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

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 $\Z \left[{i}\right]$.

This needs considerable tedious hard slog to complete it.definition of Euclidean valuation requires codomain $\N$, consider modulus squared insteadTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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.

$\blacksquare$

Although this article appears correct, it's inelegant. There has to be a better way of doing it.You can help Proof Wiki by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Example $52$