Engineering / Mathematics — Fields, Series and Factorisation
Unique Factorization of Polynomials
Unique factorisation in polynomial rings over a field, and the extension to polynomial rings over a UFD.
Executive summary
Polynomials over a field factor uniquely into irreducibles, for the same reason integers do: the ring is Euclidean, hence a principal ideal domain, hence a unique factorisation domain.
Over a general UFD the result still holds, but the proof requires Gauss's lemma to control the interaction between content and primitive parts.
Learning objectives
- Prove unique factorisation over a field via the Euclidean property.
- Define irreducibility for polynomials.
- State the extension to polynomial rings over a UFD.
01Over a field
Unique factorisation in F[X]
Every non-constant polynomial over a field factors into irreducibles, uniquely up to order and non-zero constant multiples.
The proof mirrors the integer case exactly. Division with remainder makes F[X] a Euclidean domain with the degree as the size function; every ideal is therefore principal; Bezout and Euclid's lemma follow; and unique factorisation is the consequence.
Division with remainder
Available because every non-zero constant is invertible over a field.
Every ideal is principal
Take the non-zero element of least degree in the ideal and divide.
Bezout and Euclid's lemma
An irreducible dividing a product divides one factor.
Unique factorisation
Induction on degree, cancelling common irreducibles.
Irreducible polynomial
A non-constant f ∈ F[X] is irreducible if it cannot be written as a product of two polynomials of strictly smaller degree.
Irreducibility depends on the field: X² + 1 is irreducible over R and reducible over C.
02Content and primitive parts
Over a UFD such as the integers, a polynomial splits into a constant content and a primitive part with coprime coefficients.
Content and primitive polynomial
The content of f ∈ R[X] is the gcd of its coefficients. f is primitive if its content is a unit.
Gauss's lemma
The product of two primitive polynomials is primitive. Equivalently, content is multiplicative: cont(fg) = cont(f) · cont(g) up to units.
The proof is a reduction modulo a prime dividing the content of the product: both factors would have to reduce to zero, forcing the prime to divide the content of one of them.
03Over a UFD
Polynomial rings over a UFD
If R is a unique factorisation domain, so is R[X].
By induction, R[X₁, ..., Xₖ] is a UFD whenever R is. So Z[X, Y] and F[X, Y] both have unique factorisation.
| Ring | Euclidean | PID | UFD |
|---|---|---|---|
| Z | Yes | Yes | Yes |
| F[X] | Yes | Yes | Yes |
| Z[X] | No | No | Yes |
| F[X, Y] | No | No | Yes |
| Z[√−5] | No | No | No |
04Frequently asked questions
Why is irreducibility field dependent?
Because factoring requires the factors to have coefficients in the field. Enlarging the field admits more possible factors, so a polynomial irreducible over a small field may split over a larger one — which is precisely how field extensions are constructed.
Is Z[X] a principal ideal domain?
No. The ideal generated by 2 and X consists of polynomials with even constant term and is not generated by any single element. Z[X] is a UFD without being a PID, showing the implications are strict.
How is a polynomial over Z factored in practice?
By factoring modulo a well-chosen prime, lifting the factorisation to a high power of that prime by Hensel's method, and recombining the lifted factors into integer factors.
Sources and method
Structural reference: Victor Shoup, A Computational Introduction to Number Theory and Algebra, Version 1, Cambridge University Press, 2005 — book pages 366-371.
This page carries the durable method layer only: definitions, constructions, algorithms, complexity results and selection criteria, authored originally for KEVOS. No text is transcribed or paraphrased from the source, and no numeric tables or benchmark data are reproduced — these are routed to live authoritative sources instead.
Author: Kevin Jogin. Last reviewed 2026-08-07.