An example where GCD depends on the domain - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:50:56Z http://mathoverflow.net/feeds/question/16899 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16899/an-example-where-gcd-depends-on-the-domain An example where GCD depends on the domain Alfonso Gracia-Saz 2010-03-02T19:18:27Z 2010-03-03T02:35:42Z <p>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 <I>one</I> gcd of $a$ and $b$ in $R$.</p> <p>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 $x \neq gcd(a,b)_S$.</p> <p>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.</p> <p>ETA: As suggested below, I also wanted $S$ to be a GCD-domain.</p> http://mathoverflow.net/questions/16899/an-example-where-gcd-depends-on-the-domain/16900#16900 Answer by t3suji for An example where GCD depends on the domain t3suji 2010-03-02T19:35:57Z 2010-03-03T02:35:42Z <p>(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?</p> <p>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 $S$.</p> http://mathoverflow.net/questions/16899/an-example-where-gcd-depends-on-the-domain/16931#16931 Answer by Arturo Magidin for An example where GCD depends on the domain Arturo Magidin 2010-03-02T22:23:15Z 2010-03-02T22:23:15Z <p>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 $S$. </p>