8
$\begingroup$

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.)

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ Dec 6, 2011 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. $\endgroup$ Dec 6, 2011 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, 2011 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. $\endgroup$ Dec 7, 2011 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 $\endgroup$ Dec 7, 2011 at 13:30

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.