Throughout this section, $F$ always denotes a field. Before carrying over the results of Section 1.3 on unique factorization in $\mathbb{Z}$ to the ring $F[x]$, we must first examine an area in which $\mathbb{Z}$ differs significantly from $F[x]$. In $\mathbb{Z}$ there are only two units,* namely $\pm 1$, but a polynomial ring may have many more units (see Corollary 4.5).

An element $a$ in a commutative ring with identity $R$ is said to be an associate of an element $b$ of $R$ if $a=bu$ for some unit $u$. In this case $b$ is also an associate of $a$ because $u^{-1}$ is a unit and $b=a{u}^{-1}$. In the ring $\mathbb{Z}$, the only associates of an integer $n$ are $n$ and $-n$ because $\pm 1$ are the only units. If $F$ is a field, then by Corollary 4.5, the units in $F[x]$ are the nonzero constants. Therefore,

$$f(x)\text{ is an associate of }g(x)\text{ in }F[x]\text{ if and only if }f(x)=cg(x)\text{ for some nonzero }c\in F.$$

Recall that a nonzero integer $p$ is prime in $\mathbb{Z}$ if it is not $\pm 1$ (that is, $p$ is not a unit in $\mathbb{Z}$) and its only divisors are $\pm 1$ (the units) and $\pm p$ (the associates of $p$). In $F[x]$ the units are the nonzero constants, which suggests the following definition.

"Unit" is defined just before Example 3.2.4.

Definition

Let $F$ be a field. A nonconstant polynomial $p(x)\in F[x]$ is said to be irreducible if its only divisors are its associates and the nonzero constant polynomials (units). A nonconstant polynomial that is not irreducible is said to be reducible.

Example 1

The polynomial $x+2$ is irreducible in $\mathbb{Q}[x]$ because, by Theorem 4.2, all its divisors must have degree 0 or 1. Divisors of degree 0 are nonzero constants. If $f(x)\mid (x+2)$, say $x+2=f(x)g(x)$, and if $\mathrm{deg}(f(x))=1$, then $g(x)$ has degree 0, so that $g(x)=c$. Thus $c(x+2)=f(x)$, and $f(x)$ is an associate of $x+2$. A similar argument in the general case shows that

every polynomial of degree 1 in $F[x]$ is irreducible in $F[x]$.

The definition of irreducibility is a natural generalization of the concept of primality in $\mathbb{Z}$. In most high-school texts, however, a polynomial is defined to be irreducible if it is not the product of polynomials of lower degree. The next theorem shows that these two definitions are equivalent.

Theorem 4.11

Let $F$ be a field. A nonzero polynomial $f(x)$ is irreducible in $F[x]$ if and only if $f(x)$ cannot be written as the product of two polynomials of positive degree in $F[x]$.

Proof

First, assume that $f(x)$ is reducible. Then it must have a divisor $g(x)$ that is neither an associate nor a nonzero constant, say $f(x)=g(x)h(x)$. Either $g(x)$ or $h(x)$ has the same degree as $f(x)$, then the other must have degree 0 by Theorem 4.2. Since a polynomial of degree 0 is a nonzero constant in $F$, this means that either $g(x)$ is a constant or an associate of $f(x)$, contrary to hypothesis. Therefore, both $g(x)$ and $h(x)$ have lower degree than $f(x)$.

Now assume that $f(x)$ can be written as the product of two polynomials of lower degree, and see Exercise 8. $◼$

Various other tests for irreducibility are presented in Sections 4.4 to 4.6. For now, we note that the concept of irreducibility is not an absolute one. For instance, $x^2+1$ is reducible in $\mathbb{C}[x]$ because $x^2+1=(x+i)(x-i)$ and neither factor is a constant or an associate of $x^2+1$. But $x^2+1$ is irreducible in $\mathbb{Q}[x]$ (Exercise 6).

The following theorem shows that irreducibles in $F[x]$ have essentially the same divisibility properties as do primes in $\mathbb{Z}$. Condition (3) in the theorem is often used to prove that a polynomial is irreducible; in many books, (3) is given as the definition of "irreducible".

You could just as well call such a polynomial "prime," but "irreducible" is the customary term with polynomials.

Theorem 4.12

Let $F$ be a field and $p(x)$ a nonconstant polynomial in $F[x]$. Then the following conditions are equivalent:

1. $p(x)$ is irreducible.
2. If $b(x)$ and $c(x)$ are any polynomials such that $p(x)\mid b(x)c(x)$, then $p(x)\mid b(x)$ or $p(x)\mid c(x)$.
3. If $r(x)$ and $s(x)$ are any polynomials such that $p(x)=r(x)s(x)$, then $r(x)$ or $s(x)$ is a nonzero constant polynomial.

