Irreducible Polynomials and Unique Factorization

The goal for this section is to state a version of fundamental theorem of arithmetic for polynomials. We will first need the concept of “prime numbers” in $F[x]$. Conveniently, the prime elements in $F[x]$ will be called the irreducible polynomials.

Irreducible Polynomials

Definition

A non-constant polynomial $f(x)$ in $F[x]$ is called irreducible if its only divisors are the units and the constant multiples of $f(x)$. A non-constant, non-irreducible polynomial is called reducible.

This definition should be compared with prime numbers. Where we stated the number $p$ is prime iff its only divisors are $\pm 1$ and $\pm p$. The $\pm 1$ are the units in $\mathbb{Z}$, so the definition above is a consistent generalization.

The irreducible polynomials are the prime elements in $F[x]$, the terminology is justified by the following proposition.

Proposition

A non-constant polynomial $f(x)$ is reducible if and only if there exist non-constant polynomials $g(x)$ and $h(x)$ such that $f(x)=g(x)h(x)$.

proof of the proposition

If a non-constant polynomial $f(x)$ is reducible, then there exist polynomials $g(x)$ and $h(x)$ such that $f(x)=g(x)h(x)$, we may assume $g(x)$ is non-constant and not a constant multiple of $f(x)$.

Then in this case, we will have:

$$\mathrm{deg}f(x)=\mathrm{deg}g(x)+\mathrm{deg}h(x),0<\mathrm{deg}g(x)<\mathrm{deg}f(x).$$

This implies $\mathrm{deg}h(x)>0$, so both $g(x)$ and $h(x)$ are non-constant. $■$

The proposition tells us reducible polynomials can be thought as the ones that can be factored into two smaller degree polynomials. So in particular, when we have the minimal possible degree $1$ for non-constant polynomials, it is automatically irreducible.

Proposition

A non-constant polynomial $f(x)$ of degree $1$ is irreducible.

Irreducibility depends on the field $F$. For example:

1. $x^2-2$ is irreducible in $\mathbb{Q}[x]$ but reducible in $\mathbb{R}[x]$.
2. $x^2+1$ is irreducible in $\mathbb{R}[x]$ but reducible in $\mathbb{C}[x]$.

Recall we have the following property for prime numbers:

If $p$ is prime, and $p\mid ab$, then $p\mid a$ or $p\mid b$.

We will have a similar property for irreducible polynomials.

Proposition

Let $p(x)$ be a non-constant polynomial in $F[x]$. TFAE:

1. $p(x)$ is irreducible.
2. If $p(x)\mid g(x)h(x)$, then $p(x)\mid g(x)$ or $p(x)\mid h(x)$.
3. If $p(x)=r(x)s(x)$, then $r(x)$ or $s(x)$ is a unit.

The generalization of the second property to arbitrary number of polynomials is also correct:

Proposition

Let $p(x)$ be an irreducible polynomial in $F[x]$. For any polynomials $f_1(x),f_2(x),\dots ,f_n(x)$, if $p(x)\mid f_1(x)f_2(x)\cdots f_n(x)$, then $p(x)\mid f_i(x)$ for some $i$.

Unique Factorization

We finally state the fundamental theorem of arithmetic for polynomials.

Theorem

Let $f(x)$ be a non-constant polynomial in $F[x]$. Then $f(x)$ can be factored into a product of irreducible polynomials in $F[x]$ in the following way:

$$f(x)=\prod _{i=1}^r{p}_i(x).$$

The factorization is unique in the following sense:

$$f(x)=\prod _{i=1}^r{p}_i(x)=\prod _{j=1}^s{q}_j(x).$$

Then $r=s$, and after relabeling, we have $p_i(x)=c\cdot q_i(x)$ for some non-zero constant $c$.

When we have factorizations into irreducible polynomials, we can find the greatest common divisor of two polynomials by looking for common factors in the factorizations and take the minimum of the exponents.

Example

Let $f(x)=(x^2+1{)}^2(x-1{)}^{3}x$ and $g(x)=(x^2+1)(x-1{)}^2{x}^2$ be two polynomials in $\mathbb{Q}[x]$. Then we have:

$$\gcd (f(x),g(x))=(x^2+1)(x-1{)}^{2}x.$$