News: Grigory Perelman wins Fields Medal

Grigory Perelman, a Russian mathematician who may have solved Poincare Conjecture, has won the Fields Medal for his work on the Ricci Flow. The Fields Medal is one of the most prestigious prizes in mathematics and is awarded every four years by the International Congress of Mathematicians. This year, the awards were presented by King Juan Carlos of Spain. Interestingly, Perelman has made news by not showing up to the ceremony to receive the Fields Medal. The New Yorker has recently done a story on him.
Perelman has also gained international notoriety for posting a generalized solution to the Poincare Conjecture on a web site in November, 2002. Many mathematicians believe that he may have successfully solved it. It is a statement to Perelman's reputation that mathematicians have been willing to come to his web site as opposed to the standard process where Perelman would submit his papers to a peer-reviewed math journal.

The Poincare Conjecture is a theorem about the nature of multidimensional space. It is the most famous open problem in topology. The Poincare Conject concerns the problem of transforming a torus into sphere.

If Perelman has indeed proven the Poincare Conjecture, then he will be eligible for the $1 million from the Clay Mathematics Institute. The Institute has identified seven open math problems which it has called the Millenium Problems and offers a $1 million prize for the solution of any of them. Perelman has said that if he wins the $1 million dollars, he plans to talk to the Institute.

Kummer's Proof for Regular Primes: αr

In today's blog, I continue the proof of Fermat's Last Theorem for regular primes. In today's blog, I go over a lemma which is used in the proof for Case I. For context and definitions, please start here at the beginning of this proof.

The details of today's content is taken from Harold M. Edwards Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Lemma 1: a unit is only divisible by a unit

Proof:

(1) Assume that a unit g(α) is divisible by a nonunit h(α).

(2) Ng(α) = 1

(3) Nh(α) ≠ 1 (see Definition 1, here for definition of unit) so Nh(α) ≥ 2.

(4) But then Nh(α) ≥ 2 must divide 1 (see Lemma 6, here) which is impossible.

(5) So we reject our assumption.

QED

Lemma 2: α - α^-1 is divisible by α - 1.

Proof:

(1) α - α^-1 = α - α^λ-1

(2) (α - α^λ-1) = (α - 1)(α^λ-2 - α^λ-3 + ... ± α)

QED

Lemma 3: αⁱ - α^-i is divisible by α - α^-1

Proof:

(αⁱ - α^-i) = (α - α^-1)(α^i-1 - α^i-3 + ... ± α^1-i)

QED

Examples:

(α² - α^-2) = (α - α^-1)(α + α^-1)

(α³ - α^-3) = (α - α^-1)(α² - 1 + α^-2)

(α⁴ - α^-4) = (α - α^-1)(α³ - α + α^-1 + α^-3)

Lemma 4: if e is a unit then e/e = α^r for some r.

Proof:

(1) Let E(α) = e/e.

(2) E(α) = a₀ + a₁α + ... + a_λ-1α^λ-1 (See Lemma 1, here for details)

(3) Then, E(X) = a₀ + a₁X + ... + a_λ-1X^λ-1

(4) Using the Division Algorithm for Polynomials to divide E(X^λ-1)E(X) by (X^λ-1 - 1), we know that there exists Q(X),R(X) such that:

E(X^λ-1)E(X) = Q(X)(X^λ-1 - 1) + R(X)

where R(X) is a polynomial of degree less than λ.

(5) So, we know that there exists A_i such that:

R(X) = A₀ + A₁X + ... + A_λ-1X^λ-1

(6) If we set X=1, then the equation in step #4 gives us:

E(1^λ-1)E(1) = Q(1)*(1^λ-1-1) + R(1) =

= E(1)*E(1) = (a₀ + a₁ + ... + a_λ-1)² =

= Q(1)*0 + R(1) = A₀ + A₁ + ... + A_λ-1.

(7) If we set X=α, then the equation in step #4 gives us:

E(α^-1)*E(α) = [e(α^-1)/e(α^-1)][e(α)/e(α)] = [e/e][e/e] = 1 =

= Q(α)*(α^λ-1-1) + R(α) = Q(α)*(1-1) + R(α) = R(α) = A₀ + A₁α + ... + A_λ-1α^λ-1

(8) This gives us that:

(A₀ - 1) + A₁α + ... + A_λ-1α^λ-1 = 0.

(9) Since α^λ-1 + α^λ-2 + ... + α + 1 = 0 [See Lemma 2, here], we can conclude that:

A₀ - 1 = A₁ = A₂ = ... = A_λ-1.

(10) Let k = A₀-1 = A₁ = ... = A_λ-1

(11) Using the equation from step #6, we have:

(a₀ + a₁ + ... + a_λ-1)² = A₀ + A₁ + ... + A_λ-1 = λ*k + 1.

