Polynomial Irreducibility

Rational root theorem & Eisenstein's criterion.

The idea

Definition (Irreducible polynomial).

A polynomial with coefficients in a field $F$ is irreducible over $F$ when it cannot be written as a product of two polynomials of lower degree with coefficients in $F$.

The field matters: $x^{2} - 5$ factors as $(x - \sqrt{5})(x + \sqrt{5})$ over $\mathbb{R}$, but not over $\mathbb{Q}$, because $\sqrt{5}$ is irrational. Enlarging the field can make an irreducible polynomial reducible, never the reverse.

Irreducible polynomials are the primes of the polynomial ring $F[x]$: every polynomial factors into irreducibles, essentially uniquely, and quotienting $F[x]$ by an irreducible polynomial produces a field extension in which that polynomial has a root.

Exhibiting a factorization proves a polynomial reducible, but proving it irreducible means ruling out every possible factorization, and for that we use criteria. Over $\mathbb{Q}$, two are the standard tools.

Theorem (Rational root theorem).

Every rational root $p/q$ of a polynomial with integer coefficients, written in lowest terms, has $p$ dividing the constant term and $q$ dividing the leading coefficient.

This leaves a short list of candidate roots, and when every candidate fails the polynomial has no linear factor over $\mathbb{Q}$.

Theorem (Eisenstein's criterion).

If a prime $p$ divides every coefficient of an integer polynomial except the leading one, and $p^{2}$ does not divide the constant term, then the polynomial is irreducible over $\mathbb{Q}$.

Ways to work on it

• Walkthrough. Decide whether a polynomial is irreducible, and why the answer depends on the field.
• Practice. Is x^2 - D irreducible over ℚ?
• Hardest. Decide whether a quartic is irreducible over ℚ.

Not sure where to start? Take the ten-question placement test.