Unramified at certain places and Selmer groups

Question

Consider the $p$-Selmer group of an elliptic curve $E/\mathbb{Q}$ denoted by $\operatorname{Sel}_{p}(E/\mathbb{Q})$. Why does showing that $E(\mathbb{Q}_{\ell})/pE(\mathbb{Q}_{\ell}) = 0$ (for $\ell \neq p$) imply that $\operatorname{Sel}_{p}(E/\mathbb{Q})$ is unramified at the place $\ell$?

Answer 1

I think I can now give an answer to this. Can anyone check my reasoning?

Recall that by definition, $\xi \in \operatorname{Sel}_{p}(E/\mathbb{Q})$ is unramified at a place $v$ if it is trivial in $H^{1}(I_{v}, E[p])$ where $I_{v}$ is the inertia subgroup of $G_{\overline{\mathbb{Q}}_{v}/\mathbb{Q}_{v}}$. For a prime $\ell \neq p$, we have the Kummer sequence for $E/\mathbb{Q}_{\ell}$, $$0 \longrightarrow E(\mathbb{Q}_{\ell})pE(\mathbb{Q}_{\ell}) \overset{\phi}{\longrightarrow} H^{1}(G_{\overline{\mathbb{Q}}_{\ell}/\mathbb{Q}_{\ell}}, E[p]) \overset{\psi}{\longrightarrow} H^{1}(G_{\overline{\mathbb{Q}}_{\ell}/\mathbb{Q}_{\ell}}, E(\overline{\mathbb{Q}}_{\ell}))[p] \longrightarrow 0$$ which is a short exact sequence. In particular as $E(\mathbb{Q}_{\ell}/pE(\mathbb{Q}_{\ell}) = 0$, by the exactness of the sequence, $\ker\psi = \operatorname{im}\phi = 0$. Then by the definition of unramified at the beginning of this paragraph, we have that $\operatorname{Sel}_{p}(E/\mathbb{Q})$ is unramified at $\ell$.