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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

1.1basegivenL1L4

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

2.1step 1.1ihL2L4algebra

For uniqueness, compare $p_{1} \dots p_{m} = q_{1} \dots q_{n}$ ; 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 $m = n$ .

3.1step 1.1step 2.1L3discharge-induction∎

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

Depends on

Dependency tree · 2 levels

Sources

Neil Donaldson, Math 120B Notes, Theorem 23.11

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