Torsion points of an elliptic curve over infinite extension

Question

How can I see that, if $E$ is an elliptic curve over $\mathbb{Q}$, than $$E(\overline{\mathbb{Q}})_{tors}=E(\mathbb{C})_{tors}=(\mathbb{Q}/\mathbb{Z})^2,$$ and that $E(\overline{\mathbb{Q}})/E(\overline{\mathbb{Q}})_{tors}$ has infinite rank?

Answer 1

If $P$ is a torsion point in $E(\mathbb{C})$, then there exists $n$ such that $nP = 0$. But the multiplication by $n$ map is expressable in terms of rational functions in the Weierstrass coefficients of $E$, so the coordinates of $P$ must be algebraic numbers, i.e. $P$ in $E(\overline{\mathbb{Q}})$.

Now to see that the torsion group is isomorphic to $(\mathbb{Q}/\mathbb{Z})^2$ is easy if you are familiar with the complex uniformization $E(\mathbb{C}) = \mathbb{C}/\Lambda$, since the torsion in this latter group is $\varinjlim \frac{1}{n} \Lambda / \Lambda = (\Lambda \otimes \mathbb{Q}) / \Lambda$.

Answer 2

The infinitude of the rank is somewhat subtle. You can see a proof in this article of Frey and Jarden (see Section 2).