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单词 MultiplicationRuleGivesInverseIdeal
释义

multiplication rule gives inverse ideal


Theorem.

Let R be a commutative ring with non-zero unity.  If an ideal  (a,b)  of R, with a or b regularPlanetmathPlanetmathPlanetmath (http://planetmath.org/RegularElement), 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

(a,b)2=(a,-b)(a,b)=(a2,ab-ba,b2)=(a2,b2).

Thus the product ab may be written in the form

ab=ua2+vb2,

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 (http://planetmath.org/FractionalIdealOfCommutativeRing)); 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.

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更新时间:2025/5/4 21:36:25