# Gaussian Integers form Principal Ideal Domain

## Theorem

The ring of Gaussian integers:

- $\struct {\Z \sqbrk i, +, \times}$

forms a principal ideal domain.

## Proof

From Gaussian Integers form Integral Domain, we have that $\struct {\Z \sqbrk i, +, \times}$ is an integral domain.

Let $a, d \in \Z \sqbrk i$ such that $d \ne 0$.

Suppose $\cmod a \ge \cmod d$.

Reference to an Argand diagram shows that one of:

- $a + d, a - d, a + i d, a - i d$

is closer to the origin than $a$ is.

So it is possible to subtract Gaussian integer multiples of $d$ from $a$ until the square of the modulus of the remainder drops below $\cmod d^2$.

That remainder can only take integer values.

Thus a Division Theorem result follows:

- $\exists q, r \in \Z \sqbrk i: a = q d + r$

where $\cmod r < \cmod d$.

Let $J$ be an arbitrary non-null ideal of $\Z \sqbrk i$.

Let $d$ be an element of minimum modulus in $J$.

Then the Division Theorem can be used to prove that $J = \ideal d$.

This needs considerable tedious hard slog to complete it.In particular: The above is the outline only.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 `{{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. |

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $9$: Rings: Exercise $23$