Eisenstein’s Irreducibility Criterion

We present Eisenstein’s Irreducibility Criterion which gives a sufficient condition for a polynomial over a unique factorization domain to be irreducible. This is followed by a famous application: for any prime p, the polynomial

ϕ(X) = X^{p−1} + X^{p−2} + … + 1

is irreducible in ℤ[X]. We conclude by explaining why it is possible to then deduce that ϕ(X) is irreducible in ℚ[X].

Theorem 1 (Eisenstein’s Irreducibility Criterion ). Let R be a unique factorization domain. Suppose

0 ≠ f(X) = X^{n} + a_{n−1}X^{n−1} + … + a_{0}

is a monic polynomial in R[X], and p ∈ R is a prime such that p|a_{n−1}, …, a_{0} but p^{2} ∤ a_{0}. Then f(X) is irreducible.

Proof. Suppose not, so

f(X) = (X^{r} + b_{r−1}X^{r−1} + … + b_{0})(X^{s} + c_{s−1}X^{s−1} + … + c_{0})

with 0 < r, s < n. Take i least such that p ∤ b_{i} (where we allow the case i = r and b_{r} = 1), and j least such that p ∤ c_{j} (where we allow the case j = s and c_{s} = 1). Now the (i + j)th coefficient a_{i+j} of the product is

a_{i+j} = b_{i}c_{j} + (b_{i−1}c_{j+1} + …) + (b_{i+1}c_{j−1} + …)

and so is a sum of b_{i}c_{j} and terms divisible by p. As p ∤ b_{i}c_{j}, p ∤ a_{i+j}. But p^{2} ∤ a_{0} = b_{0}c_{0} and so p cannot divide both b_{0} and c_{0}, and so either i = 0 or j = 0 and so i + j < n. So we contradict p|a_{i+j}.

Corollary 2. If p is a prime then the polynomial

ϕ(X) = X^{p−1} + X^{p−2} + … + 1

is irreducible in ℤ[X].

Proof. ϕ(X) is irreducible iff ϕ(Y + 1) is irreducible. But

ϕ(Y + 1) = (Y + 1)^{p−1} + (Y + 1)^{p−2} + … + 1
= ∑_{i=0}^{p−1} (\binom{p − 1}{i} + \binom{p − 2}{i − 1} + … + \binom{p − 1 − i}{0}) Y^{p−1−i}

= ∑_{i=0}^{p−1} \binom{p}{i}Y^{p−1−i}.

So p divides all the coefficients but the first and p^{2} does not divide the last and so by Eisenstein, ϕ(Y + 1) is irreducible.

Proposition 3. Let R be a unique factorization domain with field of fractions F. Suppose f(X) ∈ R[X] is such that f factorizes in F [X] into a product of two polynomials of lower degree. Then it does so in R[X].

Proof. Suppose f ∈ R[X] and f = gh for some g, h ∈ F [X]. Write f = af̃, g = \frac{b}{c} g̃, h = \frac{d}{e} h̃ where f̃, g̃, h̃ ∈ R[X] are primitive. Then af̃ = \frac{bd}{ce} g̃h̃ so acef̃ = bdg̃h̃. But f̃ and g̃h̃ are primitive (by Gauss) and so bd = uace for some unit u ∈ R. Then f̃ = ug̃h̃ and so f = (uag̃)h̃ is factorized in R[X].

Note that we say “it does so” in the statement of this proposition to take account of the possibility of factorization in R[X] which is irrelevant from the point of view of F [X]. For example, 2X^{2} + 2X + 2 = 2(X^{2} + X + 1) factorizes in ℤ[X] but it is irreducible in ℚ[X] (2 is a unit there).

Now, suppose we have a polynomial f(X) ∈ ℚ[X] and we want to know whether or not it is irreducible. Write f(X) = \frac{r}{s} f̃(X) with f̃ primitive in ℤ[X]. Then f is irreducible in ℚ[X] iff f̃ is irreducible in ℚ[X] iff f̃ is irreducible in ℤ[X].

Corollary 4. If p is a prime then the polynomial

ϕ(X) = X^{p−1} + X^{p−2} + … + 1

is irreducible in ℚ[X].