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Homemultiplication rule gives inverse ideal

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# 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, 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); 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 $(a\!+\!b)(c\!+\!d)\!-\!ac\!-\!bd=ad+bc$.

## Mathematics Subject Classification

13A15*no label found*16D25

*no label found*

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