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(α^p) 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 α^λ = 1.

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

(4) Let τ be mapping such that τα = α^p

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

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

(a) p^λ-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, σ^eg(α) = g(α) [See Lemma 4 here for details.]

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

τ = σ^k

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 α^γ where γ is a primitive root, see here]

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

τ^f = σ^kf

Since τ^f ≡ 1 (mod λ), we know that σ^kf ≡ 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 σ^e

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

(15) From #14, we have:

τ = (σ₁^e)(σ₂^e)...(σ_k'^e)

and therefore:

τg(α) =(σ₁^e)(σ₂^e)...(σ_k'^e)g(α) = g(α)

(16) Assume that τg(α) = g(α)

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

τ = σ^k [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,

τ^f' = σ^kf' = σ^qdf' = σ^{(λ - 1)q} which is identity[from step #17, #21a, and since σ^λ-1 is the identity, see here]

(c) So that:

f' ≥ f [Since f is least value where p^f ≡ 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,

σ^e = σ^akσ^b(λ-1) = [from step #22]

= σ^ak = [Since σ^b(λ-1) is identity]

= τ^a [Since τ = σ^k from step #17]

(24) From this, we see that:

σ^eg(α) = τ^ag(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:

x^p-1 ≡ 1 (mod p)

So that:

p divides x^p-1 - 1

So that:

x(x^p-1 - 1) = x^p - 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:

x^p - 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(α^p) (mod p), we have:

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

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

g(α^p) - 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(α^p) = g(α) [See Lemma 1 above]

So that:

g(α^p) - 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 η_i be a period of length f. Then, there exists an integer u_i such that η_i ≡ u_i (mod h(α))

Proof:

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

g(α) = (0)η₀ + ... + (1)η_i + ... + (0)η_e-1 = η_i

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

(3) So, we see that for each η_i, there exists u_i such that:

η_i ≡ u_i (mod h(α))

QED

Cyclotomic Integers: g(α)p ≡ g(αp) (mod p)

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: if p is a prime, then (x + y)^p ≡ x^p + y^p

Proof:

(1) Using the Binomial Theorem, we know that:




(2) Now, this means that each of the coefficients is equal to:



(3) But since p is a prime, p divides all coefficients.

For all values where k is between 1 and p-1, p, as a prime, will not be divisible by either k! or by (p-k)!.

(4) This means that p will divides (x + y)^p - x^p - y^p.

QED

Lemma 2: x^p ≡ x (mod p)

Proof:

(1) x^p - x = x(x^p-1 - 1)

(2) if gcd(x,p)=1, then by Fermat's Little Theorem, x^p-1 ≡ 1 (mod p) so p divides x^p-1-1.

(3) if p divides x, then of course p divides x(x^p-1-1)

QED

Lemma 3: g(α) ≡ g(α^p) mod p

Proof:

(1) Let g(α) = a₀ + a₁α + ... + α_λ-1α^λ-1.

(2) Then g(α)^p = (a₀ + [a₁α + ... + α_λ-1α^λ-1])^p

(3) Now using Lemma 1 above, we can break #2 out in the following ways:

(a₀ + [a₁α + ... + α_λ-1α^λ-1])^p ≡ (a₀)^p + (a₁α + [... + α_λ-1α^λ-1])^p (mod p).

(4) We could keep breaking out each element until we get:

(a₀ + [a₁α + ... + α_λ-1α^λ-1])^p ≡ (a₀)^p + (a₁α)^p + ... + (a_λ-1α^λ-1)^p (mod p)

(5) Now for each coefficient a_i, using Lemma 2 above, we have:

a_i^p ≡ a_i (mod p)

So this gives us:

(a₀)^p + (a₁α)^p + ... + (a_λ-1α^λ-1)^p ≡ a₀ + a₁α^p + ... + a_λ-1(α^λ-1)^p (mod p)

(6) And we are done since this shows that:

g(α)^p ≡ g(α^p) (mod p)

QED

Cyclotomic Integers: Periods

In today's blog, I will review the concept of periods for cyclotomic integers 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.

I will continue to use Edward's σg(α) notation that I reviewed in a previous blog.

