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 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 |
2 Answers
2
active
oldest
votes
$begingroup$
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 4 hours ago
reunsreuns
20.7k21148
$endgroup$
add a comment |
$begingroup$
This is, indeed, easy after you have calculated the number of points over the prime field. It is straightforward to list them
$$
E(Bbb{F}_3)={(0,1),(0,-1),(1,1),(1,-1),(-1,0),infty}.
$$
In other words $|E(Bbb{F}_3)|=6.$ This piece of information gives us the complex numbers
$alpha,overline{alpha}$
(see reuns's post for their interpretation as eigenvalues of Frobenius on the Tate module) as they are known to safisfy the equations $|alpha|^2=3$ and
$$
alpha+overline{alpha}=3+1-|E(Bbb{F}_3)|=-2.
$$
The real part of $alpha$ is thus equal to $-1$, so $alpha=-1pm isqrt2$.
The formula for the number of rational poinst on the extension field then reads
$$
|E(Bbb{F}_{3^k})|=3^k+1-alpha^k-overline{alpha}^k=3^k+1-2operatorname{Re}(-1+isqrt2)^k.
$$
For example, when $k=2$, $alpha^2=(-1+isqrt2)^2=-1-2isqrt2$
implying that $|E(Bbb{F}_9)|=9+1+2=12$. This passes the litmus test of being divisible by $|E(Bbb{F}_3)|$ (Lagrange's theorem from elementary group theory), possibly adding to our confidence in the correctness of the result.
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
$endgroup$
add a comment |
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2 Answers
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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 4 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 4 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 4 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 4 hours ago
reunsreuns
20.7k21148
edited 1 hour ago
answered 4 hours ago
reunsreuns
20.7k21148
answered 4 hours ago
reunsreuns
20.7k21148
answered 4 hours ago
reunsreuns
20.7k21148
20.7k21148
add a comment |
add a comment |
$begingroup$
This is, indeed, easy after you have calculated the number of points over the prime field. It is straightforward to list them
$$
E(Bbb{F}_3)={(0,1),(0,-1),(1,1),(1,-1),(-1,0),infty}.
$$
In other words $|E(Bbb{F}_3)|=6.$ This piece of information gives us the complex numbers
$alpha,overline{alpha}$
(see reuns's post for their interpretation as eigenvalues of Frobenius on the Tate module) as they are known to safisfy the equations $|alpha|^2=3$ and
$$
alpha+overline{alpha}=3+1-|E(Bbb{F}_3)|=-2.
$$
The real part of $alpha$ is thus equal to $-1$, so $alpha=-1pm isqrt2$.
The formula for the number of rational poinst on the extension field then reads
$$
|E(Bbb{F}_{3^k})|=3^k+1-alpha^k-overline{alpha}^k=3^k+1-2operatorname{Re}(-1+isqrt2)^k.
$$
For example, when $k=2$, $alpha^2=(-1+isqrt2)^2=-1-2isqrt2$
implying that $|E(Bbb{F}_9)|=9+1+2=12$. This passes the litmus test of being divisible by $|E(Bbb{F}_3)|$ (Lagrange's theorem from elementary group theory), possibly adding to our confidence in the correctness of the result.
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
$endgroup$
add a comment |
$begingroup$
This is, indeed, easy after you have calculated the number of points over the prime field. It is straightforward to list them
$$
E(Bbb{F}_3)={(0,1),(0,-1),(1,1),(1,-1),(-1,0),infty}.
$$
In other words $|E(Bbb{F}_3)|=6.$ This piece of information gives us the complex numbers
$alpha,overline{alpha}$
(see reuns's post for their interpretation as eigenvalues of Frobenius on the Tate module) as they are known to safisfy the equations $|alpha|^2=3$ and
$$
alpha+overline{alpha}=3+1-|E(Bbb{F}_3)|=-2.
$$
The real part of $alpha$ is thus equal to $-1$, so $alpha=-1pm isqrt2$.
The formula for the number of rational poinst on the extension field then reads
$$
|E(Bbb{F}_{3^k})|=3^k+1-alpha^k-overline{alpha}^k=3^k+1-2operatorname{Re}(-1+isqrt2)^k.
$$
For example, when $k=2$, $alpha^2=(-1+isqrt2)^2=-1-2isqrt2$
implying that $|E(Bbb{F}_9)|=9+1+2=12$. This passes the litmus test of being divisible by $|E(Bbb{F}_3)|$ (Lagrange's theorem from elementary group theory), possibly adding to our confidence in the correctness of the result.
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
$endgroup$
add a comment |
$begingroup$
This is, indeed, easy after you have calculated the number of points over the prime field. It is straightforward to list them
$$
E(Bbb{F}_3)={(0,1),(0,-1),(1,1),(1,-1),(-1,0),infty}.
$$
In other words $|E(Bbb{F}_3)|=6.$ This piece of information gives us the complex numbers
$alpha,overline{alpha}$
(see reuns's post for their interpretation as eigenvalues of Frobenius on the Tate module) as they are known to safisfy the equations $|alpha|^2=3$ and
$$
alpha+overline{alpha}=3+1-|E(Bbb{F}_3)|=-2.
$$
The real part of $alpha$ is thus equal to $-1$, so $alpha=-1pm isqrt2$.
The formula for the number of rational poinst on the extension field then reads
$$
|E(Bbb{F}_{3^k})|=3^k+1-alpha^k-overline{alpha}^k=3^k+1-2operatorname{Re}(-1+isqrt2)^k.
$$
For example, when $k=2$, $alpha^2=(-1+isqrt2)^2=-1-2isqrt2$
implying that $|E(Bbb{F}_9)|=9+1+2=12$. This passes the litmus test of being divisible by $|E(Bbb{F}_3)|$ (Lagrange's theorem from elementary group theory), possibly adding to our confidence in the correctness of the result.
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
$endgroup$
This is, indeed, easy after you have calculated the number of points over the prime field. It is straightforward to list them
$$
E(Bbb{F}_3)={(0,1),(0,-1),(1,1),(1,-1),(-1,0),infty}.
$$
In other words $|E(Bbb{F}_3)|=6.$ This piece of information gives us the complex numbers
$alpha,overline{alpha}$
(see reuns's post for their interpretation as eigenvalues of Frobenius on the Tate module) as they are known to safisfy the equations $|alpha|^2=3$ and
$$
alpha+overline{alpha}=3+1-|E(Bbb{F}_3)|=-2.
$$
The real part of $alpha$ is thus equal to $-1$, so $alpha=-1pm isqrt2$.
The formula for the number of rational poinst on the extension field then reads
$$
|E(Bbb{F}_{3^k})|=3^k+1-alpha^k-overline{alpha}^k=3^k+1-2operatorname{Re}(-1+isqrt2)^k.
$$
For example, when $k=2$, $alpha^2=(-1+isqrt2)^2=-1-2isqrt2$
implying that $|E(Bbb{F}_9)|=9+1+2=12$. This passes the litmus test of being divisible by $|E(Bbb{F}_3)|$ (Lagrange's theorem from elementary group theory), possibly adding to our confidence in the correctness of the result.
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
answered 49 mins ago
Jyrki LahtonenJyrki Lahtonen
109k13170377
109k13170377
add a comment |
add a comment |
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