Proof of the Weak Mordell-Weil Theorem

Suppose $ E$ is an elliptic curve over a number field $ K$, and fix a positive integer $ n$. Just as with number fields, we have an exact sequence

$\displaystyle 0 \to E[n] \to E \xrightarrow{n} E \to 0.

Then we have an exact sequence

$\displaystyle 0 \to E[n](K) \to E(K) \xrightarrow{n} E(K) \to \H^1(K,E[n]) \to \H^1(K,E)[n] \to 0.

From this we obtain a short exact sequence

$\displaystyle 0 \to E(K)/n E(K) \to \H^1(K,E[n]) \to \H^1(K,E)[n] \to 0.$ (12.1)

Now assume, in analogy with Section 12.1, that $ E[n]\subset E(K)$, i.e., all $ n$-torsion points are defined over $ K$. Then

$\displaystyle \H^1(K,E[n]) = \Hom (\Gal (\overline{K}/K),(\mathbf{Z}/n\mathbf{Z})^2),$

and the sequence (12.2.1) induces an inclusion

$\displaystyle E(K)/n E(K) \hookrightarrow \Hom (\Gal (\overline{K}/K),(\mathbf{Z}/n\mathbf{Z})^2).$ (12.2)

Explicitly, this homomorphism sends a point $ P$ to the homomorphism defined as follows: Choose $ Q \in E(\overline{K})$ such that $ nQ = P$; then send each $ \sigma \in \Gal (\overline{K}/K)$ to $ \sigma(Q)-Q\in E[n]\cong
(\mathbf{Z}/n\mathbf{Z})^2$. Given a point $ P\in E(K)$, we obtain a homomorphism $ \varphi : \Gal (\overline{K}/K) \to (\mathbf{Z}/n\mathbf{Z})^2$, whose kernel defines an abelian extension $ L$ of $ K$ that has exponent $ n$. The amazing fact is that $ L$ can be ramified at most at the primes of bad reduction for $ E$ and the primes that divide $ n$. Thus we can apply theorem 12.1.1 to see that there are only finitely many such $ L$.

Theorem 12.2.1   If $ P\in E(K)$ is a point, then the field $ L$ obtained by adjoining to $ K$ all coordinates of all choices of $ Q=\frac{1}{n}P$ is unramified outside $ n$ and the primes of bad reduction for $ E$.

Proof. [Sketch of Proof] First one proves that if $ \mathfrak{p}\nmid n$ is a prime of good reduction for $ E$, then the natural reduction map $ \pi: E(K)[n] \to
\tilde{E}(\O_K/\mathfrak{p})$ is injective. The argument that $ \pi$ is injective uses ``formal groups'', whose development is outside the scope of this course. Next, as above, $ \sigma(Q)-Q \in E(K)[n]$ for all $ \sigma \in \Gal (\overline{K}/K)$. Let $ I_\mathfrak{p}\subset \Gal (L/K)$ be the inertia group at $ \mathfrak{p}$. Then by definition of interia group, $ I_\mathfrak{p}$ acts trivially on $ \tilde{E}(\O_K/\mathfrak{p})$. Thus for each $ \sigma \in
I_\mathfrak{p}$ we have

$\displaystyle \pi(\sigma(Q) - Q) = \sigma(\pi(Q)) - \pi(Q) = \pi(Q) - \pi(Q) = 0.

Since $ \pi$ is injective, it follows that $ \sigma(Q) = Q$ for $ \sigma \in
I_\mathfrak{p}$, i.e., that $ Q$ is fixed under all $ I_\mathfrak{p}$. This means that the subfield of $ L$ generated by the coordinates of $ Q$ is unramified at $ \mathfrak{p}$. Repeating this argument with all choices of $ Q$ implies that $ L$ is unramified at  $ \mathfrak{p}$. $ \qedsymbol$

Theorem 12.2.2 (Weak Mordell-Weil)   Let $ E$ be an elliptic curve over a number field $ K$, and let $ n$ be any positive integer. Then $ E(K)/nE(K)$ is finitely generated.

Proof. First suppose all elements of $ E[n]$ have coordinates in $ K$. Then the homomorphism (12.2.2) provides an injection of $ E(K)/nE(K)$ into

$\displaystyle \Hom (\Gal (\overline{K}/K), (\mathbf{Z}/n\mathbf{Z})^2).$

By Theorem 12.2.1, the image consists of homomorphisms whose kernels cut out an abelian extension of $ K$ unramified outside $ n$ and primes of bad reduction for $ E$. Since this is a finite set of primes, Theorem 12.1.1 implies that the homomorphisms all factor through a finite quotient $ \Gal (L/K)$ of $ \Gal (\overline{\mathbf{Q}}/K)$. Thus there can be only finitely many such homomorphisms, so the image of $ E(K)/nE(K)$ is finite. Thus $ E(K)/nE(K)$ itself is finite, which proves the theorem in this case.

Next suppose $ E$ is an elliptic curve over a number field, but do not make the hypothesis that the elements of $ E[n]$ have coordinates in $ K$. Since the group $ E[n](\mathbf{C})$ is finite and its elements are defined over  $ \overline{\mathbf{Q}}$, the extension $ L$ of $ K$ got by adjoining to $ K$ all coordinates of elements of $ E[n](\mathbf{C})$ is a finite extension. It is also Galois, as we saw when constructing Galois representations attached to elliptic curves. By Proposition 11.3.1, we have an exact sequence

$\displaystyle 0 \to \H^1(L/K, E[n](L))\to \H^1(K,E[n])\to \H^1(L,E[n]).

The kernel of the restriction map $ \H^1(K,E[n])\to \H^1(L,E[n])$ is finite, since it is isomorphic to the finite group cohomology group $ \H^1(L/K, E[n](L))$. By the argument of the previous paragraph, the image of $ E(K)/nE(K)$ in $ \H^1(L,E[n])$ under

$\displaystyle E(K)/n E(K) \hookrightarrow \H^1(K,E[n]) \xrightarrow{\res } \H^1(L,E[n])

is finite, since it is contained in the image of $ E(L)/n E(L)$. Thus $ E(K)/nE(K)$ is finite, since we just proved the kernel of $ \res $ is finite. $ \qedsymbol$

William Stein 2012-09-24