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I found this page: http://en.wikipedia.org/wiki/Twists_of_curves#Quadratic_twist

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)?

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  • $\begingroup$ 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 $K(\sqrt{d})$. $\endgroup$
    – fretty
    Commented Oct 25, 2013 at 13:51
  • $\begingroup$ 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. $\endgroup$ Commented Oct 25, 2013 at 13:55
  • $\begingroup$ Oh sorry, too many symbols floating around :p. My bad... $\endgroup$
    – fretty
    Commented Oct 25, 2013 at 13:58
  • $\begingroup$ @fretty No sweat! :) $\endgroup$ Commented Oct 25, 2013 at 14:03

1 Answer 1

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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 $d^3$.

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  • $\begingroup$ That's a nice and easy solution, thanks! $\endgroup$
    – user100659
    Commented Oct 26, 2013 at 9:22
  • $\begingroup$ @user100659 You are welcome. $\endgroup$ Commented Oct 26, 2013 at 14:25

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