TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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Every Euclidean domain is a principal ideal domain

Statement

Every Euclidean domain is a principal ideal domain.

Facts & Assumptions

Given: A Euclidean domain $R$ with Euclidean function $δ$ , and an ideal $I ⊴ R$ .

[L1]

Euclidean division gives $a = b q + r$ with $r = 0$ or $δ (r) < δ (b)$ whenever $b \neq 0$ (Euclidean domain and Euclidean function).

[L2]

An ideal is an additive subgroup closed under multiplication by arbitrary ring elements (Left, right and two-sided ideals).

[L3]

The principal ideal $(d)$ is the ideal generated by $d$ (The ideal generated by a subset and principal ideals).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L5]

A PID is an integral domain whose every ideal is principal (Principal ideal domain).

Proof

technique · direct

1.1L3

If $I = {0}$ , then $I = (0)$ and is principal.

1.2L4givenchoose

Suppose $I \neq {0}$ . The set ${δ (x) : x \in I ∖ {0}}$ is nonempty, so choose $d \in I ∖ {0}$ whose $δ$ -value is least.

2.1step 1.2L1L2given

For $a \in I$ , divide by $d$ : $a = d q + r$ with $r = 0$ or $δ (r) < δ (d)$ . Since $r = a - d q \in I$ , minimality in step 1.2 excludes a nonzero $r$ ; hence $r = 0$ .

3.1step 2.1L2L3given

Step 2.1 gives $a = d q \in (d)$ for every $a \in I$ , so $I \subseteq (d)$ . Conversely $d \in I$ and ideal closure give $d q \in I$ for every $q \in R$ , so $(d) \subseteq I$ . Thus $I = (d)$ .

4.1step 1.1step 3.1L5∎

Every ideal is principal by step 1.1 or step 3.1; therefore $R$ is a PID.

Depends on

Dependency tree · 2 levels

Sources

Sharifi, Abstract Algebra, Advanced Ring Theory