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?

Recall that $d$ is a greatest common divisor of $u$ and $v$ if $d$ divides both $u$ and $v$, and any common divisor of $u$ and $v$ divides $d$.
Note that $2$ and $1+\sqrt{-5}$ are common divisors of $6$ and $2(1+\sqrt{-5})$. Suppose that $6$ and $2(1+\sqrt{-5})$ have a greatest common divisor $d$. 
Then $2$ divides $d$, and therefore $d$ has shape $2(a+b\sqrt{-5})$ for some ordinary integers $a$ and $b$.  
Since $d$ is a common divisor of $6$ and $2(1+\sqrt{-5})$, it follows that $d$ must divide $6$. Thus $2(a+b\sqrt{-5})$ divides $6$, and therefore $a+b\sqrt{-5}$ divides $3$.
Let's try to find $\dfrac{3}{a+b\sqrt{-5}}$. By multiplying top and bottom by $a-b\sqrt{-5}$, we find that
$$\frac{3}{a+b\sqrt{-5}}=\frac{3a}{a^2+5b^2}-\frac{3b}{a^2+5b^2}\sqrt{-5}.$$
But $\dfrac{3b}{a^2+5b^2}$ can only be an integer if $b=0$. For if $b \ne 0$, then the denominator $a^2+5b^2$ has absolute value greater than $|3b|$. 
Thus $d=2a$. But $d$ divides $2(1+\sqrt{-5})$, so the ordinary integer $a$ divides $1+\sqrt{-5}$. Thus $a=\pm 1$.
So our only candidates for $d$ are $\pm 2$. It is straightforward to check that neither of these is divisible by $1+\sqrt{-5}$.  Just note that $\dfrac{2}{1+\sqrt{-5}}=\dfrac{2(1-\sqrt{-5})}{6}$.
Remark: As an exercise, we deliberately avoided mentioning the norm, though it does show up, unnamed, in the calculation.  But the norm is very important, and should be used.  
