My question concerns a comment in Silverman-Tate's book "Rational Points on Elliptic Curves" (second printing). On page 76 they begin their proof of 'lemma 4' to the proof of Mordell's theorem, which is to say that the subgroup $2C(\mathbb{Q})$ in $C(\mathbb{Q})$, the group of rational points for some elliptic curve $E/\mathbb{Q}$, has finite index. They make the comment
"Unfortunately, we do not know how to prove Lemma 4 for all cubic curves without using some algebraic number theory, and we want to stick to the rational numbers. So we are going to make the additional assumption that the polynomial $f(x)$ has at least one rational root, which amounts to saying that the curve has at least one rational point of order two. The same method of proof works in general if you take a root of the equation $f(x) = 0$ and work in the field generated by that root over the rationals. But ultimately we would need to know some basic facts about the unit group and the ideal class group of the field, topics which we prefer to avoid."
My question is has this situation been improved? Can Mordell's theorem be proved in generality without resorting to algebraic number theory?
Best Answer
Not sure if this answers your question, but here's a thought.
The known proofs that $E(K)/mE(K)$ is finite are non-effective because they embed $E(K)/mE(K)$ into a larger group that is shown to be finite. Let's call that larger group $S^{(m)}$. The proof that $S^{(m)}$ is finite is effective, and indeed it yields a very nice upper bound for the order of $S^{(m)}$. However, as far as I know, this bound always involves (at least) the $m$-part of the class number of $K(P)$, where $P$ is an $m$-torsion point of $E$. So the proof requires algebraic number theory in the sense that one needs to know that the class group (or at least the $m$ part) of some extension field of $K$ is finite, unless there is a rational $m$ torsion point. So proofs along these lines would seem to require algebraic number theory and ideal class groups, either explicitly or disguised in some way.