## Tuesday, May 09, 2006

### Basic Properties of Cyclotomic Integers

Today's blog continues the discussion of Kummer's proof of Fermat's Last Theorem for regular primes. If you would like to review the historical context for this proof, start here.

Today, I will review the basic properties of cyclotomic integers. Today's content comes directly from Chapter 4 of Harold M. Edwards' Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

1. Notation

For Kummer's notation, he used λ to represent the odd prime number and α to represent the root of unity so that we have:

Definition 1:
αλ = 1

2. Standard Form of Cyclotomic Integers

Lemma 1:
If a0, a1, ... aλ-1 are integers, then all cyclotomic integers for a given value of λ can be represented in the following form:

a0 + a1α + a2α2 + ... + aλ-1αλ-1

Proof:

(1) Let's assume that we have cyclotomic integer = a0 + a1α + a2α2 + ... + aλ-1αλ-1 + aλαλ

(2) By definition 1 above, αλ = 1

(3) So that we have:

(a0 + aλ) + a1α + a2α2 + ... + aλ-1αλ-1

(4) We can do the same thing for any power of αi where i ≥ λ

(5) So we can conclude that all values can be reduced to the form required.

QED

Lemma 2: For any given value of λ, 1 + α+α2 + ... + αλ-1 = 0

Proof:

(1) Since αλ = 1, we have:

1 + α+α2 + ... + αλ-1λ + α+α2 + ... + αλ-1 =

= α(αλ-1 + 1 + α+α2 + ... + αλ-2)

(2) Now, we know that α ≠ 0 since 0λ = 0 which contradicts with definition 1.

(3) We also know that α ≠ 1 since α is a λth root of unity [using Euler's Identity, see here], we know that α = e2iπ/λ

(4) So, therefore, 1 + α+α2 + ... + αλ-1 = 0

QED

Corollary 2.1: for any given integer c, a0 + a1α + a2α2 + ... + aλ-1αλ-1 = (a0 + c) + (a1 + c)α + (a2 + c)α2 + ... + (aλ-1 + c)αλ-1.

Proof:

(1) 1 + α + α2 + ... + αλ-1 = 0 [From Lemma 2 above]

(2) c + cα + cα2 + ... + cαλ-1 = c*0 = 0

(3) So that:

a0 + a1α + a2α2 + ... + aλ-1αλ-1 = a0 + a1α + a2α2 + ... + aλ-1αλ-1 + 0 =

= a0 + a1α + a2α2 + ... + aλ-1αλ-1 + c + cα + cα2 + ... + cαλ-1 =

= (a0 + c) + (a1 + c)α + (a2 + c)α2 + ... + (aλ-1 + c)αλ-1.

QED

3. Conjugates

Since each cyclotomic value can be represented as:

a0 + a1α + a2α2 + ... + aλ-1αλ-1

Kummer used the following shorthand to represent a cyclotomic integer:

f(α), g(α), φ(α), F(α), etc.

One important point that we find is that if f(α) = g(α), then f(α2) = g(α2) and so on up until λ - 1.

Lemma 2.5: Conjugates preserve relations between equations

That is, if f(α) = g(α), then f(αi) = g(αi) where i is a positive number less than λ, αi ≠ 1 and αλ = 1.

Proof:

(1) Let f(α) = a0 + a1α + ... + aλ-1αλ-1

(2) For any value f(αi) we see that:

f(αi) = a0 + a1αi + ... + aλ-1αi*(λ-1)

(3) In step #1, let j be the possible values ranging from 1 to λ -1. Combining this with step #2, we get:

f(αi) = ∑ ajαj*i

(4) To prove this lemma, we need to show each element j*i is congruent to a unique value of i modulo λ

In other words, we are trying to prove that each element of the f(αi) is distinct.

(5) This turns out to be the case from Lemma 1 here.

QED

For this reason, we say that f(α), f(α2), ...., and f(αλ-1) are conjugates of each other.

4. Norm

Definition 2: Norm of a cyclotomic integer f(α)

Nf(α) = f(α)*f(α2)*...*f(αλ-1)

I will now use this definition in the following proofs.

Lemma 3: Nf(α) = Nf(αi) for all values of i between 1 and λ-1.

Proof:

(1) Nf(αi) = f(αi)*f(α2*i)*...*f(αi(λ-1))

(2) Now, each value i, 2*i, 3*i, ... (λ-1)*i maps to a distinct value of 1,2,3,...,(λ-1) modulo λ (see Lemma 1 here)

(3) So in each case, i,2*i, etc. maps to a1*λ+1, a2*λ+2, etc.

(4) So we get Nf(αi) = f(αa0*λ+1)*f(αa1*λ+2)*...*f(αaλ-1*λ+λ-1) where ai is a nonnegative integer.

