I was reading the following question: [About isogeny theorem for elliptic curves](https://mathoverflow.net/questions/41931/about-isogeny-theorem-for-elliptic-curves) and was interested in the following statement at the end of Torsten Ekedahl's answer:


"Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection."


Could someone please clear up exactly what could be meant by `algebraically they cannot be put into bijection...'?



Another question related to the isogeny theorem for elliptic curves
I think that maybe what is meant is that there is no *functorial* way to define the bijection in the category of algebraic geometry. So suppose that we let $\hbox{Ell}(R)$ denote the set of elliptic curves with $\hbox{End}(E)\cong R$, where for simplicity $R$ is the maximal order of an imaginary quadratic field. (The isomorphism with $R$ is over $\overline{\mathbb{Q}}$.) The ideal class group $H\_R$ is, as its name proclaims, a group. So it has a preferred element, namely the identity element. But $\hbox{Ell}(R)$ does not have a preferred element in any natural sense. The right way to think of this is that there is a natural action of $H\_R$ on $\hbox{Ell}(R)$, and this action is simply transitive. In particular, $H\_R$ and $\hbox{Ell}(R)$ have the same number of elements. But if you want to identify them $\hbox{Ell}(R)\leftrightarrow H\_R$, you need to choose an element of $\hbox{Ell}(R)$ to be distinguished.


On the other hand, if you fix an embedding $R\subset\mathbb{C}$, then every ideal $\mathfrak{a}$ in $R$ is a lattice in $\mathbb{C}$, so you can identify the ideal class $\overline{\mathfrak{a}}$ with the complex torus $\mathbb{C}/\mathfrak{a}$. This is analytically isomorphic to an elliptic curve $E\_{\mathfrak{a}}$ in $\hbox{Ell}(R)$.


