Calculate the number of points of an elliptic curve in medium Weierstrass form over finite field












5












$begingroup$


Let $E$ be the elliptic curve over $mathbb{F}_3$ in medium Weierstrass form $E:y^2=x^3+x^2+x+1$. How to compute the number of points $|E(mathbb{F}_{3^k})|$? I read that there are some formulas for computing number of points for short Weierstrass form by Frobenius endomorphism. But they don't work in this case.






number-theory elliptic-curves







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$endgroup$

















    5












    $begingroup$


    Let $E$ be the elliptic curve over $mathbb{F}_3$ in medium Weierstrass form $E:y^2=x^3+x^2+x+1$. How to compute the number of points $|E(mathbb{F}_{3^k})|$? I read that there are some formulas for computing number of points for short Weierstrass form by Frobenius endomorphism. But they don't work in this case.






    number-theory elliptic-curves







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      Let $E$ be the elliptic curve over $mathbb{F}_3$ in medium Weierstrass form $E:y^2=x^3+x^2+x+1$. How to compute the number of points $|E(mathbb{F}_{3^k})|$? I read that there are some formulas for computing number of points for short Weierstrass form by Frobenius endomorphism. But they don't work in this case.






      number-theory elliptic-curves







      share|cite|improve this question









      $endgroup$




      Let $E$ be the elliptic curve over $mathbb{F}_3$ in medium Weierstrass form $E:y^2=x^3+x^2+x+1$. How to compute the number of points $|E(mathbb{F}_{3^k})|$? I read that there are some formulas for computing number of points for short Weierstrass form by Frobenius endomorphism. But they don't work in this case.






      number-theory elliptic-curves




      number-theory elliptic-curves






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      asked 5 hours ago









      NickyNicky

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          1 Answer
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          $begingroup$

          Let $phi^k(x,y)= (x^{3^k},y^{3^k})$ then $#E(mathbb{F}_{3^k}) =deg_s(phi^k-1)$. Is the endomorphism $phi^k-1$ separable ? Yes because inserapable endomorphisms are of the form $rho circ phi$. Then $$deg_s(phi^k-1) = deg(phi^k-1)=((phi^*)^k-1)(phi^k-1)\= (phi^*phi)^k+1-(phi^*)^k-phi^k = 3^k+1-alpha^k-(alpha^*)^k$$ where $phi^*$ is the dual isogeny such that $phi^* phi = deg(phi) = 3$ and $phi+phi^* = t = 3+1-#E(mathbb{F}_{3})$ and $alpha$ is the root of the minimal polynomial $X^2-t X + 3 = 0$ of the Frobenius



          magma code



               F := FiniteField(3); A<x,y> := AffineSpace(F,2);
          
     C := Curve(A,y^2-x^3-x^2-x-1);
          
     t :=3+1- #Points(ProjectiveClosure(C));
          
     P<z> := PolynomialRing(Integers()); K<a> := NumberField(z^2-t*z+3); aa := Norm(a)/a;
          

          
     for k in [2..10] do
          
       Ck := BaseChange(C,FiniteField(3^k));
          
       Ek := #Points(ProjectiveClosure(Ck));
          
       [Ek,3^k+1-a^k-aa^k];
          
     end for;


          To obtain the minimal polynomial of endomorphisms :



          Write that $E(overline{mathbb{F}_3}) $ is a subgroup of $mathbb{Q}/mathbb{Z}times mathbb{Q}/mathbb{Z}$ so any group homomorphism acts as a matrix
          $A=pmatrix{a & b \c & d} in M_2(widehat{mathbb{Z}})$ (matrix of profinite integers). Then the dual homomorphism is $A^*=pmatrix{d & -b \-c & a}$ so that $A^* A = pmatrix{ad-bc& 0 \ 0 & ad-bc}$ and $A + A^* = pmatrix{a+d & 0 \0 & a+d}$, so they both act as direct multiplication by an element in $widehat{mathbb{Z}}$. If $A$ is an endomorphism (defined by polynomial equations) then so are $A^*,A + A^*,A^*A$ so the latter must act as multiplication by elements in $mathbb{Z}$.






