Irreducibility of polynomials in ℚ
We were able to characterize irreducible polynomials on ℝ and ℂ, but we won't able to to the same in ℚ. The factorisation in ℚ is strictly related to that on ℤ. So we are going to study polynomials on ℤ. We already studied different properties of polynomials on a field K. We can as well define polynomials with coefficients in a communtative ring with unit as ℤ, even though some of the properties that hold true for fields do not hold as well for commutative rings.
Lemma 4.6.1. Let f(x) ∈ ℚ[x]. Then
f(x) = i/m f*(x)
where ∑ni=0 aixi ∈ ℤ[x], (a0,a1, ..., an) = 1 and d,m ∈ ℤ, (d,m) = 1.
Proof. Let f(x) = q0 + q1 x + q2x2 + ... + qnxn, qi = bi/ci ∈ ℚ ∀i and bi,ci ∈ ℤ. Thus,
f(x) = b0/c0 + b1/c1x + ... + bn/cn xn
Let m' the lcm(c0, c1,... , cn), the polynomial m'f(x) = b'0 + b1'x + ...+ b'nxn is a polynomial with integer coefficients. Let d' = GCD(b'0, b'1,... , b'n) then m'f(x) = d' (a0 + a1x + ... + anxn) with ai ∈ ℤ and GCD(a0 + a1x + ...+ anxn) = 1. If necessary we divide by GCD(d',m') both sides and get
mf(x) = d ⋅ ∑ni=0 aixi
which is the claim. □
Definition 4.6.2. Let f(x) ∈ ℤ[x]. The GCD of the coefficients of f(x) is known as content or divisor of f(x).
Definition 4.6.3. A polynomial f(x) = ∑ni = 0 aixi ∈ ℤ[x] is said primitive if the gcd of its coefficient is 1 (i.e. its content is 1).
That is, a polynomial is primitive if and only if there is no prime integer that divides all of its coefficients.
The lemma above states that if f(x) ∈ ℚ[x] then there exists in ℤ[x] a primitive polynomial f*(x) associate to f(x).
Example 4.6.4. Consider the following polynomial with coefficients in ℚ
f(x) = 5/7 + 5/8 ⋅x − 10/3 ⋅ x2
m = 7 ⋅ 8 ⋅ 3 = 168. Then
168 f(x) = 24 ⋅ 5 + 21 ⋅ 5x − 56 ⋅ 10x2 = 5(24 + 21x − 112x2)
as GCD(24, 21, 112) = 1, f*(x) = 24 + 21x − 112x is primitive and associate to f(x).
What presented shows that when we want to factorize a polynomial f(x) ∈ ℚ[x], we can always suppose that the polynomial is in ℤ[x] primitive or not depending on cases.
Lemma 4.6.5 (Gauss's Lemma). The product of two primitive polynomials is again a primitive polynomial.
Proof. Let
f(x) = a0 + a1x + ⋅⋅⋅ + anxn, g(x) = b0 + b1x + ⋅⋅⋅ + bnxm, ai,bi ∈ ℤ
be the two primitive polynomials. Assume by contradiction that f(x)g(x) is not primitive. Then there exists a prime number p dividing all the coefficients of f(x)g(x). This element p cannot divide all the coefficients of f(x) and of g(x) by the hypothesis of primitivity. Let ah and bk be the smallest-indexed coefficients of f(x) and g(x), respectively, that are not divided by p. Look at the coefficient with index h + k in f(x)g(x). It is
ch+k = ah bk + (ah−1 bk + 1 + ··· + a0 bh+k) + (ah+1 bk−1 + ··· + ah+k b0).
Now, p divides ch+k and both summands in the parentheses, and so it divides ah bk. Thus, it divides one of the two factors, contradicting the hypothesis.□
Corollary 4.6.6. The content of the product of two polynomials equals the product of the contents of the two polynomials.
Proof. Let f(x) = g(x)h(x), with f(x), g(x), h(x) ∈ ℤ[x]. Let's extract the content of the polynomials
f(x) = df*(x) = d1g*(x) d2h*(x) = d1d2 · g*(x)h*(x)
Since g*(x)h*(x) is primitive, we have d = d1d2, which means that the content of the product is the product of the content of the single polynomials.□
4.6.7 (Gauss's theorem). If a polynomial f(x) ∈ ℤ[x] factors as a product of two polynomials g(x), h(x) with rational coefficients, then it also factors as the product of two polynomials of the same degree with integer coefficients.
Proof. Assume first that f(x) is primitive. So let f(x) = g(x)h(x), with g(x), h(x) ∈ ℚ[x]. By lemma 4.6.1, we have
g(x) = d1/m1 g*(x), h(x) = d2/m2 h*(x)
with g*(x) and h*(x) ∈ ℤ[x] primitive polynomials in ℤ[x] and di, mi integers. As a consequence, f(x) = (d/m)g*(x)h*(x), d = d1 d2, m = m1 m2. By Gauss lemma, the polynomial f*(x)= g*(x)h*(x) is primitive and we have
mf(x) = df*(x) with f(x) and f*(x) primitive
The left-hand-side polynomial and the right-hand-side one have the same content. But as f(x) and f*(x) are primitive, we have m = d and so
f(x) = g*(x)h*(x),
which is a factorisation over ℤ, with with associate factors of same degree as the original factorisation in ℚ.
