Tuesday, July 25, 2006

Cyclotomic Integers: Periods mod p

Today's blog continues the details that are used by Ernst Kummer in his theory of ideal numbers. He then uses ideal numbers in his proof for Fermat's Last Theorem for regular primes.

The contents of today's blog are taken from Harold M. Edwards Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

In previous blogs, I have talked about cyclotomic integers and cyclotomic periods derived from cyclotomic integers. In today's blog, I will focus on a single, very important result. That for any prime p where p ≠ λ, there exists an integer f = exponent mod λ (see definition 2, here) and an integer e = (λ - 1)/f (see definition 3, here).

This means that for any prime p, there exist a set of e cyclotomic periods that each have a length of f elements (see here for review of the basic properties of cyclotomic periods).

Lemma 1: xp - x ≡ (x-1)(x-2)*...*(x-p) ≡ 0 (mod p)


(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) ≡ 0 (mod p)


Lemma 2: 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 (see Lemma 1, here)

(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)


Theorem 3: For any cyclotomic integer g(α) made up of periods of length f where f is the exponent mod λ for p, there exists an integer u such that g(α) ≡ u (mod p)


(1) From Lemma 1 above, we have:

xp - x ≡ (x-1)(x-2)*...*(x-p) (mod p)

(2) Let x = g(α)

(3) Then:

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

(4) 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)

(5) Now, since g(α) is made up of periods of length f, we know that g(αp) = g(α) [See Lemma 2 above]

So that:

g(αp) - g(α) = 0 ≡ 0 (mod p)

(9) This gives us:

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

(10) So at least one of these values must be divisible by p (otherwise, the product of each g(α) - u could not be 0.)

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

Which means that:

g(α) ≡ u (mod p)


Corollary 3.1: For each of the e cyclotomic periods ηi for a given odd prime p, there exists an integer ui such that ηi ≡ ui (mod p)


(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 Theorem 3 above, we know that there exists an integer ui such that g(α) ≡ ui (mod p)

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

ηi ≡ ui (mod p)


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