ and $s \\in S$.\nThe *left Ore condition* asks instead for $b \\in R$ and $t \\in S$ with $t a = b s$.\nA *right Ore domain* is a domain $R$ such that $R \\setminus \\set{0}$ satisfies the right Ore condition.\n\n## Remarks\n\nExample: if $R$ is commutative, every multiplicative set $S$ satisfies both Ore conditions: take $t = s$ and $b = a$.\nEvery right Noetherian domain is a right Ore domain.\n\nNon-example: let $k$ be a field and $R = k\\langle x, y \\rangle$ the free algebra, with $S = R \\setminus \\set{0}$.\nEach 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 $y$.\nSo $xS \\cap yR = \\set{0}$, while $0 \\notin xS$ since $R$ is a domain.\nHence $xS \\cap yR = \\emptyset$ and the right Ore condition fails for $a = x$ and $s = y$.\n\nThe 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.\nA 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*.\nIf $S$ contains no [[zero divisors|0NNRDW10RZ]], reversibility is automatic.\n",
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