Lemma 1:

Let e be a factor of λ - 1 where λ is an odd prime and α is a primitive root of unity where α^λ = 1.

Then, for a given cyclotomic integer g(α) there exists a cyclotomic integer G(α) such that:

Ng(α) = G(α)*σG(α)*...*σ^e-1G(α)

Proof:

(1) Let G(α) = g(α)*σ^eg(α)*σ^2eg(α)*...*σ^-eg(α) where σ^-e denotes σ^λ-1-e

(2) With this definition we can see that:

G(α) = g(α)*g(α^γe)*g(α^γ2e)*...*g(α^γ(λ-1-e))

σG(α) = g(α^γ)*g(α^γ(e+1))*g(α^γ2e+1)*...*g(α^γλ-e)

...

σ^e-1G(α) = g(α^γ(e-1))*g(α^γ(2e-1))*...*g(α^γ(λ-2))

(3) Putting this all together, we see that:

Ng(α) = g(α)*g(α^γ)*g(α^γ*γ)*...*g(α^γ(λ-2))

(4) Since γ is a primitive root mod λ, we know that (3) is the same as:

Ng(α) = g(α)*g(α²)*g(α³)*..*g(α^λ-1)

QED

Example 1: λ = 13, γ = 2, and e = 4

In this case, Ng(α) = G(α)G(α)G(α⁴)G(α⁸)

Each term G(α) = g(α)*g(α²⁴)*g(α^22*4) = g(α)*g(α³)*g(α⁹)

Since:

2 ≡ 2 (mod 13)
2² ≡ 4 (mod 13)
2³ ≡ 8 (mod 13)
2⁴ ≡ 16 ≡ 3 (mod 13)
2⁵ ≡ 32 ≡ 6 (mod 13)
2⁶ ≡ 64 ≡ 12 (mod 13)
2⁷ ≡ 128 ≡ 11 (mod 13)
2⁸ ≡ 256 ≡ 9 (mod 13)
2⁹ ≡ 512 ≡ 5 (mod 13)
2¹⁰ ≡ 1024 ≡ 10 (mod 13)
2¹¹ ≡ 2048 ≡ 7 (mod 13)
2¹² ≡ 4096 ≡ 1 (mod 13)

Corollary 1.1: σ^eG(α) = G(α)

Proof:

(1) G(α) = g(α)*g(α^γe)*g(α^γ2e)*...*g(α^γ(λ-1-e))

(2) And we find that:

σ^eG(α) = g(α^γe)*g(α^γ2e)*g(α^γ3e)*...*g(α^γλ-1)

(3) And since γ^λ-1 ≡ 1 (mod λ) (By Fermat's Little Theorem), we see that the result in step #2 is the same as the result in step #1.

QED

Corollary 1.2: σ^eG(α) = G(α) implies that there exists a set of e cyclotomic integers: η₀, η₁, ... η_e-1 such that:

(a) η₀ = α + σ^eα + σ^2eα + ... + σ^λ-1-eα

(b) η_i+1 = ση_i

(c) G(α) = a₀ + a₁η₀ + a₂η₁ + ... + a_eη_e-1 where a_i are integers.

Proof:

(1) σ^eG(α) = G(α)

(2) Since G(α) is a cyclotomic integer, we know that (see Lemma 1 here):

G(α) = a₀ + a₁α + ... + a_λ-1α^λ-1

(3) Now, (#1) implies that the coefficient for any α^j, α^j = σ^eα^j

(4) In order for #3 to be true, then G(α) must have the following form:

G(α) = a₀ + a₁(α + σ^eα + σ^2eα + ... + σ^λ-1-e) + a₂(σα + σσ^eα + ...) + ... + a_e(σ^e-1α + σ^e-1σ^eα + ...)

The reason being that each time all the values G(α) shifts by e, the new value must also have had the same coefficient and further, all new resulting values must be the same as values that existed before the shift. For each coefficient, the set of possible α^j values must be closed.