(12) If we subtract k from each A_i we get:

A₀ + A₁ + ... + A_λ-1 = 1

Under this situation, we get:

(a₀ + a₁ + ... + a_λ-1)² = 1

which means that

a₀ + a₁ + ... + a_λ-1= ± 1

Further we have the following:

k=0
A₀=1
A₁ = A₂ = ... = A_λ-1 = 0.

(13) We can now show that A₀ = a₀² + a₁² + ... + a_λ-1² since:

(a) Each term of E(X^λ-1)E(X) is of the form (a_iX^λi-i)(a_jX^j) = a_ia_jX^(λi-i+j)

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

λi - i + j = qλ + r where r is less than λ.

(c) Since λ divides λ*i, we know that r ≡ j - i (mod λ).

(d) We now have:

a_ia_jX^(λi-i+j) = a_ia_jX^qλ+r =

= a_ia_jX^r*(X^qλ) =

= a_ia_jX^r*(X^qλ - 1) + a_ia_jX^r

(e) We can now define a function Q_i,j(X) = a_ia_jX^r*(X^qλ - 1)/(X^λ-1)

(f) We can see that using Q_i,j(X), each term has the following value:

a_ia_jX^(λi-i+j) = a_ia_jX^r*(X^qλ - 1) + a_ia_jX^r = Q_i,j(X)(X^λ-1) + a_ia_jX^r

(g) if q is greater than 1, we have:

Q_i,j(X) = a_ia_jX^r*(X^qλ - 1)/(X^λ-1) = a_ia_jX^r*(X^qλ-λ + ... + X^λ + 1)

(h) if q = 1, then Q_i,j(X) = a_ia_jX^r.

(i) if q = 0, then Q_i,j(X) = 0.

(j) Since E(X^λ-1)E(X) = Q(X)(X^λ-1 - 1) + R(X) is the sum of all of these terms we can conclude that:

R(X) = ∑ (r=0,λ-1) [∑ (j-i ≡ r) a_ia_j]X^r.

(k) For A₀ where r=0, we have the following:

A₀ = (∑(j-i ≡ 0) a_ia_j)X⁰ = a₀*a₀ + a₁*a₁ + ... + a_λ-1*a_λ-1 = a₀² + a₁² + ... + a_λ-1².

(14) So A₀ = 1 implies that one a_i = ±1 and the rest = 0.

(15) Since E(α) = a₀ + a₁α + ... + a_λ-1α^λ-1, step #14 implies that:

E(α) = 0 + 0 + (±1)α^r + 0 + 0 + 0 +...

So that:

E(α) = ± α^r for some r.

(16) Finally, we can show that E(α) = α^r

(a) Assume that E(α) = -α^r

(b) Since λ is odd, either r is even or r+λ is even (since odd + odd = even) so there exists an s such that E(α) = -α^2s

NOTE: Since α = 1, -α^r = -α^r*1 = -*α^r*α^λ = -α^λ+r

(c) Since E(α) = e/e, we have:

e/e = -α^2s

NOTE: -α^2s is a unit since e is a unit and e is a unit. [See Lemma 5, here]

This implies that:

eα^-s = -eα^s.

NOTE: By Lemma 1 above, α^-s and α^s are units since they divide -α^2s which is a unit.

Further, eα^-s, -eα^s are units because the product of units is a unit (see Lemma 3, here)

(d) Since eα^-s is a cyclotomic integer, we have:

eα^-s = b₀ + b₁α + ... + b_λ-1 [See Lemma 1, here for details]

(e) Let f(α) = b₀ + b₁α + ... + b_λ-1α^λ-1

(f) Then we can see that f(α) = eα^-s and f(α^-1) = e^s.

where f(α^-1) = b₀ + b₁α^-1 + ... + b_λ-1α

(g) So that, f(α) = -f(α^-1) [from step #16c]

(h) 2*f(α) = f(α) + -f(α^-1) = (b₀ - b₀) + b₁(α - α^-1) + b₂(α² - α^-2) + ... + b_λ-1(α^-1 + α)

(i) Now it is clear that (α - α^-1) divides 2*f(α) [See Lemma 3 above]

(j) We know further (α - 1) divides 2*f(α) since α - 1 divides (α - α^-1) [See Lemma 2 above]

(k) But (α - 1) does not divide 2 since Norm(α-1) = λ and λ doesn't divide Norm(2) = 2^λ-1 (see Lemma 6, here for details)

(l) But if (α - 1) divides f(α) then we have a contradiction by Lemma 1 above since a unit is only divisible by a unit and Norm(α - 1) = λ is not a unit (see Definition 1, here).

(m) Therefore, we reject our assumption at step #16a.

QED

Kummer's Proof for Regular Primes: Case I

In today's blog, I continue the proof of Fermat's Last Theorem for regular primes. Case I makes the assumption that x,y,z,λ are relatively prime and that each of the factors are relatively prime. For context and definitions, please start here at the beginning of this proof.

