Elliptic curves with complex multiplication by an order $\mathcal{O}$ I am reading a paper about elliptic curves with CM, and since I'm new to this I'm having troubles understanding the essence of some things.
Let $E$ be an elliptic curve with complex multiplication by an order $\mathcal{O}$, let $K$ be the field of fractions of $\mathcal{O}$, and $$K_{\Delta}=K(j(E)),$$ $$F=\mathbb{Q}(j(E)),$$ $$w=\#\mathcal{O}_K^{\times},$$ $$h(\Delta)=\# Pic(\mathcal{O})=[K_{\Delta}:K].$$
I saw an enormous amount of results with expressions above. I do understand what CM by an order means, but namely, I don't understand why $K_{\Delta}$ or $F$ would be worth considering? Does it have anything to do with twists?
I am also interested in the interpretation of the numbers $w$ and $h(\Delta).$ I feel like these things are fairly common in this topic, I am just failing to understand why.
 A: There are lots of interesting things to say about them. To start, assume $\mathcal O$ is the maximal order.
A very interesting exact sequence in number theory is
$$0\rightarrow \mathcal O_K^\times \rightarrow K^\times \rightarrow  I_K \rightarrow H_K\rightarrow 0$$
where $I_K$ is the group of fractional ideals and $H_K$ is the class group.
Lots of the quantities you listed give information about members of this sequence. When $K$ is quadratic imaginary, $\mathcal O_K^\times$ is finite by dirichlet, and its order is your $w$. The degree of $K_\Delta$ over $K$ is the class number $h_K = |H_K|$.
In fact, the latter follows from the fact that $K_\Delta$ is the hilbert class field of $K$ (which is itself very interesting). This is the start of explicit class field theory for quadratic imaginary extensions.
Also, $[F:\mathbb Q]$ also coincides with $h_K$ for maximal orders.
I don't remember much about CM in the case of nonmaximal orders - it is more complicated (hence more interesting :) ). For instance, the notion of "fractional ideal" has to be tweaked so the ideal and ideal class groups will actually be groups (e.g. the latter is replaced by the Picard group, which is exactly what appears in your question). Hopefully someone who knows more can weigh in.
