Cyclotomic Integers: The Fundamental Theorem

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.

Theorem: The Fundamental Theorem for Cyclotomic Integers

Let λ be a given prime greater than 2 and let g(α), h(α) be nonzero cyclotomic integers built up from a λth root of unity α ≠ 1.

Then g(α) divides h(α) if and only if every prime divisor which divides g(α) also divides h(α) with multiplicity at least as great.

Proof:

(1) Let h(α) be divisible by g(α) such that h(α) = q(α)g(α)

(2) Certainly every prime divisor which divides g(α) also divides h(α) with multiplicity at least as great.

(3) Now g(α) divides h(α) if and only if Ng(α) = g(α)*g(α²)*...*g(α^λ-1) divides h(α)*g(α²)*...*g(α^λ-1)

(4) If every prime divisor which divides g(α) also divides h(α) with multiplicity at least as great then every prime divisor which divides Ng(α) also divides h(α)*g(α²)*...*g(α^λ-1) with the multiplicity at least as great.

(5) Since Ng(α) is an integer, from this point on we can assume that g(α) is a rational integer and that h(α) = h(α)*g(α²)*...*g(α^λ-1).

NOTE: The justification is that every cyclotomic integer can be made to depend on this equation where we show that every prime divisor that divides its norm necessarily divides the cyclotomic integer h(α)*g(α²)*...*g(α^λ-1)

(6) Now, we will proceed to prove this theorem for 3 cases:

Case I: g(α) is a prime integer p ≠ λ

Case II: g(α) is a prime integer p = λ

Case III: g(α) is not a prime integer

(7) Assume g(α) is a prime integer p ≠ λ

(8) From a previous result (See Theorem 4, here), we know that if h(α) is divisible by each of the e prime divisors of p, then h(α) is divisible by p.

(9) Assume g(α) is a prime integer p = λ

(10) From a previous result (See Proposition 2, here), we know that if h(α) is divisible by (α - 1)^{λ - 1}, then it is divisible by p = λ

(11) To prove case III, we will show that if the thereom is true for g₁(α) and g₂(α), then it is true for the product, g₁(α)*g₂(α)

(12) So, let us assume that all the prime divisors which divide g₁(α)*g₂(α) also divide h(α)

(13) Since all the prime divisors g₁(α) divide h(α), then it follows that g₁(α) divides h(α) and there exists h₁(α) such that:

h(α) = g₁(α)*h₁(α)

(14) By assumption, every divisor of g₂(α) divides h₁(α).

(15) Since we are assuming that the theorem holds true for g₂(α), then it follows that g₂(α) divides h₁(α) and there exists h₂(α) such that:

h₁(α) = g₂(α)*h₂(α)

(16) Therefore h(α) = g₁(α)g₂(α)h₂(α) we have shown that h(α) is divisible by g₁(α)g₂(α)

QED

Corollary 1.1: Unique Factorization is "saved"

If two cyclotomic integers g(α) and h(α) are divisible by exactly the same prime divisors with exactly the same multiplicities, then they differ only by a unit multiple, g(α) = unit * h(α)

Proof:

(1) By the fundamental theorem above, g(α) divides h(α) and h(α) divides g(α)

(2) Therefore, h(α)/g(α) and g(α)/h(α) are cyclotomic integers.

(3) Since their product is 1, they must both be units. (See here for more details if needed)

QED

NOTE: On the Fundamental Theorem and Unique Factorization.

The use of the fundamental theorem "saves" the property of unique factorization for primes as shown by the corollary but there is a price to pay. With the integers, we can arbitrarily organize any combination of primes and know that we have numbers. For prime divisors, this is no longer the case. For example, for λ = 23, there is no cyclotomic integer which is divisible once by one prime divisor of 47 and not divisible by any other prime divisor.