## Thursday, August 10, 2006

### Ideal Numbers: Distinct Prime Divisors - Part III

As stated previously, ideal numbers are generalizations of cyclotomic integers. Every time that a cyclotomic prime exists for a given rational prime, there is a mapping between one of the prime divisors and that the real cyclotomic prime. Still, this raises a question. Is this construction mathematically valid? Can we assume that if all the prime divisors for a given prime p divide a cyclotomic integer, then p is also divides it and if all but one of the prime divisors divide a cyclotomic integer, that p does not divide it?

In today's blog, I will show that the set of e distinct prime divisors that divide a rational prime are mathematically equivalent. In other words, I will show that a cyclotomic integer g(α) is divisible by a rational prime p if and only if g(α) is divisible by the e prime divisors that divide p.

The content in today's blog is taken from Harold M. Edwards Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Theorem: a cyclotomic integer g(α) is divisible by a rational prime p if and only if g(α) is divisible by each of the prime divisors of p.

Proof:

(1) In the case of p = λ, we know this is true since λ = (α - 1)(α2 - 1)*...*(αλ-1 - 1). [See Lemma 2, here]

(2) So, for the rest of the proof, we will assume that p ≠ λ.

(3) So, we need to show:

(a) Division by p implies division by all e of the prime divisors.

(b) Division by all e of the prime divisions implies division by p.

(4) Since p is divisible by all e of them, divisions by p implies division by all e of the prime divisors since congruence modulo a prime divisor is consistent with multiplication.

(5) Divisibility of g(α) by the prime divisor of p corresponding to u1, u2, ..., ue is equivalent to g(α)ψ(η) ≡ 0 (mod p) where ψ(η) is the product of ep-e factors j - ηi in which j ≠ ui. [See Definition 7, here for definition of division by the prime divisor of p]

(6) Divisibility of g(α) by the prime divisor of p corresponding to uk+1, uk+2, ... , uk is equivalent to g(α)σ-kψ(η) ≡ 0 mod p because:

σ-kψ(η) = σ-k[ (i=1,e)∏ (j=1,p)∏ (j - ηi)/[(i=1,e)∏ (ui - ηi)]] =

= (i=1,e)∏ (j=1,p)∏ (j - ηi-k)/(i=1,e)∏(ui - ηi-k) =

= (i=1,e)∏ (j=1,p)∏ (j - ηi)/(i=1,e)∏(ui+k - ηi)

(7) That is, σ-kψ(η) is the ψ(η) that results from replacing u1, u2, ..., ue with u1+k, u2+k, ..., uk.

(8) Therefore, if g(α) is divisible by all e prime divisors of p, then g(α)σ-kψ(η) ≡ 0 (mod p) for all k = 0, 1, ..., e-1.

(9) The sum of these congruences gives g(α)*M ≡ 0 (mod p) which, because M ≠ 0 mod p (see Lemma 1, here for details on M) implies g(α) ≡ 0 (mod p) as was to be shown.

QED