The details of today's content is taken from Harold M. Edwards Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Lemma 1: If n is odd, then xⁿ + yⁿ = zⁿ implies that (x)ⁿ + (y)ⁿ + (-z)ⁿ = 0.

Proof:

This works because n is odd. n is odd implies that (-z)ⁿ = -(z)ⁿ

QED

Lemma 2: Every cyclotomic integer is congruent mod (α - 1) to a rational integer

Proof:

(1) Let g(α) be any cyclotomic integer.

(2) g(α) = a₀ + a₁α + ... + a_λ-1α^{λ - 1} [See Lemma 1, here]

(3) α ≡ 1 (mod α - 1)

(4) g(α) ≡ g(1) ≡ a₀ + a₁ + ... + a_λ-1 = rational integer

QED

Lemma 3: For any given cyclotomic integer g(α) and any given integer k greater than 0, there exists integers a₀, a₁, ..., a_k-1 such that:

g(α) ≡ a₀ + a₁(α - 1) + ... + a_k-1(α - 1)^k-1 (mod (α - 1)^k)

Proof:

(1) Let a₀ ≡ g(α) (mod α - 1) [See Lemma 2 above]

(2) Let h(α) = [g(α) - a₀]/(α - 1)

(3) Let a₁ ≡ h(α) (mod α -1 ) [See Lemma 2 above]

(4) We can see that g(α) ≡ a₀ + a₁*(α - 1) mod (α - 1)²

(5) Using step #4, let a₂ ≡ [g(α) - a₀ - a₁(α - 1)]/(α - 1)² (mod α -1)

(6) Then g(α) ≡ a₀ + a₁(α - 1) + a₂(α - 1)² (mod (α - 1)³).

(7) We can continue this process up to any integer k so we are done.

QED

Lemma 4: Coefficients of corresponding powers of (α - 1) must be congruent mod λ provided all powers are less than the (λ - 1)st.

if:

a₀ + a₁(α - 1) + a₂(α-1)² + ... + a_λ-2(α-1)^λ-2 ≡ b₀ + b₁(α - 1) + b₂(α-1)² + ... + b_λ-2(α-1)^λ-2 (mod λ)

then:

a₀ ≡ b₀ (mod λ)
a₁ ≡ b₁ (mod λ)
...
a_λ-2 ≡ b_λ-2 (mod λ)

Proof:

(1) If a₀ + a₁(α - 1) + ... + a_λ-2(α - 1)^λ-2 ≡ b₀ + b₁(α -1) + ... + b_λ-2(α-1)^λ-2 (mod (α - 1)^λ-1), then:

c₀ + c₁(α-1) + ... ≡ 0 (mod (α - 1)^λ-1) where c_i = a_i - b_i

(2) Then c₀ ≡ 0 (mod α - 1) which also gives us that c₀ ≡ 0 (mod λ)  [See Theorem here for explanation of λ and see Lemma 2, here for properties of cyclotomic integers, and Corollary 3.2 here]

(3) Then c₁(α-1) ≡ 0 (mod (α-1)²) so that (α-1) divides c_1.   Also, from step(1) and Corollary 3.2, c₁ ≡ 0 (mod λ).

(4) We can use the same logic to show that all c_i ≡ 0 (mod λ)

(5) So, a_i - b_i = 0, this implies that a_i = b_i

QED

Theorem: x,y,z,λ are coprime and all factors (x + α^jy) are relatively prime, then x^λ + y^λ = z^λ has no integer solutions except for where xyz=0.

Proof:

(1) z^λ = x^λ + y^λ = (x + y)(x + αy)(x + α²y)*...*(x + α^λ-1y). [See Lemma 1 here for details on this refactoring]

(2) From the given, we know that all the factors (x + α^jy) are relatively prime, so we can conclude that the divisor of each (x + α^jy) is λth power. [See Lemma 1, here]

(3) Further, we can conclude that each factor (x + α^jy) is a unit*λth power. [This follows from Corollary, here]

(4) So, there exists a unit e and a cyclotomic integer t such that:

x + αy = et^λ

(5) If permute α to α^-1 on each side, we get:

x + α^-1y = e*t^λ where e is the inverse of e and t is the inverse of t.

(6) Now, there exists an r such that e/e = α^r. [See Lemma 4, here]

(7) t^λ ≡ C ≡ C ≡ t^λ mod λ because mod λ of every λth power is a rational integer (see Theorem 6, here) and rational integers are invariant under α → α^-1.

