---
title: "Polynomial Rings, Irreducibility and Unique Factorisation"
description: "Polynomials over a field divide with remainder, and that single fact forces unique factorisation into irreducibles. Deciding whether a given polynomial is irreducible over the rationals needs tests ra"
canonical: https://lightmysky.com/learn/mathematics/polynomial-rings-irreducibility-and-unique-factorisation-mt_m2G9PIHj6y
source: https://lightmysky.com/learn/mathematics/polynomial-rings-irreducibility-and-unique-factorisation-mt_m2G9PIHj6y.md
retrieved: 2026-11-12
---

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# Polynomial Rings, Irreducibility and Unique Factorisation

Polynomials over a field divide with remainder, and that single fact forces unique factorisation into irreducibles. Deciding whether a given polynomial is irreducible over the rationals needs tests rather than inspection.

Subject: Mathematics · Area: Abstract Algebra · Ages 22 to 24
Page: https://lightmysky.com/learn/mathematics/polynomial-rings-irreducibility-and-unique-factorisation-mt_m2G9PIHj6y

## Ready when they can

- Divide polynomials with remainder and read off what it says about roots
- Apply Eisenstein's criterion and reduction modulo a prime
- Trace the chain from division with remainder to principal ideals to unique factorisation

## Lesson: Dividing polynomials and factoring uniquely

Over a field F, divide f by a nonzero polynomial g by repeating four actions: divide the leading terms, add that term to the quotient, multiply g by it, then subtract. Stop when the remainder is zero or has smaller degree than g. The result f = qg + r is unique. Field coefficients let you divide by g's nonzero leading coefficient.

**Example.** Divide x^2 + 6x + 8 by x + 3. Leading terms give x. Subtract x(x + 3), leaving 3x + 8. The next quotient term is 3. Subtract 3(x + 3), leaving -1. Thus the quotient is x + 3 and the remainder is -1. A nonzero remainder means the divisor is not a factor.

Division by x - a leaves a constant remainder. Substituting a into f = q(x)(x - a) + r gives r = f(a), the Remainder Theorem. Hence x - a is a factor exactly when f(a) = 0. A quadratic over a field is reducible exactly when it has a root there. Over the reals, a quadratic with no real root is irreducible; every real cubic has a real root and therefore a real linear factor.

*(drawing: Cancel the leading term, subtract and repeat until the remainder degree is below the divisor degree.)*

**Example.** Grouping reverses distributivity. For x^3 + 2x^2 + 6x + 12, the pairs become x^2(x + 2) + 6(x + 2). Factor the common x + 2 to obtain (x^2 + 6)(x + 2). Expanding a proposed factorization checks every coefficient. State the coefficient field when claiming that the remaining factors are irreducible.

For an integer-coefficient polynomial, a rational root u/v in lowest terms has u dividing the constant term and v dividing the leading coefficient. Test all such candidates. For degree 2 or 3 over Q, no rational root means irreducible; higher degrees can split without a linear factor. Irreducibility depends on the field: a rationally irreducible cubic still has a real root and factors over R.

An integer polynomial is primitive when its coefficients have gcd 1. For a positive-degree integer polynomial, Eisenstein's criterion applies if a prime p does not divide the leading coefficient, divides every other coefficient, and p^2 does not divide the constant coefficient. Then the polynomial is irreducible over Q. Missing coefficients are zero, which is divisible by p. Check all three divisibility conditions.

*(drawing: These checks establish Eisenstein's criterion for this primitive cubic.)*

For primitive f in Z[x], reduce coefficients modulo a prime p. If the degree is preserved and the reduction is irreducible over F_p, then f is irreducible over Q. For reduced degree 2 or 3, test all field elements for roots. Higher degrees need other factor tests. For example, x^2 + x + 1 has value 1 at both 0 and 1 modulo 2, so its reduction is irreducible. A degree drop invalidates this test.

*(drawing: For a quadratic, the complete root check proves irreducibility of the reduction.)*

A principal ideal domain is a nonzero commutative domain in which every ideal has one generator. F[x] is one: in a nonzero ideal choose a nonzero g of least degree. Divide any f in the ideal by g. Its remainder also belongs to the ideal, so minimality forces it to be zero. Thus the ideal is (g); the zero ideal is (0). Repeated division also computes gcds, and the gcd generates the ideal (f, g).

In F[x], gcd division gives Bezout identities. If irreducible p does not divide a, gcd(p, a) = 1, so up + va = 1. If p divides ab, multiply this identity by b to see that p divides b. Thus irreducibles are prime. Repeated positive-degree splitting terminates, giving each nonzero polynomial a nonzero constant times irreducible factors. Primeness makes the factors unique up to order and nonzero constant multiples. Zero has no such factorization.

*(drawing: Division proves the principal-ideal and prime-factor properties used in unique factorization.)*

**Recap.** Division gives remainders for testing factors and leads to unique factorisation of polynomials.

*Lesson and questions written for this dataset.*

## Practice

16 questions on this page, each with its working shown.

## Needs first

- [Ideals and Quotient Rings](https://lightmysky.com/learn/mathematics/ideals-and-quotient-rings-mt_uQBA9E1JIP)

## Optional connections

- [Roots of Polynomials and Roots of Unity](https://lightmysky.com/learn/mathematics/roots-of-polynomials-and-roots-of-unity-mt_XAISygqdQr)

## Opens up

- [Modules: Linear Algebra Over a Ring](https://lightmysky.com/learn/mathematics/modules-linear-algebra-over-a-ring-mt_0GMVp-s8wD)
- [Field Extensions and Their Degrees](https://lightmysky.com/learn/mathematics/field-extensions-and-their-degrees-mt_CjOMNt5_i0)
