Korean, Edit

Proof of Fermat’s Last Theorem

Recommended posts: 【Modern Mathematics】 Group Theory

1. Overview

2. Step 1. Kummer’s attempt

3. Step 2. Ribet’s theorem (Serre’s \(\varepsilon\)-conjecture)

4. Step 3. Wiles’s proof

5. Subsequent research

1. Overview

⑴ Infinite descent

① Problem : Prove that \(\sqrt{2}\) is irrational.

② Proof : Proven by the Pythagorean school

Let \(\sqrt{2} = \frac{b}{a}\) with \(a,b \in \mathbb{Z}\).

∴ \(2a^2 = b^2 \Rightarrow 2 \mid b \Rightarrow 2 \mid a \Rightarrow 2^2 \mid b \Rightarrow 2^2 \mid a \Rightarrow \cdots\)

∴ \(2^\infty \mid a, b \Rightarrow\) impossible

∴ \(\sqrt{2} \notin \mathbb{Q}\)

⑵ Fermat’s infinite descent

① Problem : Prove that there are no integer solutions \(x, y, z\) to \(x^4 + y^4 = z^2\).

② Proof : Proven by Fermat

Even if an arbitrary \((x, y, z)\) is given, after reducing appropriately we may assume \(\gcd(x, y) = 1\). If \(\gcd(x, y) = 1\), then \((x^2, y^2, z)\) are pairwise coprime, so they form a Pythagorean triple. We may assume without loss of generality that \(x^2 = m^2 - n^2, y^2 = 2mn, z = m^2 + n^2\). From this we see that \(y\) is even, so we may assume without loss of generality that \(m = 2a, n = 2b+1\). Looking at \(x^2 = m^2 - n^2 = (m+n)(m-n)\), we have \(\gcd(m+n, m-n) = \gcd(m+n, 2n) = 1\) (∵ \(\gcd(x, y) = 1\)). Therefore we can set \(m+n = v^2, m-n = w^2\), and since \(v, w\) are odd, the remainders of \((m+n)\) and \((m-n)\) upon division by \(4\) are \(1\). We have \(2m = (m+n) + (m-n) \equiv 1 + 1 = 2 \pmod{4}\), but \(2m = 4a \equiv 0 \pmod{4}\), a contradiction. Of note, this solution did not use the method of infinite descent.

⑶ Fermat’s right-triangle problem: Claimed in a letter to Digby and Carcavi

① Problem: Show that among right triangles with rational side lengths, none has area \(1\).

⑷ Fermat’s Last Theorem (Fermat’s Last Theorem, FLT; 1637)

① Problem: Let the Fermat equation be \(x^n + y^n = z^n\) (\(n \ge 3\)). Prove that there is no integer solution \(x, y, z \in \mathbb{Z}\) with \(xyz \ne 0\).

② Proof: To solve the difficult FLT, prove an even more difficult elliptic-curve problem (see below).

③ For \(n = 4\) there is a separate proof by Fermat, and for \(n = 3\) Euler proved it.

④ “The margin is too small to contain the proof”: A famous note Fermat left in the margin next to a particular problem in Diophantus’s Arithmetica

2. Step 1. Kummer’s attempt (1840s)

⑴ Summary: FLT, a purely integer Diophantine problem, is equivalent to a problem about certain algebraic objects (rings/curves/groups).

⑵ Group-theoretic concepts

① Units: You must know what a unit is to distinguish being a \(p\)-th power from being a unit times a \(p\)-th power

② Ideal: An additive subset closed under multiplication inside the ring of integers \(\mathbb{Z}[\zeta]\)

○ Mathematical expression: \(M\cdot\mathbb{Z}[\zeta] \subset \Omega \subset \mathbb{Z}[\zeta]\) (some \(M \in \mathbb{Z}\)) s.t. \(\alpha\Omega \subset \Omega\) if \(\alpha \in \mathbb{Z}[\zeta]\)

○ Type 1. Principal ideal: \(\Omega = \beta\cdot\mathbb{Z}[\zeta]\), etc.

