Gaussian Integers form Euclidean Domain/Proof 2

Proof
We have by definition that $\Z \sqbrk i \subseteq \C$.

Let $a, b \in \Z \sqbrk i$.

We have from Modulus of Product that:


 * $\cmod a \cdot \cmod b = \cmod {a b}$

From Complex Modulus is Norm we have that:


 * $\forall a \in \C: \cmod a \ge 0$


 * $\cmod a = 0 \iff a = 0$

Suppose $a = x + i y \in \Z \sqbrk i$ and $a \ne 0$.

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

So $\cmod a^2 \ge 1$.

If also $b \in \Z \sqbrk i, b \ne 0$, then:


 * $\map \nu {a b} = \cmod {a b}^2 \ge \cmod a^2 = \map \nu a$

Now, consider $x, y \in Z \sqbrk i$ with $y \ne 0$.

We want to find $q, r \in Z \sqbrk i$ such that $x = q y + r$ and $\map \nu r < \map \nu y$.

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

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

We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as:
 * $\forall z \in \C: \map \nu z = \cmod z^2$

Let $q = q_r + i q_i$ be the Gaussian integer such that:


 * $\map \nu {p - q} = \cmod {p - q}^2 = \paren {p_r - q_r}^2 + \paren {p_i - q_i}^2$

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:
 * $(1): \quad \map \nu {p - q} \le \paren {\dfrac 1 2}^2 + \paren {\dfrac 1 2}^2 = \dfrac 1 2$

Now by Complex Modulus is Norm, for any two complex numbers $z_1, z_2$ we have:
 * $\map \nu {z_1 z_2} = \map \nu {z_1} \, \map \nu {z_2}$

Thus we obtain:

On the other hand:

So letting $r = x - y q$ we have $\map \nu r < \map \nu y$.

Moreover we trivially have $x = q y + r$.

Thus $\Z \sqbrk i$ is a Euclidean domain.