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 = Xt + ct-1Xt-1 + ... + c1X + c0 ∈ Z[X]
If there is a prime number p which divides ci for i = 0, ..., t-1 but p2 does not divide c0
P is irreducible over Q
(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 + an-1Xn-1 + ... + a1X + a0 ∈ Z[X]
g = Xm + bm-1Xm-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/an is monic and an*g is monic since P = (f/an)*(g*an)]
(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 ai. It is clear that 0 ≤ i ≤ n-1. So, we can conclude that for any i, p divides a0, a1, ..., ai.
(10) Let k = i+1, it is clear that 1 ≤ k ≤ n and p does not divide ak.
ci+1 = ai+1b0 + aib1 + ai-1b2 + ... + a0bi+1.
where bj = 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 ci+1. [From the given in the theorem that p divides c0, ..., ct-1]
(14) So, from step #13, it follows that since p divides ci+1 and p divides a0, a1, ..., ai [step #9] that p must also divide ai+1b0.
(15) But this is impossible since p doesn't divide b0 (step #8) and p doesn't divide ai+1 [step #10]
(16) So we have a contradiction and we must reject our assumption in step #1.
Corollary 1.1: For every prime p, the cyclotomic polynomial
Φp(x) = xp-1 + xp-2 + ... + x + 1
is irreducible over the field of rational numbers
(1) From the definition of the cyclotomic polynomial (see Definition 1, here):
Φp(x) = (xp - 1)/(x - 1)
(2) Setting x = y + 1, we get:
Φp(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.)
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001