Given a GCD domain $D$:
$D$ is a GCD domain if for every two elements $a,b\in D$, $\exists d \in D\setminus \{0_D\} $ such that:
(GCD.I) $d\mid a$ and $d\mid b$
(GCD.II) If $\exists c\in D\setminus \{0_D\}$ such that $c\mid a$ and $c\mid b$ then $c\mid d$
Note that all not all GDC domains are UFD
I have defined $A:=\{t\in D\setminus \{0_D\} : \exists n\in D$ such that $nt=a \}$ the set with every divisor of a
This same definitions works for the set $AB$, the set that has every divisor of $a \cdot b$.
What I want to prove is that the set
$AB=\{t\in D\setminus \{0_D\} : \exists n\in D$ such that $nt=ab \}$
and the set
$AB^*=\{ x\cdot y : x\in A, y \in B \}$
Are the same set.
To do this I've first proven that $AB^* \subset AB$. This is easy as for every $xy \in AB^*, \exists n$ such that $xn=a$ and $\exists m$ such that $ym=b$ therefore $\exists p=mn \in D$ such that $xyp=ab$ so $AB^* \subset AB$.
However, to prove that $AB \subset AB^*$ I'm missing one detail. This is what I've done so dar.
$t\in AB \Rightarrow \exists n\in D$ such that $tn=ab$
If $t\in A$, then $\exists 1_D \in B$ such that $t\cdot 1_D=t \Rightarrow t\in AB^*$
Same logic applies when $t \in B$
However I'm missing the case when $t$ is neither in $A$ or $B$, when $t$ isn't a divisor of $a$ or $b$.
I'd like some help with this last part of the problem as I'm now considering that this might not even be true as it's being so hard to prove.
Thanks in advance
