PlanetMath: multiplication rule gives inverse ideal

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multiplication rule gives inverse ideal

(Theorem)

Theorem 1 Let

be a commutative ring with non-zero unity. If an ideal $(a,\,b)$ of

, with

$\displaystyle (a,\,b)(c,\,d) = (ac,\,ad\!+\!bc,\,bd)$

(1)

with all ideals $(c,\,d)$ of

, then $(a,\,b)$ is an invertible ideal.

Proof. The rule gives

$\displaystyle (a,\,b)^2 = (a,\,-b)(a,\,b) = (a^2,\,ab\!-\!ba,\,b^2) = (a^2,\,b^2).$

Thus the product

may be written in the form

$\displaystyle ab = ua^2\!+\!vb^2,$

where

and

are elements of

. Let's assume that e.g.

is regular. Then

. Again applying the rule yields

$\displaystyle (a,\,b)(va,\,a-vb)(a^{-2}) = (va^2,\,a^2-vab+vab,\,ab-vb^2)(a^{-2}) = (va^2,\,a^2,\,ua^2)(a^{-2}) = (v,\,1,\,u) = R.$

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 theorem may be replaced with the rule

$\displaystyle (a,\,b)(c,\,d) = (ac,\,(a\!+\!b)(c\!+\!d),\,bd)$

(2)

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

"multiplication rule gives inverse ideal" is owned by pahio.

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AMS MSC:	16D25 (Associative rings and algebras :: Modules, bimodules and ideals :: Ideals)
	13A15 (Commutative rings and algebras :: General commutative ring theory :: Ideals; multiplicative ideal theory)

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