(5) Now, we can define η₀ in the following way:

η₀ = α + σ^eα + σ^2eα + ... + σ^λ-1-e

(6) We can likewise define η_i+1 as:

η_i+1 = ση_i

(7) Putting (5) and (6) into the equation in step #4 gives us:

G(α) = a₀ + a₁η₀ + ... + a_eη_e-1

QED


Definition 1: Cyclotomic Period

Let λ be a prime integer. Let α be a primitive root of unity such that α^λ=1. Let γ be a primitive root modulo λ. Let e be a factor of λ - 1.

Let G(α) = g(α)*g(α^γe)*g(α^γ2e)*...*g(α^γ(λ-1-e))

Then there exists a set of e cyclotomic integers: η₀, η₁, ... η_e-1 such that:

(a) η₀ = α + σ^eα + σ^2eα + ... + σ^λ-1-eα

(b) η_i+1 = ση_i

(c) G(α) = a₀ + a₁η₀ + a₂η₁ + ... + a_eη_e-1 where a_i are integers.

The cyclotomic integers η₀, η₁, ... η_e-1 that meet conditions (a), (b), and (c) are called cyclotomic periods.

Note:

As a convention, η_e = η₀ and η_-1 = η_e-1. So one can think of η_i as an ever repeating pattern of e values such that η_i+1 = ση_i

In this way, η_i is defined for all integers i.

Example 1: period for λ = 13, μ = 2, γ = 4

In this case, the cyclotomic periods are:

η₀ = α + σ⁴α + σ⁸α = α + α³ + α⁹

η₁ = σα + σα³ + σα⁹ = α² + α⁶ + α⁵

η₂ = σ²α + σ²α³ + σ²α⁹ = α⁴ + α¹² + α¹⁰

η₃ = σ³α + σ³α³ + σ³α⁹ = α⁸ + α¹¹ + α⁷

Lemma 2: σ^eη_i = η_i

Proof:

(1) For Case 0, this is given by the note in definition 1 above.

σ^eη₀ = η_e = η₀

(2) Let's assume that this is true up to i so that:

σ^eη_i = η_i where 0 ≤ i.

(3) Then, this is also true for η_i+1 since:

σ^e(η_i+1) = σ^e(ση_{i) [From Definition 1 above]}

Now σ^eσ = σ^e+1 = σσ^e

So that we get:

σ^eη_i+1 = σ(σ^eη_i)

By assumption at step #2, σ^eη_i = η_i

So that we end up with:

σ^eη_i+1 = ση_i = η_i+1

(4) So, by the principle of induction, we are done.

QED


Lemma 3: Each period η consists of (λ-1)/e terms.

Proof:

(1) Let f = (λ - 1)/e.

(2) σ^efα = σ^λ-1α = α

(3) σ^{(λ - 1 - e)}α = σ^e(f-1)α = (σ^e)^f-1α

(4) So we can see that each term in α + σ^eα + σ^2eα + ... + σ^λ-1-eα is σⁱα where i is 0*e, 1*e, 2*e, ... , (f-1)*e.

QED

Definition 2: Made up of periods of length f

A cyclotomic integer is made up of periods of length f if it can be put in the form:

a₀ + a₁η₁ + a₂η₂ + ... + a_eη_e

where a_i are integers and η_i are periods of length f.

Lemma 4: A cyclotomic integer g(α) is made up of periods of length if and only if σ^eg(α) = g(α)

Proof:

(1) Assume that g(α) is a cyclotomic integer made up of periods of length f.

(2) Then, there exists a_i and η_i such that:

g(α) = a₀ + a₁η₀ + a₂η₁ + ... + a_eη_e-1

(3) Applying σ^e to g(α) gives us:

σ^eg(α) = a₀ + a₁σ^eη₀ + ... + a_eσ^eη_e-1

(4) Applying Lemma 2 above gives us:

σ^eg(α) = a₀ + a₁η₀ + ... + a_eη_e-1

(5) Assume that σ^eg(α) = g(α)

(6) Then by Corollary 1.2 above, g(α) consists of e periods.

(7) And by Lemma 3, each period will be of length f.

QED

Corollary 4.1: The product of two cyclotomic integers made up of periods of length f is itself made up of periods of length f.

