Elliptic curve with CM and image of Galois representation in normalizer of nonsplit Cartan

Question

I am trying to understand the following. Let $E/\mathbb{Q}$ be an elliptic curve with complex multiplication given by the ring of integers $\mathcal{O}_K$. We are given a fixed rational prime $p$ which is inert in $\mathcal{O}_K $. Then the image of the mod $p$ Galois representation $\overline{\rho}_{p,E}(G_\mathbb{Q})$ is conjugate to a subgroup of the normalizer of non-split Cartan.

The way I think about it is the following. We know End$(E)\cong \mathcal{O}_K$, hence there is an isogeny: $$ [\sqrt{-d}]: E \to E. $$ I would like to make a choice of basis for $E[p]$ such that the Galois action with respect to this basis gives an element of the normalizer of a non-split Cartan group.

Any help is much appreciated.

Answer 1

By cm theory, $E[p]$ is isomorphic to $\mathcal{O}_K/(p)$ as an $\mathcal{O}_K$-module. Elements in $\operatorname{Gal}(\bar{\mathbb{Q}}/K)$ act $\mathcal{O}_K$-linearly on $E[p]$ since they commute with endomorphisms, which means they act by multiplication with an element in $\bigl(\mathcal{O}_K/(p)\bigr)^{\times}$. When viewing $E[p]$ as a $\mathbb{F}_p$-vector space, this means that the representation from $\operatorname{Gal}(\bar{\mathbb{Q}}/K)$ has image in a non-split Cartan subgroup of $\operatorname{GL}\bigl(E[p]\bigr)$.

The image of $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ might at most be twice as large, therefore it is contained in the normaliser of this non-split Cartan subgroup. By the way, it cannot be contained in the non-split Cartan subgroup itself as complex conjugation is not equal to the multiplication by a unit modulo $p$. For large enough $p$, the image should be the full normaliser.