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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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For every field FF, F[x]F[x] is a principal ideal domain

Statement

For every field FF, every ideal of F[x]F[x] is generated by one polynomial; equivalently, F[x]F[x] is a principal ideal domain.

Facts & Assumptions

Given: A field FF.

[L1]

The ring F[x]F[x] is a Euclidean domain with degree as Euclidean function (For every field FF, F[x]F[x] is a Euclidean domain with degree as Euclidean function).

[L2]

Every Euclidean domain is a principal ideal domain (Every Euclidean domain is a principal ideal domain).

[L3]

A principal ideal domain is an integral domain in which every ideal is principal (Principal ideal domain).

Proof

technique · direct
1.1

By [L1] and [L2], F[x]F[x] is a principal ideal domain.

givenL1L2
2.1

Unfolding [L3], every ideal of F[x]F[x] therefore has the form (d)(d) for some polynomial dd.

step 1.1L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 6 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