## Wednesday, January 02, 2008

### Cyclotomic Equation

Lemma 1:

Let ζ = the n-th root of unity where ζ ≠ 1

Then:

1 + ζ + ζ2 + ... + ζn-1 = 0

Proof:

(1) Since ζn = 1, we have:

1 + ζ + ζ2 + ... + ζn-1 = ζ(1 + ζ + ζ2 + ... + ζn-1)

(2) Now ζ ≠ 0 [since 0n ≠ 1] and ζ ≠ 1

(3) Let x = 1 + ζ + ζ2 + ... + ζn-1

(4) Assume that x ≠ 0

(5) So, x = ζx [from step #1]

(6) Since we assume x ≠ 0, we can divide both sides by x to get 1 = ζ

(7) But this contradicts step #2 so we reject our assumption in step #3 and conclude that:

1 + ζ + ζ2 + ... + ζn-1 =0

QED

Corollary 1.1:

if ζ is a primitive n-th root of unity, then:

a = (-a) ζ + (-a)ζ2 + ... + (-a)ζn-1

Proof:

(1) From Lemma 1 above:

-1 = ζ + ζ2 + ... + ζn-1

(2) So, a = (-a)(ζ + ζ2 + ... + ζn-1 ) = (-a) ζ + (-a)ζ2 + ... + (-a)ζn-1

QED

Lemma 2:

if i ≡ j (mod q)

and

ζ is a primitive qth root of unity

Then:

ζi = ζj

Proof:

(1) i ≡ j (mod q) implies there exists integers r,s such that:

i +r*q = j + s*q

(2) So, it follows that:

ζi*(ζq)r = ζj*(ζq)s

(3) Since ζq = 1, we have:

ζi*1r = ζj*1s

which means that:

ζi = ζj

QED