The norm of an ideal number is more than a curiosity. It has an interesting property. The norm for any ideal number is also the number of incongruent classes modulo that ideal number. I use this result in my proof of the existence of the class number for any set of cyclotomic integers. I go over this property of norms of ideal numbers here.

Today's content is once again based on Harold M. Edwards' Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Definition 1: Norm of an Ideal Number

For an ideal number A, the norm N(A) = A*σA*σ

^{2}A*...*σ

^{λ-2}A.

For details on what σA means please see the Definition 1, here.

Lemma 1: N(AB)=N(A)*N(B)

Proof:

N(AB) = AB*σ(AB)*σ

^{2}(AB)*...*σ

^{λ-2}(AB) =

= A*B*σA*σB*σ

^{2}(A)*σ

^{2}(B)*...*σ

^{λ-2}(A)*σ

^{λ-2}(B) =

= A*σA*σ

^{2}(A)*...*σ

^{λ-2}(A)*B*σB*...*σ

^{λ-2}B =

= N(A)*N(B)

QED

Lemma 2: if p ≠ λ, then P*σP*σ

^{2}P*...*σ

^{e-1}P = p

NOTE: e = (λ - 1)/f where f = exponent mod λ for p. [See Definition 2, here]

Proof:

(1) Let P be a prime divisor that divides a rational integer p.

(2) From a previous result (see Lemma 1, here), we know that P, σP, σ

^{2}P, ..., σ

^{e-1}P are the e distinct divisors that divide p.

(3) We know that if any cyclotomic integer g(α) is divisible by all the prime divisors of p, then it is also divisible by p. Likewise, we know that if g(α) is divisible by p, then it is divisible by all the prime divisors. (See Theorem, here)

(4) In other words, the product of all prime divisors is principal and it is equal to p.

QED

Lemma 3: For a Prime Divisor P that divides p, the N(P) is a rational integer.

Proof:

(1) If p = λ, then P = α - 1 and N(P) = λ [See Lemma 2, here]. So now, we can assume that p ≠ λ in order to complete the proof.

(2) If p ≠ λ, then N(P) = P*σP*σ

^{2}P*...*σ

^{λ-2}P. [See Definition above]

(3) Since σ is a permutation and there are only e distinct prime divisors, we get the following:

N(P) = (P*σP*σ

^{2}P*...*σ

^{e-1}P)*(P*σP*σ

^{2}P*...*σ

^{e-1}P)*...*(P*σP*σ

^{2}P*...*σ

^{e-1}P)

(4) Applying Lemma 2 above gives us:

N(P) = p*p*...*p = p

^{f}[since ef = λ - 1, from the definition of exponent mod λ, see Definition 2, here]

QED

Lemma 4: For any ideal number A, the N(A) is a rational integer.

Proof:

(1) Every ideal number A is composed of powers of prime divisors. [See here for definition of ideal number]

(2) Let us assume that the powers of prime divisors of A consist of: P

_{1}

^{a}*P

_{2}

^{b}*...*P

_{n}

^{c}

(3) Then using Lemma 1 above, N(A) = N(P

_{1}

^{a}*P

_{2}

^{b}*...*P

_{n}

^{c}) =

= N(P

_{1}

^{a})*N(P

_{2}

^{b})*...*N(P

_{n}

^{c}) =

= N(P

_{1})*...*N(P

_{1})*N(P

_{2})*...*N(P

_{2})*...*N(P

_{n})*...*N(P

_{n})

(4) Using Lemma 3 above, we know that each N(P

_{i}) is a rational integer. This then gives us our result since the product of a set of rational integers is itself a rational integer.

QED

Lemma 5: if g(α) is a cyclotomic prime that divides p where p ≠ λ and P is the divisor for g(α), then p divides g(α)*σg(α)*σ

^{2}g(α)*...*σ

^{e-1}g(α) with a multiplicity of 1.

Proof:

(1) If a prime divisor P is the divisor for g(α), then p divides g(α)*ψ(η) [See Definition 7, here for definition of divisible by a prime divisor]

(2) This means that σp divides σ(g(α)*ψ(η)) and since p is a rational integer σ p = p and we have p divides σg(α)*σψ(η)

(3) Using the definition for divisibility of a prime divisor, we see that σP divides σg(α).

