How fast can elliptic curve rank grow in towers of number fields?

Question

Fix $E/K$ an elliptic curve over a number field $K$. For various towers of finite field extensions $K=K_0 \subset K_1 \subset K_2\subset\cdots$ how fast can $\operatorname{rank}(E(K_n))$ grow in terms of $|K_n:K|$?

If $E:y^2=f(x)$ then there exist $x_n \in \mathbb{Q}$ such that $\sqrt{f(x_n)} \notin K_{n-1} = K(\sqrt{f(x_1)},\sqrt{f(x_2)}, \ldots, \sqrt{f(x_{n-1})})$ and $P_n=(x_n,\sqrt{f(x_n)})$ is independent of $E(K_{n-1})$, and thus $\operatorname{rank}(E(K_n)) \ge \log_2(|K_n:K|) + \operatorname{rank}(E(K))$, and if $E/K$ has complex multiplication then $\operatorname{rank}(E(K_n)) \ge 2\log_2(|K_n:K|) + \operatorname{rank}(E(K))$.

What function of $|K_n:K|$ is an upper bound to $\operatorname{rank}(E(K_n))$ for all towers $\{K_n\}$? Is there an example with $\limsup \operatorname{rank}(E(K_n))/\log_2(|K_n:K|) \gt 1$ resp.${} \gt 2$ for fixed $E/K$ without resp. with complex multiplication?

Answer 1

I think it is a good idea to compare the growth of the rank to the degree. I would say that we have excessive growth in an extension $F/K$ if $$\DeclareMathOperator{\rank}{rank}\rank E(F) - \rank E(K) > [F:K].$$ There is a conjecture, called the minimalistic conjecture for twists in here that implies that if $K=\mathbb{Q}$ then for 100% of curves $E$ there is no excessive growth in a fixed $F/\mathbb{Q}$. To find growth of rank is therefore not so easy, but a well studied question. Maybe this article by Mazur and Rubin is a good place to start.

Of course it is easy to construct large excessive growth for quadratic extensions. If $F/K$ is quadratic then $\rank E(F) - \rank E(K) = \rank \breve{E}(K)$ where $\breve{E}/K$ is the quadratic twist of $E$ corresponding to $F/K$. So over $\mathbb{Q}$, we can have growth of 28 by taking $E$ a twist of a curve of rank at least 28. It should not be hard to prove that for a given $E/\mathbb{Q}$ and $n$ there are $n$ quadratic twists with coprime discriminant all of positive rank. That would give a multiquadratic extension with rank growth at least equal to the degree. For general successive quadratic extensions, i.e., a general $2$-group, this might be harder. However the methods coming from root numbers often allow to find local conditions, especially in even order groups, which force the rank to grow as a positive proportion of the rank. See the first link above.

Another famous example in which the rank grows at least as fast as the degree are anti-cyclotomic $\mathbb{Z}_p$-extensions. Let $p>2$ be a prime and $K$ a quadratic imaginary field. There is a unique extension $F/K$ such that its Galois group is isomorphic to $\mathbb{Z}_p$, it is Galois over $\mathbb{Q}$, but the group $F/\mathbb{Q}$ is not abelian (but infinite dihedral). This is formed of extensions $F_n/K$ which are cyclic of degree $p^n$. Under suitable conditions on an elliptic curve $E$ defined over $\mathbb{Q}$, e.g. the Heegner hypothesis, one can show that the rank of $E(F_n)$ is equal to $[F_n:K]+C$ for a constant $C$ if $n$ is sufficiently large. This growth is produced by Heegner points that can be shown to be independent. This was done by Cornut and Vatsal as explained in these videos. This gives a proven example with $\bigl(\rank E(F_n) - \rank E(K) \bigr)/[F_n:K]$ converging to $1$. By taking the non-Galois subfield of degre $p^n$, we also get this limit equal to $1$ over $K=\mathbb{Q}$.

I don't think I have ever seen an infinite tower such that $$\frac{\rank E(F_n) - \rank E(K)}{[F_n:K]}$$ has a $\limsup$ larger than $1$. There are examples of finite extensions other than the quadratic case above in which the rank grows quicker, but they are relatively rare. My favourite is this curve of conductor 5692 and rank $2$ over $\mathbb{Q}$. It has rank $6$ over the cubic cyclic field with $\alpha^2-3\alpha+1=0$. For cyclic fields of prime degree larger than $5$, it should be very hard if at all possible.