# Euclidean Domain is Principal Ideal Domain

## Theorem

A Euclidean domain is a principal ideal domain.

## Proof

Let $\left({D, +, \times}\right)$ be a Euclidean domain whose zero is $0$ and whose Euclidean valuation is $\nu$.

We need to show that every ideal of $\left({D, +, \times}\right)$ is a principal ideal.

Let $U$ be an ideal of $\left({D, +, \times}\right)$ such that $U \ne \left\{{0}\right\}$.

Let $d \in U$ such that $d \ne 0$ and $\nu \left({d}\right)$ is as small as possible for elements of $U$.

By definition, $\nu$ is defined as $\nu : D \setminus \left\{{0_R}\right\} \to \N$, so the codomain of $\nu$ is a subset of the natural numbers.

By the Well-Ordering Principle, such an element $d$ exists as an element of the preimage of the least member of the image of $U$.

Let $a \in U$.

Let us write $a = d q + r$ where either $r = 0$ or $\nu \left({r}\right) < \nu \left({d}\right)$.

Then $r = a - d q$ and so $r \in U$.

Suppose $r \ne 0$.

That would mean $\nu \left({r}\right) < \nu \left({d}\right)$ contradicting $d$ as the element of $U$ with the smallest $\nu$.

So $r = 0$, which means $a = q d$.

That is, every element of $U$ is a multiple of $d$.

So $U$ is the principal ideal generated by $d$.

This deduction holds for all ideals of $D$.

Hence the result.

$\blacksquare$

## Sources

- C.R.J. Clapham:
*Introduction to Abstract Algebra*(1969)... (previous)... (next): $\S 6.27$: Theorem $53$