(4) We can make the same argument to show that σ

^{2}P divides σ

^{2}g(α) and so on up until σ

^{e-1}P divides σ

^{e-1}g(α).

(5) This gives us that the rational prime p divides g(α)*σ(gα)*...*σ

^{e-1}g(α). [See Theorem here, since division by all e of the prime divisors implies that p divides a given cyclotomic integer]

(6) Assume that p divides g(α)*σg(α)*...*σ

^{e-1}g(α) with a multiplicity greater than 1.

(7) Then the prime divisor P which divides p exactly once (from the theorem mentioned in step #5) would have to divide g(α)*σg(α)*...*σ

^{e-1}g(α) with the same multiplicity.

(8) But since P is the divisor of g(α), this would imply that g(α) must divide g(α)*σg(α)*...*σ

^{e-1}g(α) with a multiplicity greater than one which is not the case.

(9) So we reject the assumption in step #6 and conclude that p divides g(α)*σg(α)*...*σ

^{e-1}g(α) exactly once.

QED

Lemma 6: If a prime divisor P for a prime p ≠ λ is principal such that it is the divisor of a cyclotomic prime g(α), then N(P) = Ng(α)

Proof:

(1) From Lemma 3 above, N(P) = p

^{f}

(2) By definition of a divisor, we know that g(α) divides p [since P divides p and P is the divisor of g(α)]

(3) Ng(α) = g(α)*g(α

^{2})*...*g(α

^{λ-1}) [See definition of norm for cyclotomic integers]

(4) Np = p

^{λ-1}[See definition of norm for cyclotomic integers]

(5) We know that Ng(α) must be equal to a power of p since:

(a) N(p) = p

^{λ-1}[See definition of norm for cyclotomic integers]

(b) Ng(α) divides N(p) since g(α) divides p [See Lemma 6, here]

(6) Ng(α) ≡ g(1)

^{λ-1}≡ 0 or 1 (mod α - 1) since:

(a) α ≡ 1 (mod α -1)

(b) Ng(α) = g(α)*g(α

^{2})*...*g(α

^{λ-1,}) [Definition of Ng(α), see here]

(c) Using step #6a, we have:

g(α)*g(α

^{2})*...*g(α

^{λ-1,}) ≡ g(1)*g(1)*...*g(1) = g(1)

^{λ-1}= (a

_{0}+ a

_{1}+ ... + a

_{λ-1})

^{λ-1}= rational integer r (since a

_{i}are all rational integers).

(d) Now, since N(α - 1) = (α - 1)(α

^{2}- 1)*...*(α

^{λ-1}-1) = λ (see Lemma 2, here), we know that (α-1) divides λ.

(e) Since λ is a prime, we know that either λ divides r or it does not. If it does, then g(1)

^{λ-1}≡ 0 (mod α -1 ).

(f) If λ does not divide g(1)

^{λ-1}, then gcd(λ,g(1)

^{λ-1}) = 1 [Since λ is a prime]

(g) Using Bezout's Identity, there exists a,b such that a*λ + b*g(1)

^{λ-1}= 1.

(h) This gives us that b*g(1)

^{λ-1}- 1 = (-1)(aλ) so that b*g(1)

^{λ-1}≡ 1 (mod λ).

(7) So, from step #6, we see that g(1)

^{λ-1}≡ 0 or 1 (mod α - 1) if and only if g(1)

^{λ - 1}≡ 0 or 1 (mod λ)

(8) We know that λ does not divide Ng(α) since Ng(α) = p

^{x}(from step #5 above) and since gcd(p,λ)=1.

(9) So, we are left with Ng(α) ≡ 1 (mod λ) which means that if Ng(α) = p

^{x}then x is divisible by f where f is the exponent mod λ for p. [See Lemma 1, here]

(10) Finally, we show that Ng(α) = p

^{f}since:

(a) We can divide up Ng(α) into [g(α)*σg(α)*...*σ

^{e-1}g(α)]*[σ

^{e}g(α)* σ

^{e+1}g(α)*...*σ

^{2e-1}g(α)]*...*[σ

^{(f-1)*e}g(α)σ

^{(f-1)*e+1}g(α)*...*σ

^{(f-1)*e+e-1}g(α)]

(b) By the reasoning in Lemma 5 above, each of these f groupings of e elements is divisible by at most once by p. So that all f of these groups is divisible at most by p

^{f}.

