First some notation. Given a domain $R$ and $x,a,b \in R$, I write $x=gcd(a,b)\_R$ to mean that $x$ is *one* gcd of $a$ and $b$ in $R$.


I want to find an example of an GCD-domain $R$, a subdomain $S \subseteq R$, and two elements $a, b \in S$ such that there isn't any $x \in S$ such that $x=gcd(a,b)\_R$ and $x=gcd(a,b)\_S$. Notice that it is not enough to find one element $x \in S$ such that $x=gcd(a,b)\_R$ but $x \neq gcd(a,b)\_S$.


I can prove that this is impossible in as little as a Bezout domain, but I cannot prove that this is impossible in a mere GCD-domain. I do not know that many examples of GCD-domains which are not Bezout domains in the first place.


ETA: As suggested below, I also wanted $S$ to be a GCD-domain.



An example where GCD depends on the domain
(Edit: first version was about lcm rather than gcd).
Take $R=k[u,v,w]$, $a=uv$, $b=vw$. Then $gcd\_R(a,b)=v$ (times constant). Now let $S=k[a,b]$. Since $a$ and $b$ are independent, $gcd\_S(a,b)=1$ (times constant). Right?


Edit: here's an even simpler example: $R=k[u,v]$, $a=u$, $b=uv$, $S=k[a,b]$. Then $a|b$ in $R$, but $a$ and $b$ are both irreducible in $S$.


