## Friday, August 11, 2006

### Ideal Numbers: Class Number

In a previous blog, I have talked about Kummer's theory of ideal numbers in relation to his concepts of principal divisors and equivalent divisors. In today's blog, I will show how these ideas lead to the concept of the class number.

The idea of class number comes directly from the idea of equivalent divisors. In my previous blog, I showed how certain divisors can be said to be equivalent to each other. In other words, if A,B are equivalent divisors, then for a given divisor C, AC is principal if and only if AB is principal.

Kummer's further insight is that for each type of ideal number (that is, derived from an odd prime λ), there is a finite set of equivalence classes. I will show the proof for this in today's blog. The class number is the size of this minimum set of equivalence classes.

Lemma 1: If n is an odd integer, we can order 1, ..., n-1 into (n-1)/2 pairs that each add up to n.

Proof:

(1) We know this is true where n = 3

In this case, there is (3-1)/2 = 1 pair (1,2) which adds up to 3.

(2) Assume that this lemma is true up to n.

(3) So that we can divide up 1, .., n-1 into (n-1)/2 pairs that each add up to n so that we have:

(1,n-1)
(2,n-2)
...
([n-1]/2,[n-1]/2+1)

(4) If we add 2 to each the second element of each pair, we have (n-1)/2 pairs that each add up to n+2.

(1,n+1)
(2,n)
...
([n-1]/2,[n-1]/2+3)

(5) Since every first element of the pairs is between 1 and [n-1]/2 and since every second pair is between [n-1]/2+3 and n+1, we can see that [n-1]/2+1 and [n-1]/2+2 are not included in any of the pairs.

(6) So we can add the following pair ([n-1]/2+1,[n-1]/2+2) to get:

(1,n+1)
(2,n)
...
([n-1]/2,[n-1]/2+3)
([n-1]2+1,[n-1]/2+2)

(7) Now, we can see that we have (n+1)/2 pairs that each add up to n+2. In other words, we have shown that if this lemma is true for n, then it is also true for n+2.

(8) Using the principle of induction (see Theorem, here), we are done.

QED

Lemma 2: g(αj)*g(α-j) ≤ [abs(a0) + abs(a1) + .. + abs(aλ-1)]2.

Proof:

(1) g(α) = a0 + a1α + ... + aλ-1αλ-1. [See Lemma 1, here]

(2) g(αj) = a0 + a1αj + ... + aλ-1αj(λ-1).

(3) g(α-j) = a0 + a1α-j + ... + aλ-1α-j(λ-1).

(4) We can assume that each αj cancels out (since each Norm is a positive rational integer, see Lemma 5, here), so that we have:

g(αj)g(α-j) is less than [abs(a0) + abs(a1) + .. abs(aλ-1)]2

QED

Lemma 3: For any divisor A, there exists a cyclotomic integer g(α) such that A divides g(α) and the Ng(α) is less than or equal to Kn where K = λλ-1 and n = N(A)

Proof:

(1) Let g(α) = a0 + a1α + ... + aλ-1αλ-1. [See Lemma 1, here]

(2) Ng(α) = g(α)*g(α2)*...*g(αλ-1) [See definition of norm, here]

(3) Using Lemma 1 above, we can order all the elements of Ng(α) into the following pairs:

Ng(α) = [g(α)g(α-1)][g(α2)g(α-2)]...

(4) Using Lemma 2 above, we can assume that each g(αj)g(α-j) ≤ [abs(a0) + abs(a1) + ... + abs(aλ-1)]2

(5) Therefore, if each abs(ai) ≤ c, then

g(αj)g(α-j) ≤ [λ*c]2 = λ2c2

and

Ng(α) ≤ (λ2c2)(λ-1)/2 = λλ-1cλ-1

(6) The set of all g(α) in which a0 = 0 and 0 ≤ ai ≤ c for i greater than 0 contains more than n elements provided that (c+1)λ-1 is greater than n.

If we set c = 1, then we have a total of (1+1)λ-1 possible elements.

For example if λ = 3, these would include (1+1)2 = 4 elements which consist of:

0
α
α2
α + α2

(7) Then, the difference of two of them is divisible by A (see explanation below) and has the norm at most λλ-1cλ-1 (from step #5 above).

We know that there are at most Norm(A) distinct congruence classes modulo A (see Theorem here for details). For this reason, if we have a set of distinct cyclotomic integers that are greater than Norm(A), we know that the difference of two of them must be divisible by A.

The proof on this is very straight forward.

(a) Let n be the set of distinct congruence classes modulo A.

(b) Let m be a set of distinct cyclotomic integers where m is greater than n.

(c) Let's assume that we go through the first n of the m distinct cyclotomic integers. If any of two of them are in the same congruence class modulo A, then A will divide the difference. Let's assume that none are.

(d) Now, when we get to the (n+1)th distinct cyclotomic integer out of the m, we know that it must match one of the existing congruence classes since there are only n of them.

