Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-11
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For every field F, F[x] is a unique factorisation domain

Statement

For every field F, the polynomial ring F[x] is a unique factorisation domain.

Facts & Assumptions

Given: A field F.

[L1]

Every nonzero nonunit polynomial over F factors into irreducibles (Every nonzero nonunit polynomial over a field factors into irreducible polynomials).

[L2]

Every irreducible polynomial over F is prime (Every irreducible polynomial over a field is prime).

[L3]

A UFD is an integral domain with existence and uniqueness, up to order and associates, of irreducible factorizations of every nonzero nonunit (Unique factorisation domain).

[L4]

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

Proof

technique · induction
1.1

Fact [L4] makes F[x] a domain, and [L1] supplies existence of irreducible factorizations.

basegivenL1L4
2.1

For uniqueness, compare p1⋯pm=q1⋯qn; by [L2], p1 divides some qj, and irreducibility makes p1 associate to qj; after reordering and cancelling these nonzero associates in the domain, induction on m pairs all remaining factors and gives m=n.

step 1.1ihL2L4algebra
3.1

The existence and uniqueness established in steps 1.1 and 2.1 are exactly the conditions of [L3], so F[x] is a UFD.

step 1.1step 2.1L3discharge-induction∎

Depends on

Used by

Dependency tree · two levels

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Sources