Given an order $\mathscr{O}$of imaginary quadratic field, is there an elliptic curve $E/\mathbb{C}$ s.t. $\operatorname{End}(E)\cong \mathscr{O}$?

Question

Sorry for my bad English.

For a CM elliptic curve $E/\mathbb{C}$, the endomorphism ring $\operatorname{End}(E)$ is isomorphic to an order of an imaginary quadratic field.

I want to know if it is true this inverse proposition.

Answer 1

The ring $\mathscr O$ is a lattice in $\mathbf C$, so $\mathbf C/\mathscr O$ is a complex elliptic curve.

For a lattice $\Lambda$ in $\mathbf C$, the endomorphisms of the complex elliptic curve $\mathbf C/\Lambda$ are the maps $f \colon \mathbf C/\Lambda \to \mathbf C/\Lambda$ where $f(z \bmod \Lambda) = \alpha z\bmod \Lambda)$ for $\alpha$ in $\mathbf C$ such that $\alpha\Lambda \subset \Lambda$. Taking $\Lambda = \mathscr O$, which contains $1$, we have $\alpha\mathscr O \subset \mathscr O$ if and only if $\alpha \in \mathscr O$.

Therefore we get a mapping $\mathscr O \to {\rm End}(\mathbf C/\mathscr O)$ by $\alpha \mapsto m_\alpha$, where $m_\alpha$ is multiplication by $\alpha$ on $\mathbf C/\mathscr O$. That mapping $\mathscr O \to {\rm End}(\mathbf C/\mathscr O)$ is additive and surjective. If $m_\alpha$ is the zero mapping then $\alpha\mathbf C \subset \mathscr O$, which forces $\alpha = 0$, so we get injectivity. Thus ${\rm End}(\mathbf C/\mathscr O) \cong \mathscr O$.