There are no elliptic curves over $mathbb{F}_8$ with $7$ or $11$ points












4












$begingroup$


This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).



One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.



However, the question advises to use the previous part, which states:



Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.



For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.



It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.



Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?






elliptic-curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
    $endgroup$
    – sdf
    Jul 9 '14 at 3:41












  • $begingroup$
    I have tried out all possible examples and the case of $q=8$ has not failed.
    $endgroup$
    – Haikal Yeo
    Jul 9 '14 at 11:55












  • $begingroup$
    what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
    $endgroup$
    – sdf
    Jul 9 '14 at 14:31
















4












$begingroup$


This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).



One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.



However, the question advises to use the previous part, which states:



Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.



For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.



It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.



Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?






elliptic-curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
    $endgroup$
    – sdf
    Jul 9 '14 at 3:41












  • $begingroup$
    I have tried out all possible examples and the case of $q=8$ has not failed.
    $endgroup$
    – Haikal Yeo
    Jul 9 '14 at 11:55












  • $begingroup$
    what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
    $endgroup$
    – sdf
    Jul 9 '14 at 14:31














4












4








4


1



$begingroup$


This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).



One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.



However, the question advises to use the previous part, which states:



Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.



For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.



It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.



Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?






elliptic-curves







share|cite|improve this question











$endgroup$




This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).



One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.



However, the question advises to use the previous part, which states:



Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.



For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.



It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.



Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?






elliptic-curves




elliptic-curves






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 6:16









Did

248k23226465




248k23226465










asked May 1 '14 at 21:34









Haikal YeoHaikal Yeo

1,027823




1,027823












  • $begingroup$
    I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
    $endgroup$
    – sdf
    Jul 9 '14 at 3:41












  • $begingroup$
    I have tried out all possible examples and the case of $q=8$ has not failed.
    $endgroup$
    – Haikal Yeo
    Jul 9 '14 at 11:55












  • $begingroup$
    what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
    $endgroup$
    – sdf
    Jul 9 '14 at 14:31


















  • $begingroup$
    I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
    $endgroup$
    – sdf
    Jul 9 '14 at 3:41












  • $begingroup$
    I have tried out all possible examples and the case of $q=8$ has not failed.
    $endgroup$
    – Haikal Yeo
    Jul 9 '14 at 11:55












  • $begingroup$
    what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
    $endgroup$
    – sdf
    Jul 9 '14 at 14:31
















$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41






$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41














$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55






$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55














$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31




$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31










1 Answer
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$begingroup$

You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
$$
rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
$$
(Here, $lceil k/2rceil$ is the ceiling function.)



In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.






share|cite|improve this answer









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    $begingroup$

    You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
    $$
    rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
    $$
    (Here, $lceil k/2rceil$ is the ceiling function.)



    In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
      $$
      rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
      $$
      (Here, $lceil k/2rceil$ is the ceiling function.)



      In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
        $$
        rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
        $$
        (Here, $lceil k/2rceil$ is the ceiling function.)



        In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.






        share|cite|improve this answer









        $endgroup$



        You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
        $$
        rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
        $$
        (Here, $lceil k/2rceil$ is the ceiling function.)



        In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 5:10









        GispsanGispsan

        463




        463






























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