The Weak Mordell-Weil Theorem

Let A be an abelian variety over a field K. A basic object to investigate is the group A(K). Let K 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 A(K) 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 A(K) is finite as can bee seen by looking at some examples of elliptic curves but it is true that A(K) 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: A(K)/nA(K) is finite for any integer n.
  • Descent using a Height Function: Deduce the full theorem from the above using a measure of size on the points of A(K).

I will focus on the first part in this section and prove it in a motivated (but sophisticated) fashion. This proof will also differ from the standard proofs in trading in for classical algebraic number theory  results (finiteness of class group and finite generation of unit group) for class field theory. This will greatly simplify the second half of the proof.

I will make free use of 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 K is a number field and A is an abelian variety over it. I will denote by A[n] the kernel of the multiplication by n map on A. I will often identify A,A[n] with the \overline K points of A,A[n] respectively. Also denote by G_K the Galois group of \overline K over K.

Now, we can consider the exact sequence of G_K modules:

0 \to A[n](\overline K) \to A(\overline K) \xrightarrow{\times n} A(\overline K) \to 0

Taking Galois invariants, we get the long exact sequence:

0\to A[n](K) \to A(K) \xrightarrow{\times n} A(K) \to H^1(G_K, A[n](\overline K)) \to \dots

and truncating the series, we have an injection:

0 \to A(K)/nA(K) \to H^1(G_K,A[n](\overline K)).

So, we now see that to show that A(K)/nA(K) is finite, we simply need to show that H^1(G_K, A[n](\overline K)) is finite. Unfortunately, this is not true! However, we will show that A(K)/nA(K) 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 S be a finite set of primes of K and G_{K,S} the Galois group of the extension of K that is unramified away from S. In other words, for a prime \mathfrak p not in S with inertia group I_{\mathfrak p} \subset G_K, I_{\mathfrak p} is in the kernel of the quotient map G_K \to G_{K,S} and this characterizes G_{K,S}.

We want to show that there is a finite set of primes S such that the image of A(K)/nA(K) lands in H^1(G_{K,S}, A[n](\overline K)) (Note that this is a canonical subgroup of H^1(G_{K}, A[n](\overline K)) 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 x \in A(K), the boundary map takes x to a cocycle f_x in the following way:

Pick a y \in A(\overline K) such that ny=x and define f_x(g) = gy - y. One can check that this is independent of the choice of y and that it is indeed a cocycle.

Now we see that for the image to land in H^1(G_{K,S}, A[n](\overline K)), we need to be able to find, for each x \in A(K), a y \in A(K), ny = x such that the inertia groups at primes away from S fix y. Equivalently, we want y to lie in an extension of K unramified away from S. However, this follows immediately from the following observation:

Away from S, the multiplication by n map on a model of A over \mathcal O_{K,v} is unramified. This can be checked over the special and generic fiber of \mathcal O_{K,v} since the multiplication by n map acts on the tangent spaces by literally multiplying by n (which is a unit away from S).

In particular, the fiber of a point P in A(K)  lies in an extension unramified outside of S and the fiber consists precisely of points Q such that nQ = P. This is precisely what we were required to prove.

We are done with harder part and only need to show:


Finiteness of H^1(G_{K,S}, A[n](\overline K)):


Since there are only finitely many points in A[n](K) and A[n] is etale over a residue field away from S, we can find a finite, unramified away from S, extension L of K that splits H^1(G_{K,S}, A[n](\overline K)). That is, H^1(G_{L,S}, A[n](\overline K)) = Hom(G_{L,S}, A[n](\overline K)). Moreover, this is finite by Class field theory since we are looking for abelian extensions with bounded degree and ramification only at a finite number of predetermined points (and hence bounded ramification at these points).

Now, the rest of the proof follows easily on considering the inflation-restriction sequence:

 0 \to  H^1(G_{L,S}, A[n](\overline K)) \to H^1(G_{K,S}, A[n](\overline K)) \to H^1(Gal(L/K), A[n](\overline K))

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.



Closing Thoughts:


I find that this proof clarifies the role of class field theory versus the Elliptic Curve machinery. Furthermore, the usual proofs proceed by using both the finiteness of the class group and finite generation of the unit group in an essential way. This makes the analysis fairly complicated but it does have the advantage of not using any class field theory.

The proof here instead trades some complexity by using class field theory instead of more elementary algebraic number theory. This makes the final part of the proof almost trivial and the idea clearer.




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