I have the following elliptic curves:
$$E:y^2=x^3-\sqrt{2}x \quad \quad E':y^2=x^3-2x$$
over $\mathbb{Q}(\sqrt{2})$. I want to determine whether they are isogenous.
I have a few strategies to do this when the field I am working on is finite (for example counting points) or when the field is the complex numbers (mainly looking at the lattices). However, when looking at number fields like this one I am completely lost. I am only able to prove that they are isomorphic over the algebraic closure of my field since they have the same $j$-invariant, but this does not tell me that they are isomorphic over the number field itself. I have tried manually constructing an isogeny but this took me nowhere. But I also don't know how to show that such an isogeny does not exist. Can someone show me how to proceed in this case?
Best Answer
In this case, you can exploit the fact that the curves become isomorphic over $K=\mathbb Q(\sqrt[8]2)$. Let $\psi\colon E\to E'$ be an isomorphism over $K$, and suppose that $\phi\colon E\to E'$ is an isogeny defined over $\mathbb Q(\sqrt2)$. Then, over $K$, the map $\psi^{-1}\circ\phi$ would have to be an endomorphism of $E$, defined over $K$.
But, even though $\mathrm{End}_{\mathbb C}(E) = \mathbb Z[i]$ (via the isomorphism $(x, y)\mapsto (-x,iy)$), crucially, the extra endomorphisms are not defined over $K$. Hence, $\psi^{-1}\circ\phi$ must be multiplication by $n$ for some integer $n$. In particular, over $K$, $\ker\phi = \ker([n])$, which would imply that $\phi = [n]$... but that would mean that $E$ and $E'$ are already isomorphic over $\mathbb Q(\sqrt2)$, which they are not. (For example, they have different $2$-torsion, because the polynomial $x^3 - 2x$ splits into linear factors, while $x^3 - \sqrt 2x$ does not.)
It might be helpful to see how this argument can go wrong with $E$ replaced with $y^2 = x^3 +8x$ and both $E, E'$ viewed over $\mathbb Q$. This time, $E$ and $E'$ become isomorphic over $K=\mathbb Q(\sqrt[4]{-4})$, which contains $\mathbb Q(i)$. In particular, if $\phi\colon E\to E'$ is an isogeny over $\mathbb Q$, and $\psi\colon E\to E'$ is an isogeny over $K$, then it's still the case that $\psi^{-1}\circ\phi\in \mathrm{End}_K(E)$, but this time, it could be any element of $\mathbb Z[i]$. In particular, it can be $1 + i$. Note that $(1+i)^2 = 2i$, so $(1+i)$ is an endomorphism which applied twice gives you multiplication by $2i$. It turns out that this endomorphism over $K$ descends to a cyclic $2$-isogeny over $\mathbb Q$.