(8) If c is as small as possible subject to the assumption that (c+1)λ-1 is greater n, then cλ-1 ≤ n.

(9) Then, a cyclotomic integer has been found that is divisible by A and has a norm at most λλ-1n.

QED

Lemma 4: Let k be a positive, rational integer. Then there is only a finite set of divisors whose norm is less than k.

Proof:

(1) There is no maximum prime. [See Theorem, here]

(2) Let p* be a prime that is greater than k.

(3) The norm of a prime divisor that divides a rational integer p is pf where f ≥ 1. [Details to be added here]

(4) So, no prime divisor of p* has a norm less than k.

(5) Further, no prime divisor of a prime greater than p* has a norm less than k.

(6) Finally, the total number of primes less than p* is less than p*.

(7) For each prime, there are at most e distinct prime divisors (see Lemma 1, here), and since e ≤ λ - 1, we know that there are at most (p*)*(λ-1) prime divisors that divides rational primes less than p*.

(8) The norm for each prime divisor is pf where f ≥ 1 so from this, we know that the maximum number of prime divisors that have a norm less than p* is (p*)*(λ-1).

(9) To finish this proof, we need to show that there are only a finite number of ideal numbers (composed of prime divisors) that have a norm less than p*.

(10) Now, for any prime divisor P, the norm(P*P) = norm(P)*norm(P) [Detail to be added later], so norm(Px) = (norm(P)x).

(11) Let l = log2(k) [See here for an explanation of logarithm if needed; this gives a solution to 2l = k.]

(12) We know for all divisors whose norm is less than k, they cannot be divisible by a prime divisor with a power greater than l.

If they are divisible by a prime divisor with a power greater than l, then the norm of this divisor will necessarily be greater than k.

(13) This means then that the maximum number of divisors with a norm less than k is floor(l)*(p*)*(λ-1) [where floor(x) is the maximum integer that is less than or equal to x]

This is the answer because we can think of each divisor as consisting of all prime divisors with each prime divisor having a value between 0 and floor(l).

QED

Theorem: For all cases of cyclotomic integers (where λ is a prime > 2), there exists a finite set of divisors A1, ..., Ak such that every divisor is equivalent to one of the Ai

Proof:

(1) Let A be any divisor.

(2) Let K = λλ-1

(3) Let A1, ..., An be the the set of divisors whose N(Ai) is less than K.

(4) We can see that there are only a finite set of divisors that make up Ai [ See Lemma 4 above]

(5) There exists a divisor B such that N(B) is less than K such that AB is principal. [This follows from Lemma 3 above]

(6) For B, there exists a divisor C such that N(C) is less than K and BC is principal. [This also follows from Lemma 3 above]

(7) We can see that A ~ C [See Lemma 8, here]

(8) It also follows from step #3, that C = Ai where i is one of the elements 1 ... n.

(9) In this way, we have shown that for any A, there exists an Ai such that A ~ Ai and further, that this set A1, ..., An is a finite set.

QED

Definition: Class Number

For a given set of cyclotomic integers based on an odd prime λ, the class number is the number of elements in the finite set of divisors A that is described in the theorem above.

Example: cyclotomic integers for λ = 3

In this case, the cyclotomic integers are characterized by unique factorization. This means that all combinations of prime divisors result in real cyclotomic integers (in other words, all divisors are principal). In this case, all divisors are equivalent to I which means that the class number is 1.

Note:

S0 far, we have shown that a class number exists for each value of λ and that in the case of unique factorization, the class number is 1. Ernst Kummer was able to use the work of Johann Dirichlet to establish a method for determinining class number. I will talk about this more in a future blog.

## Thursday, August 10, 2006

### Ideal Numbers: Divisors

Not all cyclotomic integers are characterized by unique factorization. In addressing this situation, Ernst Kummer proposed his theory of ideal numbers which attempts to save unique factorization. While unique factorization is saved, there is a price to pay.

The result of Kummer's theory is that it is no longer possible to arbitrarily create a cyclotomic integer based on a set of primes. When using rational primes, you can take any set of primes, multiply them together and get a rational integer. Unfortunately, this is not the case with prime divisors. Not all combinations of prime divisors result in a cyclotomic integer (even though, all cyclotomic integers can be broken down to a unique set of prime divisors).

To better analyze this situation, let me offer some terms taken from Harold M. Edwards' Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory:

Definition 1: Divisor

A divisor is a list of prime divisors including their powers.

For all purposes, you can think about a divisor as a product of prime divisors. A divisor divides a given cyclotomic integer g(α) if and only if the list of prime divisors associated with the divisor divide g(α).

NOTE: Divisor is the same thing as an ideal number. As a convention (following Edwards), I will represent divisors (including prime divisors) as capital letters.

Definition 2: Principal

A divisor D is said to be principal if it can be uniquely associated with a cyclotomic integer g(α) such that g(α) divides another cyclotomic integer h(α) if and only if D divides h(α).

