4
$\begingroup$

I've just finished the proof of Mordell's Theorem given in the book "Rational Points on Elliptic Curves " by Silverman.

One of the key lemmas used in the proof of the theorem is:

Let $C(\mathbb{Q})$ denote the group of rational points of $C$ then $[C(\mathbb{Q}):2C(\mathbb{Q})]$ is finite.

But in the book the lemma is proved under an additional assumption saying that $C$ has a rational point of order 2. I'd like to know how much algebraic number theory is needed to avoid that assumption and some references to see if I could try to look at a more general proof.

$\endgroup$
0
$\begingroup$

The proof of that result, usually called (an explicit version of) the Weak Mordell-Weil theorem, can be found in Silverman's Arithmetic of Elliptic Curves book.

The proof uses Galois cohomology and some minor arithmetic that can be followed with the knowledge of a few of the main theorems of global class field theory.

$\endgroup$
  • $\begingroup$ thanks! Could you recommend me any book on global class field theory? $\endgroup$ – Abellan Jul 2 '15 at 20:20
  • $\begingroup$ J.S. Milne's notes are a great resource for both local and global class field theory. $\endgroup$ – Brandon Carter Jul 2 '15 at 20:24
  • $\begingroup$ The two results needed are the finiteness of the ideal class group and DIrichlet's unit theorem?! You certainly do not need the class field theory for this... $\endgroup$ – sdf Jul 2 '15 at 21:26
  • $\begingroup$ @sdf: Where do you think that Dirichlet's unit theorem comes into this? The usual proof of the finiteness of the Selmer group comes from showing that the $n$ Selmer group lies inside of the set of continuous homomorphisms from the absolute Galois group to $E[n]$ that are unramified outside of a finite number of places. But then you can use that such a homomorphism factors through the Galois group of the maximal abelian extension of exponent $n$ which is unramified outside of those specified places. This extension is finite by CFT. $\endgroup$ – Brandon Carter Jul 2 '15 at 21:36
  • $\begingroup$ The proof can be simplified some where the base field is $\mathbf{Q}$ and $n = 2$, even to the point of only needing to understand some basic algebraic number theory (e.g. the possible discriminants of quadratic fields). Even in that case, I don't see where the ideal class group or Dirichlet's unit theorem play a role. $\endgroup$ – Brandon Carter Jul 2 '15 at 21:39
1
$\begingroup$

I hope you can find this useful.

All elliptic curve $C(\mathbb {Q})$ is canonically write as $y^2=4x^3-g_2x-g_3$... (1) in which the three roots of the right side are distinct; by a birational transformation $(x,y)\to (\frac x4, \frac y4)$ you have $y^2=x^3-h_2x-h_3$... (2) where $h_2$ and $h_3$ can be supposed rational integers. Being $e_1, e_2, e_3$ the three distinct roots of (2) one has $y^2=(x-e_1)(x-e_2)(x-e_3)$ in which there are three possibilities: $e_1, e_2, e_3$ are cubic numbers; one of them is rational and the other two quadratic conjugate irrational; the three are rational. Now one take the norm $y^2=N(x-e_i)$ (where the norm N is properly a norm just in the first possibility and in the other two not properly) and one has to see about the rational $x$ such that $N(x-e_i)$ is a square in $\mathbb {Q}$.

Always the number of $\mathbb{Q}(e_i)$ whose norms are perfect squares in $\mathbb {Q}$ are distributed in an infinite set of classes modulo the squares of $\mathbb{Q}(e_i)$ but among them, the binomial numbers $x-e_i$ are fortunately distributed in just a finite number of such classes. More precisely, the set $K^2$ of squares of $K=\mathbb {Q}(e_i)$ is a multiplicative subgroup of $K^*$ and the quotient group $G=K^*/K^2$ formed by the classes $zK^2$ is clearly infinite but for fixed $e_i$ ; i=1, 2, 3, there is a finite set $z_1K^2, z_2K^2,…,z_rK^2$ in $G^{(r)}$ and a partition in $r$ subsets of $C(\mathbb {Q})$, say, $R_1, R_2,…,R_r$ where $(x,y)\in R_j \Rightarrow (x-e_i)\in z_jK^2$.

Finally you have to prove that if $(x,y)\in C(\mathbb {Q})$ then $x-e_i=\nu\alpha^2$ where $\nu$ and $\alpha$ are in $\mathbb {Q}(e_i)$, $\nu$ being capable of taking only a finite number of values.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.