# Gaussian Integers form Euclidean Domain

## 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 1

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 Non-Negative:

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

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.

$\blacksquare$

## Proof 2

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:

\(\ds \map \nu {y \paren {p - q} }\) | \(=\) | \(\ds \map \nu y \, \map \nu {p - q}\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \dfrac {\map \nu y} 2\) | by $(1)$ | |||||||||||

\(\ds \) | \(<\) | \(\ds \map \nu y\) | since $\map \nu y \ne 0$ |

On the other hand:

\(\ds \map \nu {y \paren {p - q} }\) | \(=\) | \(\ds \map \nu {y \paren {\frac x y - q} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map \nu {x - y q}\) |

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.

$\blacksquare$