The form of inverse ideals in a GCD-domain

Question

In my lecture I encountered the following reasoning:

Let $R$ be a GCD-domain and let $I$ be an invertible ideal in $R$. Since $I$ is finitely generated as an $R$-module we can write $I=(a_1/b_1,...,a_n/b_n)R$, where $a_i, b_i$ are elements in $R$. (Recall that by hypothesis $R$ is also an LCM-domain.) Let $c$ be the least common multiple of the $b_i$'s, $d$ be the greatest common divisor of the $a_i$'s. We see that $I^{-1}=(c/d)R$.

I do not understand why $I^{-1}=(c/d)R$ should hold. I understand the trivial direction $I^{-1} \supseteq (c/d)R$, but I do not get why $I^{-1} \subseteq (c/d)R$ should be true. Could you please explain this to me?

Answer 1

As I remarked in a comment on the same deleted question last week, this is an immediate consequence of $\,\rm\color{#c00}U$ = universal gcd property for fractions over a gcd domain, viz.

$$\begin{align} &\!a/b \in (a_1/b_1,\ldots a_n/b_n)^{-1}\\[.2em] \iff\ & b/a\mid a_1/b_1,\ldots,a_n/b_n\\[.2em] \smash[t]{\overset{\rm\color{#c00}U}\iff}\ \ \, & b/a\mid d/c,\ \ d= \gcd\{a_i\}, c={\rm lcm}\{b_i\} \\[.2em] \iff\ &\!a/b \in (c/d) \end{align}\qquad$$