Editing Localization (commutative algebra) (section)

=== Saturation of a multiplicative set ===
Let <math>S \subseteq R</math> be a multiplicative set. The ''saturation'' <math>\hat{S}</math> of <math>S</math> is the set 
:<math>\hat{S} = \{ r \in R \colon \exists s \in R, rs \in S \}.</math>
The multiplicative set {{mvar|S}} is ''saturated'' if it equals its saturation, that is, if <math>\hat{S}=S</math>, or equivalently, if  <math>rs \in S</math> implies that {{mvar|r}} and {{mvar|s}} are in {{mvar|S}}.

If {{mvar|S}} is not saturated, and <math>rs \in S,</math> then <math>\frac s{rs}</math> is a [[multiplicative inverse]] of the image of {{mvar|r}} in <math>S^{-1}R.</math> So, the images of the elements of <math>\hat S</math> are all invertible in <math>S^{-1}R,</math> and the universal property implies that <math>S^{-1}R</math> and <math>\hat {S}{}^{-1}R</math> are [[canonical isomorphism|canonically isomorphic]], that is, there is a unique isomorphism between them that fixes the images of the elements of {{mvar|R}}.

If {{mvar|S}} and {{mvar|T}} are two multiplicative sets, then <math>S^{-1}R</math> and <math>T^{-1}R</math> are isomorphic if and only if they have the same saturation, or, equivalently, if {{mvar|s}} belongs to one of the multiplicative sets, then there exists <math>t\in R</math> such that {{mvar|st}} belongs to the other.

Saturated multiplicative sets are not widely used explicitly, since, for verifying that a set is saturated, one must know ''all'' [[unit (ring theory)|units]] of the ring.