invertible ideals in semi-local rings


Let R be a commutative ring in which there are only finitely many maximal idealsMathworldPlanetmath. Then, a fractional idealMathworldPlanetmathPlanetmath over R is invertible if and only if it is principal and generated by a regular elementPlanetmathPlanetmath.

In particular, a semi-local ( Dedekind domainMathworldPlanetmath is a principal ideal domainMathworldPlanetmath and every finitely generatedMathworldPlanetmathPlanetmath ideal in a semi-local PrΓΌfer domain is principal.


First, if a is regular then (a) is invertible, with inverseMathworldPlanetmathPlanetmath (a-1), so only the converse needs to be shown.

Suppose that π”ž is invertible, and π”žβ’π”Ÿ=R. Then let the maximal ideals of R be π”ͺ1,…,π”ͺn. As π”žβ’π”ŸβŠˆπ”ͺk, there exist akβˆˆπ”ž,bkβˆˆπ”Ÿ such that ak⁒bk∈Rβˆ–π”ͺk.

By maximality, π”ͺj⊈π”ͺk whenever jβ‰ k, so we may choose Ξ»j⁒k∈π”ͺjβˆ–π”ͺk. Setting Ξ»k=∏jβ‰ kΞ»j⁒k gives Ξ»k∈π”ͺj for all jβ‰ k and, as π”ͺk is prime (, Ξ»kβˆ‰π”ͺk. Then, writing


we can expand the product to get

a⁒b=βˆ‘i,jΞ»i⁒λj⁒ai⁒bj. (1)

However, ai⁒bjβˆˆπ”žβ’π”ŸβŠ†R so Ξ»i⁒λj⁒ai⁒bj is in π”ͺk whenever either i or j is not equal to k. On the other hand, Ξ»k⁒λk⁒ak⁒bkβˆ‰π”ͺk and, consequently, there is exactly one term on the right hand side of (1) which is not in π”ͺk, so a⁒bβˆ‰π”ͺk.

We have shown that a⁒b is not in any maximal ideal of R, and must therefore be a unit. So a is regular and,


as required. ∎

Title invertible ideals in semi-local rings
Canonical name InvertibleIdealsInSemilocalRings
Date of creation 2013-03-22 18:36:17
Last modified on 2013-03-22 18:36:17
Owner gel (22282)
Last modified by gel (22282)
Numerical id 6
Author gel (22282)
Entry type Theorem
Classification msc 11R04
Classification msc 13F05
Related topic PruferDomain