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  "body": "\n## Summary\n\nThe Ore condition is a condition on a [[multiplicative set|AN9SKEZ3VY]] $S$ of a noncommutative [[ring|0B1261B5ED]].\nIt allows one to form fractions $a s^{-1}$ with denominators in $S$.\nIt asks that a \"left fraction\" $s^{-1} a$ can always be rewritten as a \"right fraction\" $b t^{-1}$.\nTogether with a mild cancellation condition, it is exactly what is needed for the [[Ore localization|SG7AQ9T8HD]] to exist.\nFor commutative rings it always holds, so it explains why ordinary [[localization|4C1AE84N5N]] needs no extra hypothesis.\n\n## Explanations\n\n### L1 human\n\nIn a commutative ring, $a / s$ means the same thing whether you divide on the left or on the right.\nIn a noncommutative ring, $s^{-1} a$ and $a s^{-1}$ are different, and products of fractions must again be fractions of one kind.\nTo multiply $a s^{-1}$ by $b t^{-1}$ one needs to move $s^{-1}$ past $b$, that is, to write $s^{-1} b$ as $c u^{-1}$.\nThe Ore condition says that this is always possible.\nFor example, in the ring of polynomials $k[x]$ every nonzero polynomial can serve as a denominator, but in the free algebra $k\\langle x, y \\rangle$ the nonzero elements cannot.\n",
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