(5) Since αn*λ=1, we get:

Nf(αi) = f(α)*f(α2)*...*f(αλ-1)

QED

Lemma 4: αj = αλ-j

Proof:

(1) From roots of unity and Euler's Formula, we know that:

α = e(i2π/λ) = cos(2π/λ) + isin(2π/λ)

(2) We also know that the complex conjugate of a + bi is a - bi, so the complex conjugate for α is:

α = cos(2π/λ) - isin(2π/λ)

(3) Likewise, we know that the complex conjugate for αj is:

αj = cos(2jπ/λ) - isin(2jπ/λ)

(4) Using Euler's Formula, we see that:

e-2jπ/λ = cos(-2jπ/λ) + isin(-2jπ/λ)

(5) Since cos(-x) = cos(x) and sin(-x) = -sin(x) [see here], we can use (#4) to get:

e-2jπ/λ = cos(2jπ/λ) - isin(2jπ/λ)

which is from #3, the complex conjugate for αj

(6) Now, e-2jπ/λ = (e2π/λ)-j =

= α-j = α-jλ =

= αλ - j

QED

Corollary 4.1: f(αj) = f(αλ-j)

Proof:

(1) From Lemma 1, we have:

f(α) = a0 + a1α + a2α2 + ... + aλ-1αλ-1

(2) From this,

f(αj) = a0 + a1αj + a2α2*j + ... aλ-1αj*(λ-1)

(3) Now, from Lemma 4, we know that:

f(αj) = a0 + a1αλ-j + a2αλ - 2*j + ... aλ-1αλ - j*(λ-1)

(4) And, we know that:

f(αλ-j) = a0 + a1αλ-j + a2α(λ-j)*2 + ... aλ-1α(λ - j)*(λ - 1)

(5) Now,

n*λ - j*n ≡ λ - j*n (mod λ) [See here if you need a review of modular arithmetic]

(6) So that we see that step #3 and step #4 are equal so that:

f(αj) = f(αλ-j)

QED

Corollary 4.2: f(αj)*f(αλ-j) is a nonnegative real number

Proof:

(1) f(αj) * f(αλ-j) = f(αj)* f(αj) [From Corollary 4.1 above]

(2) So that:
f(αj) * f(αλ-j) = (a0 + a1αj + ... + aλ-1αj*(λ-1))(a0 + a1αj + ... + aλ-1αj*(λ-1)) =

= (a0)2 + (a1)2j*αj) + ... + (aλ-1)2j*(λ-1)*αj*(λ-1))

(3) Since each α*α is a nonnegative number, the conclusion follows.

QED

Lemma 5: For any cyclotomic integer f(α), its norm is a nonnegative rational integer.

Proof:

(1) Using Lemma 1 above, we know that:

Nf(α) = a0 + a1α + a2α2 + ... + aλ-1αλ-1

(2) By Lemma 3 above, we can substitute any conjugate αj and get the same norm so that:

Nf(αj) = Nf(α)

(3) But by changing to a conjugate, we keep the same coefficients but get the following:

Nf(αj) = a0 + a1αj + a2αj*2 + ... + aλ-1α(λ-1)*j

(4) Combining the two equations gets us:

a0 + a1αj + a2αj*2 + ... + aλ-1α(λ-1)*j = a0 + a1α + a2α2 + ... + aλ-1αλ-1

(5) Subtracting one from the other gives us:

a0 - a0 + (a1 - ajj + ... = 0

(6) Since we know that each of these j,2*j,...,(λ-1)*j matches up with a value 1,2,...,λ-1, we know that:

a1 = aj

(7) Further, since j can be any value from 2 thru λ-1, we can conclude the following:

a1 = a2 = a3 = ... = aλ-1

(8) So that:
Nf(α) = a0 + a1(α + α2 + ... + αλ-1)

(9) From Lemma 2, we know that:

1 + α+α2 + ... + αλ-1 = 0

so that:

α+α2 + ... + αλ-1 = -1

(10) So, we apply (#9) to (#8) to give us:

Nf(α) = a0 - a1

(11) We know that it is nonnegative since:

Nf(α) = [f(α1)*f(αλ-1)]*[f(α2)*f(αλ-2)]*...

(12) From Corollary 4.2 above, we know that multiplication of (λ-1)/2 pairs of nonnegative values will result in a nonnegative value.

QED

Lemma 6: f(α)g(α) = h(α) → Nf(α)*Ng(α) = Nh(α)

Proof:

(1) Let f(α)g(α) = h(α)

(2) By Definition 2 above:

Nf(α) = f(α)*f(α2)*...*f(αλ-1)

Ng(α) = g(α)*g(α2)*...*g(αλ-1)

Nh(α) = h(α)*h(α2)*...*h(αλ-1)

(3) Using step #1 gives us:

Nh(α) = f(α)*g(α)*f(α2)*g(α2)*...*f(αλ-1)*g(αλ-1) =

= f(α)*f(α2)*...*f(αλ-1) *g(α)*g(α2)*...*g(αλ-1) =

= Nf(α)*Ng(α)

QED