multiplication rule gives inverse ideal


Let R be a commutative ring with non-zero unity.  If an ideal  (a,b)  of R, with a or b regularPlanetmathPlanetmathPlanetmath (, obeys the multiplicationPlanetmathPlanetmath rule

(a,b)(c,d)=(ac,ad+bc,bd) (1)

with all ideals (c,d)  of R, then  (a,b) is an invertible ideal.

Proof.  The rule gives


Thus the product ab may be written in the form


where u and v are elements of R.  Let’s assume that e.g. a is regular.  Then a has the multiplicative inverse a-1 in the total ring of fractionsMathworldPlanetmath R.  Again applying the rule yields

(a,b)(va,a-vb)(a-2)=(va2,a2-vab+vab,ab-vb2)(a-2)=(va2,a2,ua2)(a-2)=(v, 1,u)=R.

Consequently the ideal  (a,b)  has an inverse ideal (which may be a fractional ideal (; this settles the proof.

Remark.  The rule (1) in the theoremMathworldPlanetmath may be replaced with the rule

(a,b)(c,d)=(ac,(a+b)(c+d),bd) (2)

as is seen from the identical equation(a+b)(c+d)-ac-bd=ad+bc.

Title multiplication rule gives inverse ideal
Canonical name MultiplicationRuleGivesInverseIdeal
Date of creation 2013-03-22 15:24:16
Last modified on 2013-03-22 15:24:16
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 5
Author pahio (2872)
Entry type Theorem
Classification msc 13A15
Classification msc 16D25
Related topic PruferRing
Related topic CharacterizationMathworldPlanetmath