(8) Since e = e*α^-r, we have:

x + α^-1y = α^-ret^λ

(9) From t^λ ≡ t^λ (mod λ) [See step #7], we have:

α^-ret^λ ≡ α^-ret^λ (mod λ)

(10) Applying x + αy = et^λ [see step #4] give us:

α^-ret^λ ≡ α^-r(x + αy) (mod λ)

(11) This then gives us:

α^r(x + α^-1y) ≡ α^r(α^-ret^λ) ≡ et^λ (mod λ) [from step #8]

Further, from t^λ ≡ t^λ (mod λ) [See step #7], we have:

et^λ ≡ et^λ (mod λ)

Combining this result with x + αy = et^λ in step #4 gives us:

α^r(x + α^-1y) ≡ x + αy (mod λ)

(12) We know that r not ≡ 0 (mod λ) since:

(a) Assume r ≡ 0 (mod λ)

(b) Then from step #11, we have:

(x + α^-1y) ≡ x + αy (mod λ)

(c) Subtracting (x + α^-1y) to both sides gives us:

0 ≡ (x + αy) - (x + α^-1y) ≡ αy - α^-1y (mod λ)

(d) Multiplying both sides by (α) gives us:

0 ≡ α²y - y ≡ (α² - 1)y (mod λ)

(e) Since λ = (α-1)^λ-1*unit (see Corollary 3.2, here), we have:

0 ≡ (α² - 1)y (mod (α - 1)^{λ - 1})

(f) Since (α-1)^{λ - 1} divides (α² - 1)y, it follows that (α - 1) must divide y which contradicts our assumption that y and λ are coprime.

If (α - 1) divides y, then λ must divide y since α - 1 divides λ and y is a rational integer.

(g) So we reject our assumption at #12a.

(13) From step #12, we can assume that r is greater than 0 and less than λ.

NOTE: It is less than λ because it is a power of α and α^λ = 1 = α⁰.

(14) Since r is greater than 0, we can rearrange step #11 to give us:

α^r-1(αx + y) ≡ x + αy (mod λ)

(15) Since α = [1 + (α - 1)], we have:

α^r-1(αx + y) ≡ [1 + (α - 1)]^r-1[x(1 + [α - 1]) + y] ≡ [1 + (α - 1)]^r-1[x + y + x(α - 1) ] (mod λ)

x + αy ≡ x + (1 + [α - 1])y ≡ x + y + y(α - 1) (mod λ)

(16) Putting the result of step #15 together with the fact that λ = (α - 1)^{λ - 1}*unit (see Corollary 3.2, here) gives us:

[1 + (α - 1)]^r-1[x + y + x(α - 1) ] ≡ x + y + y(α - 1) (mod (α - 1)^{λ - 1})

(17) If we carry out this equation using the Binomial Theorem (see here), we get:

[1 + (α - 1)]^r-1 = 1 + (r-1)(α - 1) + [(r-1)(r-2)/(2!)](α - 1)² + ... + (α - 1)^r-1

Multiplying this with [x + y + x(α - 1)] gives us:

[x + y + x(α-1)] + (r-1)[x(α - 1) + y(α - 1) + x(α-1)²] + ... + [x(α-1)^r-1 + y(α-1)^r-1 + x(α-1)^r]

(18) Coefficients of corresponding powers of (α - 1) must be congruent mod λ provided all powers are less than the (λ - 1)st. [See Lemma 4 above]

(19) The congruence in step #16 is impossible when 2 ≤ r ≤ λ - 2 because then the highest order term on the left is x(α - 1)^r and step #18 gives us 0 ≡ x (mod λ) which contradicts x,λ being relatively prime.

(20) r = λ - 1 is also impossible since:

The last two terms are:

(λ-2)[x(α-1)^λ-3 + y(α-1)^λ-3 + x(α-1)^λ-2] + [x(α-1)^λ-2 + y(α-1)^λ-2 + x(α-1)^λ-1]

We only need to consider (α-1)^λ-2 which gives us:

(λ-2+1)x(α-1)^λ-2

Using step #18, we have:

(λ-1)x(α-1)^λ-2 ≡ 0 (mod λ)

This implies that:

x ≡ 0 (mod λ)

Which is impossible.

(21) This leaves r = 1 which gives us:

[1 + (α-1)]⁰[x + y + x(α -1)] ≡ x + y + y(α - 1) (mod λ)

Since [1 + (α-1)]⁰ = 1, we have:

[x + y + x(α-1)] ≡ x + y + y(α-1) (mod λ)

Subtracting x+y from both sides gives us:

x(α-1) ≡ y(α-1) (mod λ)

Appying step #18 gives us:

x ≡ y (mod λ)

(22) We can put x^λ + y^λ = z^λ into a symmetric form (see Lemma 1 above) where:

x^λ + y^λ + (z')^λ = 0.

(23) By the fact that x,y,z,λ are relatively prime (from the given above), we know that:

x not ≡ 0 (mod λ)
y not ≡ 0 (mod λ)
z' not ≡ 0 (mod λ)

(24) We further know that x ≡ y (mod λ) (from step #21) but because the equation is symmetric we can further conclude that x ≡ z' (mod λ) and y ≡ z' (mod λ)

(25) Using Fermat's Little Theorem (see here), we can conclude that:

x^λ ≡ x (mod λ)
y^λ ≡ y (mod λ)
z' ^λ ≡ z' (mod λ)

since:

(a) Fermat's Little Theorem gives us:

gcd(a,p)=1 → a^p-1 ≡ 1 (mod p)

(b) So we have:

x^λ-1 ≡ 1 (mod λ)

(c) Multiplying x to both sides gives us:

x^λ ≡ x (mod λ)

(d) We can make the same argument for y and z.

