【정수론】 페르마의 마지막 정리

여러 곳에서 자료를 모으는 중입니다.

현재 아래 자료는 페르마의 마지막 정리를 증명한 앤드류 와일즈 교수님의 강의록을 발췌한 것입니다.

 

#1

Fermat equation: 

xⁿ + yⁿ = zⁿ.

Problem: to show that there are no solutions x, y, z ∈ ℤ with

xyz ≠ 0  if  n ≥ 3.

 

#2. Fermat's results 

n = 3:

equivalent to showing there is no right-anged triangle with rational length sides and area 1 

claimed in letters to Digby and Carcavi 

n = 4:

no solution to x⁴ + y⁴ = z² 

proved by Fermat

Fermat uses method of infinite descent (무한강하법). 

무한강하법은 예를 들면 루트 2가 무리수인 것을 증명할 때 사용됨

 

#3. Kummer (1840's)

z^p = x^p + y^p = (x + ζ⁰ y) × ··· × (x + ζ^p-1 y), ζ = exp(2πi / p)

Idea: use (x + ζy)(1 + ζ) - (x + ζ² y) = ζ(x + y)

Need arithmetic of ℤ[ζ]:

(i) units;

(ii) ideal class groups.

Ideals: M·ℤ[ζ] ⊂ Ω ⊂ ℤ[ζ] (some M ∈ ℤ) such that αΩ ⊂ Ω  if  α ∈ ℤ[ζ].

Example: principal ideals Ω = β·ℤ[ζ].

Ideal class group: Cℓ_ℤ[ζ] = Ideals / Principal ideals

Kummer proved 

(i) 페르마의 마지막 정리는 참 if p ∤ #Cℓ_ℤ[ζ] 

(ii) p | #Cℓ_ℤ[ζ]  ⇔  p | B₂ × ··· × B_p-3 (단, B_i는 베르누이 계수)

where t / (exp(t) - 1) = 1 + B₀ t⁰ / 0! + ··· + B_n tⁿ / n! + ···

Mazur-W: fine structure of Cℓ_ℤ[ζ] is described by B_i's  

 

#4.Abelian extensions

ζ_p satisfies (x^p - 1) / (x - 1) = x^p-1 + ... + 1 = 0

Gal(ℚ(ζ_p)/ℚ) ⭇ (ℤ / pℤ)^x 

{σ_ℓ : ζ_p → ζ_p^ℓ}  ↔  ℓ mod p

The Galois group is abelian.

Call σ_ℓ the Frobenius at ℓ (Frob_ℓ). 

Theoren (Kronecker-Weber)

Get all abelian extensions of ℚ from roots of 1.

 

#5. Class field theory (Hilbert, Takagi, Artin, Hasse, ...)

Describes abelian extensions of anu number field.

Example: F = ℚ(2^1/3)

∃ F_p such that 

Gal(F_p / F) ≃ (O_F / pO_F)^x / (image of O_F,+^x)

where O_F = ℤ [2^1/3], and O_F,+^x = units of O_F

More complicated when ideal class group of O_F is not 1. 

 

#6. Non-abelian approaches

Frey (1985): New idea! Suppose a^p + b^p = c^p, p prime ≥ 3, a, b, c ∈ ℤ

Consider y² = x(x - a^p)(x + b^p).

Discriminant is (a^pb^pc^p)².

Such a curve should not exist!

Discriminants should not be p^th powers. 

New way to prove F.L.T. (페르마의 마지막 정리): show there is no such elliptic curve.

 

#7. Elliptic curves

E : y² = x³ + Ax + B   (A, B ∈ ℚ) 

E(ℂ) ≃ ℂ / Λ is a group, Λ = ℤ + ℤ.τ 

 

E(ℂ)[pⁿ] ≃ ℤ/p^nℤ⊕ℤ/pⁿℤ

Gal(ℚ(E[pⁿ])/ℚ) ↪ GL₂(ℤ/pⁿℤ)

 

Algebraically: these are the 8 points of inflection of 

y² = x³ + Ax + B plut the 9^th one at ∞.

The coordinates generate a field ℚ(E[3])

Gal(ℚ(E[3])/ℚ) ↪ GL₂(ℤ/3ℤ). 

 

#8. Modular forms

Modur forms: functions

 

 

satisfying 

 

 

Theorem (Modularity conjecture of Taniyama & Shimura)

 

 

#9. Rough over view of proof

어려운 페르마의 마지막 정리를 풀기 위해 더 어려운 타원곡선 문제를 증명

 

(Frey &) Ribet: Modularity conjecture → F. L. T

Problem : prove modularity conjecture

First step : consider

 

 

Langlands (& Tunnell) : this is connected with a modular form. Based on his theory of cyclic base change: note S₄ is a solvable group. 

 

#10. Langlands' programme

Generalisation of class field theory:

Describe all Galois extensions of ℚ (or any number field).

Generalises modular forms to automorphic representations:

roughly speaking we use functions invariant under subgroups of GL_n(O_F) to classify n-dimensional representations of 

 

 

(Better to use adeles and adelic representation theory).

 

#11. A little progress...

Theorem (Khare-Wintenberger)

 

 

Assume det ρ is odd, ρ irreducible.

Then ρ is associated to a modular form.

 

Proof uses similar techniques with an ingenious induction on ℓ.

Does not seem to work in the general case (GLn and any F in place of ℚ).