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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.

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# An example where GCD depends on the domain

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.