[Math] Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD

elliptic-curveslo.logicnt.number-theory

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 $IIIT_2$).

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?

Best Answer

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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