# Gaussian Integers form Euclidean Domain/Proof 1

## Theorem

Let $\struct {\Z \sqbrk i, +, \times}$ be the integral domain of Gaussian Integers.

Let $\nu: \Z \sqbrk i \to \R$ be the real-valued function defined as:

$\forall a \in \Z \sqbrk i: \map \nu a = \cmod a^2$

where $\cmod a$ is the (complex) modulus of $a$.

Then $\nu$ is a Euclidean valuation on $\Z \sqbrk i$.

Hence $\struct {\Z \sqbrk i, +, \times}$ with $\nu: \Z \sqbrk i \to \Z$ forms a Euclidean domain.

## 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 $\cmod a \cdot \cmod b = \cmod {a b}$.

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

and:

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

Let $a = x + i y$.

Suppose $a \in \Z \sqbrk i \ne 0$.

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

So:

$\cmod a \ge 1$

Similarly, $b \in \Z \sqbrk i, b \ne 0 \implies \cmod b \ge 1$.

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

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

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

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

We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as:

$\forall z \in \C: \map \nu z = \cmod z$

Then we have:

$\map \nu r = \map \nu y \cdot \map \nu {\dfrac x y - q}$

Thus if $\map \nu {\dfrac x y - q} < 1$ we have:

$\map \nu r < \map \nu y$

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:

$\map \nu {\dfrac x y - q} = \cmod {\dfrac x y - q} \le \dfrac {\sqrt 2} 2 = \dfrac 1 {\sqrt 2} < 1$

This element $q$ and the element $r$, where $\map \nu r < \map \nu y$, are the required values.

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

$\blacksquare$