This is a theorem in Section 1, Chapter 8, Arithmetic of Elliptic Curves by J. Silverman. (Theorem 1.5(b))
$K$ be a number field. $E(K)$ the set of $K$-rational points on elliptic curve $E$, $[m]$ is multiplication-by-$m$ map: $E \to E$. Let $L = K([m] ^{-1} (E(K)))$ be a compositum of fields $K(Q)$ where $Q$ is such that $[m]Q =P$.
I am having trouble understanding what we want to prove in part (b) of this theorem. It should be as simple as proving the action trivial of inertia group on $L$ but I don't know how to go on about that and tthe proof given in the book doesn't make sense to me.
Can someone help in summarizing what Silverman is trying to do here? I'll really appreciate it.
Thank you very much.
They're doing basically what you're describing, with a reduction to make things simpler: since the compositum of extensions unramified outside $S$ is unramified outside $S$, it suffices to show that $K(Q)/K$ is unramified outside $S$ for any $Q\in[m]^{-1}(E(K))$. To show this, they take a place $v\in M_K\smallsetminus S$, take any extension of it (which they call $v'$) to $K':=K(Q)$, and then they show that the inertia group $I_{v'/v}$ is trivial by showing it acts trivially on $K'$. More precisely, to show an element $\sigma\in I_{v'/v}$ actually acts trivially on $K'=K(Q)$ it suffices to show that $Q^\sigma=Q$, and this is exactly what they show.
I'm not sure if this will help or if I've missed what was confusing you, let me know if it's the latter.