○ Type 2. Non-principal ideal: The case where unique factorization (UFD) breaks

③ Ideal class group: \(\mathrm{Cl}_{\mathbb{Z}[\zeta]} = {\text{all ideals}}/{\text{principal ideals}}\)

○ That is, the ideal group modulo the subgroup of principal ideals.

⑶ 1-1. If a counterexample to FLT exists, then \(z^p = x^p + y^p = (x + \zeta^0 y) \times \cdots \times (x + \zeta^{p-1} y)\), where \(\zeta = \exp\left(\frac{2\pi i}{p}\right)\).

① For example, the fact that \(\sum_{k=0}^{p-1}\zeta^k = 0\) can be seen from the fact that the vector sum on the complex plane is \(0\).

⑷ 1-2. Between each factor \((x + \zeta^i y)\) and \((x + \zeta^j y)\), they are “almost coprime.”

① For example, when computing the gcd of \(x + \zeta y\) and \(x + \zeta^2 y\) via the Euclidean algorithm, we obtain \((x + \zeta y)(1 + \zeta) - (x + \zeta^2 y) = \zeta(x + y)\).

② That is, for \((x + \zeta y)\) and \((x + \zeta^2 y)\) to share a common divisor, that divisor must also appear on the \((x+y)\) side, imposing many constraints on \((x+y)\).

③ Hence it is likely that \((x + \zeta y)\) and \((x + \zeta^2 y)\) are coprime (this is the meaning of “almost coprime”).

⑸ 1-3. From the “almost coprime” condition and the fact that the left-hand side of FLT is \(z^p\), it suggests each factor must be a \(p\)-th power of some number.

⑹ 1-4. If \(p \nmid\) ＃\(\mathrm{Cl}(\mathbb{Z}[\zeta])\), then one can lift an ideal \(p\)-th power to an element \(p\)-th power or its unit multiple, yielding a contradiction → prove FLT by contradiction

① \((a) = \mathfrak{I}^p\) means \([\mathfrak{I}]^p = 1\) in the class group.

② Since \(p \nmid\) ＃\(\mathrm{Cl}\), there is no \(p\)-torsion in \(\mathrm{Cl}\), so \([\mathfrak{I}] = 1\), i.e., \(\mathfrak{I}\) is principal.

③ Therefore \(\mathfrak{I} = (b)\) and \((a) = (b^p)\), so \(a = u b^p\) (unit: \(u\)).

④ Conrad’s proof

Under this condition one obtains \(x + y \equiv 0 \pmod{p}\) or \(x \equiv y \equiv -z \pmod{p}\), and in each concrete case one can easily find a contradiction.

⑺ 1-5. The case \(p \mid\) ＃\(\mathrm{Cl}_\mathbb{Z}[\zeta] \Leftrightarrow p \mid B_2 \times \cdots \times B_{p-3}\) (where \(B_i\) are Bernoulli numbers) is outside the range where Kummer’s attempt applies.

① Here \(\frac{t}{\exp(t) - 1} = 1 + \frac{B_0 t^0}{0!} + \cdots + \frac{B_n t^n}{n!} + \cdots\) holds.

② Mazur–Wiles: Analyze the entire structure of how the part related to \(p\) grows and which function exactly generates that structure.

3. Step 2. Ribet’s theorem (Serre’s \(\varepsilon\)-conjecture)

⑴ Summary: If the Frey curve is (under the assumption) ‘modular’, then no counterexample to FLT can exist.

⑵ Group-theoretic concepts

① \(\mathrm{Frob}_\ell\): A handle for reading information at each prime

○ \(\zeta_p\): A primitive \(p\)-th root of unity. It satisfies \(\frac{x^p - 1}{x - 1} = x^{p-1} + \cdots + 1 = 0\). e.g. \(e^{2\pi i/p}\)

○ \(\mathbb{Q}(\zeta_p)\): A number field containing \(\zeta_p\) (cyclotomic field, cyclotomic character)

○ \(\sigma \in \mathrm{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})\): A map fixing \(\mathbb{Q}\) and sending \(\mathbb{Q}(\zeta_p)\) to itself. In particular, \(\sigma(\zeta_p)\) satisfies the condition of being another primitive \(p\)-th root of unity.

