octonion's profile - activity

Thanks John!

Thanks! I found a slightly different method involving calculating the short Weierstrass form, then manually building and finding the roots of the cubic from the a2, a4 and a6 coefficients. But your method is much nicer and more general, and is greatly appreciated!

Say we have an elliptic curve:

R.<a,b,c> = QQ[]
cubic = a*(a+c)*(a+b)+b*(b+c)*(a+b)+c*(b+c)*(a+c)-4*(a+b)*(a+c)*(b+c)
E = EllipticCurve_from_cubic(cubic, morphism=False)
print(E)
print(E.minimal_model())

f = EllipticCurve_from_cubic(cubic, morphism=True)
finv = f.inverse()
generators = E.gens()

print
print("generators = %s" %(generators))

Output:

Elliptic Curve defined by y^2 + x*y = x^3 + 69*x^2 + 1365*x + 8281 over Rational Field
Elliptic Curve defined by y^2 + x*y + y = x^3 - 234*x + 1352 over Rational Field

generators = [(-39 : 52 : 1)]

How do we determine which of the two elliptic curve components a particular point is on? If we had this elliptic curve in the form $y^2 = x^3 + 109 x^2 + 234 x$ it'd be quite simple (if $x<0$ it's on the egg). Is there a nice way to automate this process in SageMath? For example, in this case the generator is on the egg.