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?
 A: 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.
A: One way you can do this is using some Norm arguments.
Now first start by writing 6 as the difference of two squares: 
$$6 = (1 + i\sqrt{5})(1 - i\sqrt{5})$$
Now write suppose the gcd of $6$ and $2(1 + \sqrt{-5})$ is some number $d$. Write $x = 6$ and $y = 2(1 + \sqrt{5})$ and write $x = ad$ and $y = bd$.
Now by standard properties of the norm (which you should be familiar with), we have that $N(x) = 36 = N(a)N(d)$ and $N(y) = 24 = N(b)N(d)$.
Hence $N(d)|36$ and $N(d)|24$ so that $N(d) = 1,2,3,4,6$ or $12$. This is one key ingredient you need to derive the contradiction because $N(d)$ is a positive integer and there are very few (as I have listed above) that satisfy the above condition.
Now I leave the rest for you to do:


(1) What positive integer divides both $x$ and $y$?
(2) Prove that $a|b$ in $\mathbb{Z}[\sqrt{-5}]$ iff $N(a)|N(b)$.
(3) Deduce that $12|N(d)$ (Hint: There is a complex number that divides both $x$ and $y$).
(4) Obtain the desired contradiction.


$\textbf{Edit:}$ emiliocba has a nice idea. If $\mathbb{Z}[\sqrt{-5}]$ is an Euclidean domain it is a PID. Hence if you can show that the ideal $\mathfrak{a}$ that he has written above is not principal you have the desired contradiction.
A: 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.  
A: 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.
