Ideal Numbers: Distinct Prime Divisors - Part I

In a previous blog, I showed how each cyclotomic period η_i is congruent to a rational integer u_i 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 u_i that are congruent to them modulo p.

What makes this set of integers u_i 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, ..., u_e, 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, ..., u_e; 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 u_i (mod p) [see Corollary 3.1, here] so that you have set of them: u₁, u₂, ..., u_e.

There are two imporant ideas about cyclotomic periods. First, they don't change value for the permutation σ^e: [σ^e(η_i) = η_i]. Second, each cyclotomic period η_i is congruent to a rational integer u_i 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, ..., u_e be the integers congruent with the cyclotomic periods η₁, η₂, ..., η_e 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 u_i ≡ u_(i+k) (mod p), then k ≡ 0 (mod e)

Proof:

(1) Let u₁, u₂, ..., u_e be the integers that are congruent mod p to the cyclotomic periods η₁, η₂, ..., η_e [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:

η₁η_i = c₀ + c₁η₁ + c₂η₂ + ... + c_eη_e

where c_i 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 η₁η_i is:

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

NOTE: We can do this since c_i 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 α^μ of η_i is the inverse of the term α^-μ in η₁.

In other words, c₀ is the a count of combinations where one element of η₁ is the inverse of one element of η_i such that α^μ*α^λ-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 η_i contains the inverse σ^νeα^μ of every term σ^νeα^-μ in η₁ [Since each period corresponds to a set of elements which stay the same after each σ^e. See Lemma 2, here for more information on σ^e]

In other words, when if one of the elements of η₁ is the inverse of an element of η_i, 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 η_i 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 + ... + c_e = f[c₁ + c₂ + ... + c_e]/f = (f² - f)/f = f-1.

For all others c₀ = 0 giving us:

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

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

η₁η_i + η₂η_i+1 + ... + η_eη_i-1 = η₁η_i + σ(η₁η_i) + ... + σ^e-1(η₁η_i)

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

η1η_i + σ(η₁η_i) + ... + σ^e-1(η₁η_i) =

= c₀ + c₁η₁ + c₂η₂ + ... + c_eη_e + σ(c₀ + c₁η₁ + c₂η₂ + ... + c_eη_e) + ... + σ^e-1(c₀ + c₁η₁ + c₂η₂ + ... + c_eη_e) =

= ec₀ + c₁(η₁ + η₂ + ... + η_e) + ... + c_e(η_e + η₁ + ... + η_e-1) =

= ec₀ + (c₁ + c₂ + ... + c_e)(η_e + η₁ + ... + η_e-1)

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

η₁ + η₂ + ... + η_e = -1.

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

η₁η_i + η₂η_i+1 + ... + η_eη_i-1 = ec₀ - (c₁ + c₂ + ... + c_e)

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

η₁η_i + η₂η_i+1 + ... + η_eη_i-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 + η₁η_f + η₂η_j+1 + ... + η_eη_j-1 = λ (in one case) or 0 (in all other cases).

(15) Now, using the fact that η_i ≡ u_i (mod p) gives us:

f + u₁u_i + u₂u_i+1 + ... + u_eu_i-1 not ≡ 0 (mod p) in one case and ≡ 0 (mod p) in all other cases.

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

f + u₁u_i + u₂u_i+1 + ... + u_eu_i-1 ≡ f + u₁u_i+k + u₂u_i+k+1 + ... + u_eu_i+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 u_i+e = u_e since periods are unchanged by σ^e.

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

QED