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

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

Then we have an exact sequence

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

From this we obtain a short exact sequence

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

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

and the sequence (12.2.1) induces an inclusion

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

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

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

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

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

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

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

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

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. $\qedsymbol$