invertible ideal is finitely generated

Theorem.โ€‰ Let R be a commutative ring containing regular elementsPlanetmathPlanetmathPlanetmath.โ€‰ Every invertible ( fractional idealMathworldPlanetmath ๐”ž of R is finitely generatedMathworldPlanetmathPlanetmath and regular (, i.e. regular elements.

Proof.โ€‰ Let T be the total ring of fractionsMathworldPlanetmath of R and e the unity of T.โ€‰We first show that the inverse ideal of ๐”ž has the unique quotient presentationMathworldPlanetmathPlanetmath (โ€‰ [Rโ€ฒ:๐”ž]โ€‰ whereโ€‰ Rโ€ฒ:=R+โ„คโขe.โ€‰ If ๐”ž-1 is an inverse ideal of ๐”ž, it means thatโ€‰ ๐”žโข๐”ž-1=Rโ€ฒ.โ€‰ Therefore we have


so that


This implies thatโ€‰ ๐”ž๐”ž-1=๐”ž[Rโ€ฒ:๐”ž],โ€‰ and because ๐”ž is a cancellation ideal, it must thatโ€‰ ๐”ž-1=[Rโ€ฒ:๐”ž], i.e. [Rโ€ฒ:๐”ž] is the unique inverseMathworldPlanetmathPlanetmath of the ideal ๐”ž.

Sinceโ€‰ ๐”ž[Rโ€ฒ:๐”ž]=Rโ€ฒ,โ€‰ there exist some elements a1,โ€ฆ,an of ๐”ž and the elements b1,โ€ฆ,bn ofโ€‰ [Rโ€ฒ:๐”ž]โ€‰ such thatโ€‰ a1โขb1+โ‹ฏ+anโขbn=e.โ€‰ Then an arbitrary element a of ๐”ž satisfies


because every biโขa belongs to the ring Rโ€ฒ.โ€‰ Accordingly,โ€‰ ๐”žโŠ†(a1,โ€ฆ,an).โ€‰ Since the converse inclusion is apparent, we have seen thatโ€‰ {a1,โ€ฆ,an}โ€‰ is a finite of the invertible ideal ๐”ž.

Since the elements bi belong to the total ring of fractions of R, we can choose such a regular element d of R that each of the products biโขd belongs to R.โ€‰ Then


and thus the fractional ideal ๐”ž contains a regular element of R, which obviously is regular in T, too.


  • 1 R. Gilmer: Multiplicative ideal theory.โ€‰ Queens University Press. Kingston, Ontario (1968).
Title invertible ideal is finitely generated
Canonical name InvertibleIdealIsFinitelyGenerated
Date of creation 2015-05-06 14:44:03
Last modified on 2015-05-06 14:44:03
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 10
Author pahio (2872)
Entry type Theorem
Classification msc 13B30
Related topic InvertibilityOfRegularlyGeneratedIdeal