Let p be a prime integer, and let C(p,i) denote the binomial coefficient. We can use EIC to prove that the polynomial f(x) = xp-1 + ... + x + 1 is irreducible over Q - i.e. that it has no non-trivial factorisation.


Suppose that f(x) is reducible; that is, it can be written as f(x)=r(x)*s(x) such that both r(x) and s(x) have degrees less than f(x). Define f'(x), r'(x) and s'(x) by f'(x) = f(x+1), r'(x) = r(x+1) and s'(x) = s(x+1). It is clear that the degree of these new polynomials are the same as the originals, and furthermore:

    f'(x) = f(x+1) = r(x+1)*s(x+1) = r'(x)*s'(x)
So now f'(x) must also be reducible. Noticing that
            xp - 1
    f(x) = --------
             x - 1
it follows that:
                      (x+1)p - 1
    f'(x) = f(x+1) = ------------
                       (x+1) - 1
Using the binomial theorem to expand the numerator and simplifying, we get that:
    f'(x) = C(p,1) + C(p,2)*x + ... + C(p,p-1)*xp-2 + xp-1
Now since p divides all those coeffients except for the coefficient of xp-1, and C(p,1) = p, we can use EIC (see Noether's writeup above) to deduce that f'(x) is irreducible, which gives us a contradiction!

Hence f(x) = xp-1 + ... + x + 1 is irreducible over Q.