A question about proving that there is no greatest common divisor I have to answer this question:
Prove that in the ring $\mathbb Z[\sqrt{-5}]$ there's no gcd to $6$ and $2\cdot (1+\sqrt{-5})$.
I have no clue how to do this but however I've tried to prove that their sum $$6+2\cdot (1+\sqrt{-5}) = 8+2\cdot\sqrt{-5}$$ is not in $\mathbb Z[\sqrt{-5}]$ but I was not able to do this.
How should I proceed?

I'm assuming a greatest common divisor of $a$ and $b$ means something that divides both $a$ and $b$, such that every divisor of $a$ and $b$ divides the gcd.
Use the norm function $N(r + s\sqrt{5}i) = r^2 + 5s^2$, and show that $N(ab) = N(a)N(b)$.
This leads to the important fact that if $c$ divides $d$ in the ring, then $N(c)$ divides $N(d)$ as integers. This places restrictions on the possible norms of a gcd $r + s\sqrt{5}i$: you need $r^2 + 5s^2$ to divide $N(2(1 + \sqrt{5}i)) = 24$ and $N(6) = 36$. Thus  $r^2 + 5s^2$ divides $12$. Going through the possibilities, this can only happen if $(r,s) = (\pm2,0)$, $(1,1)$, $(-1,-1)$, and $(\pm 1,0)$. So the only possibilities for the greatest common divisor are $\pm 2$, $\pm(1 + \sqrt{5}i)$, and $\pm 1$.
To see $\pm 2$ can't be a gcd: Note that $6 = (1 + \sqrt{5}i)(1 - \sqrt{5}i)$. So $1 + \sqrt{5}i$ divides both $6$ and $2(1 + \sqrt{5}i)$. Thus it would have to divide the gcd. Since $1 + \sqrt{5}i$ has norm $6$ and $\pm 2$ has norm $4$, $\pm 2$ can't be  gcd: $6$ does not divide $4$.
To see $\pm(1 + \sqrt{5}i)$ can't be gcd: Since $2$ divides $6$ and $2(1 + \sqrt{5}i)$, $2$ divides the gcd. If $\pm(1 + \sqrt{5}i)$ were the gcd, then $N(\pm2) = 4$ would have to divide $N(1 + \sqrt{5}i) = 6$, again a contradiction.
Lastly, we consider the cases $\pm 1$. In this case, since $2$ divides the gcd, one would have to have that $N(2)$ divides $N(\pm 1) = 1$, or that $4$ divides $1$. Once again we have a contradiction.
Thus we see all potential gcd's don't work: we conclude that $6$ and $2(1 + \sqrt{5}i)$ have no gcd.
