Gaussian Integers form Euclidean Domain/Proof 2

Theorem
Let $\left({\Z \left[{i}\right], +, \times}\right)$ be the integral domain of Gaussian Integers.

Let $\nu: \Z \left[{i}\right] \to \R$ be the real-valued function defined as:
 * $\forall a \in \Z \left[{i}\right]: \nu \left({a}\right) = \left \vert{a}\right \vert^2$

where $\left \vert{a}\right \vert$ is the (complex) modulus of $a$.

Then $\nu$ is a Euclidean valuation on $\Z \left[{i}\right]$.

Hence $\left({\Z \left[{i}\right], +, \times}\right)$ with $\nu: \Z \left[{i}\right] \to \Z$ forms a euclidean domain.

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 Modulus is Norm we have that:
 * $\forall a \in \C: \left \vert{a}\right \vert \ge 0$


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

Suppose $a = x + i y \in \Z \left[{i}\right] \backslash \left\{{ 0 }\right\}$.

Then either $x \ne 0$ or $y \ne 0$, so either $x^2 \ge 1$ or $y^2 \ge 1$.

So $\left \vert{a}\right \vert^2 \ge 1$.

Similarly, if $b \in \Z \left[{i}\right] \backslash \left\{{ 0 }\right\}$ then $\left \vert{b}\right \vert^2 \ge 1$.

Thus it follows that
 * $\nu\left({ab}\right) = \left \vert{a b}\right \vert^2 \ge \left \vert{a}\right \vert^2 = \nu\left({a}\right)$

Now, consider $x, y \in \Z \left[{i}\right]$ with $y \neq 0$.

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

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

Consider the complex number $p = \dfrac x y = p_r + ip_i$.

We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as:
 * $\forall z \in \C: \nu \left({z}\right) = \left \vert{z}\right \vert^2$

Let $q = q_r + iq_i$ lie be the Gaussian integer be such that:
 * $\nu\left({p - q}\right) = \left\vert p - q \right\vert^2 = \left({p_r - q_r}\right)^2 + \left({p_i - q_i}\right)^2$

is minimal.

That is, $q_r$ is the nearest integer to $p_r$ and $q_i$ is the nearest integer to $p_i$.

A given real number can be at a distance at most $1/2$ from an integer, so it follows that
 * $(1)\qquad\nu\left({p - q}\right) \leq \left({\dfrac 1 2}\right)^2 + \left({\dfrac 1 2}\right)^2 = \dfrac 1 2$

Now by Modulus is Norm, for any two complex numbers $z_1,z_2$ we have:
 * $\nu\left({z_1z_2}\right) = \nu\left({z_1}\right)\nu\left({z_2}\right)$

Thus we obtain

On the other hand,

So letting $r = x - yq$ we have $\nu\left({r}\right) < \nu\left({y}\right)$.

Moreover we trivially have $x = qy + r$.

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