○ \((\cdot)^\times\): The elements in the set that have multiplicative inverses (units)

○ \(\mathbb{Z}/p\mathbb{Z}\): The residual classes mod \(p\), i.e. \(\{0, 1, 2, \cdots, p-1\}\)

○ \((\mathbb{Z}/p\mathbb{Z})^\times = \{a \in \mathbb{Z}/p\mathbb{Z}\mid \exists b \text{ s.t. } ab \equiv 1 \pmod{p}\}\), i.e. the set of elements with inverses mod \(p\)

○ \(\mathrm{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \simeq (\mathbb{Z}/p\mathbb{Z})^\times\): Describes that, as in Kummer’s attempt, the number field corresponds 1:1 with residual classes.

○ Concretely, we can write the above isomorphism as \({\sigma_\ell : \zeta_p \mapsto \zeta_p^\ell} \leftrightarrow \ell \bmod p\)

○ If \(\ell\) is coprime to \(p\) (in particular, a prime \(\ell \ne p\)), then \(\ell \bmod p \in (\mathbb{Z}/p\mathbb{Z})^\times\)

○ If \(\ell \equiv 1 \pmod{p}\) then \(\sigma_\ell(\zeta_p) = \zeta_p\), so the element becomes close to the identity.

○ If \(\ell \equiv -1 \pmod{p}\) then \(\zeta_p \mapsto \zeta_p^{-1}\) (over the complex numbers this is linked to complex conjugation).

○ We call \(\sigma_\ell\) the Frobenius at \(\ell\) (i.e. \(\mathrm{Frob}_\ell\))

○ Why it is called Frobenius: A prime \(\ell \ne p\) is unramified in \(\mathbb{Q}(\zeta_p)/\mathbb{Q}\) (ramification occurs only at \(p\)).

② Kronecker–Weber theorem

○ Every abelian extension of \(\mathbb{Q}\) is generated by roots of unity. The simplest success case.

③ Class field theory (Hilbert, Takagi, Artin, Hasse, etc.)

○ Extend the Kronecker–Weber theorem to a general number field \(F\): Complete classification of abelian extensions (1-dimensional representations)

○ Example: \(F = \mathbb{Q}(2^{1/3})\)

○ \(\exists F_{\mathfrak{p}} \text{ s.t. } \mathrm{Gal}(F_{\mathfrak{p}}/F) \simeq (\mathcal{O}_F/\mathfrak{p}\mathcal{O}_F)^\times / (\text{image of } \mathcal{O}_{F,+}^\times)\) where \(\mathcal{O}_F = \mathbb{Z}[2^{1/3}]\), and \(\mathcal{O}_{F,+}^\times =\) units of \(\mathcal{O}_F\)

○ \(F_{\mathfrak{p}}\): A ray class field that may ramify only at \(\mathfrak{p}\) (or the prime factors of \(\mathfrak{p}\mathcal{O}_F\)) and is unramified elsewhere (mod \(\mathfrak{p}\) or \(\mathfrak{p}\cdot\infty\)).

○ \((\mathcal{O}_F/\mathfrak{p}\mathcal{O}_F)^\times\) or \((\mathcal{O}_F/\mathfrak{p})^\times\): A generalization of \((\mathbb{Z}/p\mathbb{Z})^\times\) from \(\mathbb{Q}\)

○ image of \(\mathcal{O}_F^\times\): Reducing each unit \(u \in \mathcal{O}_F^\times\) modulo \(\mathfrak{p}\) gives \(u \mapsto \bar{u} \in (\mathcal{O}_F/\mathfrak{p})^\times\). The set of such reduced units inside \((\mathcal{O}_F/\mathfrak{p})^\times\) is called the image of \(\mathcal{O}_F^\times\).

