Eisenstein’s irreducibility criterion

There exists a criterion to determine whether a polynomial of arbitrary degree is irreducible over ℚ.

Proposition 5.7.1. (Eisenstein’s irreducibility criterion). Consider a polynomial f(x) = a₀ + a₁ x + ··· + a_n xⁿ ∈ ℤ[x]. Let p be a prime number such that

1. p∤ a_n

2. p| a_i,  ∀i = 0,...,n−1

3. p² ∤ a₀

Then f(x) is irreducible over ℚ.

Proof. By Gauss theorem it is sufficient to prove that f(x) is irreducible over ℤ. Assume by contradiction that f(x) = g(x)h(x), with

polynomials with integer coefficients of degree r < n and s < n respectively. Then r + s = n and b₀c₀ = a₀ Notice that p | b₀ and p | c₀ cannot hold simultaneously, or else we would have p²| a₀. So assume that p divides, say, b₀ but not c₀ Notice that p cannot divide all the b_is, or else p would divide all the a_is, contradicting the hypothesis.□

Notice that Eisenstein’s criterion gives a sufficient condition for irreducibility, but not a necessary one.

Example 5.7.2. It is quite evident that the Eisenstein criterion implies that the polynomial x⁵ + 5x + 5 is irreducible in ℤ[x], by using p = 7. But note that the criterion does not directly apply to the polynomial x⁵ + 5x + 4. However, you will discover with the next proposition how to apply the criterion to show that this is in fact an irreducible polynomial.

If Eisenstein’s criterion does not directly apply to a polynomial, it might nevertheless be possible to apply it after having suitably modified the polynomial. Given a constant α, it is for instance possible to consider, rather than the polynomial f(x), a new polynomial f(x − α) or, for α ≠ 0, or the polynomial f(x/α), by observing that f(x) ∈ R[x] is irreducible over the field R if and only if f(x − α) and f(x/α) with α ∈ 𝕂 are as well.

Lemma 5.7.3. Let p(x) in R[x] with a ∈ R. The application

is a ring homorphism. It is a ring isomorphism for p(x) = x − a and therefore,

Analogously for p(x) = x / a, with a ≠ 0,

Proof. We must prove that the mapping is a homomorphism, i.e. it preserves the operations between polynomials:

then T_p is a ring homomorphism. T_p is an isomorphism for the case p(x) = x − a but T_p are non-surjective for p constant or for deg(p) ≥ 2: say p(x) = x^d + (lower-degree terms), then for any nonconstant r_n xⁿ + ... + r₁ x + r₀ in R[x], substituting p(x) in for x in there gives r_n x^(dn) + (lower-degree terms). In particular this image can never be x so it is not surjective.  □

For example, Eisenstein’s criterion cannot be applied to x² + 1, but substituting x + 1 for x gives another proof that x² + 1 is irreducible over ℚ As an application of this idea, we prove the following result:

Example 5.7.4. If p is a prime number, the polynomial x^p−1 + x^p−2 + ··· + x² + x + 1 is irreducible over ℚ. Note first that

So, substituting x + 1 for x, we get

We may now apply Eisenstein’s criterion with respect to the prime p, and prove that the original polynomial is irreducible over ℚ. □

Completing Example 5.7.4. We are now able to show that x⁵ + 5x + 4 is irreducible over ℚ; Substituting x + 1 for x, we get

we can chose p = 5 and apply Eisenstein criterion to prove its irreducibility.  ■

Another method, which is practical only when the degree of the polynomial under consideration is not too large, amounts to looking directly for a factorization. For instance, if f(x) is a polynomial (that we may always suppose primitive and with integer coefficients) of degree 5 and if we have previously checked that the polynomial has no rational roots, then, if f(x) is reducible, it may only decompose as the product of a degree two polynomial and a degree three one. So, writing down the factors with indeterminate coefficients and equating the coefficients of the polynomial and those of the product, we obtain a system and we are interested in its integer solutions: indeed, as we know, we may always reduce to a factorization in ℤ[x]. If the system is not compatible, then the original polynomial is irreducible.

Example 5.7.5. Prove that x⁴ + 1 is irreducible over ℚ.
The polynomial has clearly no rational roots, so it may only factor into the product of two quadratic polynomials. Keeping in mind that the leading coefficient of the product is the product of the leading coefficients, and the constant term of the product is the product of the constant terms of the factors, we may write

where we pick either both plus signs or both minus signs. So by expressing the product (picking +1) we have

We must then have

which does not admit integer solutions (By Gauss lemma you can take a, b to be integers). Notice that, over ℝ, x⁴ + 1 factors as x⁴ + 1 = (x² + √2x + 1)(x² − √2x + 1). This method is not at all computationally efficient: in particular, it becomes more and more infeasible as the degree of the polynomial increases.
A more elegan method consists in employing Eisentein criterion by applying the automorphism ℚ[x] ⟶ ℚ[x] given by f(x) ⟼ f(x + 1).

