6
$\begingroup$

Let $E$ be a CM elliptic curve defined over a quadratic imaginary field $K$ with maximal order i.e., $\mathrm{End}_K(E)\cong \mathcal{O}=\mathcal{O}_K$. Let $\mathfrak{p}$ be a prime of $K$ such that the map $\mathcal{O}^\times \to(\mathcal{O}/\mathfrak{p})^\times$ is not surjective. With this situation, I tried to prove $E[\mathfrak{p}]\not \subset E(K)$.

My idea is as follows; If $E[\mathfrak{p}]\subset E(K)$, then we can consider $(\mathcal{O}/\mathfrak{p})^\times$ as the set of isomorphisms modulo an equivalence relation `$\sim$' defined by for $f,g\in \mathcal{O}$ we say $f\sim g$ when they coincide on $E[\mathfrak{p}]$. Then it seems contradict to the assumption $\mathcal{O}^\times \to(\mathcal{O}/\mathfrak{p})^\times$ is not surjective. Is it plausible? If it is, how to expand details of proof?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

See Cor. 5.18 of Rubin "Elliptic curves with complex multiplication" in LNM 1716 for a proof (that uses the main theorem of complex multiplication).

Improve this answer
$\endgroup$
4
  • $\begingroup$ Actually, I am reading that note, but I cannot understand the reason why that corollary follows. $\endgroup$
    – Kevin.lijh
    Commented Dec 12, 2013 at 17:36
  • 1
    $\begingroup$ Having read it a while ago, I don't remember the details, but it seems that the argument is structured like this: assuming the claimed implications of Thm. 5.15, Galois must act on $E[\mathfrak{p}]$ nontrivially, as one sees by taking $x \in \mathcal{O}_{\mathfrak{p}}^{\times}$ such that its reduction modulo $\mathfrak{p}$ does not lie in the image of $\mathcal{O}^{\times}$. $\endgroup$
    – Kestutis Cesnavicius
    Commented Dec 12, 2013 at 17:44
  • $\begingroup$ But then why do we need the condition $\psi(\mathcal{O}^\times_\mathfrak{p})\subset \mathcal{O}^\times$? $\endgroup$
    – Kevin.lijh
    Commented Dec 12, 2013 at 17:50
  • $\begingroup$ This condition assures that for an $x$ as above the image of $\psi(x)x^{-1}$ in $(\mathcal{O}/\mathfrak{p})^{\times}$ is not the identity. Since $E[\mathfrak{p}]$ is a free $\mathcal{O}/\mathfrak{p}$-module of rank $1$, this means that $[x, K^{\mathrm{ab}}/K]$ must act on $E[\mathfrak{p}]$ nontrivially. $\endgroup$
    – Kestutis Cesnavicius
    Commented Dec 12, 2013 at 17:53
0
$\begingroup$

This is not a nice approach, but the unit group of an imaginary quadratic field consist of roots of unity, and there are only the possibilities $|\mu(K)| \in \{2,4,6\}$. So if the homomorphism in question is surjective, you get a bound on the norm of $\mathfrak{p}$. Futhermore, $|\mu(K)| = 4$ iff $K = \mathbf{Q}(i)$ and $|\mu(K)| = 6$ iff $K = \mathbf{Q}(\sqrt{-3})$.

And if $E[n] \subseteq E(K)$, $\mu_n \subseteq K$ by the Weil pairing.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .