IF p is a prime, then there are p-1 primitive roots of unity in the complex numbers (the p-th roots of unity form a cyclic subgroup of the multiplicative group of non-zero complex numbers of order p. by an elementary result in group theory, any non-identity element of this group is a generator, that is: any p-th root of unity BESIDES 1 is primitive).

now all these primitive p-th roots of unity satisfy the polynomial: x^{p-1}+ x^{p-2}+....+ x + 1 = (x^{p}- 1)/(x - 1).

to prove these p-1 primitive p-th roots of unity are linearly independent over Q, it suffices to prove that x^{p-1}+ x^{p-2}+....+ x + 1 = (x^{p}- 1)/(x - 1) is irreducible over Q, and by Gauss' lemma it suffices to prove irreducibility over Z.

if we call this polynomial P(x), note that if P(x) = f(x)g(x) for f,g in Z[x], then surely P(x+1) = f(x+1)g(x+1) = h(x)k(x) for some h,k in Z[x].

but P(x+1) = ((x+1)^{p}- 1)/((x+1) - 1) = (1/x)((x+1)^{p}- 1) = x^{p-1}+ px^{p-2}+ [p(p-1)/2]x^{p-3}+.....+ [p(p-1)/2]x + p

by eisenstein's criterion, we have that p divides the coefficients of every x^{k}except the leading (monic) term, and p^{2}does not divide the constant term, so P(x+1) is irreducible over Z, hence P must be.

thus P(x) must be the minimal polynomial of each primitive p-th root of unity, for if some monic polynomial m(x) with deg(m) < deg(P) gave us for some primitive root ζ, m(ζ) = 0, and we choose the degree of m to be least among such polynomials, we have:

P(x) = q(x)m(x) + r(x), with deg(r) < deg(m). so that P(ζ) = q(ζ)m(ζ) + r(ζ), that is: 0 = 0 + r(ζ), contradicting the minimality of deg(m), unless r is the 0-polynomial. but that implies that P(x) is reducible, so there can be no such m for any primitive p-th root of unity ζ.

finally, let us show that the set {ζ,ζ^{2},....,ζ^{p-1}} is linearly independent over Q:

for suppose not: then there exist some a_{1},...,a_{p-1}not all 0 in Q for which:

a_{1}ζ +....+ a_{p-1}ζ^{p-1}= 0, that is: ζ satisfies the polynomial:

a_{1}x +... + a_{p-1}x^{p-1}= (x)(s(x)) where deg(s(x)) = p-2 < p-1 = deg(P).

since ζ is non-zero, and P is the minimal polynomial of ζ, we have a contradiction, so all the a_{j}must in fact be 0.

now any subset of a linearly independent set must also be linearly independent. the question might remain: what if we added 1 to our set, and subtracted some power of ζ?

it is easy to see that the set {1,ζ,....,ζ^{p-2}} is linearly independent (so we can take out ζ^{p-1}and add 1 without fear), by the same reasoning as above. so the only case left to establish is the set {1,ζ,...,ζ^{k-1},ζ^{k+1},...ζ^{p-1}} for 0 < k < p-1.

so suppose c_{0}+ c_{1}ζ +...+ c_{k-1}ζ^{k-1}+ c_{k+1}ζ^{k+1}+...+ c_{p-1}ζ^{p-1}= 0.

assume first that c_{p-1}≠ 0. letting d_{j}= c_{j}/c_{p-1}, we have that ζ is a root of:

d_{0}+ d_{1}x +...+ d_{k-1}x^{k-1}+ d_{k+1}x^{k+1}+...+ x^{p-1}= Q(x).

thus ζ is a root of P(x) - Q(x) = (1 - d_{0}) + (1 - d_{1})x +...+ (1 - d_{k-1})x^{k-1}+ x^{k}+ (1 - d_{k+1})x^{k+1}+...+ (1 - d^{p-2})x^{p-2}.

note that k ≤ deg(P - Q) ≤ p-2 < deg(P), (since the coefficient of x^{k}= 1), P - Q is not the 0-polynomial. but P is the minimal polynomial of ζ, so we again have a contradiction. thus we are forced to conclude that no such Q exists. since we only assumed c_{p-1}≠ 0, it must be the case c_{p-1}= 0, in which case the linear independence (over Q) of {1,ζ,...,ζ^{k-1},ζ^{k+1},...ζ^{p-1}} follows from that of {1,ζ,....,ζ^{p-2}}.