Let $E$ be an elliptic curve over $\mathbb Q_p$ and suppose that $E$ has good reduction at a prime $p$. I read here that if $E$ has ordinary (resp. supersingular) reduction at $p$ then the mod $p$ representation of $E$ is reducible (resp. irreducible). Why is this true? If the reduction at $p$ is ordinary, I think reducibility follows essentially because the reduction map $E[p]\rightarrow \tilde E[p]$ is $G_{\mathbb Q_p}$-equivariant, so its kernel is a $G_{\mathbb Q_p}$-stable copy of $\mathbb{Z}/p\mathbb{Z}$ in $E[p]$. But in the supersingular case, the reduction map is just zero, so it doesn't seem to be of any help… Can anyone point me in the right direction?
Elliptic curves with supersingular reduction have irreducible mod $p$ representations
elliptic-curvesgalois-representationsnumber theory
Best Answer
One way to see this is that, as a consequence of the theory of formal groups, the image of inertia via the mod-$p$ representation is a (full) non-split Cartan subgroup. Since the non-split Cartan has no fixed non-trivial submodules of order $p$, the representation itself must be irreducible. All of this is nicely shown in a well-known paper of Serre (where he proves the so-called "open image theorem"), and you can see a summary of what you need in Section 3 of this article, and in particular in Theorem 3.1.