I am self studying Silverman's arithmetic of elliptic curves book and I am trying to solve exercise 10.19. It has four parts, and I was able to do everything but the first one. Here is the problem:
Let $E/\mathbb{Q}$ be an elliptic curve such that $j(E)=0$. Prove that there is a unique integer $D$ which is $6$-th powe free such that $E$ can be written as $y^2=x^3+D$.
I know that an elliptic curve can be written as $y^2=x^3+Ax+B$ with $4A^3+27B^2\neq 0$ and that $j(E)=1728 \frac{4A^3}{4A^3+27B^2}$ hence $j(E)=0$ implies $A=0$ and so we have that $E$ can be written as $y^2=x^3+B$ where $B\neq0$. Also it is trivial to see that $B$ can be taken to be $6$-th power free by making a change of variable. Hence, there is only one thing left to prove:
- If we have two elliptic cuves $y^2=x^3+D$ and $y^2=x^3+D'$ with $D\neq D'$ both $6$-th power free integers then they are not isomorphic curves.
But I don't have any clue on how to prove this, since there might be a tricky change of variables and I don't know a general method to show when two elliptic curves are not isomorphic.
Thanks for your help.
I don't think the claim is quite true as stated, that these are non-isomorphic for $D \neq D'$; I think the requirement should be that $D \neq D'$ as elements of $\mathbb{Q}^\times / (\mathbb{Q}^\times)^6$. (EDIT: since $D$ and $D'$ are integers, though, these are the same, see comment.)
If that's right, here is a sketch of an ugly but reasonably concrete proof.
First note that the curves become isomorphic over $\mathbb{C}$. Choose and write down an isomorphism $\phi_{D, D'}$ explicitly. Then every isomorphism between them over $\mathbb{C}$ is a composition of $\phi_{D, D'}$ and an automorphism.
The automorphisms of $y^2 = x^3 + D$ are the six maps of the form $x \mapsto \zeta^i x$ and $y \mapsto \pm y$ where $\zeta$ is a primitive third root of unity and $i \in \{ 0, 1, 2 \}$.
Compute these compositions explicitly and observe that none comes from a map of elliptic curves over $\mathbb{Q}$.