invertible ideal is finitely generated

Theorem.  Let $R$ be a commutative ring containing regular elements.  Every invertible (http://planetmath.org/FractionalIdealOfCommutativeRing) fractional ideal $𝔞$ of $R$ is finitely generated and regular (http://planetmath.org/RegularIdeal), i.e. regular elements.

Proof.  Let $T$ be the total ring of fractions of $R$ and $e$ the unity of $T$. We first show that the inverse ideal of $𝔞$ has the unique quotient presentation (http://planetmath.org/QuotientOfIdeals)  $[R^{\prime }:𝔞]$  where  $R^{\prime }:=R+\mathbb{Z}e$.  If ${𝔞}^{-1}$ is an inverse ideal of $𝔞$, it means that  $𝔞{𝔞}^{-1}=R^{\prime }$.  Therefore we have

$${𝔞}^{-1}\subseteq \{t\in T\mathrm{⋮}t𝔞\subseteq R^{\prime }\}=[R^{\prime }:𝔞],$$

so that

$$R^{\prime }=𝔞{𝔞}^{-1}\subseteq 𝔞[R^{\prime }:𝔞]\subseteq R^{\prime }.$$

This implies that  $𝔞{𝔞}^{-1}=𝔞[R^{\prime }:𝔞]$,  and because $𝔞$ is a cancellation ideal, it must that  ${𝔞}^{-1}=[R^{\prime }:𝔞]$, i.e. $[R^{\prime }:𝔞]$ is the unique inverse of the ideal $𝔞$.

Since  $𝔞[R^{\prime }:𝔞]=R^{\prime }$,  there exist some elements $a_1,\mathrm{\dots },a_n$ of $𝔞$ and the elements $b_1,\mathrm{\dots },b_n$ of  $[R^{\prime }:𝔞]$  such that  $a_1{b}_1+\mathrm{\cdots }+a_n{b}_n=e$.  Then an arbitrary element $a$ of $𝔞$ satisfies

$$a=a_1(b_{1}a)+\mathrm{\cdots }+a_n(b_{n}a)\in (a_1,\mathrm{\dots },a_n)$$

because every $b_{i}a$ belongs to the ring $R^{\prime }$.  Accordingly, $𝔞\subseteq (a_1,\mathrm{\dots },a_n)$.  Since the converse inclusion is apparent, we have seen that  $\{a_1,\mathrm{\dots },a_n\}$  is a finite of the invertible ideal $𝔞$.

Since the elements $b_i$ belong to the total ring of fractions of $R$, we can choose such a regular element $d$ of $R$ that each of the products $b_{i}d$ belongs to $R$.  Then

$$d=a_1(b_{1}d)+\mathrm{\cdots }+a_n(b_{n}d)\in (a_1,\mathrm{\dots },a_n)=𝔞,$$

and thus the fractional ideal $𝔞$ contains a regular element of $R$, which obviously is regular in $T$, too.