(26) Putting this all together gives us:

0 = x^λ + y^λ + z'^λ ≡ x + y + z' ≡ 3x (mod λ)

(27) Since x not ≡ 0 (mod λ), this implies that λ = 3. For all other values we have a contradiction and case I is proved.

(28) Using Sophie's Theorem (see Theorem, here), we know that if 2λ+1 is a prime, then λ divides xyz.

(29) But 2*3+1 = 7 which is a prime so this means that λ must divide xyz.

(30) But it does not since x,y,z,λ are relatively prime.

(31) Therefore we have a contradiction and Case I is proved.

QED

Ideal Numbers: Distinct Congruence Classes Modulo an Ideal Number

Today's blog continues the proof for the existence of a class number for any set of cyclotomic integers. There is one last proof we need to complete this proof. We need to show that given a large enough set of cyclotomic integers with distinct congruence classes modulo an ideal number, that we can be certain that at least one is divisible by the ideal number. In other words, we can be sure that there exists within that the set of cyclotomic integers one that is congruent to 0 modulo the ideal number.

The answer to this question turns out to be the norm of the ideal number. For example, if we want to make sure that for a set of cyclotomic integers, a given ideal number A divides at least one, we can be sure of this if we have a set of N(A) distinct congruence classes.

The result presented in today's blog is taken from Harold M. Edwards' Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Lemma 1: Norm of a Prime Divisor is the number of incongruent cyclotomic integers

If A is a prime divisor, the Norm N(A) can be regarded as a positive integer that is equal to the number of incongruent cyclotomic integers mod A.

Proof:

(1) If A is a prime divisor, then it is either the exceptional prime divisor (α - 1) or it is one of e prime divisors that divide p.

(2) Assume A = α - 1

(3) Then ever cyclotomic integer is congruent to one and only one of the λ integers 0, 1, 2, ..., λ-1.

(4) And N(α-1) = λ (see Lemma 6, here for details)

(5) Assume A ≠ α - 1.

(6) Then its norm is p^f where p is the prime integer that it divides and f is the exponent of p mod λ.

The norm of a divisor is defined to be A*σA*σ²A*...*σ^λ-2A = (A*σA*σ²A*..*σ^e-1A)(σ^eA*...*σ^2e-1A)....(σ^λ-e-1A*..*σ^λ-2A)

Now, each of these set of e prime divisors is the same as p. Likewise, ef = λ - 1.

This gives us:

(p)*...*(p) = p^f.

(7) Using a previous result (see Theorem 3, here), we know that there are p^f incongruent elements mod A.

QED

Lemma 2: Norm of a Power of a Prime Divisor is the number of incongruent cyclotomic integers

If A is a power of a prime divisor, the norm N(A) can be regarded as a positive integer that is equal to the number of incongruent cyclotomic integers mod A.

Proof:

(1) Let A = Pⁿ where P is a prime divisor.

(2) Let ψ(α) be a cyclotomic integer which is divisible by P with multiplicity exactly 1 but not divisible at all by any of the conjugates of P. [See Proposition 1, here for details on this construction if needed]

(3) Each cyclotomic integer is congruent to one of the form a₀ + a₁ψ(α) + a₂ψ(α)² + ... + a_n-1ψ(α)^n-1 mod Pⁿ and two cyclotomic integers of this form are congruent if and only if the coefficients a₀, a₁, ..., a_n-1 are the same mod P since:

(a) When n=1, this is obvious since we are only dealing with a₀.

(b) Assume that it is proven for n-1.

(c) Then, for a given cyclotomic integer x(α), there are cyclotomic integers a₀, a₁, ..., a_n-2 for which x(α) ≡ a₀ + a₁ψ(α) + ... + a_n-2ψ(α)^n-2 (mod P^n-1) and the coefficients a₀, a₁, ..., a_n-2 are uniquely determined mod P. [By our assumption in step (b)]

(d) Let y(α) = x(α) - a₀ - a₁ψ(α) - ... - a_n-2ψ(α)^n-2.

(e) Then y(α) ≡ 0 mod P^n-1 [By combining step (c) and step (d)]

(f) Now, we need to prove that y(α) ≡ aψ(α)^n-1 mod Pⁿ for some a and that a is uniquely determined mod P.

(g) Let γ(α) denote the the product of e-1 distinct conjugates of ψ(α) so that γ(α)ψ(α) = pk where k is an integer relatively prime to p.

