How to prove that $text{End}_{mathbb{F}_p}(E)$ is commutative for a given elliptic curve E?












1












$begingroup$


Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:



$mathbb{Z}$



An order in an imaginary quadratic field



An order in a quaternion algebra



The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.



Any help on how to approach this will be appreciated, thanks.






number-theory analytic-number-theory elliptic-curves







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$endgroup$












  • $begingroup$
    In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
    $endgroup$
    – reuns
    Jan 26 at 18:05


















1












$begingroup$


Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:



$mathbb{Z}$



An order in an imaginary quadratic field



An order in a quaternion algebra



The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.



Any help on how to approach this will be appreciated, thanks.






number-theory analytic-number-theory elliptic-curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
    $endgroup$
    – reuns
    Jan 26 at 18:05
















1












1








1


1



$begingroup$


Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:



$mathbb{Z}$



An order in an imaginary quadratic field



An order in a quaternion algebra



The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.



Any help on how to approach this will be appreciated, thanks.






number-theory analytic-number-theory elliptic-curves







share|cite|improve this question











$endgroup$




Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:



$mathbb{Z}$



An order in an imaginary quadratic field



An order in a quaternion algebra



The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.



Any help on how to approach this will be appreciated, thanks.






number-theory analytic-number-theory elliptic-curves




number-theory analytic-number-theory elliptic-curves






share|cite|improve this question















share|cite|improve this question













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edited Jan 26 at 16:24







AiYagami

















asked Jan 26 at 16:18









AiYagamiAiYagami

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635












  • $begingroup$
    In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
    $endgroup$
    – reuns
    Jan 26 at 18:05




















  • $begingroup$
    In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
    $endgroup$
    – reuns
    Jan 26 at 18:05


















$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05






$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05












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$begingroup$

$text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
so $Bbb Z[F]$ is a quadratic imaginary order.



If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.



If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
quaternion order, and the elements that commute with $F$ are just those in
$Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
$Bbb Q(F)$.



In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
$Bbb Z [F]$.






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    $begingroup$

    $text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
    with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
    so $Bbb Z[F]$ is a quadratic imaginary order.



    If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
    and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.



    If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
    quaternion order, and the elements that commute with $F$ are just those in
    $Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
    $Bbb Q(F)$.



    In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
    $Bbb Z [F]$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
      with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
      so $Bbb Z[F]$ is a quadratic imaginary order.



      If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
      and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.



      If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
      quaternion order, and the elements that commute with $F$ are just those in
      $Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
      $Bbb Q(F)$.



      In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
      $Bbb Z [F]$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
        with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
        so $Bbb Z[F]$ is a quadratic imaginary order.



        If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
        and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.



        If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
        quaternion order, and the elements that commute with $F$ are just those in
        $Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
        $Bbb Q(F)$.



        In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
        $Bbb Z [F]$.






        share|cite|improve this answer









        $endgroup$



        $text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
        with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
        so $Bbb Z[F]$ is a quadratic imaginary order.



        If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
        and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.



        If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
        quaternion order, and the elements that commute with $F$ are just those in
        $Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
        $Bbb Q(F)$.



        In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
        $Bbb Z [F]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 16:35









        Lord Shark the UnknownLord Shark the Unknown

        106k1161133




        106k1161133






























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