# Monday January 27th

## Weak Mordell-Weil: Finishing the Proof

Recall that $n\geq 2$, $K$ a field with $\ch K$ not dividing $p$, $E/K$ an elliptic curve, $R$ a Dedekind domain with fraction field $K(E[n])$
such that $\pic R$ and $R\units$ are finitely generated.
Then $E(K) / nE(K)$ is finite.

We've shown:

- May take $K = K(E[n])$, so $K$ contains the $n\dash$torsion and all $n$th roots of unity $\mu_n \in K$.
- $R$ a PID with fraction field $K$,
- $E(K) / n E(K)$ is finite iff $L\definedas K([n]\inv E(K))$ is a finite degree extension over $K$.
- $L/K$ is abelian of exponent dividing $n$
- $L/K$ is only ramified over $\pr \in \spec(R)$ if $p\divides n$ or $E$ has bad reduction at $\pr$ (so bad reduction at only finitely many primes).

Last step of the proof: 

### Applying Kummer theory to show $L/K$ is finite

Lemma
: Let $K$ be a field with a discrete valuation $v: K \surjects \ZZ \union \theset{\infty}$ such that $v(n) = 0$.
  When are Kummer extensions unramified?
  Every Kummer extension $L$ is a compositum $L \definedas K(a^{1/n})$.
  Then $L/K$ is unramified at $v$ iff $n$ divides $v(a)$.

Proof (of Lemma)
:  \hfill

    $\implies$: 
    We can extend $v$ to $L$, and $v$ is unramified iff $v(L\units) = \ZZ$.
    Noting that $v(a^{1/n}) = \frac 1 n v(a)$, if we suppose $v(L\units) = \ZZ$, we're done.

    $\impliedby$:

    > Can check being unramified by passing to completion.

    Pass from $K$ to the completion $K_v$ and let $\pi$ be a uniformizer, so $v(\pi) = 1$.
    Then define $a' = a/\pi^{v(a)}$, so $v(a') = v(a) - v(a) = 0$ and thus $a \in R_v\units$.
    Writing 
    $$
    a/a' = \pi^{v(a)} \in (K\units)^n
    ,$$ 
    so $K(a^{1/n}) = K((a')^{1/n})$.
    So we reduce to the case $v(a) = 0$.

    Since $R_v$ is integrally closed, we have $x\in K_v^{\times n} \iff x\in R_v^{\times n}$, so there is an isomorphism

    \begin{align*}
    R_v\units / R_v^{\times n} \cong k_v\units / k_v^{\times n}
    \end{align*}

    which follows from Hensel's Lemma.

    Thus $K_v(a^{1/n})/K$ is unramified.

    > Standard argument from NT II.


Now let $S\subset \spec R$ be the set 
$$
S = \theset{\pr \in \spec R \suchthat \pr \divides n \text{ or } E  \text{ has bad reduction at } \pr }
,$$ 
and 
$$
T_S =  \theset{ a\in K\units / K^{\times n} \suchthat \forall \pr\in \spec(R\setminus S),~ n \divides \ord_p(a)}
.$$

We will be done if $T_S$ is finite.

Claim
: There exists a surjective group homomorphism $\psi: R\units \to T_S$.

Then since $R\units$ is finitely generated, so is its image, but if $T_S = T_S[n]$ is an $n\dash$torsion group, this forces $T_S$ to be finite.


\begin{tikzcd}
R\units \arrow[rr, hook] \arrow[rrrrdd] \arrow[rrrr, "\psi", bend left] &  & K\units \arrow[rr, hook'] &  & K\units/K^{\times n} \\
                                                                        &  &                           &  & \\
                                                                        &  &                           &  & T_S \arrow[uu, hook]
\end{tikzcd}


Take $\bar a \in T_s$ and lift it to $a\in K\units$, and consider $(a)$ as a fractional $R_S$ ideal; it is an $n$th power.
So write $(a) = I^n$ for $I$ another fractional $R_S$ ideal.

> Really only need to assume that the class group of $R_S$ does not have $n\dash$torsion.

Since $R_S$ is a PID, $I = (b)$, so $(a) = (b)^n = (b^n)$ so there is some $u\in R_S\units$ with $a = ub^n$.
But then 
$$
\bar a = \psi(a) = \psi(u) \psi(b^n)
 = \psi(u)
 ,$$ 
 so $\psi$ is surjective.

This finishes the proof of weak Mordell-Weil.
$\qed$

Remark
: We could weaken the assumptions to $R$ with fraction field $K(E[n])$ such that $(\pic R)[n]$ is finite
and $R\units / R^{\times n}$ is finite.

Exercise
: Let $k = \bar k$ and $C/k$ be a "nice" affine curve.
  Let $R$ be the affine coordinate ring of $C$, i.e. $R = k[C]$, and $K = k(C)$ with $\ch(k)$ not dividing $n$.
  Show that if $E/K$ is an elliptic curve then $E(K) / n E(K)$ is finite.

$n\dash$torsion on an abelian group is finite. 
Note that the Mordell-Weil group here may not itself be finite!
Also note that proof almost goes through for abelian varieties, since we didn't use anything particular to elliptic curves.

## Height Functions

Definition (Product Formula Field)
: A **product formula field** is a triple $(K, \Sigma_K, A)$ where $K$ is a field, $\Sigma_K$ is a set of places on $K$, and $A$ is a set of normalizing constants.

Definition (Place)
: A **place** $v$ on a field $K$ is an equivalence class of absolute values $\abs{\wait}: K \to [-, \infty)\subset \RR$, where e.g. the equivalence is given by absolute values inducing the same topology.

Here we don't mind if raising to some power invalidates the triangle inequality, as long as raising to *some* power preserves it.

> See ultrametric, Archimedean.

$\Sigma_K$ is **a** set of equivalence classes of absolute values (not necessarily all) such that 

1. **(PF1)**
  $$
  \Sigma_K^{\text{Arch}}  \subset \Sigma_K
  .$$

2. **(PF2)**
  For all $x\in K\units$, 
  $$
  \# \theset{ v\in \Sigma_K  \suchthat \abs{x}_v \neq 1} < \infty.
  $$ 

    A non-Archimedean valuation in $\Sigma_K$ satisfies $\abs{x}_v = c_v^{-v(x)}$ for some rank 1 valuation $v: K \to \RR\union\infty$ with $c_v > 1$, and $A$ is that data of a choice of $c_v$ for each non-Archimedean $v$.

3. **(PF3)**
  For all $x\in K\units$, we have 
  $$
  \prod_{v\in \Sigma_K} \abs{x}_v = 1
  .$$

Note that we don't care about the $c_v$s necessarily, but we do need to choose them to preserve this product formula.

All global fields are PFFs, so we'll use this to define height functions on projective spaces.