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) = x^{p-1} + ... + x + 1 is irreducible over Q - i.e. that it has no non-trivial factorisation.

*Proof:*
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

x^{p} - 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)*x^{p-2} + x^{p-1}

Now since p divides all those coeffients except for the coefficient of x

^{p-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) = x^{p-1} + ... + x + 1 is irreducible over Q.