Proof

(1) $\Rightarrow$ (2) Adapt the proof of Theorem 1.5 to $F[x]$. Replace statements about $\pm p$ by statements about the associates of $p(x)$; replace statements about $\pm 1$ by statements about units (nonzero constant polynomials) in $F[x]$; use Theorem 4.10 in place of Theorem 1.4.

(2) $\Rightarrow$ (3) If $p(x)=r(x)s(x)$, then $p(x)\mid r(x)$ or $p(x)\mid s(x)$. Since $F[x]$ is an integral domain, we can cancel $p(x)$ by Theorem 3.7 and conclude that $1_F=r(x)$ or $1_F=s(x)$. Thus $s(x)$ is a unit, and hence by Corollary 4.5, $r(x)$ is a nonzero constant. A similar argument shows that if $p(x)\mid s(x)$, then $r(x)$ is a nonzero constant.

(3) $\Rightarrow$ (1) Let $c(x)$ be any divisor of $p(x)$, say $p(x)=c(x)d(x)$. If $c(x)$ is not a unit, then $c(x)$ is a nonzero constant by (2). So $c(x)$ divides $p(x)$ and $d^{-1}$. Thus in every case, $c(x)$ is a nonzero constant or an associate of $p(x)$. Therefore, $p(x)$ is irreducible.

Corollary 4.13

Let $F$ be a field and $p(x)$ an irreducible polynomial in $F[x]$. If $p(x)\mid a_1(x)a_2(x)\cdots a_n(x)$, then $p(x)$ divides at least one of the $a_i(x)$.

Proof

Adapt the proof of Corollary 1.6 to $F[x]$.

For the meaning of "the following conditions are equivalent" and what must be done to prove Theorem 4.12, see page 508 of Appendix A. Example 2 there is the integer analogue of Theorem 4.12.

Theorem 4.14

Let $F$ be a field. Every nonconstant polynomial $f(x)$ in $F[x]$ is a product of irreducible polynomials in $F[x]$. This factorization is unique in the following sense: If

$$f(x)=p_1(x)p_2(x)\cdots p_r(x)\text{and}f(x)=q_1(x)q_2(x)\cdots q_s(x)$$

with each $p_i(x)$ and $q_j(x)$ irreducible, then $r=s$ (that is, the number of irreducible factors is the same). After $q_j(x)$ are reordered and relabeled, if necessary,

$$p_{(}x)\text{ is an associate of }{q}_{(}x)(i=1,2,3,...,r).$$

Proof

To show that $f(x)$ is a product of irreducibles, adapt the proof of Theorem 1.7 to $F[x]$. Let $S$ be the set of all nonconstant polynomials that are not the product of irreducibles, and use a proof by contradiction to show that $S$ is empty. To prove that this factorization is unique up to associates, suppose $f(x)=p_1(x)p_2(x)\cdots p_r(x)=q_1(x)q_2(x)\cdots q_s(x)$, with each $p_i(x)$ and $q_j(x)$ irreducible. Then $p_1(x)p_2(x)\cdots p_r(x)=q_1(x)q_2(x)\cdots q_s(x)$ so that $p_1(x)$ divides $q_1(x)q_2(x)\cdots q_s(x)$. Corollary 4.13 shows that $p_1(x)\mid q_j(x)$ for some $j$. After rearranging and relabeling the $q_j(x)$ if necessary, we may assume that $p_1(x)\mid q_1(x)$. Since $p_1(x)$ is irreducible, $p_1(x)$ must be either a constant or an associate of $q_1(x)$. However, $p_1(x)$ is irreducible, and so it is not a constant. Therefore, $p_1(x)$ is an associate of $q_1(x)$, with $p_1(x)=c{q}_1(x)$ for some constant $c$. Thus:

$$q_1(x)[p_2(x)p_3(x)\cdots p_r(x)]=p_1(x)p_2(x)\cdots p_r(x)=q_1(x)q_2(x)\cdots q_s(x)$$

Canceling $q_1(x)$ on each end, we have:

$$p_2(x)p_3(x)\cdots p_r(x)=q_2(x)q_3(x)\cdots q_s(x)$$

Complete the argument by adapting the proof of Theorem 1.8 to $F[x]$, replacing statements about $\pm q_j$ with statements about associates of $q_j(x)$.