○ In general, the ray class group \(\mathrm{Cl}_{\mathfrak{p}}\) = residual unit information mod \(\mathfrak{p}\) + the original ideal class group information

○ Because of the residual unit information, \(\mathrm{Gal}(F_{\mathfrak{p}}/F)\) equals exactly \((\mathcal{O}_F/\mathfrak{p})^\times/\mathrm{im}(\mathcal{O}_F^\times)\) only when \(\mathrm{Cl}(\mathcal{O}_F)=1\) (= ideal class group).

○ That is, \(1 \to (\mathcal{O}_F/\mathfrak{p})^\times/\mathrm{im}(\mathcal{O}_F^\times) \to \mathrm{Cl}_{\mathfrak{p}} \to \mathrm{Cl}(\mathcal{O}_F) \to 1\)

④ Elliptic curves

○ \(E/\mathbb{Q}\): An elliptic curve \(E\) given by an equation with coefficients in \(\mathbb{Q}\)

○ Weierstrass form: \(y^2 = x^3 + Ax + B \quad (A, B \in \mathbb{Q})\)

○ \(E(\mathbb{C}) \simeq \mathbb{C}/\Lambda, \Lambda = \mathbb{Z} + \mathbb{Z}_{\tau}\) (with \(\mathrm{Im}(\tau) > 0\))

○ Meaning: An elliptic curve over the complex numbers is analytically identical to a torus obtained by folding the complex plane by a lattice.

○ \(E(\mathbb{C})[p^n] \simeq (\mathbb{Z}/p^n\mathbb{Z}) \oplus (\mathbb{Z}/p^n\mathbb{Z}) = (\mathbb{Z}/p^n\mathbb{Z})^2\)

○ In \(\mathbb{C}/\Lambda\), \(n\)-torsion points are those that become \(0\) after adding \(n\) times, so \(E(\mathbb{C})[n] = (\mathbb{C}/\Lambda)[n] = (1/n)\Lambda/\Lambda\)

○ Since there are \(n\)-subdivisions in each of the two lattice-generator directions, \(E(\mathbb{C})[n] = (1/n)\Lambda/\Lambda \cong (\mathbb{Z}/n\mathbb{Z}) \oplus (\mathbb{Z}/n\mathbb{Z})\)

○ \(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\): The absolute Galois group of \(\mathbb{Q}\) = the group of all field automorphisms of \(\overline{\mathbb{Q}}\) fixing \(\mathbb{Q}\)

○ \(\mathrm{GL}_2(\mathbb{Z}/\ell\mathbb{Z})\): The group of invertible \(2\times 2\) matrices over \(\mathbb{Z}/\ell\mathbb{Z}\) (= \(\mathbb{F}_\ell\)), i.e. all \(2\times 2\) matrices with nonzero determinant (multiplication is matrix multiplication)

○ \(\mathrm{Gal}(\mathbb{Q}(E[p^n])/\mathbb{Q}) \hookrightarrow \mathrm{GL}_2(\mathbb{Z}/p^n\mathbb{Z})\)

○ \(E[n] = \{P \in E(\overline{\mathbb{Q}})\mid nP = O\}\) (the \(\ell\)-torsion points)

○ \(\rho_{E,n} : G_{\mathbb{Q}} \to \mathrm{Aut}(E[n]) \simeq \mathrm{GL}_2(\mathbb{Z}/n\mathbb{Z})\) (a 2-dimensional Galois representation)

○ \(\ker(\rho_{E,n}) = \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}(E[n]))\)

○ Meaning of \(\hookrightarrow\): Recording the action on torsion as matrices

○ A smooth plane cubic curve has exactly 9 inflection points over an algebraic closure (including one point at infinity (the identity))

○ Mathematical expression: ㅣ\(E[3]\)ㅣ \(= 9, E[3] \simeq (\mathbb{Z}/3\mathbb{Z})^2\)

○ Note: In \(y^2 = x^3 + Ax + B\), the 8 inflection points are \(\{(x, y)\mid \psi_3(x) = 3x^4 + 6Ax^2 + 12Bx - A^2 = 0, y = \pm\sqrt{x^3 + Ax + B}\}\)

⑤ Langlands / modularity

○ Definition of a modular form: \(f\left(\frac{az+b}{cz+d}\right) = (cz+d)^2 f(z)\) (including a level condition)

○ where \(f(z) = \sum_{n=0}^{\infty} a_n e^{2\pi i n z}\)

○ where \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})\), \(c \equiv 0 \pmod{N}\)

○ Langlands’ idea

○ Modular forms are special functions/forms arising from 2-dimensional Galois representations in \(\mathrm{GL}_2\).

