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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]] and let $S \\subseteq R$ be a [[multiplicative set|AN9SKEZ3VY]].\nOn $R \\times S$, let $(a, s) \\sim (b, t)$ if there is $u \\in S$ with $u(at - bs) = 0$.\nThis is an [[equivalence relation|0G8VEZ3WTK]].\nWrite $a/s$ for the class of $(a, s)$.\nThe *localization* of $R$ at $S$ is the set $S^{-1}R$ of these classes with the operations\n$$\\frac{a}{s} + \\frac{b}{t} = \\frac{at + bs}{st}, \\qquad \\frac{a}{s} \\cdot \\frac{b}{t} = \\frac{ab}{st}.$$\nThese operations are well defined and make $S^{-1}R$ a commutative ring with zero $0/1$ and identity $1/1$.\nThe *canonical map* is the [[ring homomorphism|0VTH8KPCG2]] $\\lambda \\colon R \\to S^{-1}R$, $a \\mapsto a/1$.\n\n## Remarks\n\nUniversal property: $\\lambda(s)$ is a [[unit|0XVEZDE9J8]] for each $s \\in S$.\nFor each ring homomorphism $f \\colon R \\to B$ to a commutative ring $B$ such that $f(s)$ is a unit for all $s \\in S$, there is a unique ring homomorphism $g \\colon S^{-1}R \\to B$ with $g \\circ \\lambda = f$, namely $g(a/s) = f(a) f(s)^{-1}$.\n\nThe kernel of $\\lambda$ is the set of $a \\in R$ with $ua = 0$ for some $u \\in S$.\nSo $\\lambda$ is injective when $S$ contains no [[zero divisors|0NNRDW10RZ]], and $S^{-1}R = 0$ when $0 \\in S$.\n\nExample: for $R = \\Z$ and $S = \\Z \\setminus \\set{0}$, the localization is $\\Q$, the [[field of fractions|0F2PTGX1A2]] of $\\Z$.\nFor $S = \\set{1, 2, 4, 8, \\ldots}$, it is the ring $\\Z[1/2]$ of rationals $a / 2^k$.\nFor an element $f$, the localization at $\\set{1, f, f^2, \\ldots}$ is written $R_f$ or $R[1/f]$.\n\nNon-example: the factor $u$ cannot be dropped.\nTake $R = \\Z/6\\Z$ and $S = \\set{1, 3}$, and consider the naive relation \"$at = bs$\".\nIt relates $(2, 1)$ and $(0, 3)$, because $2 \\cdot 3 = 0 = 0 \\cdot 1$ in $R$.\nIt relates $(0, 3)$ and $(0, 1)$.\nIt does not relate $(2, 1)$ and $(0, 1)$, because $2 \\ne 0$.\nSo the naive relation is not transitive.\nWith the factor $u = 3$, one has $3(2 \\cdot 1 - 0 \\cdot 1) = 0$, so $2/1 = 0/1$ in $S^{-1}R$.\n\nWhen $\\mathfrak p$ is a [[prime ideal|00MHHZ9SNE]] and $S = R \\setminus \\mathfrak p$, the localization is the [[localization at a prime ideal|X6JD4AG973]] $R_{\\mathfrak p}$.\n",
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