# Show an intersection of Galois groups is trivial

Let $L/K$ be a finite abelian extension of number fields, and for an extension of places $w/v$ consider the local Artin map $\Phi: K_v^{\ast} \rightarrow Gal(L_w/K_v)$, defined via the global Artin map on the ideles. Let $\pi$ be a uniformizer for $K_v$, $E$ the maximal unramified extension of $K_v$ in $L_w$, and $F$ the fixed field of the subgroup of $Gal(L_w/K_v)$ generated by $\Phi(\pi)$. Note that the restriction of $\Phi(\pi)$ to $E$ generates the Galois group of $E/K_v$.

It's not difficult to show that $$K_v = E \cap F$$ but I would really like to show that also $$L_w = EF$$ or in other words the intersection of $Gal(L_w/E)$ and $Gal(L_w/F)$ is $1$. I'm interested in this because this is the last step I need to do in a proof that the local Artin map is defined independently of the global fields inducing it.

This proposition should be true (it is an analogue of a similar argument for infinite abelian extensions I saw in Caessels and Frohlich). I'm having trouble using the fact that $E$ is maximal. Any ideas?

• One possible idea, use the fact that $L_w/E$ is totally ramified, i.e. the inclusion $\mathcal O_E/ \mathfrak p_E \rightarrow \mathcal O_w/ \mathfrak p_w$ is an isomorphism. – D_S Jun 5 '15 at 5:08

I didn't figure it out. Let $\sigma$ be in the intersection of those Galois groups. I want to show that $\sigma$ is the identity on $L_w$. Since $\sigma \in Gal(L_w/F)$, there is an integer $k$ such that $\sigma = \Phi(\pi)^k = (\pi,L_w/K_v)^k$. Now the restriction of $\sigma$ to $E$ is $Frob_{E/K_v}(\pi)^k$. And the only way for $\sigma$ to be trivial on $E$ is for $f := [E : K_v]$ to divide $k$. Without loss of generality assume $f = k$.
Let $e = [L_w : E]$. There is a local Artin map for $L_w/E$ induced by some global extension, compatible with our existing local map for $L_w/K_v$. We have $$\sigma = (\pi^f, L_w/K_v) = (N_{L_w/E}(\pi), L_w/K_v) = (\pi, L_w/E)$$ But I don't think that $(\pi, L_w/E)$ is necessarily $1$. Is it?
• This sounds complicated. Can't you just show that $L/EF$ is both totally ramified and unramified? Should be simple using multiplicative property of ramification index and inertia degree. – Prometheus Jun 5 '15 at 6:05
• Thanks for all your helpful comments on algebraic number theory recently. Now here, I can see how $L_w/EF$ has inertia degree $1$, but why should it have ramification index $1$? – D_S Jun 5 '15 at 13:32
• If my edit is correct, then I wonder if the proposition is true at all. It seems to imply that if $\pi$ is a uniformizer for $E$ which lies in $K_v$, then the local Artin map for $L_w/E$ applied to $\pi$ is $1$. – D_S Jun 5 '15 at 14:03