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. NotationFor 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:α^{λ} = 12. Standard Form of Cyclotomic IntegersLemma 1:If a_{0}, a_{1}, ... a_{λ-1} are integers, then all cyclotomic integers for a given value of λ can be represented in the following form:a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1}Proof:

(1) Let's assume that we have cyclotomic integer

= a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1} + a_{λ}α^{λ}(2) By definition 1 above,

α^{λ} = 1(3) So that we have:

(a_{0} + a_{λ}) + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1}(4) We can do the same thing for any power of

α^{i} 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+α^{2} + ... + α^{λ-1} = 0Proof:

(1) Since

α^{λ} = 1, we have:

1 + α+α^{2} + ... + α^{λ-1} =α^{λ} + α+α^{2} + ... + α^{λ-1} == α(α^{λ-1 } + 1 + α+α^{2} + ... + α^{λ-2})(2) Now, we know that

α ≠ 0 since

0^{λ} = 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 = e^{2iπ/λ}(4) So, therefore,

1 + α+α^{2} + ... + α^{λ-1 }= 0QED

Corollary 2.1: for any given integer c, a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1 }= (a_{0} + c) + (a_{1} + c)α + (a_{2} + c)α^{2} + ... + (a_{λ-1} + c)α^{λ-1}.Proof:

(1)

1 + α + α^{2} + ... + α^{λ-1 }= 0 [From Lemma 2 above]

(2)

c + cα + cα^{2} + ... + cα^{λ-1 }= c*0 = 0(3) So that:

a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1 }= a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1} + 0 == a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1} + c + cα + cα^{2} + ... + cα^{λ-1 }== (a_{0} + c) + (a_{1} + c)α + (a_{2} + c)α^{2} + ... + (a_{λ-1} + c)α^{λ-1}.QED

3. ConjugatesSince each cyclotomic value can be represented as:

a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-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(α^{2}) = g(α^{2}) and so on up until

λ - 1.

Lemma 2.5: Conjugates preserve relations between equationsThat is, if

f(α) = g(α), then

f(α^{i}) = g(α^{i}) where

i is a positive number less than

λ, α^{i} ≠ 1 and

α^{λ} = 1.Proof:

(1) Let

f(α) = a_{0} + a_{1}α + ... + a_{λ-1}α^{λ-1}(2) For any value

f(α^{i}) we see that:

f(α^{i}) = a_{0} + a_{1}α^{i} + ... + a_{λ-1}α^{i*(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(α^{i}) = ∑ a_{j}α^{j*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(α^{i}) is distinct.

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

here.

QED

For this reason, we say that

f(α), f(α^{2}), ...., and f(α^{λ-1}) are conjugates of each other.

4. NormDefinition 2: Norm of a cyclotomic integer f(α)Nf(α) = f(α)*f(α^{2})*...*f(α^{λ-1})I will now use this definition in the following proofs.

Lemma 3: Nf(α) = Nf(α^{i}) for all values of i between 1 and λ-1.Proof:

(1)

Nf(α^{i}) = f(α^{i})*f(α^{2*i})*...*f(α^{i(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

a_{1}*λ+1, a_{2}*λ+2, etc.

(4) So we get

Nf(α^{i}) = f(α^{a0*λ+1})*f(α^{a1*λ+2})*...*f(α^{aλ-1*λ+λ-1}) where

a_{i} is a nonnegative integer.

(5) Since

α^{n*λ}=1, we get:

Nf(α^{i}) = f(α)*f(α^{2})*...*f(α^{λ-1})QED

Lemma 4: α^{j} = α^{λ-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

α^{j} is:

α^{j} = 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

α^{j}(6) Now,

e^{-2jπ/λ} = (e^{2π/λ})^{-j} == α^{-j} = α^{-j}*α^{λ} == α^{λ - j}QED

Corollary 4.1: f(α^{j}) = f(α^{λ-j})Proof:

(1) From Lemma 1, we have:

f(α) = a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1}(2) From this,

f(α^{j}) = a_{0} + a_{1}α^{j} + a_{2}α^{2*j} + ... a_{λ-1}α^{j*(λ-1)}(3) Now, from Lemma 4, we know that:

f(α^{j}) = a_{0} + a_{1}α^{λ-j} + a_{2}α^{λ - 2*j} + ... a_{λ-1}α^{λ - j*(λ-1)}(4) And, we know that:

f(α^{λ-j}) = a_{0} + a_{1}α^{λ-j} + a_{2}α^{(λ-j)*2} + ... a_{λ-1}α^{(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(α^{j}) = f(α^{λ-j})QED

Corollary 4.2: f(α^{j})*f(α^{λ-j}) is a nonnegative real numberProof:

(1)

f(α^{j}) * f(α^{λ-j}) = f(α^{j})* f(α^{j}) [From Corollary 4.1 above]

(2) So that:

f(α^{j}) * f(α^{λ-j}) = (a_{0} + a_{1}α^{j} + ... + a_{λ-1}α^{j*(λ-1)})(a_{0} + a_{1}α^{j} + ... + a_{λ-1}α^{j*(λ-1)}) == (a_{0})^{2} + (a_{1})^{2}(α^{j}*α^{j}) + ... + (a_{λ-1})^{2}*α^{j*(λ-1)}*α^{j*(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) = a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}α^{λ-1}(2) By Lemma 3 above, we can substitute any conjugate

α^{j} and get the same norm so that:

Nf(α^{j}) = Nf(α)(3) But by changing to a conjugate, we keep the same coefficients but get the following:

Nf(α^{j}) = a_{0} + a_{1}α^{j} + a_{2}α^{j*2} + ... + a_{λ-1}α^{(λ-1)*j}(4) Combining the two equations gets us:

a_{0} + a_{1}α^{j} + a_{2}α^{j*2} + ... + a_{λ-1}α^{(λ-1)*j }= a_{0} + a_{1}α + a_{2}α^{2} + ... + a_{λ-1}αλ-1(5) Subtracting one from the other gives us:

a_{0} - a_{0} + (a_{1} - a_{j})α^{j} + ... = 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:

a_{1} = a_{j}(7) Further, since

j can be any value from

2 thru

λ-1, we can conclude the following:

a_{1} = a_{2} = a_{3} = ... = a_{λ-1}(8) So that:

Nf(α) = a_{0} + a_{1}(α + α^{2} + ... + α^{λ-1})(9) From Lemma 2, we know that:

1 + α+α^{2} + ... + α^{λ-1} = 0so that:

α+α^{2} + ... + α^{λ-1 }= -1(10) So, we apply (#9) to (#8) to give us:

Nf(α) = a_{0} - a_{1}(11) We know that it is nonnegative since:

Nf(α) = [f(α^{1})*f(α^{λ-1)}]*[f(α

^{2})*f(α^{λ-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(α^{2})*...*f(α^{λ-1})Ng(α) = g(α)*g(α^{2})*...*g(α^{λ-1})Nh(α) = h(α)*h(α^{2})*...*h(α^{λ-1})(3) Using step #1 gives us:

Nh(α) = f(α)*g(α)*f(α^{2})*g(α^{2})*...*f(α^{λ-1})*g(α^{λ-1}) == f(α)*f(α^{2})*...*f(α^{λ-1}) *g(α)*g(α^{2})*...*g(α^{λ-1}) == Nf(α)*Ng(α)QED