Gauss: Q(μp)

In today's blog, I will review some basic properties of Q(μ_p) which is the set of complex numbers which are rational expressions in the p-th roots of unity.

The content in today's blog is taken straight from Jean-Pierre Tignol's Galois Theory of Algebraic Equations. I will later use the results in today's blog in the proof by Carl Friedrich Gauss that all roots of unity are expressible as radicals.

Definition 1: μ_p

Let μ_p denote the set of p-th roots of unity.

Example:

μ₁ = {1}
μ₂ = {1, -1}
μ₃ = {1, (1/2)[-1 + √-3], 1/2)[-1 - √-3])
μ₄ = {1, -1, i, -i}

Now, let's use this to define a field (see Definition 3, here for definition of a field if needed)

Definition 2: Q(μ_p)

Let Q(μ_p) denote the set of complex numbers that are rational expressions in these p-th roots of unity.

This gives us that:



Now, we can show that Q(μ_p) = Q(ζ).

Lemma 1:

Q(μ_p) = Q(ζ)

Proof:

(1) From Defintion 1 above:

μ_p = { ρ₁, ..., ρ_p}

(2) Let ζ = a primitive p-th root of unity (see Definition 2, here for definition of a primitive p-th root of unity if needed)

(3) Using Theorem 3, here, we have:

μ_p = { 1, ζ, ζ², ..., ζ^p-1 }

(4) Then we have:



QED

Lemma 2:

Let P,Q be polynomials with coefficients in field F.

If:

P is irreducible in F[X]. (See Definition 1, here, for definition of irreducible polynomials)

P,Q have a common root u in field K which contains F

Then:

P divides Q.

Proof:

(1) Assume that P does not divide Q.

(2) Then P,Q are relatively prime polynomials since P is irreducible. [See Definition 1, here for definition of irreducible polynomials]

(3) Then (see Corollary 3.1, here), there exists polynomials U,V in F[X] such that:

P(X)U(X) + Q(X)V(X) = 1

(4) Substituting the common root u into the polynomials, we get:

P(u)U(u) + Q(u)V(u) = 1 in K

(5) Since u is the root for P(X) and Q(X), this gives us that:

P(u) = 0, Q(u) = 0 in K

(6) But then:

o*U(u) + 0*V(u) = 1 in K which is impossible.

(7) Therefore we reject our assumption in step #1.

QED

Lemma 3:

If u ∈ field K is a root of an irreducible polynomial P ∈ F[X] of degree d

and



Then:

every element in F(u) can be uniquely written in the form:

a₀ + a₁u + a₂u² + ... + a_d-1u^d-1 with a_i ∈ field F.

Proof:

(1) Let f(u)/g(u) be an arbitrary element in F(u)

(2) Since g(u) ≠ 0, it follows that g(u) is not divisible by P since:

(a) Assume that g(u) is divisible by P.

(b) Since P(u) = 0, if follows that (X - u) divides P. [See Theorem, here]

(c) But if P divides g(u), then (X -u) divides g(u).

(d) But this is impossible since this implies g(u) = 0 [See Theorem, here] but g(u) ≠ 0.

(e) So we reject the assumption in step #2a.

(3) Then, g(u) is relatively prime to P since P is irreducible. [See Definition 1, here for definition of irreducible polynomials]

(4) Then (see Corollary 3.1, here), there exists polynomials h,U in F such that:

g(X)h(X) + P(X)U(X) = 1 in F[X]

(5) Since P(u) = 0, substituting u for X gives us:

g(u)h(u) + P(u)U(u) = g(u)h(u) + 0*U(u) = g(u)h(u) = 1 in K.

(6) Since g(u) ≠ 0, we have:

h(u) = 1/g(u) in K

which gives us that:

f(u)/g(u) = f(u)h(u) in K

(7) Using the Division Algorithm for Polynomials (see Theorem, here), there exists Q, R such that:

fh = PQ + R in F[X] with deg R ≤ d - 1.

(8) Since P(u) = 0, it follows that:

f(u)h(u) = P(u)Q(u) + R(u) = 0*Q(u) + R(u) = R(u) in K

(9) Since R(u) is a polynomial of degree at most d-1, we have converted an arbitrary expression f(u)/g(u) ∈ F(u) into a polynomial expression of the type:

a₀ + a₁u + ... + a_d-1u^d-1

with a_i ∈ F

(10) Now, I will prove uniqueness.

(11) Assume that:

a₀ + a₁u + ... + a_d-1u^d-1 = b₀ + b₁u + ... + b_d-1u^d-1

(12) Let us define V(X) such that:

V(X) = (a₀ - b₀) + (a₁ - b₁)X + ... + (a_d-1 - b_d-1)X^d-1 ∈ F[X]

(13) It is clear that V(u) = 0 from step #11.

(14) Using Lemma 2 above, it is clear that P divides V.

(15) But since deg V ≤ d-1 and since P is of degree d, this is impossible unless V=0.

(16) Therefore, it follows that:

a₀ - b₀ = a₁ - b₁ = ... = a_d-1 - b_d-1 = 0

QED

Theorem 4: Every element in Q(μ_p) can be expressed in one and only one way as a linear combination with rational coefficients of the p-th roots of unity other than 1:

a₁ζ + a₂ζ² + ... + a_p-1ζ^p-1

with a_i ∈ Q.

Proof:

(1) Let ζ be a root of Φ_p [See Definition 1, here]

(2) Q(μ_p) = Q(ζ) [From Lemma 1 above]

(3) Since p is a prime, Φ_p is irreducible over Q . [See Corollary 1.1., here]

(4) Using Lemma 3 above, since the degree of ζ is p-1, it follows that every element a ∈ Q(μ_p) can be uniquely expressed in the form:

a = a₀ + a₁ζ + a₂ζ² + ... + a_p-2ζ^p-2

for some a_i ∈ Q.

(5) Using the cyclotomic equation (see Lemma 1, here), we have:

Φ_p(ζ) = 1 + ζ + ζ² + ... + ζ^p-1 = 0

which means that:

ζ + ζ² + ... + ζ^p-1 = -1

(6) This gives us that:

a₀ = -a₀(ζ + ζ² + ... + ζ^p-1)

(7) Combining step #6 with step #4 gives us:

a = (a₁ - a₀)ζ + (a₂ - a₀)ζ² + ... + (a_p-2 - a₀)ζ^p-2 + (-a₀)ζ^p-1

(8) To prove uniqueness, let's assume that:

a₁ζ + ... + a_p-1ζ^p-1 = b₁ζ + ... + b_p-1ζ^p-1

(9) From step #5, we also have:

ζ^p-1 = -1 - ζ - ζ² + ... -ζ^p-2

(10) Putting this into step #8 gives us:

a₁ζ + ... + a_p-1(-1 - ζ - ζ² + ... -ζ^p-2) = b₁ζ + ... + b_p-1(-1 - ζ - ζ² + ... -ζ^p-2)

which reduces to:

-a_p-1 + (a₁ - a_p-1)ζ + (a₂ - a_p-1)ζ² + ... + (a_p-2 - a_p-1)ζ^p-2 = -b_p-1 + (b₁ - b_p-1)ζ + (b₂ - b_p-1)ζ² + ... + (b_p-2 - b_p-1)ζ^p-2

(11) From Lemma 3 above, we can conclude that the coefficients on both sides are equal which gives us:

a_p-1 = b_p-1

which then gives us:

a₁ = b₁
...
a_p-2 = b_p-2

QED

References

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