Proof:

(1) Let f(α), g(α) be cyclotomic integers made up of periods of length f so that:

σ^ef(α) = f(α)
σ^eg(α) = g(α)

(2) But then:

σ^e[f(α)g(α)] = [σ^ef(α)][σ^eg(α)] = f(α)g(α)

(3) So that by Lemma 4 above, we have proved that f(α)g(α) must itself be made up of periods of length f.

QED

Example: λ = 13, f = 3, e = 4

We note that:

(η₀)² = η₁ + 2η₂

η₀η₁ = η₀ + η₁ + η₃

η₀η₂ = 3 + η₁ + η₃

Likewise, applying σ to these equations gives us:

(η₁)² = η₂ + 2η₀

η₁η₂ = η₁ + η₂ + η₀

And so forth.

Cyclotomic Integers: σg(α) notation

In today's blog, I will review the σg(α) notation 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.

I will use this notation later on when I go over Kummer's proof of Fermat's Last Theorem for regular primes.

Edwards offers the following notation in Section 4.5:

Definition: σg(α)

Let σ signify the conjugate α^γ for a given cyclotomic integer where γ is a primitive root mod λ and λ is a prime integer and α is a primitive root of unity where α^λ = 1.

Example 1: σg(α), σ²g(α)

For a given cyclotomic integer g(α), σg(α) = g(α^γ) and σ²g(α) = g(α^γγ)

Example 2: Ng(α) in terms of σg(α) [See here for more information on Ng(α) notation]

Using this new notation, since γ is a primitive root mod λ, we see that:

Ng(α) = g(α)*σg(α)*σ²g(α)*...*σ^λ-2g(α)

Example 3: σ^λ-1g(α)

Since σ is defined as α → α^γ, and since the order of a primitive root is λ - 1 (see Theorem 1 here), we see that:

σ^λ-1 = g(α^(λ-1)*(γ)) = g(α¹) = g(α)

Note:

The advantage of this notation is that it avoids many of the subscripts that we have been using previously.

Based on the properties of primitive roots (see here), we can see that σg(α), σ²g(α), ..., σ^λ-1g(α) are all distinct.

Cyclotomic Integers: More x + αiy

In today's blog, I continue going through lemmas from Harold Edward's Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory. I am continuing down the same section that I started in my previous blog. If you would like to start at the beginning of cyclotomic integer properties, start here.

I will use many of these properties later when I go over Kummer's proof of Fermat's Last Theorem for regular primes.

Lemma 1:

if f(α) ≡ g(α) (mod h(α)) → f(k) ≡ g(k) (mod p) with α^λ = 1 and λ is a prime.

then k^λ ≡ 1 (mod p)

Proof:

(1) Let f(α) = α^λ-1 + α^λ-2 + ... + α + 1.

(2) We know that f(α) = 0 [See Lemma 2 here for proof if needed]

(3) So we have f(α) ≡ 0 (mod h(α)) [See here if you need a review of modular arithmetic]

(4) But from assumption f(α) ≡ g(α) (mod h(α) → f(k) ≡ g(k) (mod p) so:

f(k) ≡ 0 (mod p)

(5) So k^λ-1 + k^λ-2 + ... + k + 1 ≡ 0 (mod p)

(6) Now, if we multiply (k-1) to this result:

(k-1)(k^λ-1 + k^λ-2 + ... + k + 1) =

= k^λ + k^λ-1 + ... + k² + k - k^λ-1 - k^λ-2 - ... - k - 1 =

= k^λ - 1

(7) Now combining step #5 and step #6 gives us:

k^λ - 1 ≡ 0 (mod p)

Which is the same as:

k^λ ≡ 1 (mod p)

QED

Lemma 2:

If k is an integer and p is a prime such that k^j ≡ 1 (mod p)

Then, there exists a smallest nonzero integer d such that k^j ≡ 1 (mod p) if and only if d divides j.

Proof:

(1) By the Well-Ordering Principle, we know that there must be a smallest nonzero integer where k^d ≡ 1 (mod p)

(2) Let us assume that k^j ≡ 1 (mod p) but d doesn't divide j.

