Cyclotomic Integers: Conjugations and the norm of a divisor

Definition 1: σ(p,ψ(η))

σ(p,ψ(η)) = (p,σψ(η))

NOTE: In other words, if A is a prime divisor for a prime p, then σA divides g(α) implies that g(α)σψ(η) ≡ 0 (mod p).

Proposition 1:

Let σⁱ be a conjugation of cyclotomic integers, let g₁(α), g₂(α) be cyclotomic integers, and let A be a divisor.

Then, to say g₁(α) ≡ g₂(α) mod σⁱA is equivalent to saying σ^-ig₁(α) ≡ σ^-ig₂(α) mod A.

Proof:

(1) If A is a power of a prime divisor, say A = P^ν, and if σA divides g(α), then p^ν divides g(α)*σψ(η)^ν

By definition, we say that P^ν divides g(α) if g(α)ψ(η)^ν ≡ 0 (mod p^ν)

By σA divides g(α), we mean that g(α)σψ(η)^ν ≡ 0 (mod p^ν)

(2) This implies that p^ν divides σ^-1g(α)*ψ(η)^ν.

(a) g(α)σψ(η)^ν ≡ 0 (mod p^ν) means that g(α) is divisible by a conjugate to P which is σ¹P.

(b) This is the same as saying that σ^-1g(α) is divisible by P.

(3) Therefore it shows that A divides σ^-1g(α).

This follows since A dividing a cyclotomic integer is the same as showing σ^-1g(α)ψ(η)^ν ≡ 0 (mod p).

(4) If A,B are relatively prime and σ(AB) = σ(A)σ(B) divides g(α) then both σ(A) and σ(B) divide g(α), both A and B divide σ^-1g(α), and therefore, AB divides σ^-1(α).

This is the same reasoning as in step #3.

(5) This shows that, for any A, g(α) ≡ 0 (mod σA) implies σ^-1g(α) ≡ 0 (mod A).

(6) Then, g(α) ≡ 0 (mod σⁱA) implies σ^-1g(α) ≡ 0 (mod σ^i-1A), ...., σ^-ig(α) ≡ 0 (mod A).

(7) Conversely, σ^-ig(α) ≡ 0 (mod σ^λ-1-i(σⁱA)) implies g(α) = σ^-(λ-1)+iσ^-ig(α) ≡ 0 (mod σⁱA).

(8) Thus, (g₁ - g₂) ≡ 0 (mod σⁱA) is equivalent to σ^-i(g₁ - g₂) ≡ 0 (mod A), as was to be shown.

QED

Proposition 2:

If A is any divisor, there is a positive integer whose divisor is equal to the norm of A.

Proof:

(1) The norm of a divisor is defined to be:

A*σA*σ²A*...*σ^λ-2A.

(2) From this definition, it is easy to see that the norm of a product is the product of norms.

(3) So, that this theorem is proven if we can show that it is true for prime divisors.

(4) By the nature of the prime divisor, we know that it is divisor for an odd prime p so we are done.

QED