$gcd(a,b)$ in a UFD subring is not a greatest common divisor in the ring

211 Views Asked by At

Give a counterexample that $R$ is a unique factorization domain but not a principal ideal domain, $S$ is a ring containing $R$, such that $a,b\in R$, $gcd(a,b)$ in $R$ is not a greatest common divisor of $a,b$ in $S$.

1

There are 1 best solutions below

0
On BEST ANSWER

$\gcd(2,x)=1$ in $\,\Bbb Z[x],$ but the gcd $\,= 2\,$ in $\,\Bbb Z[x/2]$