§4.5 Irreducibility in Polynomials with Rational Coefficients
The central theme of this section is that factoring in \(\mathbb{Q}[x]\) can be reduced to factoring in \(\mathbb{Z}[x]\). Then elementary number theory can be used to check polynomials with integer coefficients for irreducibility. We begin by noting a fact that will be used frequently:
\[ \text{If } f(x) \in \mathbb{Q}[x], \text{ then } cf(x) \text{ has integer coefficients for some nonzero integer } c. \]
This section is used only in Chapters 11, 12, and 15. It may be omitted until then, if desired. Section 4.6 is independent of this section.
For example, consider
\[ f(x) = x^5 + \frac{2}{3}x^4 + \frac{3}{4}x^3 - \frac{1}{6}. \]The least common denominator of the coefficients of \(f(x)\) is 12, and \(12f(x)\) has integer coefficients:
\[ 12f(x) = 12 \left(x^5 + \frac{2}{3}x^4 + \frac{3}{4}x^3 - \frac{1}{6}\right) = 12x^5 + 8x^4 + 9x^3 - 2. \]According to the Factor Theorem, finding first-degree factors of a polynomial \(g(x) \in \mathbb{Q}[x]\) is equivalent to finding the roots of \(g(x)\) in \(\mathbb{Q}\). Now, \(g(x)\) has the same roots as \(cg(x)\) for any nonzero constant \(c\). When \(c\) is chosen so that \(cg(x)\) has integer coefficients, we can find the roots of \(g(x)\) by using
Theorem 4.21: Rational Root Test
Let \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\) be a polynomial with integer coefficients. If \(r \neq 0\) and the rational number \(r/s\) (in lowest terms) is a root of \(f(x)\), then \(r | a_0\) and \(s | a_n\).
Proof
First consider the case when \(s = 1\), that is, the case when the integer \(r\) is a root of \(f(x)\), which means that
\[ a_nr^n + a_{n-1}r^{n-1} + \cdots + a_1r + a_0 = 0. \]Hence,
\[ a_0 = -a_nr^n - a_{n-1}r^{n-1} - \cdots - a_1r, \]which says that \(r\) divides \(a_0\).
In the general case, we use essentially the same strategy. Since \(r/s\) is a root of \(f(x)\), we have
\[ a_n \left(\frac{r}{s}\right)^n + a_{n-1} \left(\frac{r}{s}\right)^{n-1} + \cdots + a_1 \left(\frac{r}{s}\right) + a_0 = 0. \]We need an equation involving only integers (as in the case when \(s = 1\). So multiply both sides by \(s^n\), rearrange, and factor as before:
\[ a_nr^n + a_{n-1}s r^{n-1} + \cdots + a_1s^{n-1}r + a_0s^n = 0. \]This last equation says that \(r\) divides \(a_0s^n\), which is not quite what we want. However, since \(r/s\) is in lowest terms, we have \(\gcd(r, s) = 1\). It follows that \((r, s^n) = 1\) (a prime that divides \(r\) also divides \(s\), by Corollary 1.6). Since \(r | a_0s^n\) and \((r, s^n) = 1\), Theorem 1.4 shows that \(r | a_0\). A similar argument proves that \(s | a_n\) (just rearrange Equation (*) so that \(a_nr^n\) is on one side and everything else is on the other side).