If f(x) is not primitive let again f(x) = g(x)h(x) a factorisation in ℚ[x] and let f(x) = df*(x) (that is f(x) is the product of its content by a primitive polynomial), it will result df*(x) = g(x)h(x), then
f*(x) = d−1g(x)h(x),
is a factorisation over ℚ[x] and f*(x) is primitive. Thus, by what has already been proved, f*(x) is factorisable over ℤ as well, that is,
f(x) = ḡ(x)h̄(x), ḡ(x), h̄(x) ∈ ℤ[x]
by replacing in the original expression for f(x)
f*(x) = df*(x) = dḡ(x) h̄(x)
is a factorisation of f(x) in ℤ. □
Remark
Gauss theorem states that if a polynomial with coefficients in ℤ can be factored over ℚ, then is factorasable over ℤ as well, or equivalently if it is irreducible over ℤ then is irreducible over ℚ. Now somebody might want to venture the hypothesis that a polynomial with coefficients in ℤ is irreducible if and only if it is irreducible over ℚ, that's what happens for polynomials in ℚ[
4.6.8 Example. Consider the polynomial
f(x) = 3x2 + 6,
which is irreducible over ℚ, as it is associate to the polynomial x2 + 2 which is irreducible over ℚ. However, f(x) is reducible over ℤ, as the factorisation 3x2 + 6 = 3(x2 + 2) is a non-trivial factorisation over ℤ because 3 is not invertible over ℤ!
In other words, the two polynomials 3x2 + 6 and x2 + 2 are not associate in ℤ[x], as the invertible elements in ℤ[x] are not all non-zero constants, but only ±1. It should now be clear why reducibility of polynomials over ℤ does not imply their reducibility over ℚ. On the other hand, if a polynomial is reducible over ℚ, then it is reducible over ℝ. Indeed, both ℚ and ℝ are fields, one contained in the other, and an element of ℚ is invertible in ℚ if and only if it is invertible in ℝ. Thus, if a polynomial in ℚ[x] is irreducible over ℝ it is obviously also irreducible over ℚ.
Even if over ℤ the claim "a polynomial is irreducible if and only if it is irreducible over ℚ" is not in general valid, it is for primitive polynomials as the following lemma states:
Lemma 4.6.9. A primitive polynomial f(x) ∈ ℤ[x] is irreducible over ℤ if and only if it is irreducible over ℚ.□
Proof. In case of a primitive polynomial, it can't have a factorisation in which one of the factor is an element of ℤ ≠ ±1.
In order to decide about the reducibility of a polynomial f(x) ∈ ℤ[x] over ℚ it is useful in the first place to have a criterion to establish whether the polynomial has roots in ℚ or not. If it has roots, it is sure to be reducible. The following proposition teaches us something useful in this regard.
Proposition 4.6.10 (Rational root theorem). Let f(x) = a0 + a1 x + a2 x2 + ··· + an xn ∈ ℤ[x]. Let α = r/s ∈ ℚ be a root of f(x), with r and s relatively prime. Then r | a0 and s | an.
Proof. If α = r/s is a root of f(x), we find 0 = f(r/s) = a0 + a1 (r/s) + ··· + an (r/s)n. Multiplying by sn we get
0 = sn a0 + sn−1 a1 r + ··· + an rn =
= s(sn−1 a0 + sn−2 a1 r + ··· + an−1 rn−1) + an rn =
= sna0 + r(sn−1 a1 + ··· + an rn−1).
From the relation 0 = s(sn−1 a0 + sn−2 a1 r + ··· + an−1 rn−1) + an rn, it follows that s | an rn, and from 0 = sna0 + r(sn−1 a1 + ··· + an rn−1) it follows that r | sna0. As GCD(r, s) = 1, we may conclude that s | an and r | a0.□
Proposition 4.6.10 gives a set of elements of ℚ which are candidates to be roots of the polynomial f(x ∈ ℤ[x], these are of the form r/s, where r varies in the set of divisors of a0 and s is among the set of divisors of an.
Example 4.6.11. Determine whether the polynomial
f(x) = 3x3 − 4x2 + 2
has rational roots. By applying the algorithm described above, the possible rational roots of f(x) must lie in the set
{±1, ±2, ±1/3, ±2/3}
Corollary 4.6.12. If a monic polynomial with integer coefficients has a rational root, this is a an integer.
Proof. Monic means an = 1. s|1 i.e s = 1 and r/s = r.
As it may be easily verified, none of these rational numbers is a root of f(x), so f(x) admits no rational roots. Being of degree 3, the polynomial is irreducible over ℚ.
Exercises
Determine whether the polynomial 8x3 6x −1 is irreducible over ℚ.
Solutions
By applying the Rational root theorem, the possible rational roots of f(x) must lie in the set
{±1/1, ±1/2, ±1/4, ±1/8}
thus f is irreducible over ℚ. ■