I need to show that there is no greatest common divisor for $4$ and $2 \cdot(1 + \sqrt{-3})$ in the ring $R := \{ x + y \cdot\sqrt{-3} \,\,|\,\, x, y \in \mathbb{Z}\}$.
I found that both $2$ and $(1 + \sqrt{-3})$ divide $4$ as well as $2 \cdot(1 + \sqrt{-3})$. Therefore if $x$ was a greatest common divisor of the two, it has to be divisible by both $2$ and $(1 + \sqrt{-3})$.
How can I proceed from here?
Let the common divisor of $4$ and $2(1+\sqrt{-3})$ be $x+y\sqrt{-3}$. Then we have \begin{align} (x+y\sqrt{-3})(a+b\sqrt{-3})&=4,\\ (x+y\sqrt{-3})(c+d\sqrt{-3})&=2(1+\sqrt{-3}), \end{align} where $x,y,a,b,c,d\in\mathbb{Z}$. Multiplying $x-y\sqrt{-3}$ on both sides of the equations, one obtains \begin{align} a&=\frac{4x}{x^2+3y^2},\quad b=-\frac{4y}{x^2+3y^2},\\ c&=\frac{2(x+3y)}{x^2+3y^2},\quad d=\frac{2(x-y)}{x^2+3y^2}. \end{align} The denominators have higher degrees than the numerators, so the number of integer pairs of $(x,y)$ that makes $|a|\geq1$ or $|b|\geq 1$ or $|c|\geq 1$ or $|d|\geq 1$ is finite. One can therefore enumerate all such cases. After ruling out them all, the only possibility left is $a=b=c=d=0$, which cannot hold either (not allowed to have $x=y=0$ as they appear in the denominators), thus completing the proof.