We know that p divides γ(α)*ψ(α) exactly once so k is what is left over. Since p is a prime, we know that gcd(p,k)=1.

(h) So if we multiply γ(α)^n-1 to both sides of (f), we get:

y(α)γ(α)^n-1 ≡ aγ(α)^n-1ψ(α)^n-1 ≡ ap^n-1k^n-1 (mod Pⁿ)

(i) Since gcd(p,k)=1, there exists an integer m such that mk ≡ 1 (mod p).

NOTE: This comes straight from Bezout's Identity which says there exist m,n such that mk + np = 1. In other words (-n)p = mk - 1.

(j) Multiplying m^n-1 to both sides of (h) gives us:

y(α)γ(α)^n-1m^n-1 ≡ ap^n-1k^n-1m^n-1 ≡ ap^n-1 (mod Pⁿ)

(k) Since y(α) ≡ 0 (mod P^n-1) by assumption, then y(α)γ(α)m^n-1 is divisible by p^n-1

(l) This shows that y(α) is determined by a mod P [See here for details if needed]

(m) This completes the proof in the case A = Pⁿ.

(4) This proves that the number of classes mod Pⁿ is equal to the number of ways of choosing a₀, a₁, ..., a_n-1 mod P which is N(P)ⁿ = N(Pⁿ).

QED

Theorem 1: Norm of a Divisor is the number of incongruent cyclotomic integers

If A is a divisor, the Norm N(A) can be regarded as a positive integer that is equal to the number of incongruent cyclotomic integers mod A.

Proof:

(1) The number of incongruent elements mod AB (where A,B are relatively prime) is never more than the number of incongruent elements mod A times the number of incongruent elements mod B since:

(a) Let x ≡ x' (mod A)

(b) Let x ≡ x' (mod B)

(c) Then:

A divides x - x'

B divides x - x'

(d) Since A,B are relatively prime, this means that AB divides x - x'.

(e) So that x ≡ x' (mod AB)

(2) The Chinese Remainder Theorem for Divisors (see here) shows that all possible classes mod A and mod B occur.

(3) So that, the number of classes mod AB is equal to the number of classes mod A times the number of classes mod B.

(4) By induction, if A,B,C,D are relatively prime divisors, then the number of classes mod ABC*..*D is equal to number of classes mod A times ... times the number of classes mod D.

(5) If A,B,C,D are powers of prime divisors, by Lemma 2 above this number is equal to N(A)*N(B)*N(C)*...*N(D) = N(ABC*...*D).

(6) Since any divisor can be written in the form A*B*C*....*D where A,B,C,...,D are relatively prime powers of prime divisors, it follows that the number of classes mod any divisor is the norm of that divisor.

QED

Ideal Numbers: Norm of an Ideal Number

In my last blog, I offered a definition of an ideal number. Based on this definition, it is possible to define a norm function that maps a divisor to a rational integer by multiplying it with its conjugates.

The norm of an ideal number is more than a curiosity. It has an interesting property. The norm for any ideal number is also the number of incongruent classes modulo that ideal number. I use this result in my proof of the existence of the class number for any set of cyclotomic integers. I go over this property of norms of ideal numbers here.

Today's content is once again based on Harold M. Edwards' Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Definition 1: Norm of an Ideal Number

For an ideal number A, the norm N(A) = A*σA*σ²A*...*σ^λ-2A.

For details on what σA means please see the Definition 1, here.

Lemma 1: N(AB)=N(A)*N(B)

Proof:

N(AB) = AB*σ(AB)*σ²(AB)*...*σ^λ-2(AB) =

= A*B*σA*σB*σ²(A)*σ²(B)*...*σ^λ-2(A)*σ^λ-2(B) =

= A*σA*σ²(A)*...*σ^λ-2(A)*B*σB*...*σ^λ-2B =

= N(A)*N(B)

QED

Lemma 2: if p ≠ λ, then P*σP*σ²P*...*σ^e-1P = p

NOTE: e = (λ - 1)/f where f = exponent mod λ for p. [See Definition 2, here]

Proof:

(1) Let P be a prime divisor that divides a rational integer p.

(2) From a previous result (see Lemma 1, here), we know that P, σP, σ²P, ..., σ^e-1P are the e distinct divisors that divide p.

(3) We know that if any cyclotomic integer g(α) is divisible by all the prime divisors of p, then it is also divisible by p. Likewise, we know that if g(α) is divisible by p, then it is divisible by all the prime divisors. (See Theorem, here)

(4) In other words, the product of all prime divisors is principal and it is equal to p.

QED

Lemma 3: For a Prime Divisor P that divides p, the N(P) is a rational integer.

Proof:

(1) If p = λ, then P = α - 1 and N(P) = λ [See Lemma 2, here]. So now, we can assume that p ≠ λ in order to complete the proof.

