^{n}- 1) by (x -1) and solving the equation.

In the case of n=2, this gives us: (x

^{2}- 1)/(x - 1) = (x + 1). So, the other root of unity is x = -1.

In the case of n=3, this gives us: (x

^{3}- 1)/(x - 1) = (x

^{2}+ x + 1). Using the quadratic equation to solve for this (see Theorem, here) gives us:

(-1 ± √1 - 4)/2 = (-1 ± √-3)/2.

In light of this and a previous result (see Lemma 1, here), it is possible to generate an equation to determine the primitive nth roots of unity.

Definition 1: Cyclotomic Polynomial Φ

_{n}(x)

Φ

_{1}(x) = x - 1

Otherwise:

where d includes all divisors of n except for n itself.

Lemma 1:

If p is prime, then:

Φ

_{p}= x

^{p-1}+ x

^{p-2}+ ... + x + 1

Proof:

(1) Since p is prime, the only divisor is 1 and:

Φ

_{p}= (x

^{p}- 1)/(x - 1)

(2) Now, we know that (x

^{p}- 1)/(x - 1) = x

^{p-1}+ x

^{p-2}+ ... + x + 1 since:

(a) Let s = x

^{p-1}+ x

^{p-2}+ ... + x + 1

(b) x*s = x

^{p}+ x

^{p-1}+ ... + x

^{2}+ x

(c) x*s - s = x

^{p}- 1

(d) s(x - 1) = x

^{p}- 1

(e) s = (x

^{p}- 1)/(x - 1)

QED

Now, the next thing we need is a proof that Φ

_{n}will generate a polynomial whose solution is the primitive n-th roots of unity:

Theorem 2: Cyclotomic Polynomial

For every integer n ≥ 1, Φ

_{n}is a monic polynomial (see Definition 5, here for monic if needed) with integral coefficients and:

Φ

_{n}= ∏(x - ζ) ∈ Z[x]

where ζ runs over the set of primitive n-th roots of unity and Z is the set of integers.

Proof:

(1) For Φ

_{1}=x-1, the theorem is clearly true.

(2) Let's assume that the theorem is true up to n-1.

(3) Since d is always less than n, for each d, we have:

Φ

_{d}(x) = ∏(x - ζ) ∈ Z[x]

where ζ runs over the roots of unity with exponent d.

(4) Since each root of Φ

_{d}is a root of unity, it follows that Φ

_{d}divides x

^{n}- 1 [see Lemma 1, here]

(5) If d

_{1}, d

_{2}are distinct divisors of n, then it follows that Φ

_{d1}and Φ

_{d2}are relatively prime polynomials and distinct -- that is, they do not include any common root of unity since:

Each root of unity has a unique exponent so no root of unity included in Φ

_{n}will be found in more than one divisor Φ

_{di}.

(6) Using Corollary 2.2, here, it follows that:

is a polynomial since it is divible by a product of all Φ

_{d}

(7) Since each Φ

_{d}is monic, it follows that Φ

_{n}is monic.

(8) Since each Φ

_{d}is in Z[x] and since x

^{n}- 1 is in Z[x], using the Division Algorithm for Polynomials, it follows that Φ

_{n}is in Z[x]. [See Theorem, here]

(9) Now for each (x - ζ

_{i}) that divides Φ

_{n}, it follows that it's exponent equals n since:

(a) If its exponent is less than n and it is an n-th root of unity, then its exponent e is a divisor of n and it has been divided out.

(b) Assume that e doesn't divide n.

(c) So that n = qe + r where r ≥ 1 and r is less than e.

(d) Then:

(ζ)

^{qe+r}= 1 = (ζ)

^{qe}*(ζ

^{r})

(e) We know that (ζ)

^{qe}= 1 since:

(ζ)

^{qe}= (ζ

^{e})

^{q}

(f) But then:

ζ

^{r}= 1/(ζ

^{qe}) = 1

(g) This is impossible since r is less than e and e is the exponent of the root of unity.

(h) Therefore, we reject the assumption in 9b.

QED

References

- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001