# Mordell-Weil theorem

One version of Mordell-Weil theorem is as follows:

Suppose $K$ is a number field, $E$ is an elliptic curve in $\mathbb{P}^2_{\mathbb{K}}$, then $E$ is an Abelian group of finite type.

In other words, $E(K)=\mathbb{Z}^r\bigoplus A$ where $r$ is a non-negative integer and $A$ is a finite Abelian group. At first sight, this theorem is indeed surprising, at least for me, because I do not know, a priori, anything on the algebraic structure, except that it is an Abelian group.

The proof of this theorem relies heavily on the height function defined on the elliptic curves as we will give right now. Suppose $P=[x_0:x_1:...:x_n]\in \mathbb{P}^n_K$ is a point in the projective space, we denote $\Sigma(K)$ the set of places of $K$, finite places($\Sigma_f(K)$) and infinite places($\Sigma_i(K)$) as well. And we set $H(P)=\prod_{v\in\Sigma(K)}max_i{|x_i|_v}$. Another height function is defined on $E(K)$, that is for $P=[x:y:z]\in E(K)$, we set $h(P)=H([x:z])$ if $P\neq[0:1:0]$, and $h(P)=0$ if $P=[0:1:0]$. This height function is the Weil height function. We can verify that

There exists constants $c,d>0$ such that (1)$|h(2P)-4h(P)|;(2)$|h(P+Q)+h(P-Q)-2h(P)-2h(Q)|.

The first inequality inspires us that $h$ is perhaps a ‘polynomial function’ on $n$ in $nP$. Indeed, we can define $F(P)=\lim_{n\rightarrow +\infty}\frac{h(2^nP)}{4^n}$. Note that , due to the first inequality, this limit does exist, and in this way we have that

$F(2P)=4F(P)$.

As for the second inequality, we can also show that for $F$

$F(P+Q)+F(P-Q)=2F(P)+2F(Q)$.

So, for the present, we have shown that we can define a positive quadratic form on the Abelian group $E(K)$.

To prove the Mordell-Weil theorem, we can proceed in two steps:(1)show that $E(K)/2E(K)$ is a finite set;(2) show that $E(K)$ is an Abelian group of finite type.

Note that, the whole argument is not valide in general. It uses essentially the following fact:

The set $\{P\in \mathbb{P}^n_K|H(P) for some constant $c$ is a finite set.

Let’s first assume this fact, and then we observe that for a point $P\in E(K)$, $h(P)=H([x_P:z_P])$, and the set $\{P\in E(K)|h(P) is finite(of course, we can change here the $h(P)$ by $F(P)$, the conclusion remains unchanged since the difference between these two functions is uniformly bounded).

Now combine the first argument that $E(K)/2E(K)$ is finite and the above result, in using the descent method due to Fermat, we can show that $E(K)$ is of finite type as an Abelian group. The argument is rather classic. First we find a set of representatives $S$ of the quotient $E(K)/2E(K)$, and since it is finite, we can set $c=max_{s\in S}\sqrt{F(s)}$. For convenience, we can set $|P|=\sqrt{F(P)}$. So for any $P\in E(K)$ with $|P|>c$, there is $Q\in S$ such that $P-Q=2R$. Yet, the crucial fact is

$|R|=\frac{1}{2}|P-Q|\leq(|P|+|Q|)/2\leq(|P|+c)/2<|P|$

This is the famous descent method! So, as long as $|R|>c$, we can continue to use this method, and with the important fact that $\{P\in E(K)||P| is always finite, we get that this process will eventually get terminated, and at last we get $P=Q+2R=Q+2(Q'+2R')=...=\sum_{Q\in S}a_QQ+R_1$ with $|R_1|\leq c$. This means that $E(K)$ can be spanned by the finite set $\{P\in E(K)||P|\leq c\}$.

So, it remains to show that $E(K)/2E(K)$ is finite. This is the most difficult part of the proof. The idea is to construct a homomorphism $\phi'=(\phi_1',\phi_2',\phi_3'):E(K)\rightarrow (K^*)^3$. And we can show that the induced morphism $\phi:E(K)\rightarrow (K^*/K^{*2})^3$ has kernel exactly $2E(K)$. What is more, we can show that $Im(\phi)\subset (O_{K,S}^*/O_{K,S}^{*2})^3$ where $O_{K,S}^*$ will be defined in the following and the quotient is a finite set because $O_{K,S}^*$ can to be shown to be an Abelian group of finite type.

