Let be an abelian variety over a field . A basic object to investigate is the group $A(K)$. Let be a number field. In light of the Birch and Swinnerton-Dyer conjecture and it’s relation to the class number formula, one should think of as the analog of the class group of a global field.
Thus, one might conjecture some finiteness properties of this group. It is not true that is finite as can bee seen by looking at some examples of elliptic curves but it is true that is finitely generated as an abelian group and this is the content of the Mordell-Weil Theorem.
The proof is usually broken up into two parts:
- Weak Mordell-Weil Theorem: is finite for any integer .
- Descent using a Height Function: Deduce the full theorem from the above using a measure of size on the points of .
I will focus on the first part in this section and prove it in a motivated (but sophisticated) fashion. This proof is largely the same as the one given in Milne’s Elliptic Curves notes but I find the current presentation far easier to understand.
I will free use general theory about Abelian Varieties, Algebraic geometry and Galois Cohomology. The point of this post is not to fill in the details but to show a framework that makes the proof seem natural.
The Kummer Sequence:
Recall that is a number field and is an abelian variety over it. I will denote by the kernel of the multiplication by map on . I will often identify with the points of respectively. Also denote by the Galois group of over .
Now, we can consider the exact sequence of modules:
Taking Galois invariants, we get the long exact sequence:
and truncating the series, we have an injection:
So, we now see that to show that is finite, we simply need to show that is finite. Unfortunately, this is not true! However, we will show that actually lands in a subgroup of the cohomology group and this subgroup can easily be shown to be finite.
Ramification of Galois Cohomology groups:
A little more precisely, let be a finite set of primes of and the Galois group of the extension of that is unramified away from . In other words, for a prime not in with inertia group , is in the kernel of the quotient map and this characterizes .
We want to show that there is a finite set of primes such that the image of lands in (Note that this is a canonical subgroup of by the restriction map. ) This is by far the hardest part of the proof:
Controlling the image of the boundary map:
To do this, let us examine the first boundary map in the Kummer sequence more closely. For an element , the boundary map takes to a cocycle in the following way:
Pick a such that and define . One can check that this is independent of the choice of and that it is indeed a cocycle.
Now we see that for the image to land in , we need to be able to find, for each , a such that the inertia groups at primes away from fix .
We will pick to be the set of places of where either has bad reduction or lies above . Now, fix a valuation not in and let denote the maximal unramified extension of the local field . If the residue field of at is , then the residue field of is .
Since is henselian, the reduction map is surjective. Furthermore, we have chosen precisely so that the reduction map is a bijection on the – torsion and so the kernel of the reduction map has no – torsion. (One could also see this by looking at formal groups).
That is, we have the following diagram:
By what we have said above, the first vertical map is injective while the other two are surjective. The following easy lemma is all we need to complete the proof:
Lemma: Let with reduction . Then if and only if .
Proof: One direction is trivial. Assume then that there exists such that and lift to some element of $A(\tilde K_v)$. Then, maps to under the reduction map and so lies in the kernel . However, this is a divisible group and so we can find some such that or in other words, .
Therefore, we can define our cocycle by and the inertia group at maps to by construction. Since was arbitrary, we are done with this part of the proof. We only need to show
Finiteness of :
Since there are only finitely many points in and is etale over a residue field away from , we can find a finite, unramified away from $S$, extension of that splits . That is, . Moreover, this is finite by Class field theory since we are looking for abelian extensions with bounded degree (and hence bounded ramification).
Now, the rest of the proof follows easily on considering the inflation-restriction sequence:
The first set is finite as discussed above while the third group is finite since both the group and the module are finite. This establishes finiteness in the middle as required.
This proof is, in spirit, exactly the same as the one in standard references like Milne that proceed by bounding the Selmer group. However, they tend to use very little Cohomological machinery and do everything by hand. I find the presentation here to be much easier to understand and haven’t seen it elsewhere.