Theorem 1   Let $R$ be a commutative ring with non-zero unity. If an ideal $(a,\,b)$ of $R$ , with $a$ or $b$ regular, obeys the multiplication rule

(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); this settles the proof.

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

(2)