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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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  • $\begingroup$ The way I understand your question, it would suffice to take any subdomain of a GCD-domain that is not integrally closed, for then it cannot be a GCD-domain. But this applies equally well to Bezout domains like Z[\sqrt{-1}] (take Z[2\sqrt{-1}]), so I don't understand that part of your question. Am I missing something? $\endgroup$
    – Pete L. Clark
    Commented Mar 2, 2010 at 19:30
  • $\begingroup$ Maybe you want that $S$ is also a GCD-domain? $\endgroup$
    – Pete L. Clark
    Commented Mar 2, 2010 at 19:31
  • $\begingroup$ A fantastic modern discovery is the notion of set. It could allow to use the notation $x\in gcd(a,b)_R$ instead of $x=gcd(a,b)_R$, which gives rise to the somewhat unpleasant $1= gcd(2,3)_{\mathbb{Z}}=-1$... $\endgroup$
    – YCor
    Commented Oct 29, 2014 at 16:28

3 Answers 3

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(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 $S$.

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  • $\begingroup$ You're computing the lcm and not the gcd. But the example works for these too: $gcd_R(a,b)=v$ and $gcd_S(a,b)=1$. $\endgroup$
    – Johannes Hahn
    Commented Mar 2, 2010 at 19:48
  • $\begingroup$ Duh. That seems to work. Is $S$ a GCD-domain as well? $\endgroup$
    – Alfonso Gracia-Saz
    Commented Mar 2, 2010 at 20:18
  • $\begingroup$ Isn't S isomorphic to k[x,y]? (in which case it is even a UFD) $\endgroup$
    – Paul Yuryev
    Commented Mar 2, 2010 at 20:59
  • $\begingroup$ @Johannes Hahn: Oops, I just hate these abbreviations :) Thanks, I'll fix it. $\endgroup$
    – t3suji
    Commented Mar 2, 2010 at 22:39
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If you do not require $S$ to be a GCD domain, then here's a simple example: let $R=\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, and let $S=\mathbb{Z}[\sqrt{-3}]$. Then $R$ is a GCD domain (in fact, a PID). Let $a=2$, $b=1+\sqrt{-3}$. Since $a|b$ in $R$, any gcd of $a$ and $b$ must be an associate of $2$. However, in $S$ the only common divisors of $a$ and $b$ are $1$ and $-1$, so no gcd of $a$ and $b$ in $R$ can be a common divisor of $a$ and $b$ in $S$.

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Take $S=\mathbb{Z}[X]$ and $R=\mathbb{Z}+X\mathbb{Q}[X]$. Then $gcd_S (2,X)=1$ and $gcd_R(2,X)=2$, where both $R$ and $S$ are GCD-domains.

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