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Let $K$ be a number field and let $\mathcal O_K$ be the ring of integers. Following this paper of Cornelissen, Pheidas, and Zahidi, a key ingredient needed to show that Hilbert's tenth problem has a negative solution over $\mathcal O_K$ is an elliptic curve $E$ defined over $K$ with rank$(E(K))=1$.

Recently Mazur and Rubin have shown that such a curve exists assuming the Shafarevich-Tate conjecture for elliptic curves over number fields. They actually use a weaker, but still inaccessible hypothesis (See conjecture $ШT_2$).

If you wanted to eliminate the need for this hypothesis you would have to write a proof that simultaneously demonstrated that rank$(E(K))=1$ for infinitely many pairs $(K,E)$ where $E$ is an elliptic curve defined over $K.$ This raises (as opposed to begs) the easier question:

Can you show unconditionally that rank$(E(\Bbb Q)) = 1$ for infinitely many elliptic curves $E$ over $\Bbb Q$?

It would appear that Byeon, Jeon, and Kim have done so in this paper (probably need an institutional login). Vatsal obtains a weaker result here that still does the job. Unfortunately both of these results invoke the fact that the BSD rank conjecture is true for elliptic curves over $\Bbb Q$ with analytic rank 1. Which won't help at present working over number fields.

Can anyone do the above WITHOUT invoking the proven part of the BSD rank conjecture or assuming any conjectures?

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  • $\begingroup$ This is a nice question. A couple of quick comments: (i) the way you have stated Mazur-Rubin's result doesn't make sense: the existence of an elliptic curve cannot be conditional on the structure of its Shafarevich-Tate group! (Go back and look again at conjecture ST_2.) (ii) Full BSD is not yet known for all elliptic curves over Q of analytic rank 1; but it is known that this implies that the Mordell-Weil rank is 1, which is what you have in mind. $\endgroup$
    – Pete L. Clark
    Commented Apr 16, 2010 at 2:05
  • $\begingroup$ Sorry, I missed that you said "BSD rank conjecture" rather than just "BSD". I withdraw my second comment. $\endgroup$
    – Pete L. Clark
    Commented Apr 16, 2010 at 2:06
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    $\begingroup$ @Joel: Thanks, I will try to avoid using "begs the question" improperly. @Pete: Thanks for your comments, I've fixed the error you pointed out. Let me know if there's anything else I should do to make the question more clear. I know you've thought about this stuff a lot. $\endgroup$
    – James Weigandt
    Commented Apr 16, 2010 at 4:38
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    $\begingroup$ Last month I went to two amazing talks by Manjul Bhargava (one at IAS and one at Princeton) in which he proved (among other things) that the average rank of the 2-Selmer group of an elliptic curve over $\mathbb{Q}$, where the curves are ordered by Faltings height, is exactly 3. This is unconditional (no GRH, no BSD) obtained by analyzing rings of invariants and making precise counts of lattice points. $\endgroup$
    – Victor Miller
    Commented Apr 21, 2010 at 1:15

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I don't think that this should be too hard: take a simple family of curves, such as $y^2 = x^3 + px$ or something similar, and choose $p$ from a certain set of residue classes to guarantee that the 2-Selmer group has rank 1. You can complete the proof either by invoking rather deep constructions using Heegner points, or finding a family for which conditions such as $p = a^4 + b^2$ (there are infinitely many primes of this form) give you a global point (choose your family in such a way that $b^2 = px^4 - y^4$ occurs as a principal homogeneous space in your standard 2-descent; see e.g. Silverman's book). Sorry for being a little bit vague - a hard disk crash currently prevents me from looking at my own notes.

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    $\begingroup$ This answers the question over ${\bf Q}$ (and it's easier to use $y^2 = x^3 - p^2 x^2$ for $p \equiv 5 \bmod 8$, where the rank is at most $1$ by $2$-descent, and Heegner used what are now ironically called "mock-Heegner points" to construct a point of infinite order). But the question for arbitrary $K$ is harder. For example, no CM curve can work if $K$ contains the CM field, because then the rank is even. For example, $y^2 = x^3 + a x$ always has even rank over a field containing ${\bf Q}(i)$. $\endgroup$
    – Noam D. Elkies
    Commented May 8, 2012 at 19:25
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This is just a tiny follow up to Victor Miller's remark above. Manjul Bharghava has a second paper with Shankar entitled "Ternary cubic forms having bounded invariants, and the existence of a positive proportion of elliptic curves having rank 0", and in there he states the following theorem:

Theorem 5: Assume $Sha(E)$ is finite for all $E$. When all elliptic curves $E/\mathbf{Q}$ are ordered by height, a positive proportion of them have rank $1$.

This obviously doesn't answer your original question, because Manjul is assuming that Sha is finite. However, I believe his arguments for this theorem use results toward BSD over $\mathbf{Q}$.

EDIT: Crud -- I meant to say that I believe his arguments don't use results toward BSD, which was the whole point. Oops.

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  • $\begingroup$ I don't Theorem 5 uses anything analytic. Basically they (Bhargava/Shankar) shows an upper bound on average 3-rank of 7/6 for under their ordering (Corollary 2), and from this in Section 4 use Dokchitsers' parity result to split into even and odd in an acceptable family, and everything follows from this. So that 50% have odd 3-rank, and 3-rank is $\le 7/6$ on average, so if Sha[3] is finite, it follows (Theorem 37). The analytic part they quote the evanescent Skinner-Urban result in the even case to get trivial Sel[3] plus technical conditions implies analytic rank 0. $\endgroup$
    – Junkie
    Commented Aug 13, 2010 at 6:20
  • $\begingroup$ To be more explicit, in Theorem 37, they note that, in the odd case the 3-rank is either 1 or at least 3. By the bound of $7/6$ the 3-rank is 1 a positive proportion of the time. If Sha is finite in these cases, then 3-rank is 1 implies rank is 1. The finiteness of Sha$[3^\infty]$, correcting my above comment, is used to imply that 3-rank is odd implies rank is odd, hence positive. $\endgroup$
    – Junkie
    Commented Aug 13, 2010 at 6:26
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    $\begingroup$ The forthcoming work of Manjul Bhargava and Wei Ho on elliptic curves in families with one and two marked points shows that there are infinitely many elliptic curves over $\Bbb Q$ with exact Mordell-Weil ranks 1 and 2 respectively. I think this gives the most satisfying answer to this question, and it's believed that (with work) these results should carry over to number fields. I suppose the next natural question is wether this can be used to attack Hilbert's Tenth Problem for rings of integers of number fields. $\endgroup$
    – James Weigandt
    Commented Apr 16, 2011 at 19:09
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If $K$ is totally real, then the rank part of BSD is known in analytic rank 1 (using CM points for instance) for modular elliptic curves. As potential modularity is also known under the same hypotheses, I am quite confident that this part of Vatsal and Byeon, Jeaon and Kim generalizes. Is it OK for you to assume that $K$ is totally real?

On the other hand, their proofs also use some results on the class group of quadratic fields, so this part would not generalize (as far as I know, which is not very far).

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  • $\begingroup$ The hope is to do it for any number field $K$. I was unaware that results on the BSD rank conjecture were extended to totally real fields. This is my own ignorance because I've only really thought about curves over Q until very recently. $\endgroup$
    – James Weigandt
    Commented Apr 16, 2010 at 17:15

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