## Wednesday, August 09, 2006

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

In today's blog, I continue the discussion about ideal numbers. If you are new to ideal numbers, I suggest that you start here.

In my last blog, I demonstrated that there were at least e distinct prime divisors for a given rational prime p where p ≠ λ. In today's blog, I use the previous result to establish that there are exactly e distinct prime divisors. I also provide an example of applying these ideas to λ = 3, p=5.

The details here are taken directly from Harold M. Edwards Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Lemma 1: There are at most e distinct prime divisors for a given rational prime p where p ≠ λ

Let u1, u2, ..., ue be any set of integers such that ηi uj (mod p), then there exists an integer k such that:

u1 ≡ u(1+k) (mod p)

u2 ≡ u(2+k) (mod p)

...

ue ≡ u(e + k) (mod p)

Proof:

(1) Let u1, u2, ..., ue be any set of integers such that ηi uj (mod p).

(2) Let M = ψ(η) + σψ(η) + ... + σe-1ψ(η) where σ is the conjugation α → αγ which carries ηi to ηi+1. and ψ(η) is the cyclotomic integer constructed in Definition 4, here.

(3) Since σM = M, it is clear that M is an ordinary integer.

NOTE: One might also wonder about the possibility where M = d + cα + cα2 + ... + cαλ-1. Since α + α2 + ... + αλ-1= -1 (see Lemma 2, here), this really gives us d - c which is still an ordinary integer.

(4) Now,

Mψ(η) = ψ(η)ψ(η) + [ σψ(η)] * ψ(η) + ... + [ σe-1ψ(η)]*ψ(η) ≡

≡ ψ(u)ψ(η) + [σψ(u)]ψ(η) + ... + [σe-1ψ(u)]*ψ(η) mod p.

where σkψ(u) denotes the integer obtained by applying σk to ψ(η) and setting η1 = u1, η2 = u2, ..., ηe = ue in the result. [Taken from Corollary 3.1, here where I showed ηi ≡ ui (mod p)]

(5) In short, σkψ(u) = ∏(j - ui+k) where i = 1, 2, ..., e and j ranges over all integers from 1 to p except ui.

NOTE: This is clear based on the definition of ψ(u) = ∏(j - ui)

(6) By what was just shown, σkψ(u) contains a factor of 0 (namely a factor ui+k -ui+k) unless k=0, in which case σkψ(u) = ψ(u).

NOTE: The reason for this is when the value ψ(η) was constructed, we skipped the value ηi and in doing this, we also skipped ui. With σk, we are now creating a situation where each ui is shifted to ui+k so in this context, ui-k which we didn't skip before now becomes ui-k+k = ui.

(7) Therefore Mψ(η) ≡ ψ(u)ψ(η) ≡ (not 0) (mod p) because ψ(u) ≡ (not 0) (mod p).

NOTE: ψ(u) is a product of ep-e integers, none of them divisible by p. So in the definition of Mψ(η) all σkψ(u) = 0 so Mψ(η) ≡ (not 0) (mod p) because ψ(u)ψ(η) ≡ (not 0) (mod p).

(8) Further, M not ≡ 0 (mod p) since only ψ(η) not ≡ 0 (mod p).

(9) If u1, u2, ..., ue satisfy the condition ηi uj (mod p), then M ≡ ψ(u) + σψ(u) + ... + σe-1ψ(u) mod p since they follow the same relations as ui.

(10) Since M not ≡ 0 (mod p), this implies σkψ(u) not ≡ 0 (mod p) for at least one k.

(11) That is, ∏(j - ui+k) not ≡ 0 (mod p) for at least one k, where i = 1, 2, ..., e and j assumes all values 1, 2, ..., p except ui.

NOTE: If this mapping did not occur then M ≡ 0 (mod p) which is contradicts our assumption that ui satisfies all the equations of ui.

(12) This implies that uiui+k (mod p) and the u's are a cyclic permutation of the u's as was to be shown.

(13) The conclusion follows from applying a previous result where we showed that there are at least e distinct prime divisors. [See Lemma 1, here]

QED

Example: the prime divisors of 5 where λ = 3.

To be clear on what this means, let me end this blog with an example of constructing the e and ψ(η)p values for a given prime p. In this example, λ = 3 and p = 5.

Let f be the exponent mod λ for p; we can see that f = 2 since 52 ≡ 25 ≡ 1 (mod 3).

This means that e = (3-1)/2 = 1

So we have only one prime divisor.

In addition, there is only one cyclotomic period which consists of two terms: α + α2. Let γ be the primitive root mod λ. We can see that γ = 2 since 21 ≡ 2 (mod 3), 22 ≡ 1 (mod 3).

We can see that σe1 + α2) = α(γ*1) + α(γ*2) = α2 + α4 = α2 + α1

Now, let's determine u1. We are looking for a value such that p divides η1 - u1

It turns out u1 = 4 since:

-4 + α + α2 = -4 + α + α2 + 4(1 + α + α2) = 5α + 5α2 = 5(α + α2)

Additionally, we can see that the Norm(5) divides the Norm(α + α2 - 4) since:

Let g(α) = α + α2 - 4.

Norm(5) = 5*5 = 25

Ng(α) = (α + α2 - 4)(α2 + α4 - 4) =

= 1 + α2 - 4α + α + 1 - 4α2 -4α2 - 4α + 16 =

= 16 + 2 + (-4α + α + -4α) + (-4α2 + α2 -4α2 ) = 18 + (-7α) + (-7α2) = 18 + (-1)(-7) = 25.

So, finally, we can construct ψ(η)5 (see definition 4, here):

ψ(η)5 = (1 - η1)(2 - η1)(3 - η1)(5 - η1) = (1 - α - α2)(2 - α - α2)(3 - α - α2)(5 - α - α2) = 6*24 = 144.

See below for details on the 6 and 24.

(1 - α - α2)(2 - α - α2) = 2 - α - α2 - 2α + α2 + 1 - 2α2 + 1 + α = 4 -2α -2α2 = 4 + (-2)(α + α2) = 4 + (-2)(-1) = 6

(3 - α - α2)(5 - α - α2) = 15 - 3α -3α2 -5α + α2 + 1 -5α2 + 1 + α = 17 -7α -7α2 = 17 + (-7)(α + α2) = 17 + (-7)(-1) = 24

So, putting it all together, a cyclotomic integer g(α) is said to be congruent to another cyclotomic integer h(α) modulo the prime divisor of 5 if and only if:

g(α)144 ≡ h(α) mod 5.