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Localization (commutative algebra)
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===Module properties=== If {{mvar|M}} is a [[submodule]] of an {{mvar|R}}-module {{mvar|N}}, and {{mvar|S}} is a multiplicative set in {{mvar|R}}, one has <math>S^{-1}M\subseteq S^{-1}N.</math> This implies that, if <math>f\colon M\to N</math> is an [[injective]] [[module homomorphism]], then :<math>S^{-1}R\otimes_R f : \quad S^{-1}R\otimes_R M\to S^{-1}R\otimes_R N</math> is also an injective homomorphism. Since the tensor product is a [[right exact functor]], this implies that localization by {{mvar|S}} maps [[exact sequence]]s of {{mvar|R}}-modules to exact sequences of <math>S^{-1}R</math>-modules. In other words, localization is an [[exact functor]], and <math>S^{-1}R</math> is a [[flat module|flat {{mvar|R}}-module]]. This flatness and the fact that localization solves a [[universal property]] make that localization preserves many properties of modules and rings, and is compatible with solutions of other universal properties. For example, the [[natural transformation|natural map]] :<math>S^{-1}(M \otimes_R N) \to S^{-1}M \otimes_{S^{-1}R} S^{-1}N</math> is an isomorphism. If <math>M</math> is a [[finitely presented module]], the natural map :<math>S^{-1} \operatorname{Hom}_R (M, N) \to \operatorname{Hom}_{S^{-1}R} (S^{-1}M, S^{-1}N)</math> is also an isomorphism.<ref>{{harvnb|Eisenbud|1995|loc=Proposition 2.10}}</ref> If a module ''M'' is a [[finitely generated module|finitely generated]] over ''R'', one has :<math>S^{-1}(\operatorname{Ann}_R(M)) = \operatorname{Ann}_{S^{-1}R}(S^{-1}M),</math> where <math>\operatorname{Ann}</math> denotes [[annihilator (ring theory)|annihilator]], that is the ideal of the elements of the ring that map to zero all elements of the module.<ref>{{harvnb|Atiyah|Macdonald|1969|loc=Proposition 3.14.}}</ref> In particular, :<math>S^{-1} M = 0\quad \iff \quad S\cap \operatorname{Ann}_R(M) \ne \emptyset,</math> that is, if <math>t M = 0</math> for some <math>t \in S.</math><ref>Borel, AG. 3.1</ref>
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