---
tag: Z38EDQTCTS
kind: proof
name: Proof that Euclidean domains are principal ideal domains
msc: [13F07]
proves: 0QKVE1GD9A
uses: [0GJ2HRG95T, 0RZYDVZ475, 0MYSA6MYNA, 00A2EDFT9Q, 030TZ0DQWT, 01BX15HSRR]
sources:
  - url: https://proofwiki.org/wiki/Euclidean_Domain_is_Principal_Ideal_Domain
---

## Proof

Let $R$ be a [[Euclidean domain|0GJ2HRG95T]] with Euclidean function $\delta \colon R \setminus \set{0} \to \N$. Let $I$ be an [[ideal|0MYSA6MYNA]] of $R$.

1. For $b \in R$, the set $b R = \set{b t : t \in R}$ is an ideal that contains $b$. Every ideal containing $b$ contains $b R$. So the [[ideal generated by|030TZ0DQWT]] $b$ is $(b) = b R$.
2. If $I = \set{0}$, then $I = (0)$, so $I$ is [[principal|00A2EDFT9Q]].
3. Now let $I \neq \set{0}$. The set $\set{\delta(c) : c \in I, c \neq 0}$ is a nonempty set of natural numbers.
4. By the [[well-ordering principle|01BX15HSRR]], it has a least element. Choose $b \in I$ with $b \neq 0$ such that $\delta(b)$ is this least element.
5. Since $b \in I$, step 1 gives $(b) \subseteq I$.
6. Let $a \in I$. Since $b \neq 0$, there are $q, r \in R$ with $a = b q + r$, and $r = 0$ or $\delta(r) < \delta(b)$.
7. Since $I$ is an ideal, $b q \in I$, so $r = a - b q \in I$.
8. If $r \neq 0$, then $r$ is a nonzero element of $I$ with $\delta(r) < \delta(b)$. This contradicts the choice of $b$. So $r = 0$.
9. Hence $a = b q \in (b)$. So $I \subseteq (b)$, and with step 5, $I = (b)$.
10. Every ideal of $R$ is principal, and $R$ is an integral domain. So $R$ is a [[principal ideal domain|0RZYDVZ475]].

## Idea

In a nonzero ideal, pick a nonzero element $b$ with the smallest value of $\delta$. Dividing any element of the ideal by $b$ leaves a remainder that lies in the ideal and is smaller than $b$. So the remainder must be zero, and $b$ generates the ideal.