Let $F$ be a nonarchimedean local field of characteristic zero, and $E$ an extension of $F$ with $[E:F]=2^n$ for some $n$. Is it always possible to find a quadratic extension $M$ of $F$ such that $F\subseteq M\subseteq E$ ?



Quadratic extension of local field
* Try $f(x)=x^4+2x+2\in \Bbb{Q}\_2[x]$ and $K=\Bbb{Q}\_2[a]/(f(a))$.


Then $f(x) = (x-a)g(x)\in K[x]$ where $g$ is irreducible ($a^{-3}g(ax-a/3)$ is Eisenstein). So $f$ has only one root in $K$, $Aut(K/\Bbb{Q}\_2)$ is trivial and hence there is no subfield $L$ such that $[K:L]=2$ ie. no subfield such that $[L:\Bbb{Q}\_2]=2$.
* If $F$ is a finite extension of $\Bbb{Q}\_p$ with $p$ odd and $[E:F]=2^n,n\ge 1$ then yes there is always a quadratic subfield. Assume the opposite, take an uniformizer $\pi\_E$, the residue field of $F$ is $\Bbb{F}\_q$, that of $E$ is $\Bbb{F}\_{q^m}$, it must be that $m | 2^n$, and it must be that $m=1$ as otherwise $F(\zeta\_{q^2-1})/F$ is a quadratic subfield.


Therefore, $E = F(\pi\_E)$ where $\pi\_E^{2^n} = u \pi\_F$ for some $u\in O\_E^\times$.


So $u=\zeta\_{q-1}^r (1+c\pi\_E)$ with $c\in O\_E$,


$\varpi\_E =(1+c\pi\_E)^{-1/2^n} \pi\_E\in E$ is again an uniformizer and $\varpi\_E^{2^n}=\zeta\_{q-1}^r \pi\_F\in F$.


ie. $F(\varpi\_E^{2^{n-1}})/F$ is a quadratic subextension.


