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 \(b = au^{-1}\). 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 \(\deg(f(x)) = 1\), 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. \(\blacksquare\)

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

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

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,

\[ p_(x) \text{ is an associate of } q_(x) \quad (i=1,2,3,..., r). \]

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 \(q_j(x)\).