Prove that $R=\mathbb{Z}[i]$ is a Euclidean domain via $N(a+bi)=a^2+b^2$
We want to show that $\forall a,b\neq 0\in\mathbb{Z}[i],\exists q,r\in\mathbb{Z}[i]$ such that $a=bq +r$ and $N(r) < N(b)$ OR $r=0.$ Let $x,y\in R, y\neq 0.$ Consider $z=xy^{-1} \in \mathbb{Q}(i).$ Take $w= p +qi\in R.$ Then since $z=w+(z-w),$ we have that $yz = yw + y(z-w)\Rightarrow x = yw+y(z-w)$ (by commutativity of $\mathbb{Z}[x].$ We want to show that $y(z-w)=0$ or $N(y(z-w))<N(y).$ Since $z\in \mathbb{Q}(i), z$ is of the form $a+bi,a,b\in\mathbb{Q}(i).$ Hence $|a-p|\leq \dfrac{1}{2}$ and $|b-q|\leq \dfrac{1}{2}.$ We have two cases: $z=w$ or $z\neq w.$ If $z=w,$ then $y(z-w)=0.$ If $z\neq w,$ then we have that $N(y(z-w))= N(y)N(z-w)=N(y)N((a+bi)-(p+qi))=N(y)[(a-p)^2 + (b-q)^2]\leq N(y)\cdot \dfrac{1}{2}\Rightarrow N(y) > N(y(z-w)),$ since $z\neq w$ and $y\neq 0$ so $N(y)$ is strictly positive.
Question: Why can't we use the division algorithm here? Is it similar to assuming the hypothesis?
Also, why can one claim that $N(y(z-w)) = N(y)N(z-w)$ when $y,z,w$ are in different rings? Don't they have to be in the same ring?
Finally, I don't understand why this proof works. Why must it be the case that $(a-p)^2+(b-q)^2 \leq \dfrac{1}{2}$? I know there's a pair $(a,b)$ for every $(p,q)$ that satisfies this, but aren't $p$ and $q$ arbitrary?
I feel like the proof actually chooses the closest point to $z=xy^{-1}$ as $w=p+qi$ in the beginning.
We can claim that $N(y(z-w)) = N(y)N(z-w)$ since $N$ can be seen as the square of the norm of complex number and $N(ab)=N(a)N(b)$ holds for any $a,b\in \mathbb C$.