TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck 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 $p_1\cdots p_m=q_1\cdots q_n$ ; by [L2], $p_1$ divides some $q_j$ , and irreducibility makes $p_1$ associate to $q_j$ ; 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

Eisenstein criterion over the integers Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 9 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

Neil Donaldson, Math 120B Notes, Theorem 23.11 (standard reference, not scraped)

For every field FFF, F[x]F[x]F[x] is a unique factorisation domain