Alphabeta Math
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 Euclidean domain with degree as Euclidean function

Statement

For every field FF, the ring F[x]F[x] is a Euclidean domain with Euclidean function δ(f)=deg⁡f\delta(f)=\deg f on nonzero polynomials.

Facts & Assumptions

Given: A field FF.

[L1]

A polynomial ring over an integral domain is an integral domain (A polynomial ring over an integral domain is an integral domain).

[L2]

For f∈F[x]f\in F[x] and 0≠g∈F[x]0\ne g\in F[x], there are q,rq,r with f=qg+rf=qg+r and r=0r=0 or deg⁡r<deg⁡g\deg r<\deg g (Division algorithm for polynomials over a field).

[L3]

A Euclidean domain is an integral domain with a natural-valued function on nonzero elements satisfying exactly that division condition (Euclidean domain and Euclidean function).

[L4]

A field is an integral domain because nonzero elements are invertible and 0≠10\ne1 (Field).

Proof

technique · direct
1.1

By [L4] and [L1], F[x]F[x] is an integral domain.

givenL1L4
2.1

Degree is natural-valued on nonzero polynomials, and [L2] supplies the division condition of [L3], so F[x]F[x] is Euclidean with δ=deg⁡\delta=\deg.

step 1.1L2L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 7 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