Show that $\mathbb Z[\sqrt 2i]=\{a+bi\sqrt2|a,b \in \mathbb Z\}$, with a function $V: \mathbb Z[\sqrt 2i]/\{0\} \rightarrow \mathbb N$ defined $V(a+bi\sqrt2)=|a^2+2b^2|$ is a Euclidian domain.
I managed to prove that $V((a+b\sqrt2i)(c+d\sqrt2i))=V(a+b\sqrt2i)\cdot V(c+d\sqrt2i)$. I didn't manage to prove however EF1. Any help or hints would be appriciated!
Thank you!
Outline: It looks as if you are using $V(a+b\sqrt{-2})=a^2+2b^2$, so you are calling the norm by the name $V$.
We need to show that if $\alpha$ and $\beta$ are in our ring, with $\alpha\ne 0$, then there exist objects $\delta$ and $\rho$ in our ring such that $$\beta=\delta\alpha+\rho,$$ where $V(\rho)\lt V(\alpha)$ (the "remainder" $\rho$ has smaller norm than the divisor $\alpha$).
Divide $\beta$ by $\alpha$. By rationalizing the denominator, we can write $$\frac{\beta}{\alpha}=x+y\sqrt{-2},$$ where $x$ and $y$ are rational. Let $u$ and $v$ respectively be integers nearest to $x$ and $y$. So $|x-u|\le 1/2$ and $|y-v|\le 1/2$.
Let $\delta=u+v\sqrt{-2}$, and let $\rho=\beta-\delta\alpha$. Then $$\rho=\left(x-u +(y-v)\sqrt{-2} \right)\alpha.$$ To show that $V(rho)\lt V(\alpha)$, it is enough to show that $x-u+(y-v)\sqrt{-2}$ has norm less than $1$. But the norm of $x-u+(y-v)\sqrt{-2}$ is $\le |x-u|+2|y-v|$, and by the choice of $u$ and $v$ this is $\le 3/4$.