# 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 $ab$ 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 InvertibleIdealsInSemilocalRings 2013-03-22 18:36:17 2013-03-22 18:36:17 gel (22282) gel (22282) 6 gel (22282) Theorem msc 11R04 msc 13F05 PruferDomain