The answer to this question turns out to be the norm of the ideal number. For example, if we want to make sure that for a set of cyclotomic integers, a given ideal number A divides at least one, we can be sure of this if we have a set of N(A) distinct congruence classes.

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

Lemma 1: Norm of a Prime Divisor is the number of incongruent cyclotomic integers

If A is a prime divisor, the Norm N(A) can be regarded as a positive integer that is equal to the number of incongruent cyclotomic integers mod A.

Proof:

(1) If A is a prime divisor, then it is either the exceptional prime divisor (α - 1) or it is one of e prime divisors that divide p.

(2) Assume A = α - 1

(3) Then ever cyclotomic integer is congruent to one and only one of the λ integers 0, 1, 2, ..., λ-1.

(4) And N(α-1) = λ (see Lemma 6, here for details)

(5) Assume A ≠ α - 1.

(6) Then its norm is p

^{f}where p is the prime integer that it divides and f is the exponent of p mod λ.

The norm of a divisor is defined to be A*σA*σ

^{2}A*...*σ

^{λ-2}A = (A*σA*σ

^{2}A*..*σ

^{e-1}A)(σ

^{e}A*...*σ

^{2e-1}A)....(σ

^{λ-e-1}A*..*σ

^{λ-2}A)

Now, each of these set of e prime divisors is the same as p. Likewise, ef = λ - 1.

This gives us:

(p)*...*(p) = p

^{f}.

(7) Using a previous result (see Theorem 3, here), we know that there are p

^{f}incongruent elements mod A.

QED

Lemma 2: Norm of a Power of a Prime Divisor is the number of incongruent cyclotomic integers

If A is a power of a prime divisor, the norm N(A) can be regarded as a positive integer that is equal to the number of incongruent cyclotomic integers mod A.

Proof:

(1) Let A = P

^{n}where P is a prime divisor.

(2) Let ψ(α) be a cyclotomic integer which is divisible by P with multiplicity exactly 1 but not divisible at all by any of the conjugates of P. [See Proposition 1, here for details on this construction if needed]

(3) Each cyclotomic integer is congruent to one of the form a

_{0}+ a

_{1}ψ(α) + a

_{2}ψ(α)

^{2}+ ... + a

_{n-1}ψ(α)

^{n-1}mod P

^{n}and two cyclotomic integers of this form are congruent if and only if the coefficients a

_{0}, a

_{1}, ..., a

_{n-1}are the same mod P since:

(a) When n=1, this is obvious since we are only dealing with a

_{0}.

(b) Assume that it is proven for n-1.

(c) Then, for a given cyclotomic integer x(α), there are cyclotomic integers a

_{0}, a

_{1}, ..., a

_{n-2}for which x(α) ≡ a

_{0}+ a

_{1}ψ(α) + ... + a

_{n-2}ψ(α)

^{n-2}(mod P

^{n-1}) and the coefficients a

_{0}, a

_{1}, ..., a

_{n-2}are uniquely determined mod P. [By our assumption in step (b)]

(d) Let y(α) = x(α) - a

_{0}- a

_{1}ψ(α) - ... - a

_{n-2}ψ(α)

^{n-2}.

(e) Then y(α) ≡ 0 mod P

^{n-1}[By combining step (c) and step (d)]

(f) Now, we need to prove that y(α) ≡ aψ(α)

^{n-1}mod P

^{n}for some a and that a is uniquely determined mod P.

(g) Let γ(α) denote the the product of e-1 distinct conjugates of ψ(α) so that γ(α)ψ(α) = pk where k is an integer relatively prime to p.

We know that p divides γ(α)*ψ(α) exactly once so k is what is left over. Since p is a prime, we know that gcd(p,k)=1.

(h) So if we multiply γ(α)

^{n-1}to both sides of (f), we get:

y(α)γ(α)

^{n-1}≡ aγ(α)

^{n-1}ψ(α)

^{n-1}≡ ap

^{n-1}k

^{n-1}(mod P

^{n})

(i) Since gcd(p,k)=1, there exists an integer m such that mk ≡ 1 (mod p).

NOTE: This comes straight from Bezout's Identity which says there exist m,n such that mk + np = 1. In other words (-n)p = mk - 1.

(j) Multiplying m

^{n-1}to both sides of (h) gives us:

y(α)γ(α)

^{n-1}m

^{n-1}≡ ap

^{n-1}k

^{n-1}m

^{n-1}≡ ap

^{n-1}(mod P

^{n})

(k) Since y(α) ≡ 0 (mod P

^{n-1}) by assumption, then y(α)γ(α)m

^{n-1}is divisible by p

^{n-1}

(l) This shows that y(α) is determined by a mod P [See here for details if needed]

(m) This completes the proof in the case A = P

^{n}.

(4) This proves that the number of classes mod P

^{n}is equal to the number of ways of choosing a

_{0}, a

_{1}, ..., a

_{n-1}mod P which is N(P)

^{n}= N(P

^{n}).

QED

Theorem 1: Norm of a Divisor is the number of incongruent cyclotomic integers

If A is a divisor, the Norm N(A) can be regarded as a positive integer that is equal to the number of incongruent cyclotomic integers mod A.

Proof:

(1) The number of incongruent elements mod AB (where A,B are relatively prime) is never more than the number of incongruent elements mod A times the number of incongruent elements mod B since:

(a) Let x ≡ x' (mod A)

(b) Let x ≡ x' (mod B)

(c) Then:

A divides x - x'

B divides x - x'

(d) Since A,B are relatively prime, this means that AB divides x - x'.

(e) So that x ≡ x' (mod AB)

(2) The Chinese Remainder Theorem for Divisors (see here) shows that all possible classes mod A and mod B occur.

(3) So that, the number of classes mod AB is equal to the number of classes mod A times the number of classes mod B.

(4) By induction, if A,B,C,D are relatively prime divisors, then the number of classes mod ABC*..*D is equal to number of classes mod A times ... times the number of classes mod D.

(5) If A,B,C,D are powers of prime divisors, by Lemma 2 above this number is equal to N(A)*N(B)*N(C)*...*N(D) = N(ABC*...*D).

(6) Since any divisor can be written in the form A*B*C*....*D where A,B,C,...,D are relatively prime powers of prime divisors, it follows that the number of classes mod any divisor is the norm of that divisor.

QED