Gcd of two elements

Question

Consider the ring $\mathbb{Z}[\sqrt{2}]$. I need to find $\gcd(4, 6)$.

My try

Let $N$ be norm function defined on $\mathbb{Z}[\sqrt{2}]$ and $d$ be proper divisor of $4$ and $6$ then $d$ can't be unit and associate of $4$ and $6$. Then by property of norm function $N(d)$ divides $N(4)N(6)$. Since $N(4)= 16$ and $N(6)=36$ then $N(d)$ divides $gcd(16,36)$. Finally, N(d) divides $4$. Thus we have $N(d) =1,2$ or $4$. Now $N(d)$ can't be $1$ as $d$ is nonunit then we have the choice left for $N(d)$ is $2$ or $4$. I don't know how to proceed from here. I am totally stuck from yesterday.

Answer 1

Let $\delta$ be a gcd of $4$ and $6$ in $\mathbb Z[\sqrt 2]$.

Then $2$ divides $\delta$ because $2$ divides $4$ and $6$.

Write $\delta=2\alpha$. Then $N(\delta)=4N(\alpha)$ and so $N(\delta)\ge 4$.

You have already proved that $N(\delta) \le 4$. So $N(\delta)=4$.

But then $N(\alpha)=1$ and $\alpha$ is a unit.

Bottom line: $2$ and $\delta$ are associates and so $\gcd(4,6)=\delta \sim 2$.

Answer 2

$\gcd(6,4) = 2$ in every ring since $\,2\mid 6,4$ and $\,d\,\mid 6,4\,\Rightarrow\, d\mid 2=6-4$.

More generally gcds that are (Bezout) linear persist in extension rings or R-algebras / R-modules.