15
$\begingroup$

Let $E/\mathbb{Q}$ be an elliptic curve. The weak Birch and Swinnerton-Dyer conjecture predicts that $$\text{ord}_{s=1}L(E, s)=\text{rank} E(\mathbb{Q}).$$ Thanks to the work of Gross-Zagier and Kolyvagin, we know that this conjecture is true if $\text{ord}_{s=1}L(E, s)\le 1$.

What is known in the case $\text{rank} E(\mathbb{Q})\le 1$? Is it known that if $\text{rank} E(\mathbb{Q})=1$, then $L'(E, 1)\neq 0$?

Thank you!

Improve this question
$\endgroup$
3
  • 9
    $\begingroup$ In fairness, it's probably worth mentioning that in addition to the work of G-Z and Kolyvagin, one also needs the theorem of Wiles et al that every elliptic curve over $\mathbb Q$ is modular, since [G-Z] and [K] start with the assumption that their elliptic curve is modular. $\endgroup$
    – Joe Silverman
    Commented Aug 25, 2018 at 11:35
  • $\begingroup$ ...and a non-vanishing result for special values of L functions. $\endgroup$
    – Pasten
    Commented Aug 26, 2018 at 0:29
  • $\begingroup$ Cross-posted: math.stackexchange.com/questions/2893669 $\endgroup$
    – Watson
    Commented Oct 20, 2018 at 12:00

1 Answer 1

Reset to default
14
$\begingroup$

The following theorem is due to Chris Skinner, in this 2014 paper.

Let E/Q be an elliptic curve such that rank E(Q) = 1 and the Tate-Shafarevich group Sha(E / Q) is finite, and some other technical assumptions hold. Then $ord_{s = 1} L(E, s) = 1$, and in particular $L'(E, 1) \ne 0$.

This is, as far as I know, the best one can do at the moment; if you don't know that Sha (or at least its p-primary part for some p) is finite, then you're stuck.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .