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

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

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


Proving the condition for two elliptic curves given in Weierstrass form to be isomorphicEndomorphism Ring of an Elliptic Curve over Finite FieldComputation of the 2-torsion group of an elliptic curveHasse's Theorem for Elliptic Curves over Finite Fields + proof clarificationTopics in elliptic curves over finite fieldsElliptic curve $y^2= x^3 + x$ over the finite field $mathbb{F}_p$ with $p geq 3$.Addition of points on elliptic curves over a finite fieldAdding points on an elliptic curveDirect sum of two points on an elliptic curveWeierstrass Form of an Elliptic Curve













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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.










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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.










    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.










      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






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      asked 4 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



                share|cite|improve this answer








                edited 20 mins ago

























                answered 3 hours ago









                reunsreuns

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