# Thread: Linearly independant p-th roots of unity

1. ## Linearly independant p-th roots of unity

Lately, I have read a paper dealing with roots of unity.

It claimed that less than p p-th roots of unity were always linearly independant (p obviously are linearly dependant) but gave no proof and unfortunately no reference (probably because this result is very basic).

I have been trying to prove this myself and later to find a proof in a paper but I failed.

Does anyone know where I can find such a proof or can anyone give me the outline of such a proof?

Cheers,Thanks,

Mordin

2. ## Re: Linearly independant p-th roots of unity

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: xp-1 + xp-2 +....+ x + 1 = (xp - 1)/(x - 1).

to prove these p-1 primitive p-th roots of unity are linearly independent over Q, it suffices to prove that xp-1 + xp-2 +....+ x + 1 = (xp - 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) = xp-1 + pxp-2 + [p(p-1)/2]xp-3+.....+ [p(p-1)/2]x + p

by eisenstein's criterion, we have that p divides the coefficients of every xk except the leading (monic) term, and p2 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,ζ2,....,ζp-1} is linearly independent over Q:

for suppose not: then there exist some a1,...,ap-1 not all 0 in Q for which:

a1ζ +....+ ap-1ζp-1 = 0, that is: ζ satisfies the polynomial:

a1x +... + ap-1xp-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 aj 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,....,ζp-2} is linearly independent (so we can take out ζp-1 and add 1 without fear), by the same reasoning as above. so the only case left to establish is the set {1,ζ,...,ζk-1k+1,...ζp-1} for 0 < k < p-1.

so suppose c0 + c1ζ +...+ ck-1ζk-1 + ck+1ζk+1 +...+ cp-1ζp-1 = 0.

assume first that cp-1 ≠ 0. letting dj = cj/cp-1, we have that ζ is a root of:

d0 + d1x +...+ dk-1xk-1 + dk+1xk+1 +...+ xp-1 = Q(x).

thus ζ is a root of P(x) - Q(x) = (1 - d0) + (1 - d1)x +...+ (1 - dk-1)xk-1 + xk + (1 - dk+1)xk+1 +...+ (1 - dp-2)xp-2.

note that k ≤ deg(P - Q) ≤ p-2 < deg(P), (since the coefficient of xk = 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 cp-1 ≠ 0, it must be the case cp-1 = 0, in which case the linear independence (over Q) of {1,ζ,...,ζk-1k+1,...ζp-1} follows from that of {1,ζ,....,ζp-2}.