# November 11th, 2017

## Fermat's Last Theorem

First case, due to Kumar. Here's what we'll show:

**Theorem:**
Take a prime $p > 3$, assume $p$ is *regular* (i.e. $p\not\mid h_{\QQ(\zeta_p)}$the class number). Then $x^p + y^p = z^p \implies xyz = 0 \mod p$.

**Kummer's Criterion:**
$p$ is irregular (so $p \mid h_{\QQ(\zeta_p)})$ iff $\text{ord}_p(B_k) > 0$ for some $k = 2,3,\cdots p-3$, where $B_k$ is a Bernoulli Number.
$\frac{z}{e^z-1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!}, \abs{z} < 2\pi$.

Infinitely many irregular - known, 39%

Infinitely many regular - open, 61%

**Herbrand-Ribet:**

$A = \text{Cl}_{\QQ(\zeta_p)}, C = A/A^p$ is an $\FF_p$ vector space where $C = 0 \iff p \not\mid h$.

$G = \text{Gal}(\QQ(\zeta_p) / \QQ) \cong (\ZZ/p\ZZ)^\times$ is cyclic, so its dual group is also cyclic $\hat G = <x>$ where $X \cdot G \into \FF_p^\times$ is the cyclotomic character $X(\sigma) = [a]$ if $\sigma(\zeta_p) = \zeta_p^a$.

So fix an even $2 \leq k \leq p-3$. Then $\text{ord}_p(B_k) > 0$ $\iff$ $C(X^{1-k}) \neq 0$. (Only known to be iff this past century!)
Was known assuming Vandiver's conjecture: $p\not\mid h_{\QQ(\zeta_p + \zeta_p^{-1})}$. Ribet was able to bypass using Galois representations associated to modular forms. Under this assumption, create a cusp form congruent to an Eisenstein series$\mod p$. Move back into Galois side to recover nontriviality on RHS.

Idea: Factor both sides in the cyclotomic field, so really need to know units in these fields.

There is a natural notion (intrinsic) of conjugation on the cyclotomic field. Take $K_n = \QQ(\zeta), \text{ord}(\zeta) = n, \zeta \in \bar\QQ$.

Then $\text{Gal}(K_n/\QQ) \cong (\ZZ/n\ZZ)^\times$ by $c \mapsto [-1]$. Notate by $x \mapsto c(x)$.

Then $c(\zeta) = \zeta^{-1}$, nd for all $\sigma : K_n \into \CC, \sigma(\zeta) = e^{2\pi i k n}$ with $\text{gcd}(k,n) = 1$.

So $\sigma(\zeta^{-1}) = \overline{\sigma(\zeta)}$

and $\sigma \circ c = \bar\sigma = \text{(conjugation)} \circ \sigma$


**Kronecker's Lemma**: 
Take $\alpha \in \bar\ZZ / \theset{0}, \abs{\sigma(\alpha)} \leq 1$ for all $\sigma: \bar\QQ \into \CC$. Then $\alpha$ is a root of unity.

*Proof*: 
$f(x) = \text{Irr}(\alpha, \QQ, x) \in \ZZ[x]$. $n = \text{deg}(f) = [\QQ(\alpha): \QQ]$. Then $f(x) = \prod_{i=1}^n(x-\alpha_i) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$. Then $a_m = \pm \sum_{j \leq m} \alpha_{i_j}, \abs{a_m} \leq {n\choose m}$. But only finitely many $f(x) \in \ZZ[x]$ satisfy $\text{deg}(f) \leq n$. Thus there are only finitely many $\alpha \in \bar{\ZZ}$ that satisfy $\deg(f_\alpha) \leq n$.

Note that $\alpha^k$ satisfies the hypothesis, $f_\alpha k$ satisfies the bounds and $\deg f_\alpha k \leq n$.

**Proposition (Kummer):** 
$p > 2$ prime,$u\in \ZZ[\zeta_p]^\times$. Then $u/\bar u = \zeta_p^k$ for some $k\in \ZZ$.

**Lemma:** 
$\alpha \in \ZZ[\zeta_p]$, then $\exists a \in \ZZ$ such that $\alpha^p = a \mod (p)$.

**Proof:** 
$\alpha = a_0 + a_1\zeta + \cdots + a_{p-2}\zeta^{p-2}$, $a^p = \sum_{i=0}^{p-2} a_i \mod (p)$ where $a\in \ZZ$.

**Lemma:** 
$\mu_\infty(K) = \cup \mu_n(K) = (K^\times)_\text{torsion}$. Then $\mu_\infty (\QQ(\zeta_n)) = <(-1)^m \zeta_n>$ where $m \definedas n \mod 2$.

*Proof(Kummer):*
Take $\alpha = u/\bar u \in \ZZ[\zeta_p]$. Then $\sigma(\alpha)= \sigma(u)/\bar\sigma(u) \in \CC^d$.
By Kronecker, $\alpha = \pm \zeta_p^k$. Claim: sign = $\pm1$. Otherwise, $u^p = -\bar u^{p}$.

By other lemma, $\exists a \in \ZZ \suchthat u^p  = a\mod (p) \iff \bar u^p = a\mod (p) \iff a = -a \mod p$. But then $p\divides a$, so $p\divides u^p$ and $p$ is a unit. So $\abs{N(p)} > 1$.

**Corollary:** 
Every unit $u \in \ZZ[\zeta_p]^\times$ for $p > 2$ factors as $u = v \cdot \zeta_p^k$ where$v\in \ZZ[\zeta_p + \zeta_p^{-1}]^\times$ for some $0 \leq k \leq p$.

*Proof:* 
Know from proposition that $u/\bar u = \zeta_p^{k'}$, so find $0 \leq k \leq p$ such that $2k = k' \mod p$. Then $u \zeta^{-k} = \bar u \zeta_p^{k' - k} = \bar u \zeta_p^k$, so take $v=u$ and $\bar u \zeta_p^k = \bar v$.

Note that $\QQ(\zeta + \zeta^{-1})$ is totally real (see midterm!), so $\mathcal O_{\QQ(\zeta + \zeta^{-1})} = \ZZ[\zeta + \zeta^{-1}]$.

CM Field: $K$ over $K^+$ (s), $K^+$ over $\QQ$ totally real. Then $U_k = \mathcal O_K^\times$. So define $Q \definedas [U_k : \mu(k) U_{K^+}] \leq 2$. Why? Let $u \in U_k$, then take a complex embedding $\sigma(u/\bar u) \in \CC^1$. Then consider $U_k \into \mu(K)/ \mu(K)^2 \cong \ZZ/2\ZZ$ where$u \mapsto [u/\bar u]$, which is a homomorphism. The isomorphism follows from $\mu(K)$ being finite and cyclic.

Then $\ker \phi = \mu(K) U_{K^+}$. The LTR inclusion is from $u / \bar u = \zeta^2$, then $u \zeta^{-1} = \bar u \zeta = \overline{u\zeta^{-1}}$.

Can show $Q=1$ for $K = \QQ(\zeta_n), n=p^r$ (i.e. $n$ is any prime power), and $Q=2$ when $n$ is not a prime power and $n\neq 2$. This uses the fact that $1 - \zeta_n$ is a unit when $n\neq p^r$.