Now for any finite subset $S\subset \Sigma_f(K)$, we define $O_{K,S}=\{x\in K|ord_v(x)\geq0,\forall v\not\in S\}$. What does this set mean? It can be easily shown that $O_{K,S}$ is in fact a ring, a subring of $K$. An important fact due to Dirichlet, Chevalley and Hasse is

The group of units $O_{K,S}^*$ is of finite type.

A typical proof consists of constructing a map, $f:O_{K,S}^*\rightarrow V=\mathbb{R}^{r+t+s},x\mapsto (log|\sigma_1(x)|,...,log|\sigma_r(x)|,...,log|\sigma_{r+t}(x)|,log|x|_{v_1}...,log|x|_{v_s})$ where $\sigma_j$ are the real(complex) infinite places of $K$ and $v_j\in S$ with $s=\# S$. It is indeed a group morphism. What is more, we can show that $Im(f)$ is discret in $V$. How to intuitively show this? The last $s$ factors take discret values simple because the function $ord_v(x)$ takes discret values. And the first $r+t$ factors takes discret values because there can not be too many elements $x\in K$ all of whose conjugates are very close to $1$. Note that by multiplying by some integers, an algebraic number can become an algebraic integer. So the precedent sentence can by verified for the algebraic integers. Yet, if the conjugates of an algebraic integer are close to $1$, then the coefficients of the minimal polynomial(which is unitary, that is, the coefficients of the highest degree is $1$) are bounded, depending only on the degree $[K:\mathbb{Q}]$. Thus the finiteness. To show the kernel of $f$ is also finite, it is equivalent to showing that $\{x\in O_K^*||\sigma_i(x)=1|,\forall \sigma_i\in \Sigma_i(K)\}$ is finite. Yet the argument is the same as above. So, we have shown that $O_{K,S}^*$ is of finite type.

Then we should try to give such a homomorphism $\phi:E(K)/2E(K)\rightarrow (K^*/K^{*2})^3$. One remark is in order. Note that, in $K$, the elliptic curve is $E:y^2=x^3+ax+b$. In a larger field $K\subset L$, we can assume that $y^2=(x-x_1)(x-x_2)(x-x_3)(x_i\in L)$. And note that $E(K)\subset E(L)$. So, if we can show that $E(L)$ is of finite type, then so is $E(K)$. The reason of enlarging the field, is just to make the polynomial $x^3+ax+b$ decompose into three linear factor and so that we can define the function $\phi:E(L)\rightarrow (L^*/L^{*2})^3$. So, we can suppose directly that in $K$ we already have, $y^2=(x-x_1)(x-x_2)(x-x_3)(x_i\in K)$. For $\phi_i:E(K)\rightarrow K^*/K^{*2}$, for a point $P\in E(K)$, if $P=e_E$, the neutral element, we set $\phi_i(P)=1$, if $P=[x_i:0:1]$, then we set $\phi_i(P)=(x_i-x_{i-1})(x_i-x_{i+1})$, otherwise we set $\phi_i(P)=x_P-x_i$. We can show that, $\phi=(\phi_1,\phi_2,\phi_3)$ is indeed a homomorphism.

The difficult thing is to show that $\phi$ has a finite image. Suppose that $P=(x,y)\in E(K)$. We can assume that $x=r/t^2,y=s/t^3(r,s,t\in O_K)$ with $=<1>=$ where $$ is the ideal of $O_K$ generated by $r,s$. Then we have that $s^2=(r-x_1t^2)(r-x_2t^2)(r-x_3t^2)$. Note that $\supset \supset (i\neq j)$, so if $=\prod p^{e_p}$ where $p$ are prime ideals of $O_K$ with $e_p$ has only finite non-zero, this induces that $$ has only finite many possibilities modulo square(we should also consider the units, yet from the above we know that the group of units in $O_K$ is of finite type). Thus this shows that $Im(\phi)$ has finite image, and thus the whole Mordell-Weil theorem.