# Irreducible polynomials

**Throughout this page, we let $F$ be a field.**

**Definition of an irreducible polynomial**

Let $f(x)$ be a non-zero polynomial in $F[x]$. Then $f(x)$ is irreducible in $F[x]$ if

(1) $f(x)$ is not a unit in $F[x]$, and

(2) whenever $f(x) = g(x)h(x)$ with $g(x), h(x) \in F[x]$, then either $g(x)$ or $h(x)$ is a unit in $F[x]$.

We may also say $f(x)$ is irreducible over $F$.

**Alternative definition of an irreducible polynomial**

Let $f(x)$ be a non-zero polynomial in $F[x]$. Then $f(x)$ is irreducible in $F[x]$ if

(1) deg$\, f(x) \geq 1$, and

(2) whenever $f(x) = g(x)h(x)$ with $g(x), h(x) \in F[x]$, then either deg$\,g(x) = 0$ or deg$\,h(x) = 0$.

**Theorem (degree 1)**

Every linear polynomial in $F[x]$ is irreducible in $F[x]$.

**Proposition**

Let $f(x)$ be a polynomial in $F[x]$ with deg$\,f(x) \geq 2$. If $f(x)$ is irreducible in $F[x]$ then $f(x)$ has no roots in $F$.

$\qquad\qquad$ *Remark: Be careful when using this result as the converse is not true in general!*

**Theorem (degree 2 or 3)**

Let $f(x)$ be a polynomial in $F[x]$ of degree 2 or 3. Then $f(x)$ is irreducible in $F[x]$ if and only if $f(x)$ has no roots in $F$.

Next section on Theorems on irreducible polynomials

Back to Chapter 2: Polynomials