Proof of Fermat’s Last Theorem

Recommended posts: 【Modern Mathematics】 Group Theory

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

► Proof of Fermat’s Last Theore Video Lecture

1. Overview

In the seventeenth century, there was a man named Fermat whose hobby was mathematical research. Although he studied mathematics as a hobby, his achievements were remarkable. Fermat devoted his life to law and worked in the legal profession until his death. However, it is said that he became deeply inspired after reading a translated edition of Diophantus’s Arithmetica and began studying mathematics in his spare time. Fermat was more interested in mathematical research itself than in gaining fame. While working at the parliament, Fermat would write the results of his research, discovered during his free time, in the margins of his copy of Diophantus’s Arithmetica. The most famous of these marginal notes is the following statement, believed to have been written in 1637: for the equation xⁿ + yⁿ = zⁿ, if n is an integer greater than or equal to 3, there are no natural numbers x, y, and z that satisfy this equation. This is known as Fermat’s Last Theorem, or Fermat’s Great Theorem. Fermat left many such statements, but this particular conjecture remained unsolved despite numerous attempts by many people, including some of the most famous mathematicians in history. Another mystery, almost as great as Fermat’s Last Theorem itself, which remained unsolved for hundreds of years, was whether Fermat had truly proved it. Most scholars believe that, considering the level of mathematical development at the time, it would have been impossible for him to have done so.

Fermat’s most outstanding achievements were in the field of number theory. However, he also made important contributions to calculus and discovered the principle of least time, also known as Fermat’s principle, among many other accomplishments. In this way, Fermat achieved astonishing results through what was merely a hobby for him. Judging from this, Fermat may have been naturally gifted, but I believe that his determination and effort also played an important role. Above all, Fermat seems to have been a sincere and unpretentious person. He did not seek fame through his mathematical research, and he tried to avoid abstract mathematical disputes because he was busy. I agree with his attitude. In any case, Fermat was an indispensable figure in the history of mathematics.

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))

○ An example of the modular symmetry of an elliptic curve.

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

○ In general, if a line meets $E$ at $P, Q, R$, then $P + Q + R = O$.

○ The tangent line at $P$ gives $P + P + R = O$.

○ An inflection point gives $P + P + P = 3P = O$.

○ Inflection points $=$ 3-torsion points $=$ $E[3]$ $=$ the objects carrying the $\mod 3$ cyclic/linear/Galois-permutation symmetries.

○ 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. Modularity Assumption of Elliptic Curve: 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. Level-lowering: 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’ 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.

(push down the Galois representation side)

R_{Q_n+1}	→	R_{Q_n}
↓ (at each level) R ≃ T		↓ (at each level) R ≃ T
T_{Q_n+1}	→	T_{Q_n}

(push down the modular forms / Hecke algebra side)

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.

R_∞ := lim← R_{Q_n}	≃	T_∞ := lim← T_{Q_n}
↓ (patched isomorphism)		↓ (patched isomorphism)

(specialization: removing the auxiliary primes)

↓

R (or T) at the original level

⑼ Application 1. ECDSA: It is based on the fact that, by leveraging the group operation properties of elliptic curves defined over a finite field (modulo arithmetic), one can easily derive a public key, whereas inferring the corresponding private key in reverse is computationally intractable.

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

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Proof of Fermat’s Last Theorem

1. Overview

2. Step 1. Kummer’s attempt (1840s)

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

4. Step 3. Wiles’ proof

5. Subsequent research

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