Why there are only finitely many $\overline{\mathbb{Q}}$-isomorphism classes of elliptic curves with CM by $\mathcal{O}$?

Question

For someone who does not have a very extensive knowledge of number theory, what is a good intuitive explanation as to why there are only finitely many $\overline{\mathbb{Q}}$ isomorphism classes of elliptic curves with complex multiplication by $\mathcal{O}$?

Answer 1

Here is the standard argument. Obviously, there are details to fill in. I'm glad to fill in the ones you have trouble with, but I want to get the overview down first.

Isomorphism over $\overline{\mathbb{Q}}$ is the same as isomorphism over $\mathbb{C}$, by the usual nonsense that allows us to exchange any two algebraically closed fields of the same characteristic.

Over $\mathbb{C}$, every elliptic curve is of the form $\mathbb{C}/\Lambda$ for a discrete rank two sublattice $\Lambda$ of $\mathbb{C}$. A complex analytic homomorphism $\mathbb{C}/\Lambda_1 \to \mathbb{C}/\Lambda_2$ must lift to a map $\mathbb{C} \to \mathbb{C}$ of the form $z \mapsto a z$ for some $a \in \mathbb{C}$ such that $a \Lambda_1 \subseteq \Lambda_2$.

So $\mathbb{C}/\Lambda$ has complex multiplication by $\mathcal{O}$ if and only if, for all $a \in \mathcal{O}$ we have $a \Lambda \subseteq \Lambda$. And two lattices $\Lambda_1$ and $\Lambda_2$ give isomorphic quotients if and only if there is a scalar $a$ with $a \Lambda_1 = \Lambda_2$.

So we need to prove the following result: There are only finitely many lattices $\Lambda$ such that $\{ a \Lambda \subseteq \Lambda : \forall a \in \mathcal{O} \}$, up to the relation that $\Lambda_1 \sim \Lambda_2$ if there is a scalar $a$ with $a \Lambda_1 = \Lambda_2$.

This is the result that the class semi-group of $\mathcal{O}$ is finite. I think the shortest proof is via Minkowski's theorem on lattice points in convex bodies.