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I want to solve the following exercise:

Show that the two elliptic curves $E/ \mathbb{Q}$ and $E'/ \mathbb{Q}$ are isomorphic.
$E: y^2 = x^3+x-2$ and $E': y'^2 = x'^3-\frac{1}{3}x' - \frac{52}{27}$.

I am trying to find a change of variables $(x,y)\mapsto(x',y')$ transforming the Weierstraß equation defining $E$ to the Weierstraß equation defining $E'$.

I tried this by guesswork because I couldn't think of a clever way.
A first idea was to put $y = (\sqrt{27 y'}-\sqrt{2})$ because then I already get $27y'^2 = 27x'^3-27x'-52$ which is $y'^2 = x'^3-x'-\frac{52}{27}$ which looks a bit more like $E'$. But I don't know what to do about the $x$.

Is there a more strategic way to do this? Does anyone have a hint how to solve this exercise?

All the best!

Luca
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  • Your $y=\sqrt{27y'}-\sqrt2$ isn't defined over $\Bbb{Q}$ I think. – Jyrki Lahtonen Apr 22 '14 at 20:02
  • ups >.< thanks. That was stupid.. – Luca Apr 22 '14 at 20:05
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    Two Weierstrass equations define isomorphic curves if and only if they are related by a change of variables of the form with $u,r,s,t ∈ k$. (u not 0). $x′ =u^2x+r$
 , $y′ =u^3y+u^2sx+t$ – Alan Apr 22 '14 at 20:10
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    You may find this useful: http://math.stackexchange.com/questions/289859/proving-the-condition-for-two-elliptic-curves-given-in-weierstrass-form-to-be-is?rq=1 – Samuel Reid Apr 22 '14 at 20:23
  • Thank you for your comments! I tried to solve the exercise with this formula, but it gets very complicated.. I put the new formula for $x'$ and $y'$ in the equation and transformed the equation such that on the right is only $y^2$ so that I can put the long formula with $r,s,t,u$ $=x^3+x-2$. But this becomes so long that I don't know how to find out the right $r,s,t,u$. Is there a way to 'look at it' cleverly? – Luca Apr 23 '14 at 00:02

2 Answers2

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Typo: the equation of the first curve should be $y^2 = x^3 + x^2 - 2$. Hint: $y=y'$, so you need only find a transformation from $x$ to $x'$ that kills the quadratic term of the cubic on the right-hand side.

Noam D. Elkies
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  • Thanks a lot both of you for pointing out the typo! I tried what you said, and put the equation for the new $x$ in the other equation and then searched for $u$ and $r$ such that the quadratic term is deleted, but it doesn't realloy work out. I will ask my tutor tomorrow because I'm out of ideas and power ;) But thanks anyway. – Luca Apr 23 '14 at 01:33
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    @Luca: If $x=x'+c$, then $x^3=x'^3+3cx'^2+O(x')$ and $x^2=x'^2+O(x')$, so $x^3+x^2 = x'^3+(3c+1)x'^2+O(x')$. Now you just need to find $c$ such that the $x'^2$ term here becomes $0$. – Steven Stadnicki Apr 23 '14 at 01:49
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Well, you must have the wrong equations, because they are not isomorphic. The $j$-invariant classifies elliptic curves up to isomorphism (over $\mathbb{C}$), and the $j$-invariants of these curves are $432/7$ and $-64/25$, respectively. Since they are distinct, they are not isomorphic.

In light of Noam Elkies' answer, the $j$-invariant of $y^2=x^3+x^2-2$ is indeed $-64/25$, the same as $E'$ in the statement of the problem.

Álvaro Lozano-Robledo
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