Cyclotomic Integers: Terminology

In today's blog, I continue to review results from Harold Edward's Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory. I am continuing down the same chapter that I started in an earlier blog. If you would like to start at the beginning of cyclotomic integer properties, start here.

Thereom 1:

Let A and B be divisors. If every cyclotomic integer divisible by A is also divisible by B, then A is divisible by B.

Proof:

(1) Let p₁, p₂, ..., p_n be the prime integers that are divisible by the prime divisors which divide A.

In other words, the p's are the prime factors of the norm of A. [This is explained later in the section on norms of divisors]

(2) Let A = A₁, A₂, ..., A_n be a decomposition of A as a product of divisors A_i such that A_i contains all prime divisors in A which divide p_i.

(3) Since a sufficiently high power of p₁, p₂, ... p_n is divisible by A it must, by assumption, be divisible by B.

NOTE: By assumption, we know that ∏ p_i is divisible by each prime divisor of A. If we raise this product to a sufficiently high power, then we know that each prime divisor of A will divide at the appropriate multiplicity. Without assuming a sufficiently high power, we cannot assume that this is the case.

(4) This shows that every prime divisor in B divides one of the p_i and therefore that B can be decomposed as a product B = B₁B₂*...*B_n in which B_i contains all prime divisors in B which divide p_i

Some of the B_i may be I since it may be possible that that a given p_i is not otherwise divisible by any of the divisors of B.

(5) Since the order of the p's is arbitrary, it will suffice to prove that B₁ divides A₁

We can then make the same argument for each B_i, A_i for p_i.

(6) If p₁ = λ, then A₁ = (α - 1)^μ for μ

(7) Then the cyclotomic integer (α - 1)^μ(p₂p₃*...*p_n)^ν is divisible by A for all sufficiently large ν

(8) Therefore it is also divisible by B for large ν.

(9) This implies because B₁ does not divide (p₂p₃*...*p_n)^ν at all, that B₁ divides the cyclotomic integer (α - 1)^μ

Again, since p₁ = λ by assumption, we know that the other p_i do not.

(10) Therfore B₁ divides the divisor of (α - 1)^μ, which is A₁.

(11) If p₁ ≠ λ, let f be the exponent of p₁ mod λ and let ψ be a cyclotomic integer made up of periods of length f which is divisible by one of the e = (λ - 1)/f prime divisors of p₁ with multiplicity 1 and is not divisible at all by the remaining e-1.

(12) Then one can form a product of conjugates of ψ, call it x, with the property that the divisor of x is A₁ times prime divisors which do not divide p₁.

(13) Now x(p₂p₃*...*p_n)^ν is divisible by A for large ν

(14) Therefore it is divisible by B₁.

(15) Since B₁ does not divide (p₂p₃*...*p_n)^ν at all, it follows that B₁ divides the divisor of x.

(16) Since the divisor of x is A₁ times a factor that contains none of the prime divisors of p₁ (and a fortiori none of the prime divisors of B₁) B₁ divides A₁ as was to be shown.

QED


Corollary:

If A and B are divisors and if the relation g₁(α) ≡ g₂(α) mod A" is equivalent to "g₁(α) ≡ g₂(α) mod B" (that is, one relation holds if and only if the other does), then A=B.

Proof:

This follows form the the Theorem above since if A divides B and B divides A then A = B.

QED