invertible ideals in semi-local rings

Theorem.

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

In particular, a semi-local (http://planetmath.org/SemiLocalRing) Dedekind domain is a principal ideal domain and every finitely generated ideal in a semi-local Prüfer domain is principal.

Proof.

First, if $a$ is regular then $(a)$ is invertible, with inverse $(a^{-1})$ , so only the converse needs to be shown.

Suppose that $\mathfrak{a}$ is invertible, and $\mathfrak{a}\mathfrak{b}=R$ . Then let the maximal ideals of $R$ be $\mathfrak{m}_{1},\ldots,\mathfrak{m}_{n}$ . As $\mathfrak{a}\mathfrak{b}\not\subseteq\mathfrak{m}_{k}$ , there exist $a_{k}\in\mathfrak{a},b_{k}\in\mathfrak{b}$ such that $a_{k}b_{k}\in R\setminus\mathfrak{m}_{k}$ .

By maximality, $\mathfrak{m}_{j}\not\subseteq\mathfrak{m}_{k}$ whenever $j\not=k$ , so we may choose $\lambda_{jk}\in\mathfrak{m}_{j}\setminus\mathfrak{m}_{k}$ . Setting $\lambda_{k}=\prod_{j\not=k}\lambda_{jk}$ gives $\lambda_{k}\in\mathfrak{m}_{j}$ for all $j\not=k$ and, as $\mathfrak{m}_{k}$ is prime (http://planetmath.org/PrimeIdeal), $\lambda_{k}\not\in\mathfrak{m}_{k}$ . Then, writing

a=\lambda_{1}a_{1}+\cdots+\lambda_{n}a_{n}\in\mathfrak{a},\ b=\lambda_{1}b_{1}% +\cdots+\lambda_{n}b_{n}\in\mathfrak{b}

we can expand the product to get

ab=\sum_{i,j}\lambda_{i}\lambda_{j}a_{i}b_{j}.

(1)

However, $a_{i}b_{j}\in\mathfrak{a}\mathfrak{b}\subseteq R$ so $\lambda_{i}\lambda_{j}a_{i}b_{j}$ is in $\mathfrak{m}_{k}$ whenever either $i$ or $j$ is not equal to $k$ . On the other hand, $\lambda_{k}\lambda_{k}a_{k}b_{k}\not\in\mathfrak{m}_{k}$ and, consequently, there is exactly one term on the right hand side of (1) which is not in $\mathfrak{m}_{k}$ , so $ab\not\in\mathfrak{m}_{k}$ .

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,

(a)\subseteq\mathfrak{a}=ab\mathfrak{a}\subseteq a\mathfrak{b}\mathfrak{a}=aR=% (a)

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