6
$\begingroup$

I am currently studying mod-$\ell$ Galois representations of $CM$ elliptic curves. More precisely, let $\ell$ be a prime, $K$ a number field and $E/K$ a $CM$ elliptic curve, where all endomorphisms of $E$ are defined over $K$, then I am looking at the representation $\rho_{E,\ell}: \mathrm{Gal}(\overline{K}/K) \to \mathrm{Aut}(E[\ell])$.

I was wondering if there is an effective/computable criterion to determine for which $ \ell's $ is this representation semi-simple.

$\endgroup$
5
$\begingroup$

Let us assume that $E$ has complex multiplication by a maximal order $\mathcal{O}$. Then $E[\ell]$ is a free rank $1$ module over $\mathcal{O}/\ell\mathcal{O}$. The image $G$ of $\rho_{E,\ell}$ is contained in the $\mathcal{O}$-automorphism of $E[\ell]$, which are isomorphic to $(\mathcal{O}/\ell\mathcal{O})^{\times}$.

If $\ell$ is unramified in $\mathcal{O}$, then $G$ is contained in a group of order either $\ell^2-1$ or $(\ell-1)^2$. Since $\ell$ will not divide the group order of $G$, the action on $E[\ell]$ will be semi-simple.

Instead for ramified primes $\ell$, it can happen that $E[\ell]$ is not semi-simple. For instance for the curve $y^2 = x^3 + 1$ over $K=\mathbb{Q}(\sqrt{-3})$ and $\ell=3$. Then $K(E[3])/K$ is an extension of degree $3$ and hence the Galois group acts on $E[3]$ via the matrix $\bigr(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\bigr)$.

I believe in the ramified case one has to check by hand if $E$ admits two isogenies of degree $\ell$ defined over $K$. Of course in practice one can determine $G$ completely for these $\ell$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.