Let $E$ be an elliptic curve over $\mathbb Q$. Let's look at the group of points of this elliptic curve over $\mathbb Q(1^{1/\infty})$ which we get after adding all roots of unity to $\mathbb Q$. It is easy to prove that it is not finitely generated and the theorem of K.Ribet asserts that its torsion is finite. What else can we say about it? Does it look more like $\bigoplus \mathbb Z$ or $\mathbb C/\Gamma$? Is it divisible? (I don't think so). I apologize that the questions are not precise, I just need to know which results do we have in this direction.



Points of elliptic curves over cyclotomic extensions
$E (\mathbb Q^{ab})/\operatorname{tors}$ is a sum of countably many copies of $\mathbb Z$. 


To prove this, take a countable basis $x\_1,x\_2,\dots$ of $E (\mathbb Q^{ab})/\operatorname{tors}\otimes \mathbb Q$. Suppose we show that $\oplus\_{i=1}^n x\_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$ is finitely generated. Then it is necessarily isomorphic to $\mathbb Z^n$. Moreover, the only embeddings $\mathbb Z^n \to \mathbb Z^{n+1}$ where the elements of $\mathbb Z^n$ don’t get any more divisible are isomorphic to the standard embedding. So if $x\_1,\dots,x\_n$ form a $\mathbb Z$-basis of $\oplus\_{i=1}^n x\_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$, we can change $x\_{n+1}$ so it generates the same $\mathbb Q$-subspace and also $x\_1,\dots,x\_{n+1}$ form a $\mathbb Z$-basis of $\oplus\_{i=1}^{n+1} x\_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$. Iterating, we form a $\mathbb Z$-basis of the whole group.


Now let’s check the claim. Let $k$ be a cyclotomic field containing $x\_1,\dots,x\_n$ and all the torsion points of $E (\mathbb Q^{ab})/\operatorname{tors}$. Let $N$ be the maximum order of a torsion point. Let $y$ be any point of $E (\mathbb Q^{ab})/\operatorname{tors}$
such that $m y \in E(k)$ for some natural number $m$. We will check that $N y \in E(k)$. It will follow that$ \oplus\_{i=1}^n x\_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$ is contained in $(1/N) E(k)$, which is finitely generated, so is finitely generated.


To check this, let $y’$ be any Galois conjugate of $y$ relative to $k$. Because $\mathbb Q^{ab}/\mathbb Q$ is Galois, $y’ \in E(\mathbb Q^{ab})$, so $y’-y \in E(\mathbb Q^{ab})$. Because the Galois conjugation is relative to $k$, $m y’=my $, so $m(y’-y) =0$ and $y’-y$ is torsion, hence it is $N$-torsion, so $Ny’ =Ny$. Because this is true for all $y’$, $Ny$ is Galois-invariant and thus is in $E(k)$.


