Polynomial Factorization over Fields

Overview

Because F[x] is a Euclidean domain (and hence a unique factorization domain) whenever F is a field, every nonzero polynomial over F factors uniquely, up to order and unit scalars, into a product of irreducible polynomials over F. This mirrors unique factorization of integers into primes, with irreducible polynomials playing the role of primes. What changes dramatically from one field to another is which polynomials count as irreducible: the same polynomial can be irreducible over ℚ, split into distinct linear factors over ℝ or ℂ, or factor in yet another pattern over a finite field 𝔽_p. Comparing a polynomial's factorization pattern across ℚ, ℝ, ℂ, and 𝔽_p (via reduction mod p) is one of the most practical tools for proving irreducibility and for computing Galois groups.

Intuition

Factoring a polynomial over a field is exactly like factoring an integer into primes, except the 'primes' available depend on which number system (field) you allow coefficients to live in. Over ℂ, the Fundamental Theorem of Algebra guarantees every polynomial splits completely into linear factors — there is nowhere left to hide any irreducibility. Over ℝ, complex roots come in conjugate pairs α, ᾱ, and multiplying (x−α)(x−ᾱ) always produces a real quadratic, so irreducibles over ℝ have degree at most 2. Over ℚ, arithmetic obstructions (irrational or complex roots) can make degree-3-or-higher polynomials genuinely irreducible with no simpler factors at all. Reducing coefficients modulo a prime p and factoring over the finite field 𝔽_p is often the fastest practical irreducibility test over ℚ: if f mod p stays irreducible of the same degree for some prime not dividing the leading coefficient, f was already irreducible over ℚ.

Formal Definition

Definition

Let F be a field and f(x) ∈ F[x] a nonzero polynomial. Unique factorization in F[x] states:

f(x) = c \cdot p_1(x)^{e_1} p_2(x)^{e_2} \cdots p_k(x)^{e_k}

Factorization into irreducibles, c \in F^\times a unit

\text{the } p_i(x) \text{ are irreducible over } F \text{ and unique up to order and scalar multiples}

Uniqueness of the factorization

\mathbb{C}[x]: \ f(x) = c\prod_i (x-\alpha_i) \quad \text{(Fundamental Theorem of Algebra: only linear irreducibles)}

Over ℂ every irreducible is linear

\mathbb{R}[x]: \ f(x) = c\prod_i (x-r_i) \prod_j (x^2+b_jx+c_j),\quad b_j^2-4c_j<0

Over ℝ irreducibles have degree 1 or 2

Worked Examples

Over ℚ: difference of squares twice, then x²+1 stays irreducible (no rational/real root).
$x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)$
Over ℝ: (x−1)(x+1) are already real linear factors; x²+1 has no real root (x²=−1 impossible), so it stays irreducible of degree 2.
$x^4-1 = (x-1)(x+1)(x^2+1) \quad \text{in } \mathbb{R}[x]$
Over ℂ: x²+1 factors further as (x−i)(x+i), giving four linear factors total, matching the Fundamental Theorem of Algebra.
$x^4-1 = (x-1)(x+1)(x-i)(x+i) \quad \text{in } \mathbb{C}[x]$

Answer: Over ℚ and ℝ: (x−1)(x+1)(x²+1); over ℂ: (x−1)(x+1)(x−i)(x+i) — four distinct linear factors, the 4th roots of unity.

Practice Problems

Difficulty 5/10

Factor x³ − x over ℚ completely, and state how many roots it has in ℚ.

Difficulty 6/10

Show that x² + x + 1 is irreducible over 𝔽₂ = {0,1} by checking both elements as roots, and give its factorization over the extension field 𝔽₄.

Difficulty 7/10

Use reduction mod 5 to argue that f(x) = x³ + x + 1 is irreducible over ℚ.

Common Mistakes

Common Mistake

Assuming a polynomial's factorization pattern is the same over every field.

Factorization depends entirely on the field: x²+1 is irreducible over ℝ and 𝔽₃, but factors as (x−2)(x−3) over 𝔽₅ and (x−i)(x+i) over ℂ.

Common Mistake

Thinking the mod-p irreducibility test is an if-and-only-if criterion.

The test is one-directional: irreducibility mod p implies irreducibility over ℚ, but a polynomial can be irreducible over ℚ while factoring mod every prime p (this happens, e.g., for some quartics with Galois group V₄).

Quiz

Summary

F[x] is a UFD for any field F: every nonzero polynomial factors uniquely into irreducibles up to order and scalar units.
Over ℂ every irreducible has degree 1 (Fundamental Theorem of Algebra); over ℝ irreducibles have degree 1 or 2 (conjugate root pairs).
The same polynomial can have different factorization patterns over ℚ, ℝ, ℂ, and finite fields 𝔽_p — e.g. x²+1 is irreducible over ℝ and 𝔽₃ but splits over ℂ and 𝔽₅.
Reducing mod a prime p gives a practical sufficient test: irreducibility mod p (same degree) implies irreducibility over ℚ.