Elliptic curve over algebraically closed field of characteristic $0$ has a non-torsion point

Question

Let $E/k$ be an elliptic curve over an algebraically closed field $k$ of characteristic $0$. Can one prove that the abelian group $E(k)$ is non-torsion? Better yet, can one prove that $E(k) \otimes_\mathbb Z \mathbb Q$ is an infinite-dimensional $\mathbb Q$-vector space?

It is very tempting here to try to use the Lefschetz principle, to try to reduce the situation to $k= \mathbb C$ where both statements are obvious. However I am not sure that one can actually apply the Lefschetz principle as it would require formulating the statements in the first-order theory of fields and I am unfortunately not much of a logician.

At least one can say that if the field $k$ is uncountable then $E(k)$ is uncountable whereas $E(k)^{\text{tors}}$ is countable (so much is true over any field), so there is always a non-torsion point.

However when the field $k$ is countable, there seems to me to be no "trivial" reason why $E(k)$ should have an element of infinite order. The fact that $k$ has characteristic $0$ has to intervene somehow, as the statement is false in finite characteristic...

Answer 1

After some searching, I found the following article:

G. Frey and M. Jarden, Approximation theory and the rank of abelian varieties over large algebraic fields. Proc. London Maths. Soc. 28 (1974), 112-128.

A link is provided on the second author's website; see here for the actual article.

In it, they prove the following:

Theorem 10.1. If $A$ is an abelian variety of positive dimension defined over an algebraically closed field which is not the algebraic closure of a finite field, then the rank of $A(K)$ is equal to the cardinality of $K$.

The proof is a bit cumbersome, but in the second remark following the theorem, they provide an alternative and more direct method that does not depend on the main results of their paper. This alternative method seems useful for the weaker question you asked.

In the introduction, they also say:

Another proof was indicated by J.-P. Serre in a letter.

Since they do not include a reference for the letter, it might have been private communication. (I didn't search in Serre's Œuvres for the letter.) Or maybe Serre's proof is the alternative method they give.

Remark. Note that the theorem is trivial for $K$ uncountable:

• The cardinality of $A(K)$ is at most that of $K$, with equality if $K$ is algebraically closed;
• The torsion is countable, since the $n$-torsion has size $\leq n^{2g}$ (with equality if $\operatorname{char} K \nmid n$);
• Thus, $A(K) \otimes_{\mathbb Z} \mathbb Q$ has the same cardinality as $K$.
• Thus, if $K$ is uncountable, then $A(K) \otimes \mathbb Q$ cannot be finite-dimensional.
• Now use that for $I$ infinite, the cardinality of $I$ equals the cardinality of $\mathbb Q^{(I)}$ (but of course not that of $\mathbb Q^I$; cf. Cantor's diagonal argument).

Similarly, this argument proves for any infinite field (not necessarily algebraically closed, nor uncountable) that the dimension of $A(K) \otimes \mathbb Q$ is at most the cardinality of $K$. Thus, the only content of the theorem is exactly the question you asked: if $K$ is algebraically closed and not the algebraic closure of a finite field, does $A(K) \otimes \mathbb Q$ have infinite dimension?