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 $(r,s) = (2,0)$ or $(1,1)$. So the only possibilities for the greatest common divisor are $2$ and $1 + \sqrt{5}i$. 

To see $2$ can't be the 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 $2$ has norm $4$, $2$ can't be the gcd: $6$ does not divide $4$.

To see $1 + \sqrt{5}i$ can't be the gcd: Since $2$ divides $6$ and $2(1 + \sqrt{5}i)$, $2$ divides the gcd. If $1 + \sqrt{5}i$ were the gcd, then $N(2) = 4$ would have to divide $N(1 + \sqrt{5}i) = 6$, again a contradiction.

Thus we see both potential gcd's don't work: we conclude that $6$ and $2(1 + \sqrt{5}i)$ have no gcd.
First at all, $6$ and $2(1+\sqrt{-5})$ are in $\mathbb Z[\sqrt{-5}]$, then their sum are too becuse $\mathbb Z[\sqrt{-5}]$ is a ring.

Hint: you must prove that the ideal $\mathfrak a=\langle 6,2(1+\sqrt{-5})\rangle$ generated by $6$ and $2(1+\sqrt{-5})$ is not principal.