Finding the isomorphism type of an elliptic curveTurning an elliptic curve over C into a complex torusLet E be defined over Fq and let n ≥ 1. Show that E(Fq)[n] and E(Fq)/nE(Fq) have the same order.Isomorphism type of a finite group with respect to multiplication modulo 65Elliptic Curve Group and Multiplicative Inverse of an element.Chinese Remainder theorem on Elliptic Curve groupDetermine Isomorphism typeIsomorphism Type of $mathbbZ_8timesmathbbZ_6timesmathbbZ_4 /langle (2,2,2) rangle$Exercise 3.8b) of Silverman's “Arithmetic of Elliptic Curves”Rank of Elliptic Curve over Finite FieldDistribution of torsion subgroups of elliptic curve
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Finding the isomorphism type of an elliptic curve
Turning an elliptic curve over C into a complex torusLet E be defined over Fq and let n ≥ 1. Show that E(Fq)[n] and E(Fq)/nE(Fq) have the same order.Isomorphism type of a finite group with respect to multiplication modulo 65Elliptic Curve Group and Multiplicative Inverse of an element.Chinese Remainder theorem on Elliptic Curve groupDetermine Isomorphism typeIsomorphism Type of $mathbbZ_8timesmathbbZ_6timesmathbbZ_4 /langle (2,2,2) rangle$Exercise 3.8b) of Silverman's “Arithmetic of Elliptic Curves”Rank of Elliptic Curve over Finite FieldDistribution of torsion subgroups of elliptic curve
$begingroup$
Consider the elliptic curve $E: y^2=x^3+1$ over $mathbbF_q$ where $q = 15485863$ (the $1000000^textth$ prime).
I have computed (using sage) that $P=(15065540,4435916)$ has order $5160153 = 3cdot19cdot90529$
and that $|E(mathbbF_q)| = 15480459 = 3^2cdot19cdot90529$
I am asked to find the isomorphism type of $E(mathbbF_q)$, I know from the Fundamental Theorem of Finite Abelian Groups that $E(mathbbF_q) cong mathbbZ/3^2mathbbZ timesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
or that $E(mathbbF_q) cong mathbbZ/3mathbbZ times mathbbZ/3mathbbZtimesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
I would like to know if there's any way that I can discard one of this options. I guess that I would have to find a point of order 9 or at least two points of order 3, but I am not sure how to do this.
group-theory finite-groups elliptic-curves
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
$endgroup$
add a comment |
$begingroup$
Consider the elliptic curve $E: y^2=x^3+1$ over $mathbbF_q$ where $q = 15485863$ (the $1000000^textth$ prime).
I have computed (using sage) that $P=(15065540,4435916)$ has order $5160153 = 3cdot19cdot90529$
and that $|E(mathbbF_q)| = 15480459 = 3^2cdot19cdot90529$
I am asked to find the isomorphism type of $E(mathbbF_q)$, I know from the Fundamental Theorem of Finite Abelian Groups that $E(mathbbF_q) cong mathbbZ/3^2mathbbZ timesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
or that $E(mathbbF_q) cong mathbbZ/3mathbbZ times mathbbZ/3mathbbZtimesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
I would like to know if there's any way that I can discard one of this options. I guess that I would have to find a point of order 9 or at least two points of order 3, but I am not sure how to do this.
group-theory finite-groups elliptic-curves
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
$endgroup$
4
$begingroup$
sage will tell you the 3 torsion! you could use P.division_points(3) where P is the origin.
$endgroup$
– hunter
Mar 15 at 4:01
$begingroup$
And you can look at the polynomials such that $[3](x,y) = (fracu(x,y)v(x,y), fracw(x,y)z(x,y))$ then $[3](x,y) = O$ corresponds to the roots of $V(x) = v(x,sqrtx^3+1)v(x,-sqrtx^3+1) in mathbbF_q[x]$.
$endgroup$
– reuns
Mar 15 at 4:27
$begingroup$
@hunter So the 3-torsion points according to what you said and sage are [(0 : 1 : 0), (0 : 1 : 1), (0 : 102 : 1)], this means that there are two points of order 3 and therefore 𝐸(𝔽𝑞)≅ℤ/3ℤ×ℤ/3ℤ×ℤ/19ℤ×ℤ/90529ℤ, right?
$endgroup$
– McNuggets666
Mar 15 at 18:53
$begingroup$
If there's only two points of order 3, then the 3-torsion subgroup must be $mathbbZ/9mathbbZ$, not two copies of $mathbbZ/3mathbbZ$.
$endgroup$
– hunter
Mar 16 at 14:09
$begingroup$
EllipticCurve( GF(nth_prime(1000000)), [0,1] ).order().factor()delivers in sage2^2 * 3 * 37 * 43 * 811. On the other way, there is a big prime21843131744681521obtained by factorizing $y_0^2-x_0^3-1$, where $x_0,y_0$ are the coordinates of $P$.
