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.
(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)
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'
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)
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)
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))
(1) Fermat's Little Theorem gives us:
xp-1 ≡ 1 (mod p)
p divides xp-1 - 1
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)
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]
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))
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))
(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))