2
$\begingroup$

Let $E$ be an elliptic curve with CM by an order $\mathcal O$ in an imaginary quadratic field $K$. Choose a basis for $E[N]$ to get an isomorphism $\operatorname{Aut}(E[N])\cong \operatorname{GL}_2(\mathbb Z/N\mathbb Z) $. Complex multiplication on $E$ induces a homomorphism $(\mathcal O/ N\mathcal O )^\times\rightarrow \operatorname{GL}_2(\mathbb Z/N\mathbb Z)$. Let $C_N$ be the image.

Is $C_N$ equal to its centralizer in $\operatorname{GL}_2(\mathbb Z/N\mathbb Z)$?

This should occur in some cases, and should give rise to a rational point on certain modular curves. See Burcu Baran: Normalizers of non-split Cartan subgroups, modular curves, and the class number one problem, especially Proposition 4.1.

$\endgroup$

1 Answer 1

3
$\begingroup$

I believe the answer is yes. It is really a group theory question, so I am not the best person to answer this.

If $N$ is prime: The centraliser $Z_G(C)$ is included in $N_G(C)$. Since the action by conjugation of $N_G(C)$ is not trivial on $C$ (both in the split and non-split case), $Z_G(C)\neq N_G(C)$. As the index of $C$ in $N_G(C)$ is 2, we must have $Z_G(C)=C$.

For square-free $N$ use the Chinese remainder theorem to reduce to the prime case.

I have not thought about the case of a prime power, but my guess would be that it holds there too.

I don't know how this question relates to the rational point question. But yes, for CM curves, depending on the splitting behaviour of primes $p\mid N$ in the field $\operatorname{End}(E)\otimes\mathbb{Q}$, we get rational points on the (split and non-split) Cartan modular curves. The big question however is if there are any other rational points on these curves.

$\endgroup$

You must log in to answer this question.

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