TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-11

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Every irreducible polynomial over a field is prime

Statement

Let $F$ be a field. Every irreducible polynomial $p \in F [x]$ is prime: if $p$ divides $f g$ , then $p$ divides $f$ or $p$ divides $g$ .

Facts & Assumptions

Given: A field $F$ , an irreducible polynomial $p$ , and polynomials $f, g$ with $p ∣ f g$ .

[L1]

For polynomials not both zero, the monic gcd divides both inputs, every common divisor divides the gcd, and the gcd is a polynomial linear combination of the inputs (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L2]

An irreducible element is a nonzero nonunit whose every factorization has a unit factor; a prime element divides one factor whenever it divides a product (Irreducible and prime elements of an integral domain).

Proof

technique · direct

1.1

If $p ∣ f$ there is nothing to prove. Otherwise, if a common divisor $d$ of $p$ and $f$ were a nonunit, a factorization $p = d e$ and irreducibility would make $e$ a unit, so $d$ would be associate to $p$ and $d ∣ f$ would imply $p ∣ f$ , a contradiction. Thus every common divisor is a unit, and [L1] gives $A, B \in F [x]$ with $A p + B f = 1$ .

givenL1L2choosealgebra

2.1

Multiplying the identity by $g$ gives $A p g + B f g = g$ ; both terms on the left are divisible by $p$ , the second because $p ∣ f g$ , so $p ∣ g$ . Thus $p$ satisfies the prime condition in [L2].

step 1.1L2algebra∎

Depends on

Used by

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

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Neil Donaldson, Math 120B Notes, Theorem 23.10(2) (standard reference, not scraped)