Localization (commutative algebra): Difference between revisions

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=== General construction ===

In the general case, a problem arises with [[zero divisor]]s. Let {{mvar|S}} be a multiplicative set in a commutative ring {{mvar|R}}. IfSuppose that <math>s\tfracin a1S,</math> and <math>0\ne a\in R</math> is thea imagezero indivisor with <math>S^{-1}Ras=0.</math> ofThen <math>a\intfrac R,a1</math> andis ifthe {{math|1=''as''image =in 0}<math>S^{-1}R</math> withof <math>sa\in SR,</math> thenand one must havehas <math>\tfrac 0sa1 = \tfrac {as}{s} = \tfrac a1,0s = \tfrac 01.</math> and thusThus some nonzero elements of {{mvar|R}} must be zero in <math>S^{-1}R.</math> The construction that follows is designed for taking this into account.

Given {{mvar|R}} and {{mvar|S}} as above, one considers the [[equivalence relation]] on <math>R\times S</math> that is defined by <math>(r_1, s_1) \sim (r_2, s_2)</math> if there exists a <math>t\in S</math> such that <math>t(s_1r_2-s_2r_1)=0.</math>

The localization <math>S^{-1}R</math> is defined as the set of the [[equivalence class]]es for this relation. The class of {{math|(''r'', ''s'')}} is denoted as <math>\frac rs,</math> <math>r/s,</math> or <math>s^{-1}r.</math> So, one has <math>\tfrac{r_1}{s_1}=\tfrac{r_2}{s_2}</math> if and only if there is a <math>t\in S</math> such that <math>t(s_1r_2-s_2r_1)=0.</math> The reason for the <math>t</math> is to handle cases such as the above <math>\tfrac a1 = \tfrac 01,</math> where <math>s_1r_2-s_2r_1</math> is nonzero even though the fractions should be regarded as equal.

The localization <math>S^{-1}R</math> is a commutative ring with addition