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$E (\mathbb Q^{ab})/\operatorname{tors}$ is a sum of countably many copies of $\mathbb Z$. Iterating, we form a $\mathbb Z$-basis of the whole group.

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 $\oplus_{i=1}^{n+1} x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$.

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 $E(k)$.

$E (\mathbb Q^{ab})/\operatorname{tors}$ 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 $E(k)$.

$E (\mathbb Q^{ab})/\operatorname{tors}$ 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 $\oplus_{i=1}^{n+1} x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$.

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 $E(k)$.

$E (\mathbb Q^{ab})/\operatorname{tors}$ is a sum of countably many copies of $\mathbb Z$. Iterating, we form a $\mathbb Z$-basis of the whole group.

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 $\oplus_{i=1}^{n+1} x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$.

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 $E(k)$.

I claim for every nontorsion $x \in E(\mathbb Q^{ab})$, there is some $n$ such that $x$ is not $m$-divisible for any $m>n$.

Indeed, let $k$ be a cyclotomic field over which $E$ and all the torsion of $E(\mathbb Q^{ab})$ are defined. Without loss of generality, $x$ is indivisible in $E(k)$. Let $n$ be the maximal order of a torsion point of $E(\mathbb Q^{ab})$.

If $y$ is an $m$th root of $x$, then for each prime $p$ dividing $m$, $y^{m/p}$ is not defined over $k$, so $y$ has some Galois conjugate $y'$ with $y^{m/p} \neq y'^{m/p}$ but $y^m = y'^m$. Then if $y \in E(\mathbb Q^{ab})$, $y'$ is as well because it's Galois, so $(y/y')$ is a torsion point or order dividing $\gcd(m,n)$. If $m$ is not a multiple of $n$, then $\gcd(m,n)$ is a proper divisor of $m$. Taking $p$ dividing $m/\gcd(m,n)$, we see that $(y/y')^{m/p}$ is trivial, contradiction.

So in fact the group is closer to a sum of copies of $\mathbb Z$. But it is not clear from this sort of "local" argument whether it is a sum of copies of $\mathbb Z$.

$E (\mathbb Q^{ab})/\operatorname{tors}$ 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 $\oplus_{i=1}^{n+1} x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$.

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 $E(k)$.