(3) Using the Division Algorithm, we know that there exists q,r such that:

j = dq + r and 1 ≤ r ≤ d-1

(4) Then k^j ≡ 1 (mod p) implies that:

k^dq+r ≡ 1 (mod p)

and further, we note that:

k^{dq + r} ≡ (k^dq)*(k^r) ≡ [(k^d)^q]*(k^r) ≡ (1)(k^r) ≡ k^r ≡ 1 (mod p)

(5) But this is impossible since d is the lowest nonzero value such that k^d ≡ 1 (mod p) so we reject our assumption in #2.

(6) Now, let's assume that d divides j.

(7) So, then there exists a value q such that:

k^j ≡ k^dq ≡ (k^d)^q ≡ (1)^q ≡ 1 (mod p).

QED

Lemma 3:

If h(α) is a prime cyclotomic integer which divides a binomial x + α^jy (where x,y are relatively prime integers and α^j ≠ 1) and α^λ = 1 where λ is a prime.

Then there exists a prime p such that h(α) divides p and either p = λ or p ≡ 1 (mod λ)

Proof:

(1) Let h(α) be a prime cyclotomic integer which divides a binomial x + α^jy

(2) We know that there exists a prime p such that h(α) divides p. [See Lemma 1 here for proof]

(3) We know from Lemma 1 above that there exists an integer k such that:

k^λ ≡ 1 (mod p)

(4) We also know from Lemma 2 above that there exists a minimum element d such that α^d ≡ 1 (mod p) and since α^λ ≡ 1 (mod p), we know that d = 1 or d = λ

(5) Assume d = 1.

We will show that this implies that λ = p

(6) Then k¹ ≡ 1 (mod p)

(7) From Lemma 1 above, we know that:

k^λ-1 + k^λ-2 + ... + k + 1 ≡ 0 (mod p)

So applying step #6 gives us:

(1)^λ-1 + (1)^λ-2 + ... + (1)¹ + 1 ≡ λ ≡ 0 (mod p)

(8) This implies that p divides λ but since both p, λ are primes, this is only possible if λ = p.

(9) Assume d= λ.

We will show that this implies that p ≡ 1 (mod λ)

(10) Using Fermat's Little Theorem, we know that:

k^p-1 ≡ 1 (mod p)

(11) From Lemma 2 above, this tells us that λ must divide p-1.

(12) So that we conclude that p ≡ 1 (mod λ)

QED

Lemma 4: if n is an odd prime, then

(x - α)(x - α²)*...*(x - α^n-1) = x^n-1 + x^n-2 + ... + 1

Proof:

(1) From a previous result, we have:

xⁿ - 1 = (x - 1)(x - α)(x - α²)*...*(x - α^n-1)

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

(x-1)(x^n-1 + x^n-2 + ... + x + 1) =

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

= xⁿ - 1

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

x^n-1 + x^n-2 + ... + x + 1 = (x - α)(x - α²) * ... * (x - α^n-1)

QED

Lemma 5:

if h(α^j) divides h(αⁱ) where i is not congruent to j modulo λ then

Nh(α) divides N(α^j-i-1)

Proof:

(1) Let k be an integer such that k ≡ α (mod hα) [For proof that this value exists, see Lemma 3 here]

(2) Since conjugates preserve relations (see Lemma 2.5 here), we also know that:

α^j ≡ k (mod h(α^j))

αⁱ ≡ k (mod h(αⁱ))

(3) Since h(α^j) divides h(αⁱ), we also have:

αⁱ ≡ k (mod h(α^j))

(4) Combining #2 and #3 gives us:

α^j ≡ k ≡ αⁱ (mod h(α^j))

(5) So h(α^j) divides α^j - αⁱ

(6) So Nh(α^j) divides N(α^j - αⁱ) [See Lemma 6 here]

(7) But since Nh(α) = Nh(α^j) (see Lemma 3 here), this gives us that:

Nh(α) divides N(α^j - αⁱ)

(8) Now, since α^j - αⁱ = αⁱ(α^j-i - 1), it follows that (see Lemma 6 here):

N(α^j - αⁱ) = N(αⁱ)*N(α^j-i-1)

