Let y be a primitive nth root and let p be any prime not dividing n. Suppose ζ factors into g*h with h irreducible, and let h have the root y. Since the extension is integral, h is also the minimum polynomial for y.
Since p and n are coprime, yp is another primitive root. Suppose yp is a root of g(x). This means y is a root of g(xp).
Since h also has root y, and h is irreducible, h is a factor of g(xp). Say g(xp) = h*k.
Reduce all coefficients mod p, giving the following polynomial equation:
h*k = g(xp) = g(x)p
Some irreducible factor of h, possibly h itself, now divides g. Split this factor using the cyclotomic roots mod p. Now h and g have a root in common, and xn-1 has a multiple root. Since p does not divide n, the cyclotomic extension does not produce any double roots. We saw this earlier, when we embedded the cyclotomic extension in a larger finite field, and all the roots of 1 are distinct. If you want another proof, use formal derivatives. The derivative of xn-1 is nxn-1, which is nonzero, since p does not divide n. This has no factors in common with xn-1, hence there are no double roots.
This contradiction means yp is not a root of g(x). It therefore belongs to h(x), as does yp2, yp3, etc.
Let's look at a specific primitive root yj. If p2 divides j, raise y to the p, and then to the p again. This is a primitive root that lies in h. Then, if q divides j, raise this root to the q and find another primitive root in h. Continue through the primes of j, until yj lies in h.
Since h contains all the primitive roots, it is equal to ζn, and ζn is irreducible.
Adjoining an nth root of 1 to Q is an extension of order φ(n), with galois group Zn*.
By gauss' lemma, ζn(x) is irreducible over Q[x], or any subring of Q[x], such as Z[x] localized about p.
If n is odd, we can adjoin y or -y and get the same field extension. (This was discussed in the previous section.) This extension has a fixed dimension d. The irreducible polynomial associated with y has the same degree as the irreducible polynomial associated with -y. These are factors of ζn and ζ2n respectively. If their common degree is φ(n) then both cyclotomic polynomials are irreducible. If not, then both cyclotomics split. One splits iff the other one does, so we may as well assume n is even.
Next let n be even, but not divisible by 4. Extending Q by ζn does not span i = sqrt(-1). If it did, we would span all the 2n roots of 1, and the extension would have dimension φ(2n), which is impossible for a polynomial of degree φ(n).
Start with Q, adjoin i, then adjoin y, a primitive nth root of 1. This gives the aforementioned extension of dimension φ(2n). The polynomial associated with y is the same as it was before, namely ζn. This polynomial, of degree φ(n), produces an extension of dimension φ(n), hence it is irreducible over Q[i], the complex rationals. By Gauss' lemma, it is also irreducible over G, the gaussian integers.
Finally, when n is divisible by 4, ζ always splits. Adjoin i first, then y, and create an extension of dimension φ(n). The first has dimension 2, so the irreducible polynomial associated with y, relative to G, has degree φ(n)/2. In fact ζ splits precisely into two irreducible factors, each having half the degree.
Let's illustrate with n = 8. ζ(8) = x4+1. This does not split in the integers, but bring in i, and it becomes (x2+i)×(x2-i).
Let p q and r be polynomials with coefficients in a field F, and let p = q×r. If v is a nonzero constant taken from F, then substituting x → vx is a ring automorphism. In other words, it commutes with addition and multiplication. You can multiply q times r, and replace x with vx, or substitute first and then multiply. The result is the same. Therefore p(x) is reducible iff p(vx) is reducible. Turn this around and p(x) is irreducible iff p(vx) is irreducible.
We have shown that the cyclotomic polynomial, xn-1 + xn-2 + … + x3 + x2 + x + 1, is irreducible over the rationals. The same is true when x is replaced with x/d. Multiply this by dn-1 and the following is irreducible.
xn-1 + dxn-2 + d2xn-3 + … + dn-2x + dn-1
The second proof involves field theory. I'll illustrate with n = 7. Let p(x) = x6+2x5+4x4+8x3+16x2+32x+64, and let z be a root of this polynomial in the complex plane. Divide z by 2 and find a primitive 7th root of 1. So adjoining z to the rationals includes the cyclotomic extension, having dimension 6. The irreducible polynomial associated with z has degree at least 6, and divides p(x), hence p is irreducible.
Set d = -1 to show the alternating sum of powers of x is irreducible.