# 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^{} |