Let $E/K$ be a elliptic curve over a number field $K$ and let $L/K$ be a finite abelian (Galois) extension. Let $v’$ be a (finite) place of $L$ lying over a place $v$ of $K$. Let $I=I_{v’/v}$ be the inertia group. Suppose $I$ acts trivially on $E(L)$. Why is then $L/K$ unramified at $v’$, where by unramified I mean we have equality $v’=v$ of normalized valuations (or $e=1$ in $\mathfrak{p}_v\mathcal{O}_L = (\mathfrak{p}_{v’})^e \cdots$ as prime ideals).
This is basically what Silverman is using on p.212 of Arithmetic of Elliptic Curves. He defines unramified for any Galois module if restriction to the inertia group is trivial. I want to compare this to the usual definition of unramified field extension.
Best Answer
We have a number field $K$ and an extension $K' = K(Q)$ where $Q \in E(\overline K)$ for some elliptic curve $E$. Fix a finite place $v$ of $K$ and a place $v'$ of $K'$ lying above $v$.
Assume that every element of the inertia group $I_{v'/v}$ fixes $Q$. We want to deduce that $K' = K(Q)$ is unramified over $K$ at $v'$. By definition, it means that the ramification index $e(v'/v)$ equals $1$.
Recall that the inertia group $I_{v'/v}$ has order $e(v'/v)$. Moreover, any $g \in I_{v'/v}$ fixes $Q$, and also fix $K$, since $I_{v'/v} \subset \mathrm{Gal}(K'/K)$ (I think on p. 209, Silverman assumes $E[m] \subset E(K)$ so that $K'/K$ is Galois). Thus $g$ fixes $K(Q)$, that is $g = \mathrm{id}_{K'}$. This shows that $I_{v'/v}$ is the trivial group, of order $e(v'/v)=1$. This shows that $K'/K$ is unramified at $v'$, as claimed.