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It is conjectured that there are elliptic curves over $\mathbb Q$ of arbitrarily high rank. I was wondering wether someone made a similar conjecture if one restricts to a fixed $j$-invariant. If there are specific values of $j$ known such that the following question has a negative answer would also be nice to know.

Let $j \in \mathbb Q$ be fixed. Do there exist elliptic curves $E/\mathbb Q$ with $j(E)=j$ of arbitrary high rank over $\mathbb Q$?

I'm specifically interested in the value $j=0$.

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Do you mean the rank of $E$ as a curve over $\bf Q$ ? But then that would prove the conjecture... If you want to examine the rank of twists of $E$ then you need to enlarge the field of definition of the Mordell-Weil group (or did I misunderstand something ?). –  Damian Rössler Nov 26 '12 at 9:26
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Until recently, I thought all conjectures were in favour of "arbitrary high" for both situation. But recently I have seen counter-conjectures stating that all ranks are bounded and small... My own conjecture is that no one would bet a huge sum on anything in either directions. –  Chris Wuthrich Nov 26 '12 at 9:34
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The case $j=0$ is stated as an open problem in "Taxicabs and sums of two cubes" by Silverman, p. 337. It would follow from the fact that the number of integral points on the curves $x^3+y^3=A$ with $A$ cube-free is unbounded. –  François Brunault Nov 26 '12 at 9:37
    
@Damian: Yes I mean over $\Q$. A positive answer for a single $j$ would indeed prove the conjecture, but it might still be so that there are $j$ for which it is known that my question has negative answer. I also don't understand why I would have to enlarge the field of definition. For example all curves of the form $y^2=x^3+a$ with $a \in \mathbb Q$ have j-invariant zero, and this family of curves contains infinitly many non isomorphic curves. So it makes perfect sense to ask wether this family contains curves of arbitrary high rank. –  Maarten Derickx Nov 26 '12 at 10:54

3 Answers 3

up vote 17 down vote accepted

This is a very interesting question, and there was renewed interest in it lately. As Chris Wuthrich says in the comments, the situation used to be that most people believed in unboundedness of ranks for elliptic curves with a given $j$-invariant, although there was no particular evidence in favour of this. In fact, there was an old conjecture of Honda

T. Honda, Isogenies, rational points and section points of group varieties, Japan. J. Math., 30 (1960), 84-101

that asserted boundedness, but people did not really believe it. However, recently, there was work by Andrew Granville, Mark Watkins and others with some serious heuristics pointing towards boundedness as well. Mark's slides for his recent talk at the Warwick conference about this are here. In particular, the heuristics suggest that for the family $$ X^3+Y^3=A\qquad (j=0) $$ the correct answer could be that the ranks are bounded by $9$, except for finitely many curves in the family. Mark also notes, though, that all these heuristics are quite shaky, and are also very difficult to verify, so at the moment it is really not clear what to expect.

Finally, people working in random matrix theory, notably Nina Snaith, Jon Keating and their collaborators are also looking into higher ranks in the family of quadratic twists. Nina spoke about it here just two weeks ago, and these are her slides. They are working on a precise heuristic for the number of rank 2 twists, with a hope that the approach would eventually lead to a conjectural formula for the number of higher rank twists as well.

In any case, the current status is that the precise answer is not known for a single $j$-invariant.

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Thank you for your reference to the literature with a conjectural answer to my question and for the elaborate discussion of the current status of the conjecture. This was exactly the kind of answer I was hoping for. –  Maarten Derickx Nov 26 '12 at 12:57

This is not a complete answer to your question (since I think it isn't known), but you might still find this interesting.

Mestre has shown in Rang de courbes elliptiques d'invariant donné (http://arxiv.org/abs/alg-geom/9206007) that any elliptic curve over $\mathbf{Q}$ has a twist with rank at least $2$. For $j=1728$, he improves this to rank at least $4$, and for $j=0$ to rank at least $6$. I think he even gets infinitely many such twists in each case, but I'm not sure how easily this follows from what he says (see below though).

Stewart and Top have improved on Mestre's result for arbitrary elliptic curves, by giving a lower bound of the form $cT^a/(\log T)^b$ on the number of quadratic twists with rank at least $2$ and with "twist parameter" below $T$. See Theorem 3 in their On ranks of twists of elliptic curves and power-free values of binary forms (http://www.math.rug.nl/~top/StewartTop.pdf). For the $j=0$ family $y^2 = x^3 + d$, they get similar lower bounds for the number of members with rank at least $2, 3, 4, 5$ and $6$ (Theorem 9).

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He indeed shows that there are infinitely many by giving a whole family in each case. Is it a coincidence that 2,4 and 6 are exactly the numbers of $\bar k$ automorphisms of any $E$ with those $j$ invariants? –  Maarten Derickx Nov 26 '12 at 12:38

This doesn't answer your question, but I'll mention it anyway. My thesis research shows that the average rank of $j = 0$ elliptic curves, when ordered by discriminant, is bounded by 1.5. This follows from computing the average size of the 2-Selmer group, which is 3.

I'll also mention that your question is much easier to answer if you replace Mordell-Weil rank with 2-Selmer rank. In that case, Heath-Brown has two papers (The Size of the Selmer Group for the Congruent Number problem I, II) showing that for any $r$, a positive density of the curves $y^2 = x^3-D^2x$ have 2-Selmer rank $r$. Mazur and Rubin have a paper (Ranks of Twists of Elliptic Curves and Hilbert's Tenth Problem) where they consider quadratic twists of curves with no rational 2-torsion and give some conditions for the family of twists to have arbitrary 2-Selmer rank.

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