$endgroup$
– dan_fulea
yesterday
add a comment |
$begingroup$
Consider the elliptic curve $E: y^2=x^3+1$ over $mathbbF_q$ where $q = 15485863$ (the $1000000^textth$ prime).
I have computed (using sage) that $P=(15065540,4435916)$ has order $5160153 = 3cdot19cdot90529$
and that $|E(mathbbF_q)| = 15480459 = 3^2cdot19cdot90529$
I am asked to find the isomorphism type of $E(mathbbF_q)$, I know from the Fundamental Theorem of Finite Abelian Groups that $E(mathbbF_q) cong mathbbZ/3^2mathbbZ timesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
or that $E(mathbbF_q) cong mathbbZ/3mathbbZ times mathbbZ/3mathbbZtimesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
I would like to know if there's any way that I can discard one of this options. I guess that I would have to find a point of order 9 or at least two points of order 3, but I am not sure how to do this.
group-theory finite-groups elliptic-curves
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
$endgroup$
Consider the elliptic curve $E: y^2=x^3+1$ over $mathbbF_q$ where $q = 15485863$ (the $1000000^textth$ prime).
I have computed (using sage) that $P=(15065540,4435916)$ has order $5160153 = 3cdot19cdot90529$
and that $|E(mathbbF_q)| = 15480459 = 3^2cdot19cdot90529$
I am asked to find the isomorphism type of $E(mathbbF_q)$, I know from the Fundamental Theorem of Finite Abelian Groups that $E(mathbbF_q) cong mathbbZ/3^2mathbbZ timesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
or that $E(mathbbF_q) cong mathbbZ/3mathbbZ times mathbbZ/3mathbbZtimesmathbbZ/19mathbbZ times mathbbZ/90529mathbbZ$
I would like to know if there's any way that I can discard one of this options. I guess that I would have to find a point of order 9 or at least two points of order 3, but I am not sure how to do this.
group-theory finite-groups elliptic-curves
group-theory finite-groups elliptic-curves
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
asked Mar 15 at 3:55
McNuggets666McNuggets666
723411
723411
4
$begingroup$
sage will tell you the 3 torsion! you could use P.division_points(3) where P is the origin.
$endgroup$
– hunter
Mar 15 at 4:01
$begingroup$
And you can look at the polynomials such that $[3](x,y) = (fracu(x,y)v(x,y), fracw(x,y)z(x,y))$ then $[3](x,y) = O$ corresponds to the roots of $V(x) = v(x,sqrtx^3+1)v(x,-sqrtx^3+1) in mathbbF_q[x]$.
$endgroup$
– reuns
Mar 15 at 4:27
$begingroup$
@hunter So the 3-torsion points according to what you said and sage are [(0 : 1 : 0), (0 : 1 : 1), (0 : 102 : 1)], this means that there are two points of order 3 and therefore 𝐸(𝔽𝑞)≅ℤ/3ℤ×ℤ/3ℤ×ℤ/19ℤ×ℤ/90529ℤ, right?
$endgroup$
– McNuggets666
Mar 15 at 18:53
$begingroup$
If there's only two points of order 3, then the 3-torsion subgroup must be $mathbbZ/9mathbbZ$, not two copies of $mathbbZ/3mathbbZ$.
$endgroup$
– hunter
Mar 16 at 14:09
$begingroup$
EllipticCurve( GF(nth_prime(1000000)), [0,1] ).order().factor()delivers in sage2^2 * 3 * 37 * 43 * 811. On the other way, there is a big prime21843131744681521obtained by factorizing $y_0^2-x_0^3-1$, where $x_0,y_0$ are the coordinates of $P$.
$endgroup$
– dan_fulea
yesterday
add a comment |
4
$begingroup$
sage will tell you the 3 torsion! you could use P.division_points(3) where P is the origin.
$endgroup$
– hunter
Mar 15 at 4:01
$begingroup$
And you can look at the polynomials such that $[3](x,y) = (fracu(x,y)v(x,y), fracw(x,y)z(x,y))$ then $[3](x,y) = O$ corresponds to the roots of $V(x) = v(x,sqrtx^3+1)v(x,-sqrtx^3+1) in mathbbF_q[x]$.
$endgroup$
– reuns
Mar 15 at 4:27
$begingroup$
@hunter So the 3-torsion points according to what you said and sage are [(0 : 1 : 0), (0 : 1 : 1), (0 : 102 : 1)], this means that there are two points of order 3 and therefore 𝐸(𝔽𝑞)≅ℤ/3ℤ×ℤ/3ℤ×ℤ/19ℤ×ℤ/90529ℤ, right?
$endgroup$
– McNuggets666
Mar 15 at 18:53
$begingroup$
If there's only two points of order 3, then the 3-torsion subgroup must be $mathbbZ/9mathbbZ$, not two copies of $mathbbZ/3mathbbZ$.
$endgroup$
– hunter
Mar 16 at 14:09
$begingroup$
EllipticCurve( GF(nth_prime(1000000)), [0,1] ).order().factor()delivers in sage2^2 * 3 * 37 * 43 * 811. On the other way, there is a big prime21843131744681521obtained by factorizing $y_0^2-x_0^3-1$, where $x_0,y_0$ are the coordinates of $P$.
