Let $L/K$ be a finite extension of $p$-adic fields, $\pi$ a uniformizer of $K$, $\theta = (-, L/K)$ the local Artin map $K^{\ast} \rightarrow Gal(L/K)$, $E$ be maximal unramified extension of $K$ contained in $L$, and $F$ the fixed field of the subgroup generated by $\theta(\pi)$. It is easy to see that $E \cap F = K$ (use the fact that the restriction of $\theta(\pi)$ to $E$ generates $Gal(E/K)$). My question is whether it is true that $$EF = L$$ In Caessels and Frohlich, Serre section 2.3, a similar result is claimed (without proof) when $L$ is an infinite abelian extension of $K$ containing all unramified extensions. I was wondering whether the analagous result held in the finite case.
One idea I had to show $EF = L$ is to show that $Gal(L/E) \cap Gal(L/F)$ is trivial. If $\sigma$ is in this intersection, then $\sigma = \theta(\pi)^k$ for some $k$. Clearly $\theta(\pi)^k$ fixes $E$ if and only if $k$ is divisible by $[L :E] = f(L/K) = f$. Without loss of generality suppose $k = f$. Then $$\sigma = (\pi^f, L/K) = (N_{E/K}(\pi), L/K) = (\pi, L/E)$$ so we are reduced to showing that if $\pi$ is a uniformizer for $E$ which lies in $K$, then the Artin map of $L/E$ applied to $\pi$ is trivial. This is true if and only if $\pi$ is a norm (over $E$) from $L$. Would this be the case?