Every coefficient other than the leading one is divisible by 2 and the constant term isn’t divisble by 2², so Eisenstein’s criterion tells us that x⁴ + 4x³ + 6x² + 4x + 2 is irreducible and by Lemma 5.7.3 the original polynomial as well.  ■

Another useful method is the following.

Theorem 5.7.6. (Mod p Irreducibility Test). Let f(x) in ℤ[x], with degree n > 0. Let p be a prime such that p ∤ a_n. Reduce the coefficients modulo a prime number p, that is to say, consider the polynomial f(x) as having coefficients in ℤ_p. Denote by f̄(x) ∈ ℤ_p[x] the new polynomial, which we shall call the reduction of f(x) modulo p. If f(x) is reudicble over ℚ then f̄(x) is reducible over ℤ_p, or the contrapositive

Proof. If f(x) in ℤ[x] is re­ducible over ℚ, then by Gauss Theorem f(x) = g(x)h(x) with g(x), h(x) ℤ[x], with deg(g), deg (h) < n . The reduction of both g(x) and h(x) modulo p leads to ḡ(x) and h̄(x), so f̄(x) = ḡ(x) h̄(x). If p ∤ a_n, then p does not divide the leading coefficients, so f̄(x) still has degree n in ℤ_p[x], and ḡ(x) and h̄(x) have lower degree than n. So f̄(x) is reducible in ℤ_p[x].  □

The reduction modulo p of f, can be seen as the action of the homomorphism π : ℤ[x] ⟶ (ℤ/pℤ)[x] for which f(x) is not in the kernel, (f(x) is in the kernel iff every coefficient of f(x) is divisible by p.
If π(f) is irreducible in (ℤ/pℤ)[x] then f(x) is irreducible in ℤ[x]. As ℤ/pℤ has p elements, the number of polynomials of any fixed degree with coefficients in ℤ/pℤ) is finite. This can make it easy to prove that a polynomial is irreducible in (ℤ/pℤ)[x]. Be careful to note, however, that there exist polynomials that are irreducible in ℤ[x] but are reducible modulo p for all primes p, so the theorem above does not have a viable converse statement.

Nota bene.The converse of the theorem is not true, i.e. it is not the case that if, for some p, f̄(x) is reducible over ℤ_p, then f(x) is reducible over ℚ. We have seen that x⁴ + 1 is irreducible over ℚ, though reducible over ℤ₂; indeed, x⁴ + 1 = (x² + 1)(x² + 1) in ℤ₂[x].  ■

Since, the irreducibility test in ℤ_p[x] is finite, since there are only finitely many possible factors of the polynomial, thus working modulo p is convenient and not computationally expensive

Example 5.7.7. Let f(x) = 21x³ −3x² + 2x + 9. Then, over ℤ₂, we have f(x) = x³ + x² + 1 and, since f̄(0) = 1 and f̄(1) = 1, we see that f(x) is irreducible over ℤ₂. Thus, f(x) is irreducible over ℚ. Notice that, over ℤ₃, f(x) = 2x is irreducible, but we may not apply 5.7.6 as 3|21, to con­clude that f(x) is irreducible over ℚ. □

Example 5.7.8. Let 5x⁴ −2x³ + 9x − 1. If we consider it as a polynomial with coefficients in ℤ₂, the polynomial is x⁴ + x + 1. This polynomial has no roots in ℤ₂, so, if it factors, it does so into the product of two second degree factors: that is, we would get

or

which is clearly impossible. So x⁴ + x + 1 is irreducible over ℤ₂, which implies that the original polynomial is irreducible over ℚ. □

Methods to study the irreducibility of a polynomial over ℚ

• Consider without loss of generality only primitive polynomials f(x) ∈ ℤ[x];

• If f(x) is of degree 2 or 3, it is irreducible over ℚ if and only if it has no roots in ℚ; If f(x) has degree 2 then one of its factor is a 1st degree polynomial x − a; By Ruffin's theorem if f has no root a, then it cannot be divisible by x − a.

  In the case of f(x) of degree 3, if f were reducible one of its factor would have degree 1, but again if f has no root a, it cannot be divisible by x − a.

• Use the test for the existence of rational roots based upon Proposition 4.6.10 to conclude that f(x) is reducible: the existence of roots implies the reducibility of the polynomial;

• If possible, especially for low degrees, look for a factorization into polynomials with integer coefficients;

• When there is a prime p verifying the hypothesis of Eisenstein’s criterion, use it to conclude that f(x) is irreducible over ℚ;

• use transformations of the form x → x + α to be able to apply Eisenstein’s criterion;

• consider f̄(x) ∈ ℤ_p[x] rather than f(x) ∈ ℤ[x]: if there exists a p not dividing the leading coefficient of f(x) and such that f̄(x) is irreducible over ℤ_p, then f(x) is irreducible over ℚ.