(c) Thus, it follows that if Ng(α) = p

^{a}, then a = f.

QED

Lemma 7: Criteria for a Principal Divisor

An ideal number A is a principal divisor for a cyclotomic integer g(α) if for any prime divisor P

^{n}:

P

^{n}divides A if and only if P

^{n}divides g(α)

Proof:

(1) An ideal number by definition is a set of powers of prime divisors. [See Definition 3, here]

(2) So, if each power of each prime that makes up an ideal number A divide a given cyclotomic integer g(α), then A divides g(α)

(3) This gives us that if g(α) divides a second cyclotomic integer h(α), then A also divides h(α).

(4) Now, to complete this proof, we need to show that if A divides h(α), then g(α) also divides h(α).

(5) By the given, we know that A represents a complete set of prime divisors that divide g(α).

We know this since if there is any prime divisor P

^{n}that does not divide A, then it does not divide g(α).

(6) So, applying the Fundamental Theorem for Ideal Numbers, we know that if A divides a second cyclotomic integer h(α), then g(α) also divides h(α).

QED

Lemma 8: If an ideal number A is the principal divisor for a cyclotomic integer g(α), then σA is the principal divisor for a σg(α)

Proof:

(1) This lemma is established if we can use the criteria in Lemma 7. That is, we want to show that for any prime divisor P

^{n}:

P

^{n}divides σA if and only if P

^{n}divides σg(α).

(2) Assume P

^{n}divides σA

(3) Then σ

^{-1}P

^{n}divides A.

(4) And σ

^{-1}P

^{n}divides g(α) since A is the principal divisor for g(α).

(5) And P

^{n}divides σg(α).

(6) Assume P

^{n}divides σg(α)

(7) Then σ

^{-1}P

^{n}divides g(α) so σ

^{-1}P

^{n}divides A [Since, A is the principal divisor for g(α).]

(8) Which gives us that P

^{n}divides σA.

QED

Theorem: If an ideal number A is the principal divisor for a cyclotomic integer g(α), then N(A) is the principal divisor for Ng(α)

Proof:

(1) This lemma is established if we can use the criteria in Lemma 7. That is, we want to show that for any prime divisor P

^{n}:

P

^{n}divides N(A) if and only if P

^{n}divides Ng(α).

(2) Assume P

^{n}divides N(A)

(3) N(A) = A*σA*σ

^{2}A*...*σ

^{λ-2}A

(4) From our assumption in step #2, we know that: P

^{n}divides a subset of the list of ideal numbers in step #3 so that we have:

P

^{n}divides σ

^{a}A*...*σ

^{c}A.

(5) Since g(α) = g(α)*σg(α)*...*σ

^{λ-2}g(α), we can apply Lemma 8 above to conclude that:

σ

^{a}A is the principal divisor for σ

^{a}g(α)

...

σ

^{c}A is the principal divisor for σ

^{c}g(α)

(6) Finally, this gives us that P

^{n}divides Ng(α) since each P

^{i}that divides a given σ

^{i}A also must divide the given σ

^{i}g(α) and therefore divide Ng(α).

(7) Assume P

^{n}divides Ng(α)

(8) Again P

^{n}can only divide Ng(α), if its divides between 1 and n cyclotomic integers of the form σ

^{i}g(α).

(9) It would then divide each of the principal divisors σ

^{i}A for those cyclotomic integers and thereby divides N(A).

QED

Corollary: If an ideal number A is the principal divisor for a cyclotomic integer g(α), then N(A) = Ng(α)

Proof:

(1) N(A) is a rational integer (See Lemma 4 above) and Ng(α) is a rational integer (see Lemma 5, here)

(2) Since every prime divisor P

^{n}that divides N(A) also divides Ng(α) [By the Theorem above], we know that N(A) ≤ Ng(α).

(3) Since every prime divisor P

^{n}that divides Ng(α) also divides N(A), we know that Ng(α) ≤ N(A).

(4) The conclusion follows.

QED

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