(2) If p ≠ λ, then N(P) = P*σP*σ²P*...*σ^λ-2P. [See Definition above]

(3) Since σ is a permutation and there are only e distinct prime divisors, we get the following:

N(P) = (P*σP*σ²P*...*σ^e-1P)*(P*σP*σ²P*...*σ^e-1P)*...*(P*σP*σ²P*...*σ^e-1P)

(4) Applying Lemma 2 above gives us:

N(P) = p*p*...*p = p^f [since ef = λ - 1, from the definition of exponent mod λ, see Definition 2, here]

QED

Lemma 4: For any ideal number A, the N(A) is a rational integer.

Proof:

(1) Every ideal number A is composed of powers of prime divisors. [See here for definition of ideal number]

(2) Let us assume that the powers of prime divisors of A consist of: P₁^a*P₂^b*...*P_n^c

(3) Then using Lemma 1 above, N(A) = N(P₁^a*P₂^b*...*P_n^c) =

= N(P₁^a)*N(P₂^b)*...*N(P_n^c) =

= N(P₁)*...*N(P₁)*N(P₂)*...*N(P₂)*...*N(P_n)*...*N(P_n)

(4) Using Lemma 3 above, we know that each N(P_i) is a rational integer. This then gives us our result since the product of a set of rational integers is itself a rational integer.

QED

Lemma 5: if g(α) is a cyclotomic prime that divides p where p ≠ λ and P is the divisor for g(α), then p divides g(α)*σg(α)*σ²g(α)*...*σ^e-1g(α) with a multiplicity of 1.

Proof:

(1) If a prime divisor P is the divisor for g(α), then p divides g(α)*ψ(η) [See Definition 7, here for definition of divisible by a prime divisor]

(2) This means that σp divides σ(g(α)*ψ(η)) and since p is a rational integer σ p = p and we have p divides σg(α)*σψ(η)

(3) Using the definition for divisibility of a prime divisor, we see that σP divides σg(α).

(4) We can make the same argument to show that σ²P divides σ²g(α) and so on up until σ^e-1P divides σ^e-1g(α).

(5) This gives us that the rational prime p divides g(α)*σ(gα)*...*σ^e-1g(α). [See Theorem here, since division by all e of the prime divisors implies that p divides a given cyclotomic integer]

(6) Assume that p divides g(α)*σg(α)*...*σ^e-1g(α) with a multiplicity greater than 1.

(7) Then the prime divisor P which divides p exactly once (from the theorem mentioned in step #5) would have to divide g(α)*σg(α)*...*σ^e-1g(α) with the same multiplicity.

(8) But since P is the divisor of g(α), this would imply that g(α) must divide g(α)*σg(α)*...*σ^e-1g(α) with a multiplicity greater than one which is not the case.

(9) So we reject the assumption in step #6 and conclude that p divides g(α)*σg(α)*...*σ^e-1g(α) exactly once.

QED

Lemma 6: If a prime divisor P for a prime p ≠ λ is principal such that it is the divisor of a cyclotomic prime g(α), then N(P) = Ng(α)

Proof:

(1) From Lemma 3 above, N(P) = p^f

(2) By definition of a divisor, we know that g(α) divides p [since P divides p and P is the divisor of g(α)]

(3) Ng(α) = g(α)*g(α²)*...*g(α^λ-1) [See definition of norm for cyclotomic integers]

(4) Np = p^λ-1 [See definition of norm for cyclotomic integers]

(5) We know that Ng(α) must be equal to a power of p since:

(a) N(p) = p^λ-1 [See definition of norm for cyclotomic integers]

(b) Ng(α) divides N(p) since g(α) divides p [See Lemma 6, here]

(6) Ng(α) ≡ g(1)^λ-1 ≡ 0 or 1 (mod α - 1) since:

(a) α ≡ 1 (mod α -1)

(b) Ng(α) = g(α)*g(α²)*...*g(α^λ-1,) [Definition of Ng(α), see here]

(c) Using step #6a, we have:

g(α)*g(α²)*...*g(α^λ-1,) ≡ g(1)*g(1)*...*g(1) = g(1)^λ-1 = (a₀ + a₁ + ... + a_λ-1)^λ-1 = rational integer r (since a_i are all rational integers).

(d) Now, since N(α - 1) = (α - 1)(α² - 1)*...*(α^λ-1-1) = λ (see Lemma 2, here), we know that (α-1) divides λ.

(e) Since λ is a prime, we know that either λ divides r or it does not. If it does, then g(1)^λ-1 ≡ 0 (mod α -1 ).

(f) If λ does not divide g(1)^λ-1, then gcd(λ,g(1)^λ-1) = 1 [Since λ is a prime]

(g) Using Bezout's Identity, there exists a,b such that a*λ + b*g(1)^λ-1 = 1.

(h) This gives us that b*g(1)^λ-1 - 1 = (-1)(aλ) so that b*g(1)^λ-1 ≡ 1 (mod λ).

