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 - α2)*...*(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 (αj-1) is a conjugate of (α - 1).
(5) Therefore, N(αj-1 + αj-2 + ... + 1) = 1 which makes (αj-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 (αj - 1). [See Definition 4, here]
(4) But then h(α) = α - 1 since (α - 1) is the only prime that divides (αj - 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 (αj - 1) = (α - 1)*unit.
(3) Since there are λ - 1 factors of the form (αj - 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
Wednesday, August 02, 2006
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