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 $Z[√(-5)]$

Question

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.

Answer 1

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 $N(\gamma)\neq N(\delta)$.