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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]] and let $S$ be the set of elements of $R$ that are not [[zero divisors|0NNRDW10RZ]].\nSo $s \\in S$ means: $sa = 0$ implies $a = 0$, for all $a \\in R$.\nThen $S$ is a [[multiplicative set|AN9SKEZ3VY]].\nThe *total ring of fractions* of $R$ is the [[localization|4C1AE84N5N]]\n$$Q(R) = S^{-1}R.$$\n\n## Remarks\n\nThe canonical map $R \\to Q(R)$, $a \\mapsto a/1$, is injective, because no element of $S$ kills a nonzero element.\nIf $S'$ is a multiplicative set and $R \\to S'^{-1}R$ is injective, then $S' \\subseteq S$.\n\nExample: if $R$ is an [[integral domain|0QVD7WBFGN]], then $S = R \\setminus \\set{0}$ and $Q(R)$ is the [[field of fractions|0F2PTGX1A2]] of $R$.\nFor $R = \\Z \\times \\Z$, the non-zero-divisors are the pairs $(a, b)$ with $a \\ne 0$ and $b \\ne 0$, and $Q(R) \\cong \\Q \\times \\Q$.\nFor $R = \\Z/6\\Z$, the non-zero-divisors are $1$ and $5$, which are already units, so $Q(R) = R$.\n\nNon-example: in $R = \\Z \\times \\Z$, the set $T = \\set{(1,0)^n : n \\ge 0} = \\set{(1,1), (1,0)}$ is multiplicative, but $(1, 0)$ is a zero divisor.\nThe localization $T^{-1}R \\cong \\Z$ is not the total ring of fractions, and $R \\to T^{-1}R$ kills $(0, 1)$.\n",
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