Elliptic curves with supersingular reduction have irreducible mod $p$ representations?

Question

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?

Answer 1

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.