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 $𝔞$ is invertible, and $𝔞𝔟=R$.Then let the maximal ideals of $R$ be ${𝔪}_1,\mathrm{\dots },{𝔪}_n$. As $𝔞𝔟⊈{𝔪}_k$, there exist $a_k\in 𝔞,b_k\in 𝔟$ such that $a_k{b}_k\in R\setminus {𝔪}_k$.

By maximality, ${𝔪}_j⊈{𝔪}_k$ whenever $j\ne k$, so we may choose ${\lambda }_{jk}\in {𝔪}_j\setminus {𝔪}_k$.Setting ${\lambda }_k={\prod }_{j\ne k}{\lambda }_{jk}$ gives ${\lambda }_k\in {𝔪}_j$ for all $j\ne k$ and, as ${𝔪}_k$ is prime (http://planetmath.org/PrimeIdeal), ${\lambda }_k\notin {𝔪}_k$.Then, writing

$$a={\lambda }_1{a}_1+\mathrm{\cdots }+{\lambda }_n{a}_n\in 𝔞,b={\lambda }_1{b}_1+\mathrm{\cdots }+{\lambda }_n{b}_n\in 𝔟$$

we can expand the product to get

$$ab=\sum _{i,j}{\lambda }_i{\lambda }_j{a}_i{b}_j.$$(1)

However, $a_i{b}_j\in 𝔞𝔟\subseteq R$ so ${\lambda }_i{\lambda }_j{a}_i{b}_j$ is in ${𝔪}_k$ whenever either $i$ or $j$ is not equal to $k$. On the other hand, ${\lambda }_k{\lambda }_k{a}_k{b}_k\notin {𝔪}_k$ and, consequently, there is exactly one term on the right hand side of (1) which is not in ${𝔪}_k$, so $ab\notin {𝔪}_k$.

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

$$(a)\subseteq 𝔞=ab𝔞\subseteq a𝔟𝔞=aR=(a)$$

as required.∎