most recent 30 from http://mathoverflow.net 2013-06-20T12:00:52Z http://mathoverflow.net/feeds/question/118465 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118465/rank-growth-of-elliptic-curves-after-cubic-extensions Dave M da C 2013-01-09T16:40:09Z 2013-01-10T09:50:38Z

<p>Let $E/\mathbb{Q}$ be an elliptic curve and let $N_E(3,X)$ denote the number of cyclic cubic extensions $K/\mathbb{Q}$ of conductor no more than $X$ for which $rank~E(K)> ~rank~ E(\mathbb{Q})$. Then a conjecture of David, Fearnley and Kisilevsky (stemming from considerations in random matrix theory) states that</p> <p>$ \log N_E(3,X) \sim \frac{1}{2}\log X.$</p> <p>My question is what the conjecture should be if we remove the condition that $K/\mathbb{Q}$ is a $\textit{cyclic}$ extension. </p>

http://mathoverflow.net/questions/118465/rank-growth-of-elliptic-curves-after-cubic-extensions/118491#118491 Felipe Voloch 2013-01-10T00:36:51Z 2013-01-10T00:36:51Z

<p>I don't have a full answer, but if $E$ is given in Weierstrass form $y^2=f(x)$, then for most values of $c \in \mathbb{Q}$, if you look at the point with $y=c$ on $E$, you get a point in a cubic extension (usually non-cyclic) given by $f(x)-c^2=0$ which will not be in the division hull of $E(\mathbb{Q})$, i.e. the rank will grow. So the number of such cubic fields of conductor at most $X$ will be a constant times some power of $X$. The total number of cubic fields of conductor at most $X$ is a constant times some other power of $X$.</p>