Proof help - A finite field extension $L/K$ is unramified outside a finite set of primes (Silverman AEC, 8)

Question

I've been having trouble with the following proof in Silverman's ECs book.

Proposition 1.5(b), Chapter 8: ($K$ is a number field)

Let $L = K([m]^{-1}(E(K)))$ be a field

My attempts at understanding part(b) proof:

1. I understand why we need to prove the claim for just the finite extension $K'/K$ and we are actually considering $E$ to be defined over the local field $K_v$. To prove the required, we need to show that

$$ Q^{\sigma} = Q $$ for all $\sigma \in I_v$ but why did the book considered the inertia group $I_{v'/v}$ instead?

2. I think, at some point (towards the end) we'll use these two things- $Q^{\sigma}-Q$ is in $m$-torsion and it is in kernel of the reduction modulo $v$ map $ E(\bar{K}_v) \hookrightarrow \tilde{E}(k_v)$ where $k_v$ is the residue field of local field $K_v$.

But I don't know exactly which injection is being used here and why.

I would really appreciate it if someone could explain/clarify this proof a bit.

Thank you.

EDIT : Let $v \in M_K$ with $v \notin S$ and $Q, K'$ be as in the book. We can consider $E$ defined over the local field $K_v$. Let $K''$ be the smallest field extension of $K'$ that is Galois. Let $v''$ be a place of $K''$ lying above $v$. $E$ also has good reduction at $v''$ and we have the reduction map $$ E(K'') \longrightarrow \tilde{E}(k''_{v''}) $$

We will prove that for all $\sigma \in I(v''/v)$, $Q^{\sigma} = Q$. (I'm assuming that $I_v/I_v'$ you mentioned is isomorphic to $I(v'/v) =\{ \sigma \in Gal(K'/K) : v'(\sigma(\alpha)-\alpha) >0 \forall \alpha \text{ satisfying } v'(\alpha)\geq 0 \}$ )

Now by definition every element of $I(v''/v)$, acts trivially on $\tilde{E}(k''_{v''})$, so $$ \tilde{Q^{\sigma}-Q} = \tilde{O}$$

Also, $Q^{\sigma}-Q$ is in $E(K)[m]$. Thus by (VIII, 1.4) we have $Q^{\sigma}-Q=O$.

Answer 1

I think there is a small mistake; I believe one should replace $K'$ with its Galois closure first, only so that members of $I_v$ always take $K'$ to itself. This poses no problems for the proof.

(1) You are right that one would initially be led to consider $I_v$, because by definition everything in $\bar K$ fixed by all of $I_v$ is unramified over $K$. What we want to show is that $Q$ is fixed by all $\sigma \in I_v$. But $Q$ has coordinate in $K'$, any $\sigma$ fixing $K'$ pointwise fixes $Q$ ``for free''. So we really only need to consider members of $I_v$ modulo those which fix $K'$, which is exactly $I_{v}/I_{v'}$, aka the inertia subgroup of the Galois group of $K'/K$, which I think is Silverman's $I_{v'/v}$. (this is why one should tweak $K'$ to be Galois).

Or in other words, $Q$ lives in $K'$, so to be unramified over $K$ it is necessary and sufficient that $Q$ is fixed by the ``inertia subgroup'' of $K'$.

(2) At the end of the proof, we use the two things you wrote down: $Q^\sigma - Q$ is an $m$-torsion element, and that the reduction map is injective, and also that the reduction of $Q^\sigma - Q$ is $O$. So both $O$ and $Q^\sigma - Q$ are $m$-torsion elements which reduce to the same element $0$ mod $p$, which by injectivity means they must have been equal originally, i.e. $$Q^\sigma - Q = O\ \ \ \textrm{ equivalently }\ \ \ Q^\sigma = Q$$

We have injectivity exactly because of the choice of places: $v(m) = 0$ and we have good reduction, which is when VIII.1.4 applies and tells us that the reduction map is injective on $m$-torsion.