Example 1
The possible roots in \(\mathbb{Q}\) of \(f(x) = 2x^4 + x^3 - 21x^2 - 14x + 12\) are of the form \(r/s\), where \(r\) is one of \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6\), or \(\pm 12\) (the divisors of the constant term, 12) and \(s\) is \(\pm 1\) or \(\pm 2\) (the divisors of the leading coefficient, 2). Hence, the Rational Root Test reduces the search for roots of \(f(x)\) to this finite list of possibilities:
\[ 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12, \frac{1}{2}, -\frac{1}{2}, \frac{3}{2}, -\frac{3}{2}. \]It is tedious but straightforward to substitute each of these in \(f(x)\) to find that \(-3\) and \(\frac{1}{2}\) are the only roots of \(f(x)\) in \(\mathbb{Q}\). By the Factor Theorem, both \(x - (-3) = x + 3\) and \(x - \frac{1}{2}\) are factors of \(f(x)\). Division shows that
\[ f(x) = (x + 3)\left(x - \frac{1}{2}\right)(2x^2 - 4x - 8). \]The quadratic formula shows that the roots of \(2x^2 - 4x - 8\) are \(1 \pm \sqrt{5}\), neither of which is in \(\mathbb{Q}\). Therefore, \(2x^2 - 4x - 8\) is irreducible in \(\mathbb{Q}[x]\) by Corollary 4.19. Hence, we have factored \(f(x)\) as a product of irreducible polynomials in \(\mathbb{Q}[x]\).
Example 2
The only possible roots of \(g(x) = x^3 + 4x + 1\) in \(\mathbb{Q}\) are \(1\) and \(-1\) (Why?). Verify that neither \(1\) nor \(-1\) is a root of \(g(x)\). Hence \(g(x)\) is irreducible in \(\mathbb{Q}[x]\) by Corollary 4.19.
If \(f(x) \in \mathbb{Q}[x]\), then \(cf(x)\) has integer coefficients for some nonzero integer \(c\). Any factorization of \(cf(x)\) in \(\mathbb{Z}[x]\) leads to factorization of \(f(x)\) in \(\mathbb{Q}[x]\). So it appears that tests for irreducibility in \(\mathbb{Q}[x]\) can be restricted to polynomials with integer coefficients. However, we must first rule out the possibility that a polynomial with integer coefficients could factor in \(\mathbb{Q}[x]\) but not in \(\mathbb{Z}[x]\). In order to do this, we need
Lemma 4.22
Let \(f(x), g(x), h(x) \in \mathbb{Z}[x]\) with \(f(x) = g(x)h(x)\). If \(p\) is a prime that divides every coefficient of \(f(x)\), then either \(p\) divides every coefficient of \(g(x)\) or \(p\) divides every coefficient of \(h(x)\).
Proof
Let \(f(x) = a_0 + a_1x + \cdots + a_kx^k\), \(g(x) = b_0 + b_1x + \cdots + b_mx^m\), and \(h(x) = c_0 + c_1x + \cdots + c_px^p\). We prove by contradiction. If the lemma is false, then \(p\) does not divide some coefficient of \(g(x)\) and some coefficient of \(h(x)\). Let \(b_r\) be the first coefficient of \(g(x)\) that is not divisible by \(p\), and let \(c_t\) be the first coefficient of \(h(x)\) that is not divisible by \(p\). Then \(p | b_j\) for \(r < j\) and \(p | c_j\) for \(t < j\). Consider the coefficient \(a_{r+t}\) of \(f(x)\). Since \(f(x)=g(x)h(x)\),
\[ a_{r+t} = b_0c_{r+t} + \cdots + b_rc_t + \cdots + b_{r+t}c_0. \]Consequently,
\[ b_rc_t = a_{r+t} - [b_0c_{r+t} + \cdots + b_{r-1}c_{t+1} - [b_{r+1}c_{t-1} + \cdots + b_{r+t}c_0]. \]Now, \(p | a_{r+t}\) by hypothesis. Also, \(p\) divides each term in the first pair of brackets because \(b_j\) was chosen so that \(p | b_j\) for each \(j < r\). Similarly, \(p\) divides each term in the second pair of brackets because \(p | c_j\) for each \(j < t\). Since \(p\) divides every term on the right side, we see that \(p | b_rc_t\). Therefore, \(p | b_r\), or \(p | c_t\) by Theorem 1.5. This contradicts the fact that neither \(b_r\) nor \(c_t\) is divisible by \(p\).
