## Wednesday, August 02, 2006

### Ideal Numbers: α - 1

From the perspective of ideal numbers, there are two types of prime divisors. Prime divisors that divide a prime integer p where p ≠ λ (see here for review of λ and its relation to cyclotomic integers) and the prime divisor of λ.

In today's blog, I will go into detail into the prime divisor of λ. The important idea here is that λ has only a single prime divisor α - 1 which is also a cyclotomic integer.

In today's blog, I will review some basic properties of α - 1. The content of today's blog is based on Harold M. Edwards' Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Lemma 1: if n is an odd prime, then

(x - α)(x - α
2)*...*(x - αn-1) = xn-1 + xn-2 + ... + 1

Proof:

(1) From a previous result, we have:

xn - 1 = (x - 1)(x - α)(x - α2)*...*(x - αn-1)

(2) Now xn - 1 divided by x-1 = xn-1 + xn-2 + ... + x + 1 since:

(x-1)(xn-1 + xn-2 + ... + x + 1) =

= xn + xn-1 + xn-2 + .... + x2 + x -xn-1 - xn-2 - .... - x2 - x - 1 =

= xn - 1

(3) Putting #2 and #1 together gives us:

xn-1 + xn-2 + ... + x + 1 = (x - α)(x - α2) * ... * (x - αn-1)

QED

This second lemma uses the result of Lemma 1 to show that the Norm of α - 1 is λ. If you need a review of the norm of cyclotomic integers, see Section 4, here.

Lemma 2: N(α - 1) = λ

Proof:

(1) N(α - 1) = N(1 - α) since:

N(α - 1) = (α - 1)(α2 - 1)*...*(αλ-1 - 1)

N(1 - α) = (-1)λ-1[(α - 1)(α2 - 1)*..*(αλ-1 - 1)

Since λ is odd, λ-1 is even and (-1)λ-1 = 1

(2) N(1 - α) = (1 - α)(1 - α2)*...*(1 - αλ-1) [See here for definition of norm for cyclotomic integers]

(3) From Lemma 1 above:

(x - α)(x - α2)*...*(x - αλ-1) = xλ-1 + xλ-2 + ... + x + 1

(4) Setting x = 1 gives us:

(1 - α)(1 - α2)*...*(1 - αλ-1) = 1λ-1 + 1λ-2 + ... + 1 + 1 = λ

QED

Lemma 3: (αj - 1) = (α - 1)*unit

Proof:

(1) αj - 1= (α - 1)(αj-1 + αj-2 + ... + 1)

(2) Since ab=c → N(a)N(b)=N(c) (See Lemma 6, here), we know that:

N(αj - 1) = N(α-1) * N(αj-1 + αj-2 + ... + 1)

which gives us that:

N(αj-1 + αj-2 + ... + 1) = N(αj - 1)/N(α - 1)

(3) From Lemma 2 above, N(α-1) = λ

(4) We can also conclude from Lemma 2 above that N(αj-1) = λ since j-1) is a conjugate of (α - 1).

(5) Therefore, N(αj-1 + αj-2 + ... + 1) = 1 which makes j-1 + αj-2 + ... + 1) a unit (see Definition 1, here)

QED

Corollary 3.1: (α -1 ) is the only cyclotomic prime that divides λ

Proof:

(1) Let h(α) be a cyclotomic prime that divides λ

(2) Using Lemma 2 above, we know that:

(α - 1)(α2 - 1)*...*(αλ-1 - 1) = λ

(3) Since h(α) divides λ, it must divide one of the factors of j - 1). [See Definition 4, here]

(4) But then h(α) = α - 1 since (α - 1) is the only prime that divides j - 1). [From Lemma 3 above]

QED

Corollary 3.2: (α - 1)λ-1 = λ * unit

Proof:

(1) Using Lemma 2 above,

(α - 1)(α2 - 1)*...*(αλ-1 - 1) = λ

(2) Using Lemma 3 above, we know that each j - 1) = (α - 1)*unit.

(3) Since there are λ - 1 factors of the form j - 1), this gives us:

λ = (α-1)λ-1*(unit)λ-1

(4) Since a (unit)λ-1 = unit, we have

λ = (α -1)λ-1*unit.

QED

Lemma 4: For each cyclotomic prime h(α) that divides a standard prime p = λ, there exists a set of λ incongruent classes mod h(α)

Proof:

(1) There is only 1 cyclotomic prime that divides λ so h(α) = α - 1. [See Corollary 3.1 above]

(2) So, α ≡ 1 (mod h(α)) [Since h(α) = α - 1]

(3) Let g(α) be a cyclotomic integer.

(4) Then g(α) can be put in the following form (see Lemma 1, here for details):

g(α) = a0 + a1α + ... + aλ-1αλ - 1

(5) This means that g(α) ≡ a0 + a1 + ... + aλ - 1 (mod α - 1)

(6) So, g(α) ≡ an integer (mod h(α)).

(7) It is also clear that if g(α) ≡ x (mod λ) that x is an integer between 0 and λ - 1.

(8) But since h(α) divides λ, it is clear that h(α) must also divide g(α) - x (since λ divides g(α) - x).

QED