(7) So, from step #6, we see that g(1)^λ-1 ≡ 0 or 1 (mod α - 1) if and only if g(1)^{λ - 1} ≡ 0 or 1 (mod λ)

(8) We know that λ does not divide Ng(α) since Ng(α) = p^x (from step #5 above) and since gcd(p,λ)=1.

(9) So, we are left with Ng(α) ≡ 1 (mod λ) which means that if Ng(α) = p^x then x is divisible by f where f is the exponent mod λ for p. [See Lemma 1, here]

(10) Finally, we show that Ng(α) = p^f since:

(a) We can divide up Ng(α) into [g(α)*σg(α)*...*σ^e-1g(α)]*[σ^eg(α)* σ^e+1g(α)*...*σ^2e-1g(α)]*...*[σ^(f-1)*eg(α)σ^(f-1)*e+1g(α)*...*σ^(f-1)*e+e-1g(α)]

(b) By the reasoning in Lemma 5 above, each of these f groupings of e elements is divisible by at most once by p. So that all f of these groups is divisible at most by p^f.

(c) Thus, it follows that if Ng(α) = p^a, then a = f.

QED


Lemma 7: Criteria for a Principal Divisor

An ideal number A is a principal divisor for a cyclotomic integer g(α) if for any prime divisor Pⁿ:

Pⁿ divides A if and only if Pⁿ divides g(α)

Proof:

(1) An ideal number by definition is a set of powers of prime divisors. [See Definition 3, here]

(2) So, if each power of each prime that makes up an ideal number A divide a given cyclotomic integer g(α), then A divides g(α)

(3) This gives us that if g(α) divides a second cyclotomic integer h(α), then A also divides h(α).

(4) Now, to complete this proof, we need to show that if A divides h(α), then g(α) also divides h(α).

(5) By the given, we know that A represents a complete set of prime divisors that divide g(α).

We know this since if there is any prime divisor Pⁿ that does not divide A, then it does not divide g(α).

(6) So, applying the Fundamental Theorem for Ideal Numbers, we know that if A divides a second cyclotomic integer h(α), then g(α) also divides h(α).

QED

Lemma 8: If an ideal number A is the principal divisor for a cyclotomic integer g(α), then σA is the principal divisor for a σg(α)

Proof:

(1) This lemma is established if we can use the criteria in Lemma 7. That is, we want to show that for any prime divisor Pⁿ:

Pⁿ divides σA if and only if Pⁿ divides σg(α).

(2) Assume Pⁿ divides σA

(3) Then σ^-1Pⁿ divides A.

(4) And σ^-1Pⁿ divides g(α) since A is the principal divisor for g(α).

(5) And Pⁿ divides σg(α).

(6) Assume Pⁿ divides σg(α)

(7) Then σ^-1Pⁿ divides g(α) so σ^-1Pⁿ divides A [Since, A is the principal divisor for g(α).]

(8) Which gives us that Pⁿ divides σA.

QED

Theorem: If an ideal number A is the principal divisor for a cyclotomic integer g(α), then N(A) is the principal divisor for Ng(α)

Proof:

(1) This lemma is established if we can use the criteria in Lemma 7. That is, we want to show that for any prime divisor Pⁿ:

Pⁿ divides N(A) if and only if Pⁿ divides Ng(α).

(2) Assume Pⁿ divides N(A)

(3) N(A) = A*σA*σ²A*...*σ^λ-2A

(4) From our assumption in step #2, we know that: Pⁿ divides a subset of the list of ideal numbers in step #3 so that we have:

Pⁿ divides σ^aA*...*σ^cA.

(5) Since g(α) = g(α)*σg(α)*...*σ^λ-2g(α), we can apply Lemma 8 above to conclude that:

σ^aA is the principal divisor for σ^ag(α)
...
σ^cA is the principal divisor for σ^cg(α)

(6) Finally, this gives us that Pⁿ divides Ng(α) since each Pⁱ that divides a given σⁱA also must divide the given σⁱg(α) and therefore divide Ng(α).

(7) Assume Pⁿ divides Ng(α)

(8) Again Pⁿ can only divide Ng(α), if its divides between 1 and n cyclotomic integers of the form σⁱg(α).

(9) It would then divide each of the principal divisors σⁱA for those cyclotomic integers and thereby divides N(A).

QED

Corollary: If an ideal number A is the principal divisor for a cyclotomic integer g(α), then N(A) = Ng(α)

Proof:

(1) N(A) is a rational integer (See Lemma 4 above) and Ng(α) is a rational integer (see Lemma 5, here)

(2) Since every prime divisor Pⁿ that divides N(A) also divides Ng(α) [By the Theorem above], we know that N(A) ≤ Ng(α).

(3) Since every prime divisor Pⁿ that divides Ng(α) also divides N(A), we know that Ng(α) ≤ N(A).

(4) The conclusion follows.

QED