Let $R$ be UFD and denote. Show that if If c|a and c|b then $cgcd(a/c, b/c)=gcd(a, b)$.
My attempt.
Let $gcd(\frac{a}{c},\frac{b}{c})=d$. First need to show $(a,b) \subset (cd)$. This is clear since if $x \in (a,b)$ then; $$x=ar+bs=c(a/c)t+c(b/c)s$$
and since $d$ divides both $a/c$ and $b/c$ we are done.
Now need to show that if $(u)$ is another princ. ideal with $(a,b) \subset (u)$ then $(cd) \subset (u)$.
this is also immediate since if $y \in (cd)$ then for some $s,t \in R$ we have $(a/c)v + (b/c)w=d$ and ; $$y=cdr=c[(a/c)v + (b/c)w]r=avr+bwr$$. Which seems to show $(a,b)=(cd)$ but this can't possibly be true;
The Bezout identity only necessarily works in Bezout domains, which includes all PIDs, but not all UFDs. So the error is when you write $d$ as a linear combination of $a/c$ and $b/c$.
For example, $\mathbb{Z}[X,Y]$ is a UFD in which $gcd(X,Y)=1$. But we can't write $1$ as a linear combination of $X$ and $Y$.
Edit: Here is a sketch for the correct proof of that direction. Let $r=gcd(a,b)$ (mod units). It suffices to show $r$ divides $cd$. Note that $c$ divides $r$ by assumption, so it suffices to show $r/c$ divides $d$. By definition of $d$, it suffices to show $r/c$ divides $a/c$ and $b/c$. But this follows from the general property that if $x$ divides $y$, and $z$ divides both $x$ and $y$, then $x/z$ divides $y/z$.
In fact, this result holds in any GCD domain.