Calculate the number of points of an elliptic curve in medium Weierstrass form over finite field
$begingroup$
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
asked 5 hours ago
NickyNicky
736
$endgroup$
add a comment |
$begingroup$
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
asked 5 hours ago
NickyNicky
736
$endgroup$
add a comment |
$begingroup$
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
asked 5 hours ago
NickyNicky
736
$endgroup$
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
asked 5 hours ago
NickyNicky
736
asked 5 hours ago
NickyNicky
736
asked 5 hours ago
NickyNicky
736
asked 5 hours ago
NickyNicky
736
asked 5 hours ago
NickyNicky
736
736
add a comment |
add a comment |
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 $mathbb{Z}$.
answered 3 hours ago
reunsreuns
20.7k21148
$endgroup$
add a comment |
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1 Answer
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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 $mathbb{Z}$.
answered 3 hours ago
reunsreuns
20.7k21148
$endgroup$
add a comment |
$begingroup$
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 $mathbb{Z}$.
answered 3 hours ago
reunsreuns
20.7k21148
$endgroup$
add a comment |
$begingroup$
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 $mathbb{Z}$.
answered 3 hours ago
reunsreuns
20.7k21148
$endgroup$
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 $mathbb{Z}$.
answered 3 hours ago
reunsreuns
20.7k21148
edited 22 mins ago
answered 3 hours ago
reunsreuns
20.7k21148
answered 3 hours ago
reunsreuns
20.7k21148
answered 3 hours ago
reunsreuns
20.7k21148
20.7k21148
add a comment |
add a comment |
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