○ Langlands generalizes this to all of \(\mathrm{GL}_n(\mathcal{O}_F)\) and treats it in the language of automorphic representations.

○ Using functions invariant under a subgroup, collect symmetric functions and extract the representation hidden inside.

○ The resulting automorphic representations on the \(\mathrm{GL}_n\) side correspond to \(n\)-dimensional representations of the Galois group \(\mathrm{Gal}(\overline{F}/F)\).

○ Adeles (Adelic representation) are recommended even more than Langlands representations, as a tool bundling all finite primes and the infinite place at once.

⑶ 2-1. Assume FLT has a counterexample.

① Assume there exists \(a^p + b^p = c^p\) (with \(a, b, c \in \mathbb{Z}\) and \(p\) a prime \(\ge 3\)).

⑷ 2-2. Then the Frey–Hellegouarch curve (proposed in 1985) \(E\) arises.

① Frey curve \(E_{a,b,p}\): \(y^2 = x(x - a^p)(x + b^p)\) has distinct roots, hence is an elliptic curve.

② By a coordinate change (i.e. substituting \(x - (b^p - a^p)/2\) for \(x\) in the Frey curve), it can be written in Weierstrass form \(y^2 = x^3 + Ax + B\) (with \(A, B \in \mathbb{Q}\)).

③ Discriminant of the cubic = product of squared root differences \(= (a^p - 0)^2(0 - (-b^p))^2(a^p - (-b^p))^2 = (a^p b^p c^p)^2 \ne 0\).

④ The elliptic curve discriminant \(\Delta(E) = 16(abc)^{2p}\) means that (if it exists) the Frey curve has very special ramification/conductor properties (connected to 2-5).

⑸ 2-3. Extract the mod \(p\) residual representation \(\bar{\rho} = \rho_{E,p}\) from \(E\).

① If \(E/\mathbb{Q}\) is an elliptic curve and \(p\) is prime, then the Galois action on \(E[p]\) gives a mod \(p\) representation \(\rho_{E,p} : \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \mathrm{GL}_2(\mathbb{F}_p)\).

② Example: the case \(p = 3\)

○ \(\rho : \mathrm{Gal}(\mathbb{Q}(E[3])/\mathbb{Q}) \hookrightarrow \mathrm{GL}_2(\mathbb{Z}/3\mathbb{Z}) \to \mathrm{PGL}_2(\mathbb{Z}/3\mathbb{Z}) \simeq S_4\)

○ Langlands–Tunnell: When \(\mathrm{PGL}_2(\mathbb{F}_3) \simeq S_4\) (solvable), it guarantees that (under certain conditions) a mod \(3\) 2-dimensional Galois representation \(\bar{\rho}\) is modular.

○ Therefore, if \(\rho_{E,3}\) of the elliptic curve \(E\) is in this situation, it is connected to modularity at the mod \(3\) level.

③ Because of the Weil pairing, \(\det \bar{\rho}\) is fixed.

○ Related to the fact that when points on the plane are moved by a linear transformation, the new area of the moved points scales by a factor of \(\det\).

○ The Weil pairing implies that the area scale \(\det \bar{\rho}\) is constant.

⑹ 2-4. Level-lowering: If we assume \(E\) is modular, then \(\bar{\rho}\) must come from a weight \(2\) newform of some level \(N\) (= \(\bar{\rho}\) equals the Galois representation attached to a modular form of that level \(N\)).

① This assumption is called the Shimura–Taniyama–Weil conjecture or the Breuil–Conrad–Diamond–Taylor modularity theory.

