Showing DO NOT exist gcd of $6$ and $2+2 \sqrt{-5}$ in $\Bbb Z[\sqrt{-5}]$.
I tried it.
Suppose $d$ is GCD of $6$ and $2+2 \sqrt(-5)$. then
there exist $x,y \in \Bbb Z[\sqrt{-5}]$ s.t $6=dx$ and $2+2 \sqrt{-5}=dy$.
(1) $N(6)=36=N(d)N(x)$ and (2) $N(2+2\sqrt{-5})=24=N(d)N(y)$.
by (1), $N(d)=1$ or $2$ or $3$ or $4$ or $6$ or $12$.
But for $a+b\sqrt{-5} \in \Bbb Z[\sqrt{-5}]$, $N(a+b\sqrt{-5})=a^2+5b^2$
then possible case is $N(d)=1$ or $4$ or $6$.
Similar way, I find the case!
When $N(d)=4$, $N(x)=9$ and $N(y)=6$.
when $N(d)=6$, $N(x)=6$ and $N(y)=4$.
therefore $d=\pm2$ or $d=\pm1\pm\sqrt{-5}$.
But GCD of elements is greatest of the common divisors,
$d=\pm2$ and $d=\pm1\pm\sqrt{-5}$ is not associate.
so, DNE.
if associate, then there exist $r \in \Bbb Z\sqrt{-5}$ s.t $N(\pm2)=N(r)N(\pm1\pm\sqrt{-5})$
But $4=6N(r)$ so, Contradiction. (Since $N(r=s+t\sqrt{-5})=s^2+5t^2$)
it is my proof. is it ok?
Thanks.