Localization (commutative algebra): Difference between revisions

Revision as of 13:37, 15 May 2021

Revision as of 13:43, 15 May 2021

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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''^{{sup|–1}-1}''R''}}</math>-modules. In other words, localization is an [[exact functor]], and {{<math|''>S''^{{sup|–1}-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]]

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:<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|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>

==Localization at primes==