"No way to put a topology on $\mathbb{Q}^*/\mathbb{Q}^{*2}$ such that the map $\alpha$ is continuous."

Question

Let \begin{align*} \mathbb{Q}^{*2}:=\{x^2\;|\;x\in\mathbb{Q}^*\}. \end{align*} and \begin{align*} \alpha:\;E(\mathbb{Q})&\to\mathbb{Q}^*/\mathbb{Q}^{*2},\\ \alpha(x,y)&=x\mod\mathbb{Q}^{*2}\quad\text{if }x\neq0,\\ \alpha(T)&=b\mod\mathbb{Q}^{*2},\\ \alpha(\mathscr{O})&=1\mod\mathbb{Q}^{*2} \end{align*} where $E(\mathbb{Q})$ is an elliptic curve of the form \begin{align*} \{(x,y)\in\mathbb{Q}\times\mathbb{Q}\;|\;y^2=x^3+ax^2+bx,\;a,b\in\mathbb{Q}\}\cup\{\mathscr{O}\}. \end{align*} My question is about the almost last part of the proof of Mordell's theorem in the book Rational Points on Elliptic Curves, ed. 2, p.93. Can someone explain to me the notion: "There is no way to put a topology on $\mathbb{Q}^*/\mathbb{Q}^{*2}$ such that the map $\alpha$ is continuous"? I can see that this map $\alpha$ is a homomorphism completely defined in an arithmatic nature, while the elliptic curves are introduced geometrically in the projective plane. Of course, we embed the elliptic curve in the real projective plane but in the end we only are interested in the rational points. Why is there no possibility for a nice topology on $\mathbb{Q}^*/\mathbb{Q}^{*2}$? Or is it just too difficult that they just skip it?

Answer 1

This statement is not meant to be taken literally. It's not literally true that no topology exists for which $\alpha$ is continuous: for instance, the indiscrete topology on $\mathbb{Q}^*/\mathbb{Q}^{*2}$ makes any map to it continuous. Rather, the point is that there does not exist any topology for which $\alpha$ is easily seen to be continuous and such that this continuity is useful to the proof. The indiscrete topology is useless for this purpose, for instance, because knowing that $\alpha$ is continuous for the indiscrete topology tells you nothing. There are also probably some highly artificial topologies you could put on $\mathbb{Q}^*/\mathbb{Q}^{*2}$ for which continuity of $\alpha$ would tell you exactly what you need, but those would be equally useless because proving $\alpha$ is continuous would then be just as hard as proving what you need.

So, this is not a statement that has a precise mathematical meaning or can be proven. The point is just that there is not any "natural" topology for which $\alpha$ would obviously be continuous and such that you would then be able to use a typical continuity argument to finish the proof.