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 $\mathfrak{a}$ has the unique quotient presentation $[R':\mathfrak{a}]$ , where $R' := R+\mathbb{Z}e$ If $\mathfrak{a}^{-1}$ is an inverse ideal of $\mathfrak{a}$ it means that $\mathfrak{aa}^{-1} = R'$ Therefore we have $$\mathfrak{a}^{-1} \subseteq \{t\in T\,\vdots \,\,\, t\mathfrak{a}\subseteq R'\} = [R'\!:\!\mathfrak{a}],$$ so that $$R' = \mathfrak{aa}^{-1} \subseteq \mathfrak{a}[R'\!:\!\mathfrak{a}] \subseteq R'.$$ This implies that $\mathfrak{aa}^{-1} = \mathfrak{a}[R'\!:\!\mathfrak{a}]$ and because $\mathfrak{a}$ is a cancellation ideal, it must mean that $\mathfrak{a}^{-1} = [R'\!:\!\mathfrak{a}]$ i.e. $[R'\!:\!\mathfrak{a}]$ is the unique inverse of the ideal $\mathfrak{a}$

Since $\mathfrak{a}[R'\!:\!\mathfrak{a}] = R'$ there exist some elements $a_1,\,\ldots,\,a_n$ of $\mathfrak{a}$ and the elements $b_1,\,\ldots,\,b_n$ of $[R'\!:\!\mathfrak{a}]$ , such that $a_1b_1\!+\cdots+\!a_nb_n = e$ Then an arbitrary element $a$ of $\mathfrak{a}$ satisfies $$a = a_1(b_1a)\!+\cdots+\!a_n(b_na) \in (a_1,\,\ldots,\,a_n)$$ because every $b_ia$ belongs to the ring $R'$ Accordingly, $\mathfrak{a} \subseteq (a_1,\,\ldots,\,a_n)$ Since the converse inclusion is apparent, we have seen that $\{a_1,\,\ldots,\,a_n\}$ , is a finite generator system of the invertible ideal $\mathfrak{a}$

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 $b_id$ belongs to $R$ Then $$d = a_1(b_1d)\!+\cdots+\!a_n(b_nd) \in (a_1,\,\ldots,\,a_n) = \mathfrak{a},$$ and thus the fractional ideal $\mathfrak{a}$ contains a regular element of $R$ which obviously is regular in $T$ too.

Bibliography

1

R. GILMER: Multiplicative ideal theory. Queens University Press. Kingston, Ontario (1968).