(9) Finally, N(αⁱ) = 1 since:

(a) N(αⁱ) = N(α) = α*α²*...*α^λ-1 = α^{1+2+3+...+λ-1}

(b) 1 + 2 + 3 + ... + λ - 1 = (1/2)(λ - 1)λ since:

(1 + λ - 1) + (2 + λ - 2) + ... + (λ - 1 + 1) = 2 * (1 + 2 + 3 + ... + λ - 1)

Further, this result is = to (1 + λ -1) = λ so that we have (λ - 1)*(λ) = 2*(1 + 2 + 3 + ... + λ -1)

So dividing both sides by 2 gives us:

(1/2)(λ - 1)*(λ) = 1 + 2 + 3 +... + λ - 1)

(c) Since λ is odd, we know that λ - 1 is even so there exists a rational integer x such that x = (1/2)(λ - 1)

(d) So we get: 1 + 2 + 3 + ... + λ - 1 = x(λ)

(e) So putting it all together:

N(α) = α^{1 + 2 + 3 + ... + λ-1} = α^x(λ)) = (α^λ)^x = 1.

(10) So, Nh(α) divides N(α^j-i - 1)

QED

Lemma 6:

If h(α) is a prime cyclotomic integer which divides a binomial x + α^jy (where x,y are relatively prime integers and α^j ≠ 1),

then Nh(α) is a prime integer which is congruent to 0 or 1 modulo λ

Proof:

(1) Let h(α) be a prime cyclotomic integer that divides x + α^jy.

(2) Then, there exists a prime integer p such that h(α) divides p. [See Lemma 1 here for proof]

(3) Using Lemma 3 above we know that p ≡ 0 (mod λ) or p ≡ 1 (mod λ).

(4) From a previous result (Lemma 3 here), we know that there exists an integer k such that:

k ≡ α (mod h(α))

(5) Assume p = λ (see Lemma 3 above since this is the case where p ≡ 0 (mod λ))

We will now show that N(h(α)) = p.

(6) There exists an integer d such that d is the minimum positive integer where k^d ≡ 1 (mod p) [See Lemma 2 above]

(7) Then d = 1 since (from Lemma 3 above) if d ≠ 1, then d = λ which implies that p ≡ 1 (mod λ) which contradicts assumption in #4 so we conclude d=1.

(8) So k ≡ 1 (mod p) since k^d ≡ 1 (mod p) [See Lemma 3 above for details if needed]

(9) So, combining step #8 with step #4 gives us:

α ≡ 1 (mod p)

and since h(α) divides p, this also gives us:

α ≡ 1 (mod h(α))

(10) This shows that h(α) divides α - 1.

(11) This also shows that Nh(α) divides N(α - 1) [See Lemma 6 here for details if needed]

(12) N(α - 1) = N(1 - α) since:

N(α - 1) = (α - 1)(α² - 1)*...*(α^n-1 - 1)

N(1 - α) = (-1)^n-1[(α - 1)(α² - 1)*..*(α^n-1 - 1)

Since n is odd, n-1 is even and (-1)^n-1 = 1

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

(14) From Lemma 4 above:

(x - α)(x - α²)*...*(x - α^n-1) = x^n-1 + x^n-2 + ... + x + 1

(15) Setting x = 1 gives us:

(1 - α)(1 - α²)*...*(1 - α^n-1) = 1^n-1 + 1^n-2 + ... + 1 + 1 = n

(16) Since in this case n= λ, this proves that N(α - 1) = λ

(17) Now h(α) divides (α - 1) (step #10), this also gives us that Nh(α) divides N(α - 1) which means that Nh(α) divides p.

(18) But p is a prime and since Nh(α) ≠ 1 (since h(α) is a cyclotomic prime, see here for more details), we are left with concluding that Nh(α) = p.

(19) Assume p ≡ 1 (mod λ).