Theorem 4.23
Let \(f(x)\) be a polynomial with integer coefficients. Then \(f(x)\) factors as a product of polynomials of degrees \(m\) and \(n\) in \(\mathbb{Q}[x]\) if and only if \(f(x)\) factors as a product of polynomials of degrees \(m\) and \(n\) in \(\mathbb{Z}[x]\).
Proof
Obviously, if \(f(x)\) factors in \(\mathbb{Z}[x]\), it factors in \(\mathbb{Q}[x]\). Conversely, suppose \(f(x) = g(x)h(x)\) in \(\mathbb{Q}[x]\). Let \(c\) and \(d\) be nonzero integers such that \(cg(x)\) and \(dh(x)\) have integer coefficients. Then \(cdf(x) = [cg(x)][dh(x)]\) in \(\mathbb{Z}[x]\) with \(\text{deg}(cg(x)) = \text{deg}(g(x))\) and \(\text{deg}(dh(x)) = \text{deg}(h(x))\). Let \(p\) be any prime divisor of \(cd\), say \(cd = pt\). Then \(p\) divides every coefficient of the polynomial \(cdf(x)\). By Lemma 4.22, \(p\) divides either every coefficient of \(cg(x)\) or every coefficient of \(dh(x)\), say the former. Then \(cg(x) = pk(x)\) with \(k(x) \in \mathbb{Z}[x]\) and \(\text{deg}(k(x)) = \text{deg}(g(x))\). Therefore, \(ptf(x) = cdf(x) = [cg(x)][dh(x)] = [pk(x)][dh(x)]\). Canceling \(p\) on each end, we have \(tf(x) = k(x)[dh(x)]\) in \(\mathbb{Z}[x]\).
Now repeat the same argument with any prime divisor of \(t\) and cancel that prime from both sides of the equation. Continue until every prime factor of \(cd\) has been canceled. Then the left side of the equation will be \(\pm f(x)\), and the right side will be a product of two polynomials in \(\mathbb{Z}[x]\), one with the same degree as \(g(x)\) and one with the same degree as \(h(x)\). \(\blacksquare\)
Example 3
We claim that \(f(x) = x^4 - 5x^2 + 1\) is irreducible in \(\mathbb{Q}[x]\). But the Rational Root Test shows that \(f(x)\) has no roots in \(\mathbb{Q}\). (The only possibilities are \(\pm 1\), and neither is a root.) Thus if \(f(x)\) is reducible, the only possible factorization is as a product of two quadratics, by Theorem 4.2. In this case, Theorem 4.23 shows that there is such a factorization in \(\mathbb{Z}[x]\). Furthermore, there is a factorization as a product of monic quadratics in \(\mathbb{Z}[x]\) by Exercise 10, say
\[ (x^2 + ax + b)(x^2 + cx + d) = x^4 - 5x^2 + 1 \]with \(a, b, c, d \in \mathbb{Z}\). Multiplying out the left-hand side, we have
\[ x^4 + (a + c)x^3 + (ac + b + d)x^2 + (bc + ad)x + bd = x^4 + 0x^3 - 5x^2 + 0x + 1. \]Equal polynomials have equal coefficients; hence,
\[ a + c = 0 \quad ac + b + d = -5 \quad bc + ad = 0 \quad bd = 1. \]Since \(a + c = 0\), we have \(a = -c\), so that
\[ -5 = ac + b + d = -c^2 + b + d \]or, equivalently,
\[ 5 = c^2 - b - d. \]However, \(bd = 1\) in \(\mathbb{Z}\) implies that \(b = d = 1\) or \(b = d = -1\), and so there are only the two possibilities:
\[ 7 = c^2 \quad \text{or} \quad 3 = c^2. \]There is no integer whose square is 3 or 7, and so a factorization of \(f(x)\) as a product of quadratics in \(\mathbb{Z}[x]\), and hence in \(\mathbb{Q}[x]\), is impossible. Therefore, \(f(x)\) is irreducible in \(\mathbb{Q}[x]\). \(\blacksquare\)
The brute-force methods of the preceding example are less effective for polynomials of high degree because the system of equations that must be solved is complicated and difficult to handle in a systematic way. However, the irreducibility of certain polynomials of high degree is easily established by
Theorem 4.24: Eisenstein’s Criterion
Let \(f(x) = a_nx^n + \cdots + a_1x + a_0\) be a nonconstant polynomial with integer coefficients. If there is a prime \(p\) such that \(p\) divides each of \(a_0, a_1, \ldots, a_{n-1}\) but \(p\) does not divide \(a_n\) and \(p^2\) does not divide \(a_0\), then \(f(x)\) is irreducible in \(\mathbb{Q}[x]\).