② Mathematical expression

○ \(\{\text{Elliptic curves}/\mathbb{Q}\} \mapsto \{\text{modular form of weight }2,\ \text{some level}\}\)

○ \(\mathrm{Gal}(\mathbb{Q}(E[p^n])/\mathbb{Q}) \hookrightarrow \mathrm{GL}_2(\mathbb{Z}/p^n\mathbb{Z})\)

○ \(\mathrm{Frob}_\ell \mapsto \text{element with trace } a_\ell\)

⑺ 2-5. But because the Frey curve has special discriminant/reduction properties, \(\bar{\rho}\) becomes in fact less twisted (unramified/weakly ramified) at many primes \(\ell\) → repeatedly applying Ribet level lowering makes odd primes keep dropping out of the level.

① Specialness: A prime \(\ell \ne p\) is unramified in \(\mathbb{Q}(\zeta_p)/\mathbb{Q}\) (ramification occurs only at \(p\)) (finite primes).

② Proof

At some stage, by the “modularity assumption”, suppose \(\rho\) comes from a weight \(2\) newform of level \(N = 2\cdot 3\cdot 5\cdot 7\).

Now if \(\rho\) is unramified at \(7\), then \(7\) can be removed from the level. That is, \(2\cdot 3\cdot 5\cdot 7 \Rightarrow 2\cdot 3\cdot 5\)

Next, if it is also unramified at \(5\), then \(5\) can be removed from the level. That is, \(2\cdot 3\cdot 5 \Rightarrow 2\cdot 3\)

Next, if it is also unramified at \(3\), then \(3\) can be removed from the level. That is, \(2\cdot 3 \Rightarrow 2\)

⑻ 2-6. In the end, there must be a weight \(2\) newform of level \(2\), but none exists, a contradiction.

① Proof

from math import prod
from sympy import factorint, divisors, totient, gcd

def chi_minus4(p:int)->int:
    # Kronecker (-4/p): for odd p, 1 if p ≡ 1 (mod 4), -1 if p ≡ 3 (mod 4); for p = 2, 0.
    if p == 2:
        return 0
    return 1 if p % 4 == 1 else -1

def chi_minus3(p:int)->int:
    # Kronecker (-3/p): 0 if p = 3; for other primes, it is +1 if p ≡ 1 (mod 3) and -1 if p ≡ 2 (mod 3) (for p = 2, -1).
    if p == 3:
        return 0
    return 1 if p % 3 == 1 else -1

def index_gamma0(N:int)->int:
    # μ = [SL2(Z): Γ0(N)] = N * Π_{p|N} (1 + 1/p)
    ps = factorint(N).keys()
    return int(N * prod([(1 + 1/p) for p in ps]))

def num_cusps(N:int)->int:
    # c = Σ_{d|N} φ(gcd(d, N/d))
    return int(sum(totient(gcd(d, N//d)) for d in divisors(N)))

def e2_count(N:int)->int:
    # elliptic points of order 2
    if N % 4 == 0:
        return 0
    ps = factorint(N).keys()
    return int(prod([1 + chi_minus4(p) for p in ps]))

def e3_count(N:int)->int:
    # elliptic points of order 3
    if N % 3 == 0:
        return 0
    ps = factorint(N).keys()
    return int(prod([1 + chi_minus3(p) for p in ps]))

def genus_X0(N:int)->int:
    mu = index_gamma0(N)
    c  = num_cusps(N)
    e2 = e2_count(N)
    e3 = e3_count(N)
    g = 1 + mu/12 - e2/4 - e3/3 - c/2
    return int(round(g))  # Must be an integer.

# Key check
for N in [1,2,3,4,5,6,7,8,9,10,11]:
    g = genus_X0(N)
    print(N, g)

○ The dimension of the weight \(2\) cusp form space \(S_2(\Gamma_0(N))\) equals the genus of the modular curve \(X_0(N)\) : \(= 1 + \mu/12 - e_2/4 - e_3/3 - c/2 = 0\)

○ For \(N = 2\), \(\dim S_2(\Gamma_0(2)) = 0\) (∵ \(e_2 = 1,\ e_3 = 0,\ c = 2,\ \mu = 3\))

○ Of note, the first nonzero case is \(N = 11\) with \(g = 1\).

