# Monday January 13th

## Every Abelian Group is a Class Group

Theorem (Claborn - Leedham - Green - Clark)
: Any commutative group is the class group of some Dedekind domain.

Also see: partial re-proof by Rosen that uses elliptic curves.
This theorem: mostly a proof in commutative algebra, see end of Pete's commutative algebra notes.

## Proof Sketch
Let $E/k$ be an elliptic curve over a field.

### Step 1

Note that $\mathrm{End}_k(E) \cong_\ZZ  \ZZ^{a(E)}$ where $a(E) \in \theset{1,2,4}$.

> Could be $\ZZ$ as a $\ZZ\dash$module, could be an order in the imaginary quadratic field (e.g. a quaternion algebra)

There is a short exact sequence $$0 \to E(k) \to E(k(E)) \to \mathrm{End}_K(E) \to 0.$$
This splits because (as seen above), the RHS term is free and thus projective.
So $$E / k(E) \cong E(k) \oplus \ZZ^{a(E)}.$$

Note that $k(E)$ is an extension of $E_k$ to $E_{k(E)}$ the field of rational functions over $k$? (function field).
To simplify, take $a(E) = 1$ and $E(k) = \theset{0}$.

> Taking $k=\QQ$, this happens (probably, asymptotically) half of the time.
> It's easy to write down an elliptic curve that satisfies these conditions

Then $E/k(E) \cong \ZZ$.

Now pass to the field of rational functions over this field, taking $E(~k(E) ~(E/k(E))~)$.
Then 
$$
k^2(E) \definedas k(E) (E/k(E))
$$ 
and inductively define $k^n(E)$ by passing to function fields.
So $E(k^n (E)) \cong \ZZ^n$.

So we can construct elliptic curves that have any free commutative group as their Mordell-Weil group.

### Step 2

Loosely speaking, we'll iterate this process transfinitely. 
Then for any set $S$, there exists a field $k$ and an elliptic curve $E/k$ such that $E(k) \cong \bigoplus_S \ZZ$.
We now want to introduce a process that allows passing to quotients.
And $R \definedas k[E^0]$ is the affine coordinate ring of ?, remove the point at infinity (?).

### Step 3

Let $R$ be a Dedekind domain.
Note it has a fraction field with a certain ideal class group.
Let $W \subset \maxspec(R)$, then $$R^W \definedas \intersect_{\pr \in \maxspec R\setminus W} R_\pr.$$
Then $R^W$ is Dedekind (and every overring of a Dedekind domain is of this form)
and $\maxspec(R^W) = \maxspec(R\setminus W)$.

Then $$\pic R^W = \pic R / \generators{ [\pr] \suchthat \pr\in W }.$$
Note that if $(A, +)$ is a commutative group, writing $A = \bigoplus_S \ZZ/H$, we have a Dedekind domain $R = k[E^0]$ such that $\pic R = \bigoplus_S \ZZ$.

> Note: $\pic R$ is the class group.

Definition (Replete)
: A Dedekind domain $R$ is **replete** iff every element of the class group $\pic R$ is the class group $[\pr]$ of some ideal $\pr \in \maxspec(R)$.

> Is every ideal class the class of a prime ideal? 
> For $k$ a field, $R = \ZZ_k$. 
> This follows from Chebotarev Density (most important theorem for arithmetic geometers!)

Definition (Weakly Replete)
: A Dedekind domain $R$ is **weakly replete** iff every subgroup $H \subset \pic R$ is generated by classes of prime ideals.

Exercise (Easy)
: $K[E^0]$ is weakly replete, and an easy application of Riemann-Roch shows that if $0\neq p \in E(k) = \pic k[E^0]$, then $[p] \in \pic k[E^0]$ is generated by a prime ideal.

Note: most applications of Riemann-Roch to elliptic curves are easy!
In this case, it gives you an identification $E \cong \pic^1(E)$.

So there exists a subset $W \subset \maxspec k[E^0]$ such that $\generators{[p] \suchthat p\in W} = H$.
Then 
$$
\pic k[E^0]^W \cong \bigoplus_S \ZZ/H \cong A
.$$

$\qed$

Note that Dedekind domains don't have to be replete or even weakly replete.
The class group of a Dedekind domain could be $\ZZ$, and the class of every prime ideal could be $1 \in \ZZ$

Proof (Claborn) 
: Start with an arbitrary Dedekind domain $R$ and attach one that's replete.

	Can ask for a similar result for abelian varieties, there are conjectures here, few clear results.
	Need to get $\ZZ/(m) \cross \ZZ/(n)$, since these occur as Mordell-Weil groups.
	Take a modular curve and a generic point.
	Look at universal elliptic curves over elliptic curves and take their Mordell-Weil groups (?)

	If $k$ is algebraically closed and $\ch k = p$, can't have $\ZZ(p) \cross \ZZ/(p)$.
	Consider the $p\dash$primary torsion $E_k[p^\infty]$. 
	It is zero iff $E$ is supersingular (no points of order $p$).
	It is $\QQ_p/\ZZ_p = \lim_{\rightarrow_n} \ZZ/(p^n)$ iff $E$ is ordinary.

	> Can sometimes reduce to cases where $k=\CC$ and do things analytically.

## Mordell-Weil

Theorem (Mordell-Weil)
: 	Let $k$ be a global field (extension of $\QQ$ or function field over $\FF_p$) and $E/k$ and elliptic curve.
	Then $E(k) \cong \ZZ^r \oplus T$ (by classification of abelian groups) where $T$ is finite, and $T \cong \ZZ/(m) \oplus \ZZ/(n)$ for $m\divides n$.
	So $T$ is generated by at most two elements.

Proof (3 steps)
: **Step 1:**
	Weak Mordell-Weil theorem.

	Take any $n\geq 2$ and $\ch k$ not dividing $n$.
	Show that $E(k)/n E(k)$ is finite.

	**Step 2:**
	Define a height function $h: E(k) \to \RR$ satisfying 3 properties (see next time).
	This is approximately a quadratic form.

	> Decompose at places of a number field, see Number Theory II.

	**Step 3:**
	For any commutative group $A$, there is a notion of a height function $$h: A \to \RR.$$
	Show the Height Descent Theorem: if $A$ admits a height function and $A/nA$ is finite for some $n\geq 2$, then $A$ is finitely generated.

	> Also how you'd prove this theorem for abelian varieties, more difficulty defining $h$.