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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 RR with Euclidean function δ\delta, and an ideal IRI\mathrel{\trianglelefteq}R.

[L1]

Euclidean division gives a=bq+ra=bq+r with r=0r=0 or δ(r)<δ(b)\delta(r)<\delta(b) whenever b0b\ne0 (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)(d) is the ideal generated by dd (The ideal generated by a subset and principal ideals).

[L4]

Every nonempty subset of N\mathbb 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
1.1

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

L3
1.2

Suppose I{0}I\ne\{0\}. The set {δ(x):xI{0}}\{\delta(x):x\in I\setminus\{0\}\} is nonempty, so choose dI{0}d\in I\setminus\{0\} whose δ\delta-value is least.

L4givenchoose
2.1

For aIa\in I, divide by dd: a=dq+ra=dq+r with r=0r=0 or δ(r)<δ(d)\delta(r)<\delta(d). Since r=adqIr=a-dq\in I, minimality in step 1.2 excludes a nonzero rr; hence r=0r=0.

step 1.2L1L2given
3.1

Step 2.1 gives a=dq(d)a=dq\in(d) for every aIa\in I, so I(d)I\subseteq(d). Conversely dId\in I and ideal closure give dqIdq\in I for every qRq\in R, so (d)I(d)\subseteq I. Thus I=(d)I=(d).

step 2.1L2L3given
4.1

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

step 1.1step 3.1L5

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources