Calculate the number of points of an elliptic curve in medium Weierstrass form over finite fieldProving the...
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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
$begingroup$
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 4 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 4 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 4 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 4 hours ago
NickyNicky
736
asked 4 hours ago
NickyNicky
736
asked 4 hours ago
NickyNicky
736
asked 4 hours ago
NickyNicky
736
asked 4 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 16 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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