Editing Localization (commutative algebra) (section)

==Localization of a module ==
Let <math>R</math> be a [[commutative ring]], <math>S</math> be a [[multiplicative set]] in <math>R</math>, and <math>M</math> be an <math>R</math>-[[module (mathematics)|module]]. The '''localization of the module''' <math>M</math> by <math>S</math>, denoted <math>S^{-1}M</math>, is an <math>S^{-1}R</math>-module that is constructed exactly as the localization of <math>R</math>, except that the numerators of the fractions belong to <math>M</math>. That is, as a set, it consists of [[equivalence class]]es, denoted <math>\frac ms</math>, of pairs <math>(m,s)</math>, where <math>m\in M</math> and <math>s\in S,</math> and two pairs <math>(m,s)</math> and <math>(n,t)</math> are equivalent if there is an element <math>u</math> in <math>S</math> such that
:<math>u(sn-tm)=0.</math>
Addition and scalar multiplication are defined as for usual fractions (in the following formula, <math>r\in R,</math> <math>s,t\in S,</math> and <math>m,n\in M</math>):
:<math>\frac{m}{s} + \frac{n}{t} = \frac{tm+sn}{st},</math>
:<math>\frac rs \frac{m}{t} = \frac{r m}{st}.</math>
Moreover, <math>S^{-1}M</math> is also an <math>R</math>-module with scalar multiplication
:<math> r\, \frac{m}{s} = \frac r1 \frac ms = \frac{rm}s.</math>

It is straightforward to check that these operations are well-defined, that is, they give the same result for different choices of representatives of fractions.

The localization of a module can be equivalently defined by using [[tensor product of modules|tensor products]]:
:<math>S^{-1}M=S^{-1}R \otimes_R M.</math>
The proof of equivalence (up to a [[canonical isomorphism]]) can be done by showing that the two definitions satisfy the same universal property.

===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>