Isogenies of degree 3 of elliptic curves with j-invariant 0

Question

Let $E/\mathbb{Q}$ be an elliptic curve with $j(E)=0$. I.e., $E$ has Weierstrass equations $$ y^2 = x^3+ B$$ for some $B\in \mathbb{Q}$. $E$ has complex multiplication by $\mathcal{O}:= \mbox{ ring of integers of } \mathbb{Q}(\sqrt{-3})$.

I am looking to classify the cyclic $\mathbb{Q}$-isogenies of $E$, in particular the ones of degree 3 (which I suspect are the only ones). Given the Weierstrass equation, using the division polynomials we see that $E[3]$ has 4 cyclic subgroups: $$\{\mathcal{O}_E, (0,\sqrt{B}),(0,-\sqrt{B})\},\{\mathcal{O}_E, (\sqrt[3]{4B},\sqrt{3B}), (\sqrt[3]{4B},-\sqrt{3B})\},\{\mathcal{O}_E, (\omega\sqrt[3]{4B},\sqrt{3B}), (\omega\sqrt[3]{4B},-\sqrt{3B})\}, \{\mathcal{O}_E, (\omega^2\sqrt[3]{4B},\sqrt{3B}), (\omega^2\sqrt[3]{4B},-\sqrt{3B})\}$$ where $\omega$ is a 3rd root of unity.

So my first question boils down to which ones are $G_\mathbb{Q}:=\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$-invariant?

Furthermore if $H$ is one of these subgroups, what is the endomorphism ring of the quotient $E_H$? I want to know for which $H$ that $E_H$ has CM by the maximal order or not. I'm attempting to use the techniques of Products of CM Elliptic Curves but I cannot discern when $H$ is an ideal subgroup or not because I don't know the explicit action of $\mathcal{O}$ on $E$.

Answer 1

Each cyclic subgroup is Galois invariant if and only if its $x$ coordinate is Galois invariant, which happens if and only if it lies in $\mathbb Q$. So the first one is Galois invariant and then, if $4B$ is a perfect cube, one of the other three is Galois invariant.

The action of $\mathcal O$ is generated by the action of the third roots of unity, which act by multiplying $x$ by a third root of unity and fixing $y$. So $H$ is an ideal subgroup if and only if its stable under that action. In this case, that would require its $x$ coordinate to be zero.

These aren't quite the only cyclic isogenies. There's also a $2$-isogeny if $B$ is a perfect cube.