proof of invertible ideals are projective

We show that a nonzero fractional idealPlanetmathPlanetmath 𝔞 of an integral domainMathworldPlanetmath R is invertible if and only if it is projective ( as an R-module.

Let 𝔞 be an invertible fractional ideal and f:M𝔞 be an epimorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of R-modules. We need to show that f has a right inverseMathworldPlanetmathPlanetmath. Letting 𝔞-1 be the inverse ideal of 𝔞, there exists a1,,an𝔞 and b1,,bn𝔞-1 such that


and, as f is onto, there exist ekM such that f(ek)=ak. For any x𝔞, xbk𝔞𝔞-1=R, so we can define g:𝔞M by




so g is indeed a right inverse of f, and 𝔞 is projective.

Conversely, suppose that 𝔞 is projective and let (ai)iI generate 𝔞 (this always exists, as we can let ai include every element of 𝔞). Then let M be a module with free basis (ei)iI and define f:M𝔞 by f(ei)=ai. As 𝔞 is projective, f has a right inverse g:𝔞M. As ei freely generate M, we can uniquely define gi:𝔞R by


noting that all but finitely many gi(x) must be zero for any given x. Choosing any fixed nonzero a𝔞, we can set bi=a-1gi(a) so that


for all x𝔞, and bi must equal zero for all but finitely many i. So, we can let 𝔟 be the fractional ideal generated by the bi and, noting that xbi=gi(x)R we get 𝔞𝔟R. Furthermore, for any xR,


so that R𝔞𝔟, and 𝔟 is the inverseMathworldPlanetmathPlanetmath of 𝔞 as required.

Title proof of invertible ideals are projective
Canonical name ProofOfInvertibleIdealsAreProjective
Date of creation 2013-03-22 18:35:51
Last modified on 2013-03-22 18:35:51
Owner gel (22282)
Last modified by gel (22282)
Numerical id 5
Author gel (22282)
Entry type Proof
Classification msc 16D40
Classification msc 13A15