It seems to me that there ought to be elliptic curves over number fields with arbitrarily large torsion subgroups but Mordell-Weil rank zero. But I'll settle for a point of order 64. Does anyone know that such things must exist?
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3$\begingroup$ One strategy is to find an elliptic curve $E$ over $\mathbb{Q}$ with full 2-torsion and rank $0$, then consider its $64$-torsion field $K_{64}$. Since there’s no reason for $E$ to have a non-torsion rational point over $K_{64}$ other than luck, at least for some $E$ the rank will stay $0$ over $K_{64}$. The choice of $E$ to have full 2-torsion over $\mathbb{Q}$ is simply to reduce the degree of $K_{64}$ over $\mathbb{Q}$, which minimizes the chances $E$ will pick up new points. $\endgroup$– Stanley Yao XiaoCommented Nov 1, 2023 at 17:56
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$\begingroup$ @StanleyYaoXiao Don't we expect the average rank to increase with number field degree? $\endgroup$– Dror SpeiserCommented Nov 1, 2023 at 19:53
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$\begingroup$ @DrorSpeiser I am actually not sure what to believe on that front. I am not sure if the Katz-Sarnak conjecture extends to arbitrary number fields (that is, if one fixes a number field $K$, and one orders elliptic curves over $K$ with respect to some reasonable height, whether the average rank should be $1/2$). Certainly for a fixed elliptic curve, one can find a tower of number fields $K_0 \subset K_1 \subset \cdots K_j \subset \cdots$ such that the rank of $E$ increases as $j$ increases, but usually the construction of such a tower is contrived. $\endgroup$– Stanley Yao XiaoCommented Nov 1, 2023 at 20:15
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1$\begingroup$ @StanleyYaoXiao In any case, I'd expect that over a fixed number field and with elliptic curves ordered by some reasonable height, 100% of them should have rank 0 or 1. (Probably 50% of each, but since I haven't thought about it, maybe there's something that affects the sign of the functional equation to create unequal probabilities.) As for David's question, better probably to start with a curve with an 8-torsion point over Q. But even then, adjoining a 64-torsion point gives an extension of huge degree. Can we do a 2-descent in that field, even with the 2-torsion point to help? $\endgroup$– Joe SilvermanCommented Nov 1, 2023 at 22:46
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$\begingroup$ I agree, if $E$ is a curve that is not coming from a lower field, having a $n$-torsion point, I bet chances are good that the rank is $0$ or $1$. On the other hand for certain Galois groups the parity results coming from Brauer relations impose rank growth. I have not thought about $n=64$. But in any case, all I can offer so far is speculation that this should exist, but no idea how to construct an example or how to verify it. $\endgroup$– Chris WuthrichCommented Nov 1, 2023 at 23:52
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1 Answer
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This is not an answer, but an idea how one might get an answer with quite a bit of work. I am not too confident that I have not overlooked something.
Let $p$ be a prime and $k>1$. I am aiming to construct an elliptic curve $E$ over a number field $K$ with an $p^k$-torsion point in $E(K)$ but $E(K)$ has rank $0$.
Start with an elliptic curve $E$ with complex multiplication over some small field. Let $H$ be the field of definition of $E$ where the endomorphism ring has rank $2$. Discard $E$ if $E(H)$ is not of rank $0$. Now the extension $L/H$ adjoining the full $p^k$-torsion to $E$ is an abelian extension. There is a prime-to-$p$ extension $K/H$ such that $L/K$ is a product of two cyclic groups of $p$-power order. This sits inside the $\mathbb{Z}_p^2$-extension of $K$. One can now use the $2$-variable $p$-adic $L$-function and the main conjecture in Iwasawa theory to bound the rank of $E(L)$ from above. There is one sub-extension in which the Heegner points (from that curve $E$ itself) will force the rank to grow and that is linked to the issue with root numbers, but with some luck one can show that there is a point $P$ of order $p^k$ and $E(H(P))$ has rank $0$.
I hope there is enough results in Iwasawa theory for elliptic curves with complex multiplication that one could use this to work for odd primes $p$. For $p^k=64$, it may well be that one would have to work out first the detail of $2$-adic Iwasawa theory.
Maybe 49a1 over $\mathbb{Q}$ could work. We would have $H=\mathbb{Q}(\sqrt{-7})$, the rank of $E(H)$ is zero, and all $2$-torsion points are $H$-rational.
This relies on $E$ having complex multiplication. Because of parity/root number I am sceptical that a similar procedure should work for a curve without cm.
answered Nov 2, 2023 at 9:43
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$\begingroup$ This sounds neat! I will have to improve my Iwasawa fu to try it out, but that's a worthy cause in itself. $\endgroup$ Commented Nov 2, 2023 at 17:55