Isomorphic Elliptic Curves

Question

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!

Answer 1

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.

Answer 2

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.