Gaussian Integers form Euclidean Domain/Proof 1

Proof
We have by definition that $\Z \left[{i}\right] \subseteq \C$.

Let $a, b \in \Z \left[{i}\right]$.

We have from Modulus of Product that $\left \vert{a}\right \vert \cdot \left \vert{b}\right \vert = \left \vert{a b}\right \vert$.

From Properties of Complex Modulus we have that:
 * $\forall a \in \C: \left \vert{a}\right \vert \ge 0$


 * $\left \vert{a}\right \vert = 0 \iff a = 0$

Let $a = x + i y$.

Suppose $a \in \Z \left[{i}\right] \ne 0$.

Then $x \ne 0$ or $y \ne 0$ and $x^2 \ge 1$ or $y^2 \ge 1$.

So $\left \vert{a}\right \vert \ge 1$.

Similarly, $b \in \Z \left[{i}\right], b \ne 0 \implies \left \vert{b}\right \vert \ge 1$.

Thus it follows that $\left \vert{a b}\right \vert \ge \left \vert{a}\right \vert$ and so $\nu$ is a Euclidean valuation on $\Z \left[{i}\right]$.

Now, consider $x, y \in \Z \left[{i}\right]$.

We want to find $q, r \in \Z \left[{i}\right]$ such that $x = q y + r$.

Note that this means we want $r = y \left({\dfrac x y - q}\right)$ where $\dfrac x y$ is complex but not necessarily Gaussian.

We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as:
 * $\forall z \in \C: \nu \left({z}\right) = \left \vert{z}\right \vert$

Then we have:
 * $\nu \left({r}\right) = \nu \left({y}\right) \cdot \nu \left({\dfrac x y - q}\right)$

Thus if $\nu \left({\dfrac x y - q}\right) < 1$ we have:
 * $\nu \left({r}\right) < \nu \left({y}\right)$

Consider the point $P = \dfrac x y$ as a point on the complex plane.

Let $Q$ lie at a point representing the Gaussian integer $q$ which lies closest to $P$.

The distance $P Q$ is at most half the length of a diagonal of a unit square in the complex plane.

Thus:
 * $\nu \left({\dfrac x y - q}\right) = \left \vert {\dfrac x y - q} \right \vert \le \dfrac {\sqrt 2} 2 = \dfrac 1 {\sqrt 2} < 1$

This element $q$ and the element $r$, where $\nu \left({r}\right) < \nu \left({y}\right)$, are the required values.

Thus $\Z \left[{i}\right]$ is a euclidean domain.