this is my first question ever on a platform like this so please forgive me any kind of unintended misbehaving.

In Kudla, Rapoport and Yang "On the derivative of an Eisenstein series of weight one" the authors define a (fine) moduli stack $\mathfrak{M}(1,0)$ of elliptic curves with complex multiplication by $\mathcal{O}_K$ where $K$ is the reflex field. This is a stack over $\operatorname{Spec}(\mathcal{O}_K)$.

My question is now the following. What is known about the existence of smooth elliptic curves with such a complex multiplication over the scheme $\operatorname{Spec}(\mathcal{O}_K)$ itself?

Thank you!

  • 1
    $\begingroup$ Hi Sebastian. If memory serves me correctly, the main theorem of complex multiplication implies that the j-invariant of such an elliptic curve with complex multiplication generates the Hilbert class field over K, and so at a minimum one would need the class number of K to be 1. $\endgroup$ – Tyler Lawson Aug 15 '13 at 4:58
  • $\begingroup$ Right over $\operatorname{Spec}({\mathcal{O}_H})$, where $H \supset K$ is the Hilber class field, we have it all. But if there are no such elliptic curves over $\operatorname{Spec}({\mathcal{O}_K})$, I would expect $\mathfrak{M}(1,0)$ to be a stack over $\operatorname{Spec}({\mathcal{O}_H})$? $\endgroup$ – Sebastian Thyssen Aug 15 '13 at 15:00

In the paper you mentioned, $K\subset \mathbb{C}$ is quadratic imaginary so the reflex field is just $K$ itself. Indeed, $\mathcal{M}(1,0)$ is defined over Spec$(\mathcal{O}_K)$, but the smallest field of definition of CM elliptic curves can be either bigger or smaller than $K$.

For example $E: y^2 = x^3+x$ is defined over $\mathbb{Z}$, but has complex multiplication by $\mathbb{Z}[i]$. In this case the $j$-invariant is $1728$, and $H=K$. Note that by definition only the base change of this $E$ to $E\times \operatorname{Spec}(\mathcal{O}_K)$ appears in the family $\mathcal{M}(1,0)$.

If the class number is not 1, $E$ can't be defined over $K$, since $j(E)\in K$ contradicts $K \subsetneq K(j(E)) = H$. But the moduli space itself is still defined over $\operatorname{Spec}(\mathcal{O}_K)$. The right thing to compare with the field of definition of each elliptic curve in the family, is actually $\mathcal{M}$ itself, not the base of $\mathcal{M}$. This is because the universal object for the moduli space is defined over $\mathcal{M}$, and the other objects are obtained by base changing via maps to $\mathcal{M}$.

Suppose $L$ is a field, and let $E$ be an elliptic curve over $\operatorname{Spec}(L)$ belonging to the family $\mathcal{M}$. Since $\operatorname{Spec}(\mathcal{O}_H)$ is the coarse moduli space of $\mathcal{M}(1,0)$, there's a map $\mathcal{M}(1,0) \rightarrow \operatorname{Spec}(\mathcal{O}_H)$ over $\operatorname{Spec}(\mathcal{O}_K)$. Composing this with the map $\operatorname{Spec}(L) \rightarrow \mathcal{M}(1,0)$ corresponding to $E$, gives a map $\operatorname{Spec}(L) \rightarrow \operatorname{Spec}(\mathcal{O}_H)$ over $\operatorname{Spec}(\mathcal{O}_K)$. This shows $L$ has to contain $\mathcal{O}_H$ (and hence $H$.)

  • $\begingroup$ Ah...got it. Very nice answer! Thanks $\endgroup$ – Sebastian Thyssen Aug 16 '13 at 6:57

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