BSD conjecture for rank 1 elliptic curves

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Let $E/mathbbQ$ be an elliptic curve. The weak Birch and Swinnerton-Dyer conjecture predicts that
$$textord_s=1L(E, s)=textrank E(mathbbQ).$$
Thanks to the work of Gross-Zagier and Kolyvagin, we know that this conjecture is true if $textord_s=1L(E, s)le 1$.



What is known in the case $textrank E(mathbbQ)le 1$? Is it known that if $textrank E(mathbbQ)=1$, then $L'(E, 1)neq 0$?



Thank you!







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  • 8




    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.
    – Joe Silverman
    Aug 25 at 11:35










  • ...and a non-vanishing result for special values of L functions.
    – Pasten
    Aug 26 at 0:29














up vote
13
down vote

favorite
6












Let $E/mathbbQ$ be an elliptic curve. The weak Birch and Swinnerton-Dyer conjecture predicts that
$$textord_s=1L(E, s)=textrank E(mathbbQ).$$
Thanks to the work of Gross-Zagier and Kolyvagin, we know that this conjecture is true if $textord_s=1L(E, s)le 1$.



What is known in the case $textrank E(mathbbQ)le 1$? Is it known that if $textrank E(mathbbQ)=1$, then $L'(E, 1)neq 0$?



Thank you!







share|cite|improve this question


















  • 8




    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.
    – Joe Silverman
    Aug 25 at 11:35










  • ...and a non-vanishing result for special values of L functions.
    – Pasten
    Aug 26 at 0:29












up vote
13
down vote

favorite
6









up vote
13
down vote

favorite
6






6





Let $E/mathbbQ$ be an elliptic curve. The weak Birch and Swinnerton-Dyer conjecture predicts that
$$textord_s=1L(E, s)=textrank E(mathbbQ).$$
Thanks to the work of Gross-Zagier and Kolyvagin, we know that this conjecture is true if $textord_s=1L(E, s)le 1$.



What is known in the case $textrank E(mathbbQ)le 1$? Is it known that if $textrank E(mathbbQ)=1$, then $L'(E, 1)neq 0$?



Thank you!







share|cite|improve this question














Let $E/mathbbQ$ be an elliptic curve. The weak Birch and Swinnerton-Dyer conjecture predicts that
$$textord_s=1L(E, s)=textrank E(mathbbQ).$$
Thanks to the work of Gross-Zagier and Kolyvagin, we know that this conjecture is true if $textord_s=1L(E, s)le 1$.



What is known in the case $textrank E(mathbbQ)le 1$? Is it known that if $textrank E(mathbbQ)=1$, then $L'(E, 1)neq 0$?



Thank you!









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 25 at 10:40

























asked Aug 25 at 10:29









baobab

685




685







  • 8




    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.
    – Joe Silverman
    Aug 25 at 11:35










  • ...and a non-vanishing result for special values of L functions.
    – Pasten
    Aug 26 at 0:29












  • 8




    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.
    – Joe Silverman
    Aug 25 at 11:35










  • ...and a non-vanishing result for special values of L functions.
    – Pasten
    Aug 26 at 0:29







8




8




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.
– Joe Silverman
Aug 25 at 11:35




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.
– Joe Silverman
Aug 25 at 11:35












...and a non-vanishing result for special values of L functions.
– Pasten
Aug 26 at 0:29




...and a non-vanishing result for special values of L functions.
– Pasten
Aug 26 at 0:29










1 Answer
1






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up vote
14
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accepted










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.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    14
    down vote



    accepted










    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.






    share|cite|improve this answer
























      up vote
      14
      down vote



      accepted










      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.






      share|cite|improve this answer






















        up vote
        14
        down vote



        accepted







        up vote
        14
        down vote



        accepted






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 25 at 11:27









        David Loeffler

        18.7k146112




        18.7k146112



























             

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