## Tuesday, August 08, 2006

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

In a previous blog, I showed how each cyclotomic period ηi is congruent to a rational integer ui mod a rational prime p where p ≠ λ. I also showed that cyclotomic periods can be constructed based on any factor of λ - 1. If we set the rational integer f = the exponent mod λ for p (see definition 2, here) and we set e = (λ - 1)/f [since the exponent mod λ divides λ - 1, see Lemma 1, here], then we can assume that there are e cyclotomic periods and e rational integers ui that are congruent to them modulo p.

What makes this set of integers ui especially interesting is that we can construct a special cyclotomic integer ψ(η)p (see Definition 4, here for review) that enables to define division by a "prime divisor." Now, this raises a very important question. If we accept this idea of "prime divisors" that are derived in this way. How many prime divisors are there for each rational prime p?

We can assume that p ≠ λ, since all of our analysis makes this assumption. The answer in this case turns out to be e distinct prime divisors. The proof for this is complex so I am dividing it up into three parts.

I. Using u1, u2, ..., ue, it is possible to define e distinct prime divisors. In other words, there are at least e distinct prime divisors for a given prime p.

II. All prime divisors must be based on u1, u2, ..., ue; there are no other prime divisors for a given rational prime p where p ≠ λ. In other words, there are at most e distinct prime divisors for a given prime p. (See Lemma 1, here)

III. A cyclotomic integer is divisible by a rational prime p if and only if the cyclotomic integer is divisible by all e prime divisors that divide p. (See Theorem, here)

In today's blog, I will show that there are at least e distinct prime divisors that can be constructed for a given rational prime p.

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

To follow the first lemma, you should be familiar with cyclotomic periods, the permutation σ (see definition 5, here), and the idea that all cyclotomic periods are congruent to an rational integer ui (mod p) [see Corollary 3.1, here] so that you have set of them: u1, u2, ..., ue.

There are two imporant ideas about cyclotomic periods. First, they don't change value for the permutation σe: [σei) = ηi]. Second, each cyclotomic period ηi is congruent to a rational integer ui modulo a rational prime p.

Lemma 1: There are at least e distinct prime divisors that can be constructed for each rational prime p

Let u1, u2, ..., ue be the integers congruent with the cyclotomic periods η1, η2, ..., ηe modulo a rational prime p. Let f be the exponent mod λ for p. Let e = (λ - 1)/f.

Then if there exists an integer k such that ui ≡ u(i+k) (mod p), then k ≡ 0 (mod e)

Proof:

(1) Let u1, u2, ..., ue be the integers that are congruent mod p to the cyclotomic periods η1, η2, ..., ηe [See Corollary 3.1, here for proof that such integers exist]

(2) From a previous result (see Lemma 3, here), we know that periods consist of f elements where f = (λ-1)/e.

(3) The product of two cyclotomic integers made up of periods of length f is itself made up of periods of length f (see Corollary 4.1, here) so that:

η1ηi = c0 + c1η1 + c2η2 + ... + ceηe

where ci are all rational integers since this is really counting the number of times each period results from the combinition and i is any cyclotomic period 1, 2, ..., e.

(4) So the number of components for η1ηi is:

(f)(f) = f2 = c0(1) + c1(f) + ... + ce(f)

NOTE: We can do this since ci is just the number of times a given term occurs and since the product of periods is itself a sum of periods (see Corollary 4.1, here).

[NOTE: In order for this to work c0 will have to be a multiple of f. In most cases, c0 = 0; more details below]

(5) Now, c0 ≠ 0 only if at least one term αμ of ηi is the inverse of the term α in η1.

In other words, c0 is the a count of combinations where one element of η1 is the inverse of one element of ηi such that αμλ-μ = 1.

c0 = 0 when there are no such combinations [since periods always include α to some power]

c0 ≠ 0 when there is at least one such combination.

(6) But when this is the case ηi contains the inverse σνeαμ of every term σνeα in η1 [Since each period corresponds to a set of elements which stay the same after each σe. See Lemma 2, here for more information on σe]

In other words, when if one of the elements of η1 is the inverse of an element of ηi, then all elements are inverses.

This means that c0 = 0 or c0 = f. [It equals f since we know from step #4 there are f elements that make up a cyclotomic period].

(7) We can say more than this. We can for each η1, there is one value of i for which each element of ηi is the inverse of η1. For the e-1 other values, c0 = 0.

NOTE: Where this occurs depends on the size of f. If f is even, i=0, if f is odd, then i =(1/2)e.

(8) For one value of i, then c0 = f giving us:

c1 + c2 + ... + ce = f[c1 + c2 + ... + ce]/f = (f2 - f)/f = f-1.

For all others c0 = 0 giving us:

c1 + c2 + ... + ce = f[c1 + c2 + ... + ce]/f = f2/f = f.

(9) Using σ notation (see here), we note:

η1ηi + η2ηi+1 + ... + ηeηi-1 = η1ηi + σ(η1ηi) + ... + σe-11ηi)

(10) Using the result in step #9 and combining it with the result in step #3 gives us:

ηi + σ(η1ηi) + ... + σe-11ηi) =

= c0 + c1η1 + c2η2 + ... + ceηe + σ(c0 + c1η1 + c2η2 + ... + ceηe) + ... + σe-1(c0 + c1η1 + c2η2 + ... + ceηe) =

= ec
0 + c11 + η2 + ... + ηe) + ... + cee + η1 + ... + ηe-1) =

=
ec0 + (c1 + c2 + ... + ce)(ηe + η1 + ... + ηe-1)

(11)
Remembering that η1 + η2 + ... + ηe = α + α2 + ... + αλ-1 gives us (Lemma 2, here):

η1 + η2 + ... + ηe = -1.

(12) Applying the result in step #11 to step #10 gives us:

η1ηi + η2ηi+1 + ... + ηeηi-1 = ec0 - (c1 + c2 + ... + ce)

(13) So that applying step # 8 gives us:

η1ηi + η2ηi+1 + ... + ηeηi-1 =

either:

0 - f = -f [in all but one case for i]

or

ef - (f-1) = λ - 1 -f + 1 = λ - f [in one case for i]

(14) This gives us:

f + η1ηf + η2ηj+1 + ... + ηeηj-1 = λ (in one case) or 0 (in all other cases).

(15) Now, using the fact that ηi ≡ ui (mod p) gives us:

f + u1ui + u2ui+1 + ... + ueui-1 not ≡ 0 (mod p) in one case and ≡ 0 (mod p) in all other cases.

(16) If there were a k such that ui ≡ ui+k (mod p) for all i then:

f + u1ui + u2ui+1 + ... + ueui-1 ≡ f + u1ui+k + u2ui+k+1 + ... + ueui+k-1 (mod p)

(17) But, if we choose i = the one case where the equation in step #16 not ≡ 0, we have a problem since this means i+k must ≡ 0 unless k ≡ 0 (mod e) since ui+e = ue since periods are unchanged by σe.

(18) This proves that the e cyclic permutations of the u's are all distinct. That is, if ui ≡ ui+k (mod p) then k ≡ 0 (mod e).

QED