⑼ 2-7. Since the assumption in 2-1 cannot hold, FLT holds (proof by contradiction).

4. Step 3. Wiles’s proof

⑴ Summary: (In a certain range) elliptic curves are modular.

⑵ Group-theoretic concepts

① (On the representation side) A deformation ring \(R\) that classifies certain Galois representations

② (On the modular-form side) A Hecke algebra \(T\) containing modular forms satisfying the same conditions

③ Goal: \(R \simeq T\) (if equal, it follows that “the representation comes from a modular form”)

④ In general, \(R\) and \(T\) are not identical: The degrees of freedom (variables) arising from global conditions (Galois side) and the relations (constraints) on the modular-form side must match perfectly; otherwise either “freedom” remains or constraints are insufficient, making it hard to pin down \(R = T\).

⑶ 3-1. For each \(n\), attach an auxiliary prime set \(Q_n\) to form \(R_{Q_n}\) (Galois deformation ring) and \(T_{Q_n}\) (Hecke algebra).

⑷ 3-2. Adding auxiliary primes introduces additional local conditions describing how things must behave at those primes.

① That is, the raised-level problem is in fact the problem with extra conditions: those conditions serve to exactly control the leftover global degrees of freedom.

② Level \(Q_n\): (base conditions) + (conditions at auxiliary primes \(Q_n\))

③ Level \(Q_{n+1}\): (base conditions) + (conditions at auxiliary primes \(Q_n\)) + (additional conditions at a new prime)

⑸ 3-3. Since \(Q_{n+1} \supset Q_n\), there is a downward map that “forgets” conditions.

⑹ 3-4. Arrange compatibility so that these downward maps commute between the Galois side and the modular-form side.

① Compatibility on the Galois representation side: solution set satisfying more conditions → projection to a solution set where fewer conditions are required.

② Compatibility on the modular-form/Hecke algebra side: Hecke eigenvalues at higher level must descend and agree with the eigenvalues seen at \(Q_n\) without contradiction.

⑺ 3-5. Infinite descent: Thanks to this compatibility, take inverse limits/patching of \(\{R_{Q_n}\}\) and \(\{T_{Q_n}\}\) to build an infinite-level equality like \(R_\infty \simeq T_\infty\)

① In practice, it’s not just rings—we also patch the modules, which makes a meaningful comparison possible.

⑻ 3-6. Finally, specialize (send to \(0\)) the variables/degrees of freedom introduced by auxiliary primes (= remove auxiliary-prime conditions) in \(R_\infty, T_\infty\) to obtain the original target equality \(R \simeq T\) at minimal level.

5. Subsequent research

⑴ Khare–Wintenberger theorem (Serre’s modularity conjecture)

① Summary : For \(\mathrm{GL}_2/\mathbb{Q}\), every odd irreducible mod \(\ell\) representation comes from a modular form.

○ Wiles: Proved modularity for \(\ell\)-adic 2-dimensional representations coming from elliptic curves (weight \(2\), certain local conditions, semistable).

○ Serre/KW: Proved modularity for all odd irreducible mod \(\ell\) 2-dimensional representations.

② If \(\det \rho\) is odd (i.e., when inserting complex conjugation \(c^*\) into \(\rho\), \(\det(\rho(c)) = -1\)), then \(\rho\) is irreducible.

○ Irreducible = not decomposable into a sum of two 1-dimensional representations = no invariant 1-dimensional subspace

③ Such \(\rho\) is modular: more precisely, it is the reduction mod \(\ell\) of the Galois representation arising from a Hecke eigenform.

④ Proof: Wiles’s modularity lifting + level changing

⑤ For \(\mathrm{GL}_n\) (\(n > 2\)) or for a general number field \(F\) instead of \(\mathbb{Q}\), this proof strategy does not seem to remain effective.

⑵ Lean : An open-source project at Imperial College London that formalizes a modern variant of FLT (including Taylor–Wiles-type techniques) using the Lean theorem prover.

Created: 2023.08.26 00:02

Updated: 2026.01.04 00:22