1 | initial version |

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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