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.

  • $\begingroup$ 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? $\endgroup$ Mar 2, 2010 at 19:30
  • $\begingroup$ Maybe you want that $S$ is also a GCD-domain? $\endgroup$ 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
    Oct 29, 2014 at 16:28

3 Answers 3


(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 $S$.

  • $\begingroup$ You're computing the lcm and not the gcd. But the example works for these too: $gcd_R(a,b)=v$ and $gcd_S(a,b)=1$. $\endgroup$ Mar 2, 2010 at 19:48
  • $\begingroup$ Duh. That seems to work. Is $S$ a GCD-domain as well? $\endgroup$ Mar 2, 2010 at 20:18
  • $\begingroup$ Isn't S isomorphic to k[x,y]? (in which case it is even a UFD) $\endgroup$ Mar 2, 2010 at 20:59
  • $\begingroup$ @Johannes Hahn: Oops, I just hate these abbreviations :) Thanks, I'll fix it. $\endgroup$
    – t3suji
    Mar 2, 2010 at 22:39

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 $S$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.