# Friday January 24th

## Stronger Weak Mordell-Weil

Let $K$ be a field, $n\geq 2$ with $\ch k$ not dividing $n$, and $A/K$ a commutative algebraic group (includes abelian varieties, additive/multiplicative groups, etc).
Assume that $K$ contains all $n$th roots of unity (we showed that this can be assumed).

Let $$L \definedas K([n] \inv A(K)) = K(\theset{Q \in A(K\sep) \suchthat [n] Q \in A(K)}).$$
We've shown that $L/K$ is Galois, and moreover abelian of exponent dividing $n$ and $L/K$ is finite iff $A(K) / n A(K)$ is finite.

> Take $K = \QQ$, then $\QQ[p]$ for every $p$ gives infinite extensions.

We want to show that if $A=E$ is an elliptic curve (or an abelian variety) and $K$ is the fraction field of some Dedekind domain $R$ with some finiteness condition on $\pic(R)$ and $R\units$, then $L/K$ is finite.

### Step 3 

Let $\pr \in \maxspec(R)$ with $\pr$ not dividing $n$ such that $E$ has good reduction at $\pr$.

> Take an $R\dash$integral Weierstrass equation $W/R$ for $E$, then the discriminant satisfies $\Delta(W) \neq 0$.
> So just exclude the (finitely many) primes where $\pr \divides \Delta(W)$.
> For abelian varieties, reducing the equations mod $p$ can result in singularities for only finitely many $p$.

Then $L/K$ is unramified at $\pr$ (i.e. it's ramified at only finitely many primes).

*Proof:*
We have 
$$
L = \prod_{Q \suchthat n[Q] = P \in E(K)} L(Q)
$$
as a compositum of extensions, so it's enough to show that
$$
L_p \definedas \prod_{[n] Q = p, ~p \in E(K)} K(Q)
$$ 
is unramified over $K$.

Take integral closures:

![Image](figures/2020-01-24-12:34.png)

Take the inertia group 
$$
I \definedas I(\pr \mid p) = \theset{\sigma \in g_{L_p/k} \sigma(\pr) = \pr, \sigma\actson S_p/\pr = \id} \in \aut(L_p/K)
.$$

We want to show that $\forall \sigma \in I$, $\sigma(Q) = Q$.
We have 
$$
0 = \sigma(P) - P = \sigma([n]Q) = [n]Q = [n](\sigma Q - Q)
$$ 
and thus $\sigma Q - Q \in E[n] = E[n](K)$.

We now introduce the reduction map

\begin{align*}
r: E(L_p) \to \tilde E(S/\pr)
,\end{align*}

where we use the fact that we can complete at $\pr$ and then take a reduction to obtain a map $E(L_p) \to E(\hat L_p) \to \tilde E(S/p)$, which is where we use the fact that the reduction is good.

We know $\sigma Q - Q$ is $n\dash$torsion.
Then $r$ is a homomorphism, so $r(\sigma Q - Q) = r(\sigma Q) - r(Q) = 0$ by the definition of the inertia group.
So $\sigma Q - Q$ is an $n\dash$torsion point in the kernel of the reduction map.

Fact from elliptic curves I (Silverman Ch. 7), the only torsion in the kernel of the reduction map is $\ch (S/p)\dash$primary torsion.
Since $\ch(S/p)$ does not divide $n$ here, $\ker r = 0$, so $\sigma Q = Q$.

$\qed$

> In the case of abelian varieties: The kernel of a good reduction is a formal group law of $g$ dimensions.
> Reference: Prop 3.1, Clark and Xales (?) 2001.
> Here is where it doesn't work for $\GG_m$.
> Projective variety: clear denominators for the fraction field, now need to look at integral points?


### Step 4: Number Theory

Let $S \subset \maxspec(R)$ be the finite set of $p$ such that $p \divides n$ (using global characteristic assumption here) or $E$ has bad reduction at $p$.
Then take $L = K([n]\inv E(K))$.
We know $L/K$ is abelian of exponent dividing $n$, and is a compositum of extensions $L_Q$, each of bounded degree at most $n^2$, each unramified outside of $S$.

> $Q$ is an $n$ division point of something $k$ rational. Multiplication by $n$ has degree $n^2$, so $[K_Q : K] < n^2$.

Theorem (Hermike - Minkowski)
: Let $d \in \ZZ^+$, $K$ a number field, and $s \subset \maxspec \ZZ_\ell$ finite.
 Then $$\#\theset{\text{ $L/K$ of degree $d$ unramified outside of $S$}} < \infty.$$

> One of a few finiteness theorems in elliptic curves/NT, probably the hardest one.
> Galois extension like branched cover, too few preimages. Take surface with marked points (corresponding to prime) and take all branched covering spaces that fix the degree and only ramify at those points. I.e. take unramified coverings. Then $\pi_1$ is free and f.g., which is the analogy here.

*Proof:*
See Neukirch Ch. 3, 2.13, or Milne's ANT Theorem 8.42.
Use Hermite's theorem that any given $\Delta \in \ZZ$ can only be the discriminant $\Delta(K)$ for finitely many $K$.

If $K$ is a number field, then take $R = \ZZ_k$ and we're done.

Otherwise, let $R_s \definedas \intersect_{\pr \in \maxspec(R\setminus S)} R_{\pr}$.
Then $R_s$ is a Dedekind domain, $\maxspec R_s = \maxspec(R\setminus S)$.
Also $\pic(R_s)$ is a quotient $\pic(R) / \generators{[\pr] \suchthat \pr \in S}$.

Assume $\pic R$ is finitely generated, then so is $\pic R_s$.
If so, by enlarging $S$, we can make $R_s$ a PID.

> Note $R_s$ is not necessarily a localization, although it could be, so we expect the unit group to get bigger.

Theorem (Pete's CA Notes Lemma 23.4)
:  Let $R$ be a Dedekind domain and $R\subset T \subset K$.
Then

    a. $T\units/R\units$ is torsion-free.

    b. Let $\pr \in \maxspec ?$ and define $T \definedas \intersect_{q\neq \pr} R_q$. 
      TFAE:

      - $T\units / R\units = \ZZ$.
      - $T\units \supsetneq R\units$.

    c. $[p] \in \pic R[?]$.

So killing off primes either adds factors of $\ZZ$ or does nothing at all.

By the lemma, passing from $R$ to $R_s$ by killing finitely many maximal ideals, if $\Pic R$ and $R\units$ are both finitely generated, then they are still finitely generated for $R_s$.
Thus we can assume $R_s$ is a PID.

> Proof of theorem next time. Need ramification and taking $n$th roots in local fields.