Thursday, May 25, 2006

Cyclotomic Integers: More on cyclotomic periods

In today's blog, I continue to review results from Harold Edward's Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory. I am continuing down the same chapter that I started in an earlier blog. If you would like to start at the beginning of cyclotomic integer properties, start here.

Lemma 1: g(α) = g(αp) if and only if g(α) is a cyclotomic integer made up of periods of length f.

Proof:

(1) Let λ be an odd prime.

(2) Let α be a primitive root of unity such that αλ = 1.

(3) Let p be an odd prime distinct from λ

(4) Let τ be mapping such that τα = αp

(5) Let f be the least positive integer for which pf ≡ 1 (mod λ)

(6) We know that f divides λ -1 since:

(a) pλ-1 ≡ 1 (mod λ) from Fermat's Little Theorem (since p,λ are distinct primes)

(b) From a previous result (see Lemma 2 here), we know if f is the least positive integer, then it must divide λ-1.

(7) Let e= (λ-1)/f

(8) Assume that g(α) is made up of periods of length f. [See here for review of cyclotomic periods]

(9) Then, σeg(α) = g(α) [See Lemma 4 here for details.]

(10) Then, there must exist an integer k such that:

τ = σk

Since the primitive root (see here for review if needed) can take all possible values mod λ [By definition σ = a mapping between α and αγ where γ is a primitive root, see here]

(11) Using step #10, we note that:

τf = σkf

Since τf ≡ 1 (mod λ), we know that σkf ≡ 1 (mod λ)

(12) Since the order of λ is λ -1 (see here), we know that kf must be divisible by λ - 1 (see Lemma 2 here) which means it is divisible by ef since ef=λ - 1.

So, ef divides kf means that e must divide k.

(13) So, from step #10 and step #12:

τ is a power of σe

(14) So, there exists k' such that ek' = k.

(15) From #14, we have:

τ = (σ1e)(σ2e)...(σk'e)

and therefore:

τg(α) =(σ1e)(σ2e)...(σk'e)g(α) = g(α)

(16) Assume that τg(α) = g(α)

(17) We know that there exists k such that:

τ = σk [Since σ is a mapping to γ which is a primitive root and the primitive root can take on all values modulo λ]

(18) Since g(α) repeats at τg(α), we know that g(α) consists of e periods of length f. [See Corollary 1.2 here]

(19) By definition, e divides (λ - 1) [See here for definition of periods]

(20) Thus, e is a common divisor of k and λ - 1.

(21) e is the greatest common divisor of k and λ - 1 since:

(a) Let d be an integer that divides both k, λ-1 so that:

k = qd
λ - 1 = df'

(b) Now,

τf' = σkf' = σqdf' = σ(λ - 1)q which is identity[from step #17, #21a, and since σλ-1 is the identity, see here]

(c) So that:

f' ≥ f [Since f is least value where pf ≡ 1 (mod λ)]
d ≤ e [Because d = (λ-1)/f' where f' ≥ f and e=(λ - 1)/f]

(22) Using Bezout's Identity, there exists a,b such that:

e = ak + b(λ - 1)

(23) Further,

σe = σakσb(λ-1) = [from step #22]

= σak = [Since σb(λ-1) is identity]

= τa [Since τ = σk from step #17]

(24) From this, we see that:

σeg(α) = τag(α) = g(α)

QED

Lemma 2: Let h(α) be a prime cyclotomic integer. For any cyclotomic integer g(α) made up of periods of length f, there exists an integer u such that g(α) ≡ u (mod h(α))

Proof:

(1) Fermat's Little Theorem gives us:

xp-1 ≡ 1 (mod p)

So that:

p divides xp-1 - 1

So that:

x(xp-1 - 1) = xp - x ≡ 0 (mod p)

(2) We also know that:

(x -1)*(x-2)*...(x-p) ≡ 0 (mod p)

(a) There exists a r such that x ≡ r (mod p)

(b) We know that p divides x - r

(c) We also know that r is between 0 and p-1.

(d) if r is 0, then p divides x-p; otherwise, p divides x-r.

(3) Putting (#1) and (#2) together, we have:

xp - x ≡ (x-1)(x-2)*...*(x-p) (mod p)

(4) Let x = g(α)

(5) Then:

g(α)p - g(α) ≡ [g(α) - 1][g(α) - 2]*...*[g(α) - p] (mod p)

(6) From a previous result (see Lemma 3 here), where g(α)p ≡ g(αp) (mod p), we have:

g(αp) - g(α) ≡ [g(α) - 1][g(α) - 2]*...*[g(α) - p] (mod p)

(7) Since h(α) divides p, then we have:

g(αp) - g(α) ≡ [g(α) - 1][g(α) - 2]*...*[g(α) - p] (mod h(α))

(8) Now, since g(α) is made up of periods of length f, we know that g(αp) = g(α) [See Lemma 1 above]

So that:

g(αp) - g(α) = 0 ≡ 0 (mod h(α))

(9) This gives:

[g(α) - 1][g(α) - 2][g(α) - 3]*...*[g(α)-p] ≡ 0 (mod h(α))

(10) Since h(α) is a prime cyclotomic integer, one of these values must be divisible by h(α) [By the definition of a cyclotomic prime, see here]

(11) So there exists an integer u such that h(α) divides g(α) - u

Which means that:

g(α) ≡ u (mod h(α))

QED

Corollary 2.1: Let h(α) be a prime cyclotomic integer. Let ηi be a period of length f. Then, there exists an integer ui such that ηi ≡ ui (mod h(α))

Proof:

(1) Each period ηi can be thought of as a cyclotomic integer made up of periods of length f:

g(α) = (0)η0 + ... + (1)ηi + ... + (0)ηe-1 = ηi

(2) From Lemma 2 above, we know that there exists an integer ui such that g(α) ≡ ui (mod h(α))

(3) So, we see that for each ηi, there exists ui such that:

ηi ≡ ui (mod h(α))

QED