CM Elliptic Curves

In this section we state, and in some cases sketch proofs of, some basic facts about CM elliptic curves.

If $E$ is an elliptic curve over a field $K$ we let $\End (E/K)$ be the ring of all endomorphisms of $E$ that are defined over $K$.

Definition 3.1 (CM Elliptic Curve)   An elliptic curve $E$ over a subfield of $\mathbb{C}$ has complex multiplication if $\End (E/\mathbb{C})\neq \mathbb{Z}$.

Remark 3.2   If $E$ is an elliptic curve over $\mathbb{Q}$, then $\End (E/\mathbb{Q}) = \mathbb{Z}$. This is true even if $E$ has complex multiplication, in which case the complex multiplication must be defined over a bigger field than $\mathbb{Q}$. The reason $\End (E/\mathbb{Q}) = \mathbb{Z}$ is because $\End (E/\mathbb{Q})$ acts faithfully on the $1$-dimensional $\mathbb{Q}$-vector space of invariant holomorphic differentials on $E$ over $\mathbb{Q}$ and $\End (E/\mathbb{Q})$ is finitely generated as a $\mathbb{Z}$-module.

A complex lattice $\Lambda \subset \mathbb{C}$ is a subgroup abstractly isomorphic to $\mathbb{Z}\times \mathbb{Z}$ such that $\mathbb{R}\Lambda = \mathbb{C}$. Using the Weirestrass $\wp$-function associated to the lattice $\Lambda$, one proves that there is a group isomorphism

\begin{displaymath}
\mathbb{C}/\Lambda \cong E_{\Lambda}(\mathbb{C}),
\end{displaymath}

where $E_{\Lambda}$ is an elliptic curve over $\mathbb{C}$. Conversely, if $E$ is any elliptic curve over $\mathbb{C}$, then there is a lattice $\Lambda$ such that $E = E_{\Lambda}$. Explicitly, if $\omega_E$ is an invariant differential we may take $\Lambda$ to be the lattice of all periods $\int_{\gamma} \omega_E\in \mathbb{C}$, where $\gamma$ runs through the integral homology $\H_1(E(\mathbb{C}),\mathbb{Z})$.

Proposition 3.3   Let $\Lambda_1$ and $\Lambda_2$ be complex lattices. Then

\begin{displaymath}
\Hom (\mathbb{C}/\Lambda_1, \mathbb{C}/\Lambda_2)
\cong \{ \alpha \in \mathbb{C}: \alpha \Lambda_1 \subset \Lambda_2\},
\end{displaymath}

where the homomorphisms on the left side are as elliptic curves over $\mathbb{C}$. Moreover, the complex number $\alpha \in \mathbb{C}$ corresponds to the homomorphism $[\alpha]$ induced by multiplication by $\alpha$, and the kernel of $[\alpha]$ is isomorphic to $\Lambda_2 / (\alpha \Lambda_1)$.

Corollary 3.4   If $\alpha$ is any nonzero complex number and $\Lambda$ is a lattice, then $\mathbb{C}/\Lambda\cong \mathbb{C}/(\alpha\Lambda)$.


\begin{proof}
% latex2html id marker 7577Since multiplication by $\alpha$ sen...
...ha$ defines a homomorphism with $0$ kernel,
hence an isomorphism.
\end{proof}

Now suppose $E/\mathbb{C}$ is a CM elliptic curve, and let $\Lambda$ be a lattice such that $E\cong E_{\Lambda}$. Then

\begin{displaymath}
\End (E/\mathbb{C}) \cong \{\alpha \in \mathbb{C}: \alpha \Lambda \subset \Lambda\}.
\end{displaymath}

Proposition 3.5   Let $E = E_{\Lambda}$ be a CM elliptic curve. Then there is a complex number $\omega$ and a quadratic imaginary field such $K$ that

\begin{displaymath}\omega\Lambda\subset \O_K,\end{displaymath}

where $\O_K$ is the ring of integers of $K$. Moreover, $\End (E/\mathbb{C})$ is an order (=subring of rank $2$) of $\O_K$.


\begin{proof}
% latex2html id marker 7584Write $\Lambda = \mathbb{Z}\omega_1 \...
...ubset \O_K$,
where $\omega$ is chosen so that $\omega\beta\in\O_K$.
\end{proof}



Subsections
William 2007-05-25