Let $R$ be unique factorization domain (UFD) and we know that UFD is GCD domain. And I am going to prove the following fact: the gcd in UFD exists up to units. In other words, if $a,b\in R$ and $d, d'$ their gcd then $d$ associates with $d'$.
Indeed, from the definition of gcd we get that $d\mid d'$ and $d'\mid d$, so $d'=dx$ and $d=d'y$ so we derive the following: $d'=d'xy$ or $d'(xy-1)=0$.
The case when $d'\neq 0$ is easy since it follows that $xy-1=0$ since we are working in integral domain.
But what if $d'=0$? this case corresponds to the case when $a=b=0$. Equivalently, if $a=b=0\in R$ how to show that any two gcd of $0$ and $0$ associates? I know that $\text{gcd}(0,0)=0$.
Best Answer
Well, if $d'=0$, then $d=0$ by the fact that $d'\mid d$. So $d'=d$ and they are associates a fortiori. In other words, the pair $(0,0)$ has exactly one greatest common divisor, which is $0$.
The difference between the pair $(0,0)$ as opposed to all the others is that in all other cases, if $d,d'\in\operatorname{gcd}(a,b)$ and $x$ is such that $d'=xd$, then $x\in R^*$. For $\operatorname{gcd}(0,0)$, $d'$ may be written as $rd$ for some $r\notin R^*$; however, this doesn't change the fact that there is some $x\in R^*$ such that $d'=xd$.