I'm studying Bézout domains, invertible ideals and related topics so I decided to make some searching on the Internet. Looking at Wikipedia page I found this result: "Bézout domain iff Prüfer domain and GCD-domain", but there is no proof available so I tried to prove this result.
First of all, I'm using the following definition of invertible ideal as is given in Rotman's book "An Introduction to Homological Algebra", namely we say that an ideal $I$ in an integral domain $D$ with field of fractions $K$ is an invertible ideal if there are $a_1,\ldots a_n\in I$ and $q_1,\ldots, q_n\in K$ such that $q_iI\subseteq D$ for $1\le i\le n$ and $$1=\sum_{i=1}^n q_ia_i.$$
Using the above definition is easy to show that a non-zero principal ideal is invertible and thus a Bézout domain is Prüfer and since it's well-known that every Bézout domain is a GCD-domain, we have proved one implication.
For the other implication this is my approach: given an non-zero f.g. ideal $I$, as $D$ is Prüfer then $I$ is invertible, so there are $a_1,\ldots, a_n\in I$ and $q_1,\ldots, q_n\in K$ such that $q_iI\subseteq D$ and $$1=\sum_{i=1}^n q_ia_i \;\;...\; (*)$$
From the above equation is easy to show that $I=\langle a_1,\ldots, a_n\rangle$. Since $D$ is also a GCD-domain, then $d=\gcd(a_1,\ldots, a_n)$ exists. I claim that $I=\langle d\rangle$. Indeed, as $d\mid a_i$ it follows that $a_i\in \langle d\rangle$ and from this we deduce that $I\subseteq \langle d\rangle$.
On the other hand, let's set $q_i=p_i/s_i$ such that $\gcd(p_i,s_i)=1$ (this is possible because $D$ is a GCD-domain). As $q_iI\subseteq D$, then $(p_i/s_i)a_i\in D$ so this means that $s_i\mid p_ia_i$, but since $p_i$ and $s_i$ are coprime, it follows that $s_i\mid a_i$ for every $1\le i\le n$, thus by the definition of $\gcd$ we deduce that $s_i\mid d$ for every $1\le i\le n$.
Finally, multiplying $(*)$ by $d$ gives us $$d=d\Bigr(\sum_{i=1}^n \frac{p_i}{s_i}a_i\Bigl)=\sum_{i=1}^n \Bigl(\frac{d}{s_i}\Bigr)p_ia_i.$$
Since $s_i\mid d$, then $d\in \langle a_1,\ldots, a_n\rangle=I$. Hence $\langle d\rangle\subseteq I$ and we're done.
My question is this: is the above proof right? I'm aware that the definition of invertible ideal that I'm using is not the most standard one, since usually this definition is tied with the one of fractional ideal, hence my concern about the rightness of my proof.
Thanks in advance for further answers and/or comments.
