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Let $F$ be a $p$-adic field with algebraic closure $\overline{F}$, and let $F^{\operatorname{ur}}$ be the maximal unramified extension of $F$. Let $E$ be a finite extension of $F$. I'm a bit rusty on my local class field theory and was wondering: is the composite field $EF^{\operatorname{ur}}$ always a Galois extension of $F^{\operatorname{ur}}$?

If $[EF^{\operatorname{ur}} : F^{\textrm{ur}}]$ is not divisible by $p$, then I think the answer is yes. I recall that for each $n$ with $p \nmid n$, there exists a unique extension $K$ of $F^{\textrm{ur}}$ of degree $n$, equal to $F^{\operatorname{ur}}(\sqrt[n]{\varpi})$ for a uniformizer $\varpi$ of $F$. Since all $n$th roots of elements of $\mathcal O_F^{\ast}$ lie in $F^{\textrm{ur}}$, we have that $K$ is defined independently of the choice of uniformizer, and is Galois over $F^{\operatorname{ur}}$.

So a counterexample would have something to do with wild ramification.

The reason I am asking is in another of my questions, I am considering the inertia group $\operatorname{Gal}(\overline{F}/EF^{\operatorname{ur}})$ of $E$, and was hoping for it to be normalized by $\operatorname{Gal}(\overline{F}/F)$, or at least by the Weil group $W_F \subseteq \operatorname{Gal}(\overline{F}/F)$.

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  • $\begingroup$ Are you claiming the every extension of $F^{ur}$ of finite degree prime to $p$ is obtained by the above method? This seems to be incorrect, as you only get extension abelian over $F$ in this way, and there are many extensions of $F$ which are nonabelian of degree prime to $p$ (depending on what you mean by "$p$-adic field"). Maybe I will come with an explicit counterexample later. $\endgroup$ – Gal Porat Jul 25 '18 at 7:47
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Behold! $F=\Bbb Q_p$, $E=\Bbb Q_p(\sqrt[p]{p})$, then $EF^{ur}=F^{ur}(\sqrt[p]{p})/F^{ur}$ is not normal, since it doesn't contain the $p$-th roots of unity and hence doesn't contain all roots of $x^p-p$.

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