          share|cite|improve this answer











          $endgroup$













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            4












            $begingroup$

            Let $phi^k(x,y)= (x^{3^k},y^{3^k})$ then $#E(mathbb{F}_{3^k}) =deg_s(phi^k-1)$. Is the endomorphism $phi^k-1$ separable ? Yes because inserapable endomorphisms are of the form $rho circ phi$. Then $$deg_s(phi^k-1) = deg(phi^k-1)=((phi^*)^k-1)(phi^k-1)\= (phi^*phi)^k+1-(phi^*)^k-phi^k = 3^k+1-alpha^k-(alpha^*)^k$$ where $phi^*$ is the dual isogeny such that $phi^* phi = deg(phi) = 3$ and $phi+phi^* = t = 3+1-#E(mathbb{F}_{3})$ and $alpha$ is the root of the minimal polynomial $X^2-t X + 3 = 0$ of the Frobenius



            magma code



                 F := FiniteField(3); A<x,y> := AffineSpace(F,2);
            
     C := Curve(A,y^2-x^3-x^2-x-1);
            
     t :=3+1- #Points(ProjectiveClosure(C));
            
     P<z> := PolynomialRing(Integers()); K<a> := NumberField(z^2-t*z+3); aa := Norm(a)/a;
            

            
     for k in [2..10] do
            
       Ck := BaseChange(C,FiniteField(3^k));
            
       Ek := #Points(ProjectiveClosure(Ck));
            
       [Ek,3^k+1-a^k-aa^k];
            
     end for;


            To obtain the minimal polynomial of endomorphisms :



            Write that $E(overline{mathbb{F}_3}) $ is a subgroup of $mathbb{Q}/mathbb{Z}times mathbb{Q}/mathbb{Z}$ so any group homomorphism acts as a matrix
            $A=pmatrix{a & b \c & d} in M_2(widehat{mathbb{Z}})$ (matrix of profinite integers). Then the dual homomorphism is $A^*=pmatrix{d & -b \-c & a}$ so that $A^* A = pmatrix{ad-bc& 0 \ 0 & ad-bc}$ and $A + A^* = pmatrix{a+d & 0 \0 & a+d}$, so they both act as direct multiplication by an element in $widehat{mathbb{Z}}$. If $A$ is an endomorphism (defined by polynomial equations) then so are $A^*,A + A^*,A^*A$ so the latter must act as multiplication by elements in $mathbb{Z}$.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              Let $phi^k(x,y)= (x^{3^k},y^{3^k})$ then $#E(mathbb{F}_{3^k}) =deg_s(phi^k-1)$. Is the endomorphism $phi^k-1$ separable ? Yes because inserapable endomorphisms are of the form $rho circ phi$. Then $$deg_s(phi^k-1) = deg(phi^k-1)=((phi^*)^k-1)(phi^k-1)\= (phi^*phi)^k+1-(phi^*)^k-phi^k = 3^k+1-alpha^k-(alpha^*)^k$$ where $phi^*$ is the dual isogeny such that $phi^* phi = deg(phi) = 3$ and $phi+phi^* = t = 3+1-#E(mathbb{F}_{3})$ and $alpha$ is the root of the minimal polynomial $X^2-t X + 3 = 0$ of the Frobenius



              magma code



                   F := FiniteField(3); A<x,y> := AffineSpace(F,2);
              
     C := Curve(A,y^2-x^3-x^2-x-1);
              
     t :=3+1- #Points(ProjectiveClosure(C));
              
     P<z> := PolynomialRing(Integers()); K<a> := NumberField(z^2-t*z+3); aa := Norm(a)/a;
              

              
     for k in [2..10] do
              
       Ck := BaseChange(C,FiniteField(3^k));
              
       Ek := #Points(ProjectiveClosure(Ck));
              
       [Ek,3^k+1-a^k-aa^k];
              
     end for;


              To obtain the minimal polynomial of endomorphisms :



              Write that $E(overline{mathbb{F}_3}) $ is a subgroup of $mathbb{Q}/mathbb{Z}times mathbb{Q}/mathbb{Z}$ so any group homomorphism acts as a matrix
              $A=pmatrix{a & b \c & d} in M_2(widehat{mathbb{Z}})$ (matrix of profinite integers). Then the dual homomorphism is $A^*=pmatrix{d & -b \-c & a}$ so that $A^* A = pmatrix{ad-bc& 0 \ 0 & ad-bc}$ and $A + A^* = pmatrix{a+d & 0 \0 & a+d}$, so they both act as direct multiplication by an element in $widehat{mathbb{Z}}$. If $A$ is an endomorphism (defined by polynomial equations) then so are $A^*,A + A^*,A^*A$ so the latter must act as multiplication by elements in $mathbb{Z}$.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                Let $phi^k(x,y)= (x^{3^k},y^{3^k})$ then $#E(mathbb{F}_{3^k}) =deg_s(phi^k-1)$. Is the endomorphism $phi^k-1$ separable ? Yes because inserapable endomorphisms are of the form $rho circ phi$. Then $$deg_s(phi^k-1) = deg(phi^k-1)=((phi^*)^k-1)(phi^k-1)\= (phi^*phi)^k+1-(phi^*)^k-phi^k = 3^k+1-alpha^k-(alpha^*)^k$$ where $phi^*$ is the dual isogeny such that $phi^* phi = deg(phi) = 3$ and $phi+phi^* = t = 3+1-#E(mathbb{F}_{3})$ and $alpha$ is the root of the minimal polynomial $X^2-t X + 3 = 0$ of the Frobenius



