Another question related to the isogeny theorem for elliptic curves - MathOverflow

Another question related to the isogeny theorem for elliptic curves

2012-10-09T19:42:00Z

I was reading the following question: http://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...'?

Answer by Will Sawin for Another question related to the isogeny theorem for elliptic curves

2012-10-09T20:23:43Z

The class group of the endomorphism ring $\mathcal O_K$ is defined over $K$. But unless the class group of $\mathcal O_K$ is trivial, none of the CM curves are defined over $K$. Thus there is no bijection defined over $K$.

Answer by Joe Silverman for Another question related to the isogeny theorem for elliptic curves

2012-10-09T23:09:10Z

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 $\hbox{Ell}(R)$.

Answer by JSE for Another question related to the isogeny theorem for elliptic curves

2012-10-10T01:24:26Z

I just want to add to Joe's excellent answer the following much simpler example that shows the same behavior. You might ask: what is the relationship between the set of square roots of -1 in $\bar{\mathbf{Q}}$ and the group $A = \pm 1$? As mere sets, one may say they're in bijection, which is not a very rich statement; it just says there are two of each.

When we say that $\pm i$ and $A$ are not algebraically in bijection, we are saying that there is no bijection between the two sets which is commutes with the action of the Galois group $G_Q$ on the left. On the other hand, just as in the case Joe describes, A acts on $\pm i$ (by multiplication) simply transitively, and this action is compatible with Galois; it satisfies

$a t^\sigma = (at)^\sigma$

for each $t$ in $\pm 1$ and each $\sigma$ in Galois.

We say that $\pm i$ is a torsor for $A$.