Let $K$ be an imaginary quadratic field, and $\mathfrak{f}$ an integral ideal of $K$ which is stable under complex conjugation. Assume that $(1 + \mathfrak{f} ) \cap \mathcal{O}_K^\times = \{1\}$.

Then $\mathbb{C} / \mathfrak{f}$ is an elliptic curve over $\mathbb{C}$ with CM by $\mathcal{O}_K$, and $1 \in \mathbb{C}$ maps to a primitive $\mathfrak{f}$-torsion point; and it's a standard theorem that there is a pair $(E, \alpha)$, consisting of an elliptic curve with CM by $\mathcal{O}_K$ and a primitive $\mathfrak{f}$-torsion point, defined over the ray class field $K(\mathfrak{f})$ which becomes isomorphic to $(\mathbb{C} / \mathfrak{f}, 1)$ over $\mathbb{C}$, and $(E, \alpha)$ is unique up to unique isomorphism.

Here's the question: can we find a model for $(E, \alpha)$ over $K(\mathfrak{f})^+ = K(\mathfrak{f}) \cap \mathbb{R}$? I'm pretty sure we can descend $E$ to $K(\mathfrak{f})^+$, but will the torsion point $\alpha$ be rational over this smaller field too?

(If the complex conjugation on $E(\mathbb{C})$ arising from the $K(\mathfrak{f})^+$ model of $E$ coincides with the natural complex conjugation on $\mathbb{C} / \mathfrak{f}$ this is immediate, but it's not completely clear to me that this is the case.)

  • $\begingroup$ Small comment: since the ideal is stable under complex conjugation, it is of order two in the class group. Reducing it, we get a reduced ambiguous ideal, which under the assumptions isn't above $2$ (or $1$). So it is of the form $(m,\sqrt{\Delta})$, with $m|\Delta$. Normalizing, the lattice is homothetic to $(1,\sqrt{\Delta}/m)$. Hence, by V.2.1 in [AECII], the j-invariant is real. $\endgroup$ – Dror Speiser Dec 6 '11 at 21:49
  • 2
    $\begingroup$ @Dror: The j-invariant of any lattice invariant under complex conjugation is real, because the q-expansion coefficients of the j-invariant are in $\mathbb{R}$. One doesn't need the lattice to have CM for this to work. $\endgroup$ – David Loeffler Dec 6 '11 at 21:56
  • 1
    $\begingroup$ @David: Doesn't the statement in the if clause of the last sentence in your question follow from the Main Theorem of Complex Multiplication? $\endgroup$ – monodromy Dec 6 '11 at 23:31
  • 1
    $\begingroup$ @monodromy: the Main Theorem of CM is a statement about automorphisms of $\mathbb{C}$ which restrict to the identity on $K$, which complex conjugation does not. $\endgroup$ – David Loeffler Dec 7 '11 at 8:16
  • 3
    $\begingroup$ There is the main Main theorem, which actually describes the action of the whole Galois group of $\mathbb{Q}$, though it seems a little excessive here. See Section 4 of jmilne.org/math/articles/2007c.pdf $\endgroup$ – Keerthi Madapusi Pera Dec 7 '11 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.