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  "body": "\n## Definition\n\nLet $R$ be a [[ring|0B1261B5ED]] with identity and let $S \\subseteq R$ be a [[multiplicative set|AN9SKEZ3VY]].\nA *right ring of fractions* of $R$ with respect to $S$ is a ring $Q$ together with a [[ring homomorphism|0VTH8KPCG2]] $\\varphi \\colon R \\to Q$ such that:\n\n1. $\\varphi(s)$ is a [[unit|0XVEZDE9J8]] of $Q$ for every $s \\in S$;\n2. every element of $Q$ has the form $\\varphi(a) \\varphi(s)^{-1}$ with $a \\in R$ and $s \\in S$;\n3. the kernel of $\\varphi$ is $\\set{a \\in R : a t = 0 \\text{ for some } t \\in S}$.\n\nWhen it exists, it is unique up to a unique isomorphism compatible with $\\varphi$, and it is written $R S^{-1}$.\nIt is called the *Ore localization* of $R$ at $S$.\n\n## Remarks\n\nOre's theorem: $R S^{-1}$ exists if and only if $S$ is a right denominator set.\nThis means that $S$ satisfies the right [[Ore condition|TJP4ARC2X7]] and is right reversible: if $s a = 0$ with $s \\in S$, then $a t = 0$ for some $t \\in S$.\nIn that case one may build $R S^{-1}$ as the set of classes of pairs $(a, s) \\in R \\times S$, where $(a, s) \\sim (b, t)$ when there are $c, d \\in R$ with $s c = t d \\in S$ and $a c = b d$.\nIt then has the universal property: each ring homomorphism $f \\colon R \\to B$ with $f(S)$ consisting of units factors uniquely through $\\varphi$.\n\nExample: if $R$ is commutative, every multiplicative set is a right denominator set, and $R S^{-1}$ is the usual [[localization|4C1AE84N5N]] $S^{-1} R$.\nClause 3 matches the kernel description there.\n\nNon-example: for the free algebra $R = k\\langle x, y \\rangle$ over a field $k$ and $S = R \\setminus \\set{0}$, no right ring of fractions exists.\nSuppose $\\varphi$ satisfied clauses 1 to 3; then $\\varphi$ would be injective by clause 3.\nBy clause 2, $\\varphi(y)^{-1} \\varphi(x)$ would equal $\\varphi(b) \\varphi(t)^{-1}$, giving $x t = y b$ with $t \\ne 0$, which is impossible.\n\nThe left ring of fractions $S^{-1} R$ is defined in the same way with elements $\\varphi(s)^{-1} \\varphi(a)$.\nWhen both exist, they are isomorphic as rings over $R$.\n",
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