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What is the point ๐‘ƒ1=(โˆ’1,3) on the y^2=7๐‘ฅ4+๐‘ฅ3+๐‘ฅ2+๐‘ฅ+3 curve and the point corresponding to P on the x^2=x^3โˆ’250/3xโˆ’124927 curve?

asked 2020-03-26 10:08:06 -0500

Gamzeee gravatar image

What is the point ๐‘ƒ1=(โˆ’1,3) on the y^2=7๐‘ฅ^4+๐‘ฅ^3+๐‘ฅ^2+๐‘ฅ+3 curve and the point corresponding to P on the x^2=x^3โˆ’250/3xโˆ’124927 curve? (๐ธ1:๐‘ฆ2=7๐‘ฅ4+๐‘ฅ3+๐‘ฅ2+๐‘ฅ+3,๐‘ƒ1=(โˆ’1,3) can be transformed to ๐ธ2:๐‘ฃ2=๐‘ข3โˆ’250๐‘ข3โˆ’124927) Thanks

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answered 2020-03-26 10:30:24 -0500

rburing gravatar image

updated 2020-03-27 09:29:34 -0500

See Construct elliptic curves as Jacobians in the manual.

sage: P2.<x,y,z> = ProjectiveSpace(QQ,2)
sage: E1 = Curve(y^2*z^2 - (7*๐‘ฅ^4+๐‘ฅ^3*z+๐‘ฅ^2*z^2+๐‘ฅ*z^3+3*z^4))
sage: phi = Jacobian(E1, morphism=True)
sage: E2 = phi.codomain(); E2
Elliptic Curve defined by y^2 = x^3 - 250/3*x - 1249/27 over Rational Field
sage: phi(E1(-1,3,1))
(-71/9 : 296/27 : 1)

Edit: in this way you certainly get a map, but not necessarily an isomorphism; see the other answer.

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Note, though, that the map phi is not uniquely determined. You can compose phi with a translation on E1 and/or negation.

nbruin gravatar imagenbruin ( 2020-03-26 11:40:23 -0500 )edit

sage: P2.<x,y,z> = ProjectiveSpace(QQ,2)

sage: E1= Curve(y^2z^2+15x^4-1516z^4)

sage: phi = Jacobian(E1, morphism=True)

sage: E2 = phi.codomain(); E2

Elliptic Curve defined by y^2 = x^3 + 14400*x over Rational Field

sage: phi(E1(2,0,1))

(0 : 1 : 0)

But the (0 : 1 : 0) point is not on the E2 curve. What should I do?

Gamzeee gravatar imageGamzeee ( 2020-03-26 12:32:21 -0500 )edit

@Gamzeee yes it is on the curve; it is the point at infinity. Maybe @nbruin can suggest how to get a different phi.

rburing gravatar imagerburing ( 2020-03-26 12:33:30 -0500 )edit
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answered 2020-03-26 16:10:47 -0500

nbruin gravatar image

The question is, unfortunately, not well-defined. The curves mentioned, while isomorphic, are not uniquely so. First you should specify which point on E1 is supposed to be mapped to the "origin" (0:1:0) on "E2", and then there is still a choice of sign.

The routine Jacobian mentioned in another answer does not give you a (birational) isomorphism; it only expresses "E1" as a cover of "E2". There is even more choice in that.

As an example:

sage: P2.<x,y,z>=ProjectiveSpace(QQ,2)
sage: C=Curve(3*x^3+4*y^3+5*z^3)
sage: phi=Jacobian(C,morphism=True)
sage: phi.codomain()
Elliptic Curve defined by y^2 = x^3 - 24300 over Rational Field

This is Selmer's famous example: the curve C isn't isomorphic to the elliptic curve over Q, because C does not have rational points. The map phi computed is of degree larger than 1 (presumable degree 9, but I didn't check)

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Asked: 2020-03-26 10:08:06 -0500

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Last updated: Mar 27