# About Mordell's Theorem (Elliptic Curves)

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.

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.

• thanks! Could you recommend me any book on global class field theory? – Abellan Jul 2 '15 at 20:20
• J.S. Milne's notes are a great resource for both local and global class field theory. – Brandon Carter Jul 2 '15 at 20:24
• 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... – sdf Jul 2 '15 at 21:26
• @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. – Brandon Carter Jul 2 '15 at 21:36
• 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. – Brandon Carter Jul 2 '15 at 21:39

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.