# Wednesday January 15th

## Proving the Mordell-Weil Theorem

Recall that we're trying to prove the Mordell-Weil theorem.
Let $K$ be a global field, so it's the field of functions over some nice curve.
Then the Mordell-Weil group $E(K)$ is finitely generated.

**Step 1:** 
The weak Mordell-Weil theorem for all $n\geq 2$ with $\ch k$ not dividing $n$, $E(k) / n E(k)$ is finite.

**Step 2:**
Construction of a height function $h: E(K) \to \RR$ that is "trying" to be a quadratic form.

**Step 3 (Today):**
The Height Descent Theorem, i.e. if $(A, +)$ is a commutative group such that $A/nA$ is finite for some $n\geq 2$ and it admits a heigh function $h: A\to \RR$, then $A$ is finitely generated.

### Step 3: Proving the Height Descent Theorem

*Question:*
What does the weak Mordell-Weil group $E(K)/ nE(K)$ tell us about $E(K)$?

Note that we'll inject this into a larger group, which we'll show is finite, but this isn't great for learning about the size.

Example
: Consider $E/\CC$, then $E(\CC) = S^1 \cross S^1$ and $E(\CC)/nE(\CC) = 0$, so the map $x\to nx$ is a surjective map and $E(K)$ is $n\dash$divisible here.
  In general, whenever $K = \bar K$ is algebraically closed, then $x \mapsto nx$ is again surjective and the weak Mordell-Weil group is trivial.
  So knowing this is small doesn't tell us much about $E(K)$ at all.

Example
:   For $E/\RR$, $E(\RR)$ is either $S^1$ (cubic with one real root, $\Delta = 0$) or $S^1 \cross \ZZ/(2)$ (cubic with three real roots, $\Delta > 0$) are the two possible group structure.

    Then

    \begin{align*}
    ? = 
    \begin{cases}
    0 & n \text{ odd } \\
    0 & n \text{ even and } \Delta < 0 \\
    \ZZ/(2) & n \text{ even and } \Delta > 0
    \end{cases}
    .\end{align*}

Example
: Consider $E/\QQ_p$, then for all $\ell \gg 0$ $E(\QQ_p) \mapsvia{[\ell]} E(\QQ_p)$ with $E(\QQ_p)/\ell E(\QQ_p) = 0$ while $E(\QQ_p) / p E(\QQ_p)$ is not zero.

Note: here is an example of a Boolean space, that ends up being homeomorphic to a Cantor set.

Suppose $E(K)$ is finitely generated, so $E(K) \cong \ZZ^r \oplus T$ with $T$ finite.
Then knowing $E(K)/ n E(K)$ gives an upper bound on $r$.

Example
:   Take $n=2$, then $E(K) / nE(K) \cong (\ZZ/(2))^s$ for some $s\in \NN$.
    Then 
    $$
    (\ZZ^r \oplus T) / 2(\ZZ^r \oplus T) \cong (\ZZ/(2))^r \oplus T/2T
    $$ 
    for $r\leq s$.
    Then either

    - $r = 2$ and $E(K[2]) = (0)$.
    - $r=1$ and $E(K[2]) \cong \ZZ/(2)$,
    - $r = 0$ and $E(K[2]) \cong (\ZZ/(2))^2$.

Note that we don't need the Mordell-Weil theorem to compute the torsion subgroups of $E(k)$.
It is often easier to compute these directly.
For all non-archimedean places $v$ of $K$, $E(K_v)[\text{tors}]$ is finite (see Silverman?) and embeds into a number of finite things.

To compute $E(K_v)[\text{tors}]$,

1. Find $N \in \ZZ^+$ such that $E(k)[\text{tors}] \subset E[N]$.
  
  - Choose 2 different places $v_0, v_1$ of good reduction (from Weierstrass equation) with different residue characteristics $\ell_1 \neq \ell_2$
  
  - Consider the map $E(K_{v_i})[\text{tors}] \to E(\FF_{v_i})$
  
  - The kernel is a finite $p_i\dash$primary group.
  
  - Comes down to torsion and formal groups, see first course.

2. Compute $E[N](K)$ (several algorithms, just checking for rational points on a zero-dimensional variety?)

> See division polynomials, can check for roots of polynomials over any global field.
> Easy to check for rational points on finite fields.

Suppose $E(K) \cong \ZZ^r \oplus T$ is finitely generated and we know $E(K) / nE(K)$ for some $n$ and we know $T$.
Then we explicitly know $r$.

> See Tate Shafarevich group -- important! But difficult, a piece of information that helps compute the rank (?).

Definition (Height Functions)
: Fix $n\geq 2$. 
  An $n\dash$**height function** on $(A, +)$ is a map $h: A\to \RR$ satisfying

  1. For all $R\geq 0$, the set $h\inv(-\infty, R)$ is finite.
  
  2. For all $Q\in A$, there exists a $C_2 = C_2(A, Q)$ such that for all $P \in A$, 
  $$
  h(P + Q) \leq 2h(P) + C_2
  (?).$$ 
  
  3. There exists a $C_3 = C_3(A, n)$ such that for all $P \in A$, 
  $$
  h(nP) \geq n^2 h(P) - C_3
  $$
  
Note: (3) would be an equality for an honest quadratic function, so this deviates in a controlled way.

Theorem (Height Descent)
: Let $(A, +)$ be a commutative group with an $n\dash$height function $h: (A, +) \to \RR$.
  If $A/nA$ is finite, then $A$ is finitely generated.

Proof
:   Let $r$ be the size of $A/nA$.
    Choose coset representatives $Q_1, \cdots, Q_r$ of $nA$ in $A$.
    Let $p\in A$ and define a sequence $\theset{P_k}_{k=0}^\infty$ in $A$ by $P_0 = P$ and for $k\geq 1$, choose $P_k$ such that $P_{k-1} = nP_k + Q_{i_k}$.
    Then for all $k\in \ZZ^+$, it's true that $P = n^k P_k + \sum_{j=1}^k n^{j-1} Q_{i_j}$.

    Claim
    : There exists a constant $c > 0$ depending only on $A, n$ such that for all $P \in A$, there exists a $K = K(P$ such that for all $k \geq K$, we have $h(P_k) \leq 0$.

    Note that this is sufficient -- if so, $A$ is generated by $\theset{Q_1, \cdots, Q_r} \union h\inv((-\infty, C])$, which are both finite.

Next time: proof of claim.

> Note: similar setup goes through for abelian varieties, see Néron-Tate height canonical height, which yields an honest "quadratic form".