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 $b$ regular (http://planetmath.org/RegularElement), obeys the multiplication 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

$${(a,b)}^2=(a,-b)(a,b)=(a^2,ab-ba,b^2)=(a^2,b^2).$$

Thus the product $ab$ may be written in the form

$$ab=u{a}^2+v{b}^2,$$

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 fractions $R$.  Again applying the rule yields

$$(a,b)(va,a-vb)(a^{-2})=(v{a}^2,a^2-vab+vab,ab-v{b}^2)(a^{-2})=(v{a}^2,a^2,u{a}^2)(a^{-2})=(v,\mathrm{\hspace{0.17em}1},u)=R.$$

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 theorem 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	Characterization