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...

  • $\begingroup$ The claim that a particular point $P$ on a particular elliptic curve is non-torsion is the infinite sequence of claims that for every positive integer $n$, some explicit polynomials in two variables (expressing $nP$ in coordinates) are not both zero when evaluated on particular elements of $k$. In particular, it's a first-order claim. $\endgroup$ – Qiaochu Yuan Dec 16 '15 at 0:54
  • $\begingroup$ Note that $E(k)$ is clearly non-torsion if $k$ contains an element transcendental over the subfield of $k$ generated by the coefficients of $E$. So the only interesting case is the case that $k$ is the algebraic closure of this subfield, in which case $k$ is countable. $\endgroup$ – Qiaochu Yuan Dec 16 '15 at 1:21
  • $\begingroup$ Maybe you should tag number theory too; they might know about this. For $k$ the algebraic closure of a field $K$ of finite type over $\mathbb Q$, you are trying to prove that the Mordell–Weil rank of $E(L)$ for $L/K$ finite goes to infinity (or even: is nonzero for some $L$). This seems like this should be known classically. $\endgroup$ – Remy Dec 16 '15 at 2:00
  • $\begingroup$ For example, if $E$ is given by $y^2 = x^3 + Ax + B$, then choosing some value for $x$ and considering $d = x^3 + Ax + B$ gives a point of $E$ over $K(\sqrt{d})$; the question is whether we can choose $x$ such that this point is linearly independent from the $K$-points. This seems like it should be true for general $x$, but I don't know how to prove it (maybe you can use something about the height pairing). $\endgroup$ – Remy Dec 16 '15 at 2:05

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?

  • $\begingroup$ Fantastic answer, thanks for your efforts in searching the literature. It seems boggling that such an intuitively obviously result should require that much work! Anyways thanks very much again. $\endgroup$ – Bruno Joyal Dec 24 '15 at 3:13
  • 1
    $\begingroup$ Maybe it's just not so obvious. If I understand it correctly, there are examples of infinite algebraic field extensions $\mathbb Q \subseteq K$ such that every elliptic curve over $K$ has finite rank! So even for an elliptic curve over a number field, you have to do something clever to let the Mordell–Weil rank go to infinity as you extend the field. $\endgroup$ – Remy Dec 24 '15 at 5:35
  • $\begingroup$ You say "the n-torsion has size ≤$n^{2g}$ (with equality if char K∤n)"; where is this proved? $\endgroup$ – Alex Nov 8 '16 at 1:00
  • 1
    $\begingroup$ @Alex: see e.g. Mumford's Abelian Varieties, p.64 (or p.61 in the second edition). $\endgroup$ – Remy Nov 8 '16 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.