So, to be clear, a combination of prime divisors exist as a cyclotomic integer if and only if the divisor equivalent to the combination of prime divisors is principal. If a divisor is not principal, then the combination of prime divisors which make it up are not uniquely associated with a cyclotomic integer.

To define the class number (link added later), we need one more concept. This one is a bit tricky. Kummer came up with this idea as he was trying to determine under what conditions a divisor is principal.

Kummer's insight is that certain divisors have a special relationship. Let A, B be two such divisors. If AC is principal, then BC is also principal. If AD is not principal, then BD is not principal. In other words, both divisors are principal in exactly the same combinations. Kummer called these two divisors equivalent.

Definition 3: Equivalent Divisors ~

Two divisors A and B are said to be equivalent if for all cases where AC is principal, BC is also principal. That is, A ~ B if and only if AC is principal when and only when BC is principal.

This concept is the basis for the idea of class number which is very important to Kummer's proof. I will talk more about class number in a future blog.

Here are some basic properties of divisors.

Lemma 1: If A and B are both principal, then so is AB

Proof:

(1) A is principal → ∃ a(α) such that A is the divisor for a(α)

(2) B is principal → ∃ b(α) such that B is the divisor for b(α)

(3) So, AB is the divisor for a(α)b(α) which means that AB is also principal.

QED

Lemma 2: If A and B are divisors such that A and AB are both principal, then B is principal.

Proof:

(1) A is principal → ∃ a(α) such that A is the divisor for a(α)

(2) AB is principal → ∃ c(α) such that AB is the divisor for c(α)

(3) Since A divides AB, we know that a(α) divides c(α) so that there exists b(α) = c(α)/a(α) and we know that B is the divisor for b(α)

QED

Lemma 3: A is principal if and only if A ~ I where I is the empty divisor, that is the divisor of 1.

Proof:

(1) Assume A is principal

(2) Then there exists a(α) such that A is the divisor for a(α)

(3) In all cases where IC is principal, it is implies that C is principal which means that in all those cases AC is also principal [By Lemma 1 above]

(4) Assume A~I

(5) If IC is principal then C is principal since IC = C.

(6) IC is principal implies that AC principal. [From A ~ I]

(7) So if AC is principal and C is principal then A is principal [By Lemma 2 above]

QED

Lemma 4: The equivalence relation ~ is reflexive, symmetric, and transitive.

Proof:

(1) ~ is reflexive, that is, A ~ A, since A can always be replaced by itself.

(2) ~ is symmetric since multiplication of divisors is commutative. If AB is principal, then BA is also principal.

(3) ~ is transitive since if A~B and B~C, then A~C. Assume there was a divisor D where AD is principal but CD is not, then BD would not be principal (by B~C) and this would be a contradiction of A~B so this divisor D cannot exist.

QED

Lemma 5: Multiplication of divisors is consistent with the equivalence relations. That is A~B implies AC~BC for all divisors C

Proof:

(1) Assume ACD is principal

(2) Then A(CD) is principal since a divisor can be formed from C,D

(3) Then B(CD) is principal (from A~B)

(4) Assume ACD is not principal

(5) Then A(CD) is not principal since a divisor can be formed from C,D

(6) Then B(CD) is not principal (otherwise, we violate A~B)

QED

Lemma 6: Given any divisor A, there is another divisor B such that AB~I

Proof:

(1) N(A) is the divisor of an integer.

(2) Let B = the complement of N(A), that is, N(A)/A.

(3) Then AB is principal and this leads to AB~I (from Lemma 3 above)

QED

Lemma 7: A ~ B implies there exist principal divisors M and N such that AM = BN.

Proof:

(1) Assume A ~ B

(2) There exists C such that AC is principal. (from Lemma 6 above)

(3) AC ~ I (from Lemma 3 above)

(4) Let M = BC

(5) Let N = AC

(6) So AM = ABC = BN

QED

Lemma 8: A ~ B if and only if there is a third divisor C such that AC and BC are both principal.

Proof:

(1) Assume A ~ B

(2) Then there exists a divisor C such that AC is principal (from Lemma 6 above) and BC is principal (by definition of ~).

(3) Assume AC,BC, AD are principal

(4) Then, AC ~ BC ~ I and AD ~ I [By Lemma 3 above]

(5) (AD)(BC) ~ I [By Lemma 1 and Lemma 3 above]

(6) (AD)(BC) = (BD)(AC) [By Lemma 5 above]

(7) So, since (AC) is principal and (AD)(BC) is principal, it follows that (BD) must also be principal. [By Lemma 2 above]

(8) Thus, we have shown that if there exists a divisor C such that AC is principal and BC is principal, then for any divisor D, if AD is principal, then BD is necessarily also principle.

(9) A~B follows directly from step#(8).

QED

### 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

## 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.

## 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