I am currently working myself through Cassels' proof of the weak Mordell-Weil theorem in the book 'Lectures on Elliptic Curves' (chapter 15). I found myself stuck quite close to the end of the proof with a few statements that seem to be elementary number theoretic in nature.
We have an elliptic curve $$Y^{2} = F(X) = X^{3} + AX + B$$
and define $\mathbb{Q}[\Theta] := Q[T]/F(X) $ with $ \Theta$ as the image of $X$. Moreover, we define $M \subset \mathbb{Q}[\Theta]^*/(\mathbb{Q}[\Theta]^*)^{2}$ as the subgroup of all $\alpha(\mathbb{Q}[\Theta]^*)^2$ for which $Norm(\alpha)\in(\mathbb{Q}^*)^{2}$. (The Norm is defined to be the determinant of the multiplication with $\alpha$, seen as a linear map over the 3-dim. vector space $\mathbb{Q}[\Theta]$ over $\mathbb{Q}$).
The clue of the proof is then to construct a homomorphism $$\mu: E(\mathbb{Q})\to M$$ with the properties
- $\mu(O) = 1(\mathbb{Q}[\Theta]^*)^2$
- $\mu(a,b) = (a-\Theta)(\mathbb{Q}[\Theta]^*)^2$ if $b\neq0$ and
- since we can view $\mathbb{Q}[\Theta]$ as a direct sum of fields, if $(a,0)\in E(\mathbb{Q})$, then $ F(a)=0$ and one component of $a-\Theta$ is $0$. This one shall be patched with any element of $\mathbb{Q}^*$ so that the resulting Norm is in $(\mathbb{Q}^*)^2$.
Cassels then goes on to proove that $ker(\mu)=2E(\mathbb{Q})$ and the only thing left to proove is that $im(\mu)$ is finite:
I cant wrap my head around why $gcd\{(r-e_{1}t^2),(r-e_{2}t^2)\}$ divides $(e_{1} - e_{2})$ and why it follows that $r-e_{j}=d_{j}v_{j}^2$ . Also, how does this help in concluding that the image is finite? I'm not sure if this proof is leaving out a few important details or if there's just a major misunderstanding on my side.
It'd certainly be awesome if someone who knows/understands this proof could help me! Thanks in advance!
PS I'm also probably going to come up with a followup question for the general case of this last part of the proof. That one requires some algebraic number theory which I don't have a background in. But I wanted to make sure of this special case here first.
Let $d$ be $\gcd\{(r-e_{1}t^2),(r-e_{2}t^2)\}$. Recall that all variables occuring here are in $\Bbb Z$. Then $d$ divides both $(r-e_{1}t^2)$, and $(r-e_{2}t^2)$, so also there difference $(e_1-e_2)t^2$. It also divides the linear combination $$ e_2(r-e_{1}t^2)-e_1(r-e_{2}t^2)\ ,$$ so $d$ divides both $(e_1-e_2)t^2$, and $(e_1-e_2)r$. Recall now that $x=r/t^2$ was written as an irreducible fraction, there is no common factor for $r$ and $t^2$. So $d$ divides the g.c.d. of these two numbers, which is (up to sign, but this does not matter) $(e_1-e_2)$. The first point is now clear, we use $d$ for other purposes in the sequel.
Now go to the expression $$ \tag{$*$} s^2=(r-e_1t^2)(r-e_2t^2)(r-e_3t^2)\ . $$ On the R.H.S. we have three explicit factors. The product is a square. Each integer $N$ can be written uniquely in the form $N=p_1^{a_1}p_2^{a_2}\dots$ up to an order of the primes $p_1,p_2,\dots$ involved in it, and we extract from each power $a_i$ the maximal even power, thus write $a_i=2a_i'+a_i''$, where $a_i''\in\{0,1\}$ is the rest after the division with rest by $2$ of $a_i$. This leads to a representation of $N$ as $N=dv^2$, where $d=\prod p_i^{a_i'}$, $v=\prod p_i^{a_i''}$.
We extract from each factor in $( * )$ the primes that appear to odd powers as above, thus write (uniquely) $$ \begin{aligned} (r-e_1\color{red}{t^2}) = d_1v_1^2\ ,\\ (r-e_2\color{red}{t^2}) = d_2v_2^2\ ,\\ (r-e_3\color{red}{t^2}) = d_3v_3^2\ , \end{aligned} $$ and note that possible common factors of $d_i$, $d_j$ are also dividing the g.c.d. of the corresponding $(r-e_i\color{red}{t^2})$, $(r-e_j\color{red}{t^2})$, which is a divisor of $(e_i-e_j)$. Cassels writes it in one sentence, $d_j$ divides the product of all three factors $(e_1-e_2)$, $(e_2-e_3)$, $(e_3-e_1)$. And of course, since the L.H.S. $s^2$ of $(*)$ is a square, the R.H.S. is also a square, i.e. $d_1d_2d_3\; v_1^2v_2^2v_3^2$ is also a square, so $d_1d_2d_3$ is a square.
There are thus finitely many possibilities for $d_1,d_2,d_3$.
Note that Cassels gives his argument in the special case when $F$ has the three roots $e_1,e_2,e_3\in\Bbb Q$. We can thus write $$ F(T)=(T-e_1)(T-e_2)(T-e_3)\ ,$$ so the quotient ring $$ \Bbb Q[\Theta] = \Bbb Q[T]/(F) \overset\cong\longrightarrow \Bbb Q\times \Bbb Q\times \Bbb Q\ , $$ where the isomorphism above is given by mapping a polynomial $p(T)$ modulo $F$ to the triple $(p(e_1),p(e_2),p(e_3))$. On the R.H.S. the ring $\Bbb Q^{\times 3}$ has the operations (addition and multiplication) done componentwise. The units of the L.H.S. correspond to the units of the R.H.S. which are explicitly forming the group $$ G:= (\Bbb Q\times \Bbb Q\times \Bbb Q)^* \cong \Bbb Q^*\times \Bbb Q^*\times \Bbb Q^*\ , $$ with the operation of multiplication. Now we build $G/G^2$ the group of triples (of non zero rationals) taken modulo squares (on each component).
The map $\mu$ of Cassels is a map $$E(\Bbb Q)=\mathfrak E \longrightarrow \Bbb Q[\Theta]^*/(\text{squares}) \overset\cong\longrightarrow G/G^2\ ,$$ which maps a "generic" point $(a,b)$ to $a-\Theta$ modulo squares, if we stop one step before going to $G/G^2$, and going to $G/G^2$, and using the notation $(x,y)$ for the point, instead of $(a,b)$, as it is taken in the proof of Theorem 1, we have the map explicitly in this language: $$ \begin{aligned} (x,y) &\to x-\Theta\text{ modulo squares} \\ &\overset\cong\longrightarrow (x-e_1,\ x-e_2,\ x-e_3) \text{ modulo squares } \\ &= (r/t^2-e_1,\ r/t^2-e_2,\ r/t^2-e_3) \text{ modulo squares } \\ &= (r-e_1t^2,\ r-e_2t^2,\ r-e_3t^2) \text{ modulo squares } \\ &= (d_1v_1^2,\ d_2v_2^2,\ d_3v_3^2) \text{ modulo squares } \\ &= (d_1,\ d_2,\ d_3) \text{ modulo squares ,} \end{aligned} $$
and as noted in loc. cit. there are only finitely many possibilities for this final result.
It may be useful to have a comparison with the excellent didactic and illustrative exposition of Álvaro Lozano-Robledo [Elliptic Curves, Modular Forms, and Their L-Functions], same result being here Theorem 2.8.3, and things will become immediately clear.