$endgroup$
– dan_fulea
yesterday
4
4
$begingroup$
sage will tell you the 3 torsion! you could use P.division_points(3) where P is the origin.
$endgroup$
– hunter
Mar 15 at 4:01
$begingroup$
sage will tell you the 3 torsion! you could use P.division_points(3) where P is the origin.
$endgroup$
– hunter
Mar 15 at 4:01
$begingroup$
And you can look at the polynomials such that $[3](x,y) = (fracu(x,y)v(x,y), fracw(x,y)z(x,y))$ then $[3](x,y) = O$ corresponds to the roots of $V(x) = v(x,sqrtx^3+1)v(x,-sqrtx^3+1) in mathbbF_q[x]$.
$endgroup$
– reuns
Mar 15 at 4:27
$begingroup$
And you can look at the polynomials such that $[3](x,y) = (fracu(x,y)v(x,y), fracw(x,y)z(x,y))$ then $[3](x,y) = O$ corresponds to the roots of $V(x) = v(x,sqrtx^3+1)v(x,-sqrtx^3+1) in mathbbF_q[x]$.
$endgroup$
– reuns
Mar 15 at 4:27
$begingroup$
@hunter So the 3-torsion points according to what you said and sage are [(0 : 1 : 0), (0 : 1 : 1), (0 : 102 : 1)], this means that there are two points of order 3 and therefore 𝐸(𝔽𝑞)≅ℤ/3ℤ×ℤ/3ℤ×ℤ/19ℤ×ℤ/90529ℤ, right?
$endgroup$
– McNuggets666
Mar 15 at 18:53
$begingroup$
@hunter So the 3-torsion points according to what you said and sage are [(0 : 1 : 0), (0 : 1 : 1), (0 : 102 : 1)], this means that there are two points of order 3 and therefore 𝐸(𝔽𝑞)≅ℤ/3ℤ×ℤ/3ℤ×ℤ/19ℤ×ℤ/90529ℤ, right?
$endgroup$
– McNuggets666
Mar 15 at 18:53
$begingroup$
If there's only two points of order 3, then the 3-torsion subgroup must be $mathbbZ/9mathbbZ$, not two copies of $mathbbZ/3mathbbZ$.
$endgroup$
– hunter
Mar 16 at 14:09
$begingroup$
If there's only two points of order 3, then the 3-torsion subgroup must be $mathbbZ/9mathbbZ$, not two copies of $mathbbZ/3mathbbZ$.
$endgroup$
– hunter
Mar 16 at 14:09
$begingroup$
EllipticCurve( GF(nth_prime(1000000)), [0,1] ).order().factor() delivers in sage 2^2 * 3 * 37 * 43 * 811. On the other way, there is a big prime 21843131744681521 obtained by factorizing $y_0^2-x_0^3-1$, where $x_0,y_0$ are the coordinates of $P$.$endgroup$
– dan_fulea
yesterday
$begingroup$
EllipticCurve( GF(nth_prime(1000000)), [0,1] ).order().factor() delivers in sage 2^2 * 3 * 37 * 43 * 811. On the other way, there is a big prime 21843131744681521 obtained by factorizing $y_0^2-x_0^3-1$, where $x_0,y_0$ are the coordinates of $P$.$endgroup$
– dan_fulea
yesterday
add a comment |
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4
$begingroup$
sage will tell you the 3 torsion! you could use P.division_points(3) where P is the origin.
$endgroup$
– hunter
Mar 15 at 4:01
$begingroup$
And you can look at the polynomials such that $[3](x,y) = (fracu(x,y)v(x,y), fracw(x,y)z(x,y))$ then $[3](x,y) = O$ corresponds to the roots of $V(x) = v(x,sqrtx^3+1)v(x,-sqrtx^3+1) in mathbbF_q[x]$.
$endgroup$
– reuns
Mar 15 at 4:27
$begingroup$
@hunter So the 3-torsion points according to what you said and sage are [(0 : 1 : 0), (0 : 1 : 1), (0 : 102 : 1)], this means that there are two points of order 3 and therefore 𝐸(𝔽𝑞)≅ℤ/3ℤ×ℤ/3ℤ×ℤ/19ℤ×ℤ/90529ℤ, right?
$endgroup$
– McNuggets666
Mar 15 at 18:53
$begingroup$
If there's only two points of order 3, then the 3-torsion subgroup must be $mathbbZ/9mathbbZ$, not two copies of $mathbbZ/3mathbbZ$.
$endgroup$
– hunter
Mar 16 at 14:09
$begingroup$
EllipticCurve( GF(nth_prime(1000000)), [0,1] ).order().factor()delivers in sage2^2 * 3 * 37 * 43 * 811. On the other way, there is a big prime21843131744681521obtained by factorizing $y_0^2-x_0^3-1$, where $x_0,y_0$ are the coordinates of $P$.$endgroup$
– dan_fulea
yesterday