Proof
The proof is by contradiction. If \(f(x)\) is reducible, then by Theorem 4.23 it can be factored in \(\mathbb{Z}[x]\), say
\[ f(x) = (b_0 + b_1x + \cdots + b_rx^r)(c_0 + c_1x + \cdots + c_sx^s). \]where each \(b_i, c_j \in \mathbb{Z}, r \geq 1\), and \(s \geq 1\). Note that \(a_0 = b_0c_0\). By hypothesis, \(p | a_0\), and hence, \(p | b_0\) or \(p | c_0\) by Theorem 1.5, say \(p | b_0\). Since \(p^2\) does not divide \(a_0\), we see that \(c_0\) is not divisible by \(p\). We also have \(a_n = b_rc_s\), and consequently, \(p\) does not divide \(b_r\) (otherwise \(a_n\) would be divisible by \(p\), contrary to hypothesis). There may be other \(b_i\) not divisible by \(p\) as well. Let \(b_k\) be the first of the \(b_i\) not divisible by \(p\); then \(0 < k \leq r < n\).
By the rules of polynomial multiplication,
\[ a_k = b_0c_k + b_1c_{k-1} + \cdots + b_kc_0, \]so that
\[ b_kc_0 = a_k - [b_0c_k + b_1c_{k-1} + \cdots + b_{k-1}c_1]. \]Since \(p | a_k\) and \(p | b_j\) for \(i < k\), we see that \(p\) divides every term on the right-hand side of this equation. Hence, \(p | b_kc_0\). By Theorem 1.5, \(p\) must divide \(b_k\) or \(c_0\). This contradicts the fact that neither \(b_k\) nor \(c_0\) is divisible by \(p\). Therefore, \(f(x)\) is irreducible in \(\mathbb{Q}[x]\). \(\blacksquare\)
Example 4
The polynomial \(x^{17} + 6x^{13} - 15x^4 + 3x^2 - 9x + 12\) is irreducible in \(\mathbb{Q}[x]\) by Eisenstein’s Criterion with \(p = 3\).
Example 5
The polynomial \(x^9 + 5\) is irreducible in \(\mathbb{Q}[x]\) by Eisenstein’s Criterion with \(p = 5\). Similarly, \(x^n + 5\) is irreducible in \(\mathbb{Q}[x]\) for each \(n \geq 1\). Thus,
There are irreducible polynomials of every degree in \(\mathbb{Q}[x]\).
Although Eisenstein’s Criterion is very efficient, there are many polynomials to which it cannot be applied. In such cases, other techniques are necessary. One such method involves reducing a polynomial mod \(p\), in the following sense. Let \(p\) be a positive prime. For each integer \(a\), let \([a]\) denote the congruence class of \(a\) in \(\mathbb{Z}_p\). If \(f(x) = a_kx^k + \cdots + a_1x + a_0\) is a polynomial with integer coefficients, let \(\bar{f}(x)\) denote the polynomial \([a_k]x^k + \cdots + [a_1]x + [a_0]\) in \(\mathbb{Z}_p[x]\). For instance, if \(f(x) = 2x^4 - 3x^2 + 5x + 7\) in \(\mathbb{Z}_7[x]\), then in \(\mathbb{Z}_3[x]\),
\[ \bar{f}(x) = [2]x^4 - [3]x^2 + [5]x + [7] \] \[ = [2]x^4 - [0]x^2 + [2]x + [1] = [2]x^4 + [2]x + [1]. \]Notice that \(f(x)\) and \(\bar{f}(x)\) have the same degree. This will always be the case when the leading coefficient of \(f(x)\) is not divisible by \(p\) (so that the leading coefficient of \(\bar{f}(x)\) will not be the zero class in \(\mathbb{Z}_p\)).
