Eisenstein: Criteria for Irreducibility of Polynomials

In today's blog, I will go over a classical result from Ferdinand Eisenstein. The content in today's proof is taken directly from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

I will later use this result in Carl Friedrich Gauss's proof that all roots of unity are expressible as radicals.

For those who are not familiar, a monic polynomial is a polynomial whose leading coefficient is 1.

Theorem 1: Eisenstein's Criteria for Irreducibility of Polynomials

Let P be a monic polynomial with integral coefficients such that:

P = X^t + c_t-1X^t-1 + ... + c₁X + c₀ ∈ Z[X]

If there is a prime number p which divides c_i for i = 0, ..., t-1 but p² does not divide c₀

then:

P is irreducible over Q

Proof

(1) Assume that P is not irreducible over Q.

(2) Then there exist two non-constant polynomials f,g ∈ Q[X] such that:

P = fg

(3) Assume that f has degree n with n ≥ 1 and g has degree m with m ≥ 1 so that:

f = Xⁿ + a_n-1X^n-1 + ... + a₁X + a₀ ∈ Z[X]

g = X_m + b_m-1X^m-1 + ... + b₁X + b₀ ∈ Z[X]

(4) Since P is monic, we can assume that f,g are monic [if f,g are not monic, then (f/a_n is monic and a_n*g is monic since P = (f/a_n)*(g*a_n)]

(5) From step #2, it is also clear that c₀ = a₀*b₀

(6) Using Euclid's Lemma (see Lemma 2, here), it is clear that for the prime p, p divides a₀ or b₀.

(7) It is also clear that since p² does not divide c₀ that p cannot divide both a₀ and b₀.

(8) We can assume that p divides a₀ but not b₀ (if it did not, we could switch f,g since they are interchangeable with respect to p and make the same assumption)

(9) Let i be the largest index of a such that p divides a_i. It is clear that 0 ≤ i ≤ n-1. So, we can conclude that for any i, p divides a₀, a₁, ..., a_i.

(10) Let k = i+1, it is clear that 1 ≤ k ≤ n and p does not divide a_k.

(11) So:

c_i+1 = a_i+1b₀ + a_ib₁ + a_i-1b₂ + ... + a₀b_i+1.

where b_j = 0 if j is greater than m.

(12) Now i+1 ≤ n which is less than t since n+1 ≤ m+n = t [since m ≥ 1 and n ≥ 1]

(13) So it follows that i+1 ≤ t-1 and p divides c_i+1. [From the given in the theorem that p divides c₀, ..., c_t-1]

(14) So, from step #13, it follows that since p divides c_i+1 and p divides a₀, a₁, ..., a_i [step #9] that p must also divide a_i+1b₀.

(15) But this is impossible since p doesn't divide b₀ (step #8) and p doesn't divide a_i+1 [step #10]

(16) So we have a contradiction and we must reject our assumption in step #1.

QED

Corollary 1.1: For every prime p, the cyclotomic polynomial

Φ_p(x) = x^p-1 + x^p-2 + ... + x + 1

is irreducible over the field of rational numbers

Proof:

(1) From the definition of the cyclotomic polynomial (see Definition 1, here):

Φ_p(x) = (x^p - 1)/(x - 1)

(2) Setting x = y + 1, we get:

Φ_p(y+1) = ([y + 1]^p - 1)/([y + 1] - 1) =

= ([y + 1]^p - 1)/y

(3) Using the Binomial Theorem (see Theorem, here):



(4) So, combining this with step #2 gives us:












(5) Now, it is clear that for (y+1)^p all the resulting coefficients are integral (since integers are closed on additional and multiplication, see Lemma 1 and Lemma 2, here).

(6) Further, we know that for p!/[(m!)(p-m)!] where 1 ≤ m ≤ p-2, p is not divisible by (m!) or by (p-1)! [Since p is a prime and only divisible by p and m! and (p-m)! does not include p]

(7) So, we can conclude that for m=1, ..., p-2, we have p divides p!/[(m!)(p-m)!]

(8) Now, we can use Eisenstein's Criterion above to conclude that the cyclotomic polynomial is irreducible in the field Q since:

the cyclotomic polynomial is monic, p divides all coefficients except for the leading coefficient, and p² does not divide the last coefficient (p² does not divide p.)

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001