We will now show that under this condition, Nh(α) = p

(20) Since k ≡ α (mod h(α)) [from step #4] we know that:

(a) h(α) divides k - α

(b) And therefore h(α^j) divides k - α^j since conjugates preserve relations (see Lemma 2.5 here)

(21) Now we know that no conjugate of h(αⁱ) divides another conjugate of h(α^j) where i is not congruent to j modulo λ since:

(a) Assume h(α^j) divides h(αⁱ) and i is not congruent to j (mod h(α))

(b) Then Nh(α) divides N(α^j-i-1) [See Lemma 5 above]

(c) But since i is not congruent to j (mod h(α)), N(α^j-i-1) = N(α - 1) [See Lemma 5 above]

(d) So that using the reasoning from #12 thru #16 gives us that:

N(α^j-i-1) = λ

(e) But then h(α) divides λ which is impossible since p does not divide λ from step #19 so we have a contradiction and we reject our assumption in step #21a.

(22) Since h(α) divides p, there exists a cyclotomic integer q(α) such that:

p = h(α)*q(α)

(23) Now we know that each of the other conjugates of h(α) must also divide p since h(α) divides p implies Nh(α) divides N(p) so that h(α)*h(α²)*...h(α^n-1) divides p*p*...*p.

(24) Since h(α²) divides p, it divides h(α)*q(α) but it cannot divide h(α) from step #21 so it divides q(α) [See here for reason that Euclid's Lemma applies to cyclotomic integers]

(25) We can therefore conclude that there exists q₂(α) such that q(α) = h(α²)*q₂(α)

(26) We can repeat step #24 and step #25 for each conjugate of h(α) until we get:

p = h(α)*h(α²)*...h(α^n-1)*q_n-1(α)

So that we get:

p = Nh(α)*q_n-1(α)

(27) But since p is a prime integer and Nh(α) is a rational integer, q_n-1(α) must also be a rational integer.

(28) But p is prime so the only values that can divide it are p and 1. Since we know that Nh(α) ≠ 1, this means that Nh(α) = p and q_n-1(α) = 1.

QED

Corollary 6.1: If Nh(α) = λ, then h(α) and all of its conjugates are unit multiples of α - 1.

Proof:

(1) Let Nh(α) = λ

(2) From Lemma 6 above, we know that there exists an integer k such that k ≡ 1 (mod p)

(3) Since k ≡ α (mod p) (from Lemma 6 above), this also means that α ≡ 1 (mod p).

(4) This shows that p divides α - 1 which means likewise that h(α) divides α - 1 since h(α) divides p.

(5) Then, there exists a value q(α) such that α - 1 = h(α)q(α)

(6) Using Lemma 6 here, we get:

N(α - 1) = Nh(α)*Nq(α)

(7) From step #16 in Lemma 6 above, we know that N(α - 1) = λ

(8) But we also know from Lemma 6 above that Nh(α) = p = λ

(9) So we see that Nq(α) = 1 which means that q(α) is a cyclotomic unit (see here for definition of a cyclotomic unit)

(10) Since all the conjugates of h(α) divide p (see Lemma 6 above for details), we can make the same argument for each of them.

QED

Lemma 7:

If h(α) is any cyclotomic integer whose norm is a prime integer, then h(α) is prime and it divides a binomial x + α^jy (x,y relatively prime, α^j ≠ 1)

Proof:

(1) Let h(α) be a cyclotomic integer whose norm Nh(α) is a prime integer p

(2) Let k be an integer such that k^λ ≡ 1 (mod p) where λ is an odd prime.

(3) Let γ be a primitive root mod p. (See here for details on primitive roots if needed)

(4) Then every nonzero integer mod p can be written in a form of γⁱ where i = 1, 2, ..., p - 1). [See here for definition of primitive root]

(5) We also see that by Fermat's Little Theorem, (γⁱ)^λ ≡ 1 (mod p) if and only if p-1 divides iλ

(6) In Lemma 6 above, this theorem has been proven for the case p = λ so we can assume that p ≡ 1 (mod λ)

(7) Based on this assumption, there exists a value μ such that p - 1 = λμ

(8) From step #5, we know that p-1 divides iλ if and only if μ divides i since:

(a) From step #7, μ = (p-1)/λ

(b) Assume p - 1 divides iλ

(c) p-1 = λ*μ

(d) So λ*μ divides i λ

(e) Since λ cancels out, this means that μ must divide i.

