# 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

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

 $ab=ua^{2}\!+\!vb^{2},$

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

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

Consequently the ideal  $(a,\,b)$  has an inverse ideal (which may be a fractional ideal (http://planetmath.org/FractionalIdealOfCommutativeRing)); 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$(a\!+\!b)(c\!+\!d)\!-\!ac\!-\!bd=ad+bc$.

Title multiplication rule gives inverse ideal MultiplicationRuleGivesInverseIdeal 2013-03-22 15:24:16 2013-03-22 15:24:16 pahio (2872) pahio (2872) 5 pahio (2872) Theorem msc 13A15 msc 16D25 PruferRing Characterization