Quadratic twist of an elliptic curve

Question

which tells me $dy^2=x^3+a_2x^2+a_4x+a_6$ is equivalent to $y^2=x^3+da_2x^2+d^2a_4x+d^3a_6$. Why is this equivalent (for $d$ as given on that page)?

By equivalent I guess you mean they are "isomorphic"? Well this is true but also according to the wiki page there is no isomorphism defined over the ground field $K$...only over the quadratic extension $K(\sqrt{d})$. — fretty, Oct 25 '13 at 13:51
Dear @fretty: No, the curve and its twist are isomorphic over the quadratic extension. But the curves $dy^2 = x^3 + a_2x^2 +a_4x + a_6$ and $y^2 = x^3 + da_2x^2 + d^2a_4x + d^3a_6$ are isomorphic over the ground field. This is the question being asked. — Bruno Joyal, Oct 25 '13 at 13:55

score 8 · Accepted Answer · answered Oct 25 '13 at 13:23

8

Put $y = y'/d^2$ and $x=x'/d$. Then

$$d^{-3}y'^2 = d^{-3}x'^3 + a_2d^{-2}x'^2 + a_4d^{-1}x' + a_6.$$

Multiply both sides by $d^3$.

answered Oct 25 '13 at 13:23

Bruno Joyal

1 Answers1