                magma code



                     F := FiniteField(3); A<x,y> := AffineSpace(F,2);
                
     C := Curve(A,y^2-x^3-x^2-x-1);
                
     t :=3+1- #Points(ProjectiveClosure(C));
                
     P<z> := PolynomialRing(Integers()); K<a> := NumberField(z^2-t*z+3); aa := Norm(a)/a;
                

                
     for k in [2..10] do
                
       Ck := BaseChange(C,FiniteField(3^k));
                
       Ek := #Points(ProjectiveClosure(Ck));
                
       [Ek,3^k+1-a^k-aa^k];
                
     end for;


                To obtain the minimal polynomial of endomorphisms :



                Write that $E(overline{mathbb{F}_3}) $ is a subgroup of $mathbb{Q}/mathbb{Z}times mathbb{Q}/mathbb{Z}$ so any group homomorphism acts as a matrix
                $A=pmatrix{a & b \c & d} in M_2(widehat{mathbb{Z}})$ (matrix of profinite integers). Then the dual homomorphism is $A^*=pmatrix{d & -b \-c & a}$ so that $A^* A = pmatrix{ad-bc& 0 \ 0 & ad-bc}$ and $A + A^* = pmatrix{a+d & 0 \0 & a+d}$, so they both act as direct multiplication by an element in $widehat{mathbb{Z}}$. If $A$ is an endomorphism (defined by polynomial equations) then so are $A^*,A + A^*,A^*A$ so the latter must act as multiplication by elements in $mathbb{Z}$.






                share|cite|improve this answer











                $endgroup$



                Let $phi^k(x,y)= (x^{3^k},y^{3^k})$ then $#E(mathbb{F}_{3^k}) =deg_s(phi^k-1)$. Is the endomorphism $phi^k-1$ separable ? Yes because inserapable endomorphisms are of the form $rho circ phi$. Then $$deg_s(phi^k-1) = deg(phi^k-1)=((phi^*)^k-1)(phi^k-1)\= (phi^*phi)^k+1-(phi^*)^k-phi^k = 3^k+1-alpha^k-(alpha^*)^k$$ where $phi^*$ is the dual isogeny such that $phi^* phi = deg(phi) = 3$ and $phi+phi^* = t = 3+1-#E(mathbb{F}_{3})$ and $alpha$ is the root of the minimal polynomial $X^2-t X + 3 = 0$ of the Frobenius



                magma code



                     F := FiniteField(3); A<x,y> := AffineSpace(F,2);
                
     C := Curve(A,y^2-x^3-x^2-x-1);
                
     t :=3+1- #Points(ProjectiveClosure(C));
                
     P<z> := PolynomialRing(Integers()); K<a> := NumberField(z^2-t*z+3); aa := Norm(a)/a;
                

                
     for k in [2..10] do
                
       Ck := BaseChange(C,FiniteField(3^k));
                
       Ek := #Points(ProjectiveClosure(Ck));
                
       [Ek,3^k+1-a^k-aa^k];
                
     end for;


                To obtain the minimal polynomial of endomorphisms :



                Write that $E(overline{mathbb{F}_3}) $ is a subgroup of $mathbb{Q}/mathbb{Z}times mathbb{Q}/mathbb{Z}$ so any group homomorphism acts as a matrix
                $A=pmatrix{a & b \c & d} in M_2(widehat{mathbb{Z}})$ (matrix of profinite integers). Then the dual homomorphism is $A^*=pmatrix{d & -b \-c & a}$ so that $A^* A = pmatrix{ad-bc& 0 \ 0 & ad-bc}$ and $A + A^* = pmatrix{a+d & 0 \0 & a+d}$, so they both act as direct multiplication by an element in $widehat{mathbb{Z}}$. If $A$ is an endomorphism (defined by polynomial equations) then so are $A^*,A + A^*,A^*A$ so the latter must act as multiplication by elements in $mathbb{Z}$.







                share|cite|improve this answer














                share|cite|improve this answer



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