Proving that $\alpha\gamma$ and $\beta\gamma$ have no gcd where $\alpha=3$ , $\beta=(1+2\sqrt{-5})$ and $\gamma=7(1+2\sqrt{-5})$in ring $Z[√(-5)]$ I have the following question before me :
$\alpha=3$ and $\beta=1+2\sqrt{-5}$ are two elements of ring $R=Z[\sqrt{-5}]$.
$\gamma=7(1+2\sqrt{-5})$ is another element of this ring.
I have to prove that elements $\alpha\gamma$ and $\beta\gamma$ have no gcd in $R$.
In this case, it turns out that $\alpha$ and $\beta$ are relatively prime.
This exact question has been asked before as well. But that was way back in $2013$. And my query on that page went unanswered. That's why I have asked a new question here.
One of the comments on that page says that potential candidates for gcd of $\alpha\gamma$ and $\beta\gamma$ are $1$,$-1$ (units in $R$ ),$7$ and $(1+2\sqrt{-5})$.
I could not quite understand why $\gamma$ is not a potential candidate for gcd here as $\gamma$ divides both $\alpha\gamma$ and $\beta\gamma$
Infact $\gamma$ is divisible by all the common factors of $\alpha\gamma$ and $\beta\gamma$. Then why is $\gamma$ not regarded as gcd here?
Please help me out regarding this.

You are right that $\gamma=7(1+2\sqrt{-5})$ is a common divisor. If it were the greatest common divisor then any other common divisor $\delta$ would have to divide it, $\delta\mid \gamma$. Now consider the element $\delta=(1+2\sqrt{-5})^2$.
This is a common divisor of $\gamma\alpha$ and $\gamma\beta$, as

*

*$\gamma\alpha=3\cdot 7(1+2\sqrt{-5})=(1-2\sqrt{-5})(1+\sqrt{-5})^2$

*$\gamma\beta=7\cdot (1+2\sqrt{-5})^2$
Now, lets check if $\delta\mid\gamma$ or $\gamma\mid\delta$, that is if one divides the other. We have:

*

*$\frac{(1+2\sqrt{-5})^2}{7(1+2\sqrt{-5})}=\frac{1+2\sqrt{-5}}{7}\not\in\Bbb Z[\sqrt{-5}]$


*$\frac{7(1+2\sqrt{-5})}{(1+2\sqrt{-5})^2}=\frac{1-2\sqrt{-5}}{3}\not\in\Bbb Z[\sqrt{-5}]$
Also, $\gamma,\delta$ are not associated, as $N(\gamma)\neq N(\delta)$.
