# Wednesday January 22nd

## Stronger Weak Mordell Weil

Theorem (Stronger Weak Mordell-Weil)
:   Let $n\geq 2$, $k$ a field, $\ch(K)$ not dividing $n$, $E/k$ an elliptic curve $K \definedas K(E[n])$.

    Suppose there exists a Dedekind domain $R$ with fraction field $K$ such that

    - $\pic R[n]$ is finite
    - The cokernel of $x\mapsto nx$ is finite

    Then $E(K)/ n E(K)$ is finite.

Exercise
: Take $k = \bar k$, $C/k$ a "nice" affine curve, $R = k[C]$, $K = k(C)$, $\ch(k)$ not dividing $n$.
  Show that if $E/k$ is any elliptic curve, then the hypotheses of Stronger Weak Mordell-Weil hold, as does the conclusion, and in fact $E(K)$ need not be finitely generated.

> Uses ANT II.

### Step 1

Let $L/K$ be a Galois extension and consider 
$$
\iota: E(K)/n E(K) \to E(L)/ n E(L)
.$$
This is not injective in general, and in fact 
$$
\ker(\iota) = (E(K) \intersect nE(L))/ nE(K)
.$$

Proposition
: \hfill
    a. There exists a map $\ker \iota \to \maps(g_{L/K}, E[n])$, where $g_{L/K}$ is the Galois group of $L/K$.
  
    b. If $L/K$ is finite, then $\ker \iota$ is finite
  
So if $E(L) / nE(L)$ is finite, then $E(K) / nE(K)$ is finite.

Proof
: Let $p\in E(K) \intersect n E(K)$.
  Choose $Q_p \in E(L)$ (only determined up to an $n\dash$torsion point) such that $[n] Q_p = p$.
  For all $\sigma \in g_{L/K}$ we have 

  \begin{align*}
  [n] (\sigma(Q_p) - Q_p) = \sigma([n] Q_p) - [n] Q_p = \sigma(p) - p = 0
  .\end{align*}

  Thus $\sigma(Q_p) - Q_p \in E[n]$.

  > Note: this resembles a certain coboundary map in a cohomology theory.

  We then associate a map

  \begin{align*}
  \lambda_p : g_{L/K} \to E[n] \\
  \sigma \mapsto \sigma(Q_p) - Q_p
  .\end{align*}

  Suppose that for $p, p' \in E(K) \intersect n E(L)$, so $\lambda_p = \lambda_{p'}$.
  Then for all $\sigma \in g_{L/K}$, we have 

  \begin{align*}
  \sigma(Q_p - Q_{p'}) &= \sigma(Q_p) - Q_p - (\sigma(Q_{p'}) - Q_{p'} ) + (Q_p - Q_{p'}) \\
  &= \lambda_p(\sigma) - \lambda_{p'}(\sigma) + Q_p - Q_{p'} \\
  &= Q_p - Q_{p'}
  .\end{align*}

  So $Q_p - Q_{p'} \in E(K)$, and thus $p - p' = [n] (Q_p - Q_{p'}) \in nE(K)$.
  Thus $$\ker \iota = \frac{E(K) \intersect nE(L)}{nE(K)}.$$

From now on, we'll assume $K = K' = K(E[n])$.

### Step 2

Define the Kummer pairing.

Let $L = K\sep$ and take $p \in E[K]$ such that $[n] Q_p = p$, and define 

\begin{align*}
\lambda_p : g_k \leq \aut(K\sep/K) &\to E[n] \\
\sigma &\mapsto \sigma(Q_p) - Q_p
.\end{align*}

Note that since $E[n] = E[n](K)$, this no longer depends on the choice of $Q_p$.

Define a Kummer pairing

\begin{align*}
\kappa: E(k) \cross g_K &\to E[n] \\
(p, \sigma) &\mapsto \sigma Q_p - Q_p
.\end{align*}

Proposition
: This pairing satisfies the following properties:

  1. For all $p, q\in E(K)$ and $\sigma \in g_k$, $\kappa(p+q, \sigma) = \kappa(p, \sigma) + \kappa(q, \sigma)$
  2. For all $p \in E(K)$ and $\sigma, \xi \in g_k$, $\kappa(p, \sigma \xi) = \kappa(p, \sigma) + \kappa(p, \xi)$.
  3. For all $p\in E(K)$, we have $\kappa(p, \sigma) = 0$ for all $\sigma$ iff $p\in nE(K)$.
  4. For all $\sigma \in g_K$, we have $\kappa(p, \sigma) = 0$ for all $p\in E(K)$ iff $\sigma \in g_L$ where $L\definedas K([n]\inv E(K)) = K(Q_p/P \in E(K))$.
  
  So $\kappa$ induces group homomorphisms
  \begin{align*}
  g_K / g_L = g_{L/K} &\injects \hom(E(K) / nE(K), E[n]) \\
  E(K)/ nE(K) &\injects \hom(g_{L/K}, E[n])
  .\end{align*}

> Thus $E(K) / n E(K)$ is finite iff $g_{L/K}$ is finite iff $[L: K]$ is finite.

Thus $g_{L/K}$ is abelian of exponent dividing $n$.
So we can study this using Kummer theory and class field theory.

Proof (of 1)
: Can take $Q_{p + p'} = Q_p + Q_{p'}$, which is a fine choice, and then $\kappa(p + p', \sigma) = \kappa(p, \sigma) + \kappa(p', \sigma)$.

Proof (of 2)
: 
  \begin{align*}
  \kappa(p, \sigma \xi) 
  &= \sigma\xi Q_p - Q_p - \sigma \xi Q_p - \sigma Q_p + \sigma Q_p - Q_p \\
  &= \sigma(\xi Q_p - Q_p) + \kappa(p, \sigma) = \sigma(\kappa(p, \xi) + \kappa(p,\sigma)) \\
  &= \kappa(p, \xi) + \kappa(p, \sigma)
  .\end{align*}

Proof (of 3)
: For $\sigma \in g_K$, then $\kappa(E(K), \sigma) = (0)$ iff $\sigma Q_p - Q_p = 0$ for all $P \in E(K)$ iff $\sigma$ pointwise fixes $L$ iff $\sigma \in g_L$.

Exercise
: Replace $E/k$ with $A/k$ a commutative group scheme such that $[n]: A\to A$ is etale and $A[n]$ is finite.
  The proof goes through without modification if $\ch k$ doesn't divide $n$ and $A/k$ is an algebraic group, reduced, and of finite type.

Exercise
: Take $A = \GG_m$, the multiplicative group of $K$.
  Then $A(K) / nA(K) = K\units/K^{\cross n}$.
  Suppose $K$ contains $n$th roots of unity, then regular Kummer theory gives a map $K\units/K^{\cross n} \cong \hom(g_K, \mu_n)$ where $\mu_n$ are $n$th roots of unity.

  This says that $K\units/K^{\times n} = \chi(g_{L/K})$ where $L$ is the maximal abelian extension of $K$ of exponent dividing $n$, and $\chi(\wait) = \hom(\wait, \ZZ/n\ZZ)$, i.e. these are Pontryagin duals.

So far, works for any algebraic group, but we'll need properness later.