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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

$ \log N_E(3,X) \sim \frac{1}{2}\log X.$

My question is what the conjecture should be if we remove the condition that $K/\mathbb{Q}$ is a $\textit{cyclic}$ extension.

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Hi Dave, did you look at Kisilevsky's paper "Ranks of elliptic curves in cubic extensions"? (In Number Theory, Analysis and Geometry, D. Goldfeld, J. Jorgenson, P. Jones, D. Ramakrishnan, K. A. Ribet, J. Tate, eds., New York, Springer, 2012). There he proves various rank growth results in general cubic extensions. Perhaps he has some precise growth conjectures there as well? –  Tim Dokchitser Jan 10 '13 at 12:16
    
Hi Tim. I have had a look at that paper and there don't seem to be any precise conjectures, however it may be possible to adapt some of his arguements to find some lower bounds. I'll have another look! –  Dave M da C Jan 10 '13 at 14:43
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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 $X$.

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Well it's easy to get a lower bound $cX^\theta$ for some $c,\theta>0$, but the question is how big $\theta$ can get. For cyclic cubics we're told it might be $1/2 - \epsilon$. For unrestricted cubics, possibly even a positive proportion (i.e. $\theta = 1$), at least if $E$ has a place of bad reduction whose contribution to the root number can change from ${\bf Q}$ to $K$ $-$ or is there a reason that this can't happen (as it presumably doesn't for cyclic cubics if we're to believe the heuristic of David, Fearnley, and Kisilevsky)? –  Noam D. Elkies Jan 10 '13 at 2:42
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Yes, Noam, root number do not change in odd-degree Galois extensions. –  Chris Wuthrich Jan 10 '13 at 10:43
    
Following the approach that Felipe mentions one can show that there are at least $c_E X^{1/2}$ cubic extensions which show an increase in rank (at least when $j(E) \neq 0$). It would be interesting to know how many more extensions there should be where rank grows which are not of the form $f(x) - c^2$. Perhaps, as Noam suggests, the true exponent of $X$ depends on the reduction types the curve exhibits and is not uniform in $E$ unlike the conjecture above and others like Goldfeld's. –  Dave M da C Jan 10 '13 at 14:15
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