Theorem 4.25
Let \(f(x) = a_nx^n + \cdots + a_1x + a_0\) be a polynomial with integer coefficients, and let \(p\) be a positive prime that does not divide \(a_n\). If \(\bar{f}(x)\) is irreducible in \(\mathbb{Z}_p[x]\), then \(f(x)\) is irreducible in \(\mathbb{Q}[x]\).
Proof
Suppose, on the contrary, that \(f(x)\) is reducible in \(\mathbb{Q}[x]\). Then by Theorem 4.23, \(f(x) = g(x)h(x)\) with \(g(x), h(x)\) nonconstant polynomials in \(\mathbb{Z}[x]\). Since \(p\) does not divide \(a_n\), the leading coefficient of \(f(x)\, it cannot divide the leading coefficients of \(g(x)\) or \(h(x)\) (whose product is \(a_n\)). Consequently, \(\text{deg}(g(x)) = \text{deg}(\bar{g}(x))\) and \(\text{deg}(h(x)) = \text{deg}(\bar{h}(x))\). In particular, neither \(\bar{g}(x)\) nor \(\bar{h}(x)\) is a constant polynomial in \(\mathbb{Z}_p[x]\). Verify that \(\bar{f}(x) = \bar{g}(x)\bar{h}(x)\) in \(\mathbb{Z}_p[x]\) implies that \(\bar{f}(x) = g(x)h(x)\) in \(\mathbb{Z}_p[x]\) (Exercise 20). This contradicts the irreducibility of \(\bar{f}(x)\) in \(\mathbb{Z}_p[x]\). Therefore, \(f(x)\) must be irreducible in \(\mathbb{Q}[x]\). \(\blacksquare\)
Example 6
To show that \(f(x) = x^5 + 8x^4 + 3x^2 + 4x + 7\) is irreducible in \(\mathbb{Q}[x]\), we reduce mod 2. In \(\mathbb{Z}_2[x]\), \(\bar{f}(x) = x^5 + x^4 + x + 1\). It is easy to see that \(\bar{f}(x)\) has no roots in \(\mathbb{Z}_2\) and hence no first-degree factors in \(\mathbb{Z}_2[x]\). The only quadratic polynomials in \(\mathbb{Z}_2[x]\) are \(x^2, x^2 + x, x^2 + 1\), and \(x^2 + x + 1\). However, if \(x^2 + x = x(x + 1)\), or \(x^2 + 1 = (x + 1)(x + 1)\) were a factor, then \(\bar{f}(x)\) would have a first-degree factor, which it doesn’t. You can use division to show that the remaining quadratic, \(x^2 + x + 1\), is not a factor of \(\bar{f}(x)\). Finally, \(\bar{f}(x)\) cannot have a factor of degree 3 or 4 (if it did, the other factor would have degree 2 or 1, which is impossible). Therefore, \(\bar{f}(x)\) is irreducible in \(\mathbb{Z}_2[x]\). Hence, \(f(x)\) is irreducible in \(\mathbb{Q}[x]\). \(\blacksquare\)
Caution: If a polynomial in \(\mathbb{Z}[x]\) reduces mod \(p\) to a polynomial that is reducible in \(\mathbb{Z}_p[x]\), then no conclusion can be drawn from Theorem 4.25. Unfortunately, there may be many polynomials for which the reduction of \(f(x)\) is reducible in \(\mathbb{Z}_p[x]\), even when \(f(x)\) is actually irreducible in \(\mathbb{Q}[x]\). Consequently, it may take more time to apply Theorem 4.25 than is first apparent.
* When no confusion is likely, we omit the brackets for elements of \(\mathbb{Z}_p\).