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  "body": "\n## Statement\n\nLet $R$ be a [[commutative ring|0310REJJY8]], let $S \\subseteq R$ be a [[multiplicative set|AN9SKEZ3VY]], and let $\\lambda \\colon R \\to S^{-1}R$, $a \\mapsto a/1$, be the canonical map to the [[localization|4C1AE84N5N]].\nThen the maps\n$$\\mathfrak q \\mapsto \\lambda^{-1}(\\mathfrak q), \\qquad \\mathfrak p \\mapsto S^{-1}\\mathfrak p = \\set{a/s : a \\in \\mathfrak p, \\ s \\in S}$$\nare mutually inverse bijections between the [[prime ideals|00MHHZ9SNE]] $\\mathfrak q$ of $S^{-1}R$ and the prime ideals $\\mathfrak p$ of $R$ with $\\mathfrak p \\cap S = \\emptyset$.\nBoth maps preserve inclusion.\n\n## Remarks\n\nThe preimage of a prime ideal under a ring homomorphism is prime, and $\\lambda^{-1}(\\mathfrak q)$ misses $S$ because elements of $S$ become units.\nConversely, if $\\mathfrak p \\cap S = \\emptyset$, then $a/s \\in S^{-1}\\mathfrak p$ forces $ua \\in \\mathfrak p$ for some $u \\in S$, hence $a \\in \\mathfrak p$ since $u \\notin \\mathfrak p$.\nThis shows $\\lambda^{-1}(S^{-1}\\mathfrak p) = \\mathfrak p$ and that $S^{-1}\\mathfrak p$ is a proper prime ideal.\nFinally every ideal $J$ of $S^{-1}R$ is the [[extension|5AFH94P4S3]] of its contraction, which gives $S^{-1}\\lambda^{-1}(\\mathfrak q) = \\mathfrak q$.\n\nExample: for $R = \\Z$ and $S = \\set{1, 2, 4, \\ldots}$, the prime ideals of $\\Z[1/2]$ are $(0)$ and $p\\Z[1/2]$ for the odd primes $p$; the prime $2\\Z$ meets $S$ and disappears.\n\nFor $S = R \\setminus \\mathfrak p$ with $\\mathfrak p$ prime, the primes of $R_{\\mathfrak p}$ correspond to the primes of $R$ contained in $\\mathfrak p$.\nSo $R_{\\mathfrak p}$ has the unique maximal ideal $\\mathfrak p R_{\\mathfrak p}$, that is, it is a [[local ring|MZ02BKPHJA]].\n\nThe hypothesis that $\\mathfrak p$ be prime matters: for a non-prime ideal $I$ with $I \\cap S = \\emptyset$ one can have $\\lambda^{-1}(S^{-1}I) \\ne I$.\nFor instance, with $R = \\Z$, $I = 6\\Z$ and $S = \\set{1, 2, 4, \\ldots}$, one gets $\\lambda^{-1}(S^{-1}I) = 3\\Z$.\n\n## Explanations\n\n### L1 human\n\nWhen you localize, some prime ideals vanish and the others survive unchanged.\nThe ones that vanish are exactly those that contain an allowed denominator, since such an ideal would contain a unit.\nIn the language of geometry, localizing keeps an open part of the space of prime ideals, or the primes near a chosen point.\n\n### L2 human\n\nThis result is the reason why localization is a \"local\" operation on the [[spectrum|FB45BHHA2J]].\nFor $S = \\set{1, f, f^2, \\ldots}$ it identifies $\\operatorname{Spec} R_f$ with the [[basic open set|N6JX9BCKQP]] $D(f)$.\nFor $S = R \\setminus \\mathfrak p$ it identifies $\\operatorname{Spec} R_{\\mathfrak p}$ with the primes inside $\\mathfrak p$, so the [[height|87WH65FJB8]] of $\\mathfrak p$ equals the [[Krull dimension|MVGBS3J1P7]] of $R_{\\mathfrak p}$.\nCombined with the description of the primes of a quotient $R/I$, it describes the primes of rings such as $R_{\\mathfrak p}/I R_{\\mathfrak p}$ for an ideal $I$ of $R$.\n",
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