The Shafarevich-Tate Group and the Mordell-Weil Theorem

Definition 2.8 (Shafarevich-Tate Group)   The Shafarevich-Tate group of an elliptic curve $E$ over a number field $K$ is

\begin{displaymath}
{\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\...
...}}}(E/K) = \ker\left(\H^1(K,E) \to \prod_v \H^1(K_v,E)\right).
\end{displaymath}

For any positive integer $n$, we may thus add in a row to (2.2.2):

\begin{displaymath}
\xymatrix @=1.2em{
0 \ar[r]& {E(K)/n E(K)} \ar[r]\ar[d]& {\...
... \H^1(K_v,E[n])}
\ar[r]& {\prod_v H^1(K_v,E)[n]} \ar[r]& 0.
}
\end{displaymath}

The $n$-descent sequence for $E$ is the short exact sequence

\begin{displaymath}
0 \to E(K)/n E(K) \to \Sel ^{(n)}(E/K) \to {\mbox{{\fontenc...
...yr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E/K)[n] \to 0.
\end{displaymath} (2.2.3)

Theorem 2.9   For every integer $n$ the group $\Sel ^{(n)}(E/K)$ is finite.


\begin{proof}
% latex2html id marker 4433
[Sketch of Proof]
Let $K(E[n])$ denot...
...ly that there are only
finitely many such extensions of $K[n]$.
\par
\end{proof}

Exercise 2.10   Prove the that $E[n]$ is a finite Galois extension of $K$.

Theorem 2.11 (Mordell-Weil)   The group $E(\mathbb{Q})$ is finitely generated.


\begin{proof}
% latex2html id marker 4453The exact sequence \eqref{eqn:ndescen...
...te,
i.e., $S$ is finite, so $E(\mathbb{Q})$ is finitely generated.
\end{proof}

William 2007-05-25