(f) Assume μ divides i

(g) Then λ*μ divides i*λ

(h) Which means that p-1 divides i*λ

(9) Let m = γ^μ

(10) If k = m, m², m³, ..., then:

k^λ ≡ 1 (mod p) [This follows from step #8, step #9]

(11) h(m)*h(m²)*...*h(m^λ-1) ≡ 0 (mod p) since:

(a) Using the Division Algorithm for polynomials (see here), we know that there exists q(x), r(x) such that:

h(x)*h(x²)*...*h(x^λ-1) = q(x)(x^λ-1 + x^λ-2 + ... + 1) + r(x)

where the degree of r(x) is a polynomial of degree less than λ - 1.

(b) Since Nh(α) = p and since α^λ-1 + α^{λ - 2} + ... + 1 = 0 (see Lemma 2 here), we get:

p = q(α)(0) + r(α)

so that r(x) = p

(c) Using step #11b, we get:

h(m)*h(m²)*...*h(m^λ-1) = q(m)(m^λ-1 + m^{λ - 2} + ... + 1) + p.

(d) Now, from an earlier result (see step #6 in Lemma 1 above), we know that:

(m^λ - 1)/(m - 1) = m^λ-1 + m^{λ - 2} + ... + 1

(e) We also know that m is not congruent 1 (mod p) since:

μ divides p-1 from step #7 above

So μ is less than p-1.

The order of γ is p-1 [See here for definition of order of a primitive root]

So therefore m = γ^μ cannot be congruent to 1 (mod p)

(f) Since m^λ ≡ 1 (mod p) from step #10 and since m is not congruent 1 (mod p) , we get:

(m^λ - 1)/(m - 1) ≡ (1 - 1)/(m - 1) ≡ 0 (mod p)

(12) From this, we know that at least one of the values h(m^j) ≡ 0 (mod p)

(13) Using the Division Algorithm for Polynomials with h(x), we have:

h(x) = q(x)(x - m^j) + r where 1 ≤ j ≤ λ -1.

We see that r is an integer.

(13)Using h(m^j) = q(m^j)(m^j - m^j) + r shows that r = h(m^j) and from step #12, we can assume that r ≡ 0 (mod p)

(14) So h(αⁱ) ≡ q(αⁱ)(αⁱ - m^j) (mod p) for i = 1, 2, ..., λ - 1.

(15) Now, from step #14, we get:

(α - m^j)*h(α²)*h(α³)*...*h(α^λ-1) ≡

≡ (α - m^j)q(α²)(α² - m^j)q(α³)*(α³ - m^j)* ... * q(α^λ-1)(α^λ-1 - m^j) ≡

≡ N(α - m^j)q(α²)q(α³)*...*q(α^λ-1) (mod p)

(16) Now, since (x - α)*...*(x - α^λ-1) = (x^λ - 1)/(x - 1) [See here], we know that:

N(α - k) = (k^{λ - 1})/(k - 1) ≡ (1 - 1)/(k - 1) ≡ 0 (mod p)

(17) And since k can equal m^j (see step #10), we have shown that N(α - m^j) ≡ 0 (mod p).

(18) This proves that h(α) divides α - m^j since h(α) divides p.

(19) Now, we need only to prove that h(α) is prime to complete the proof.

(20) Assume that h(α) divides f(α)g(α)

(21) Then f(α)g(α) ≡ 0 (mod h(α))

(22) Since α ≡ k (mod h(α)), we also have:

f(k)g(k) ≡ 0 (mod h(α))

(23) We also have f(k)g(k) ≡ 0 (mod p) since:

(a) Assume f(k)g(k) ≡ w (mod p) and w is not congruent to 0.

(b) Then f(k)g(k) ≡ w (mod h(α)) since h(α) divides p.

(c) And then w ≡ 0 (mod h(α)) which is impossible so we reject our assumption at (#23a)

(24) Now p either divides f(k) or g(k) by Euclid's Lemma.

(25) So finally, h(α) must divide either f(k) or g(k).

(26) And since k ≡ α (mod h(α)), we also have:

h(α) must divide either f(α) or g(α).

QED