Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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For every field F, F[x] is a Euclidean domain with degree as Euclidean function

Statement

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

Facts & Assumptions

Given: A field F.

[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] and 0≠g∈F[x], there are q,r with f=qg+r and r=0 or 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≠1 (Field).

Proof

technique · direct
1.1

By [L4] and [L1], 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] is Euclidean with δ=deg⁡.

step 1.1L2L3∎

Depends on

Used by

Dependency tree · two levels

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Sources