Let $F$ be a nonarchimedean local field. Let $k$ be a global field with place $v$ such that $F = k_{v}$.
Let $E$ be a finite extension of the completion $F = k_{v}$. Does there exist a global field $K$ containing $k$, and a place $w$ of $K$ lying over $v$, such that $K_w = E$ with $[K : k] = [E : F]$? It seems like this should come from Krasner's lemma, but I'm a bit hazy on these kind of arguments.
Krasner's lemma says that if $\alpha, \beta \in \overline{F}$, and $|\alpha - \beta| < |\sigma(\alpha) - \alpha|$ for all $\sigma \in \operatorname{Hom}_{\textrm{$F$-alg}}(F, \overline{F})$, then $F(\alpha) \subseteq F(\beta)$.
I recall Krasner's lemma implies that if $f(t) = a_0 + a_1t + \cdots + a_nt^n \in F[t]$ is irreducible with roots $\alpha_1, ... , \alpha_n$, then there exists an $\epsilon > 0$ such that if $g(t) = b_0 + b_1t+ \cdots +b_nt^n \in F[t]$, and $|a_i - b_i| < \epsilon$, $g$ is also irreducible, and $F(\alpha_i) = F(\beta_i)$, where $\beta_1, ... , \beta_n$ is some ordering on the roots of $g$.
Krasner's lemma is also saying if $f \in F[t]$ is irreducible then you can approximate it with a polynomial $g\in k[t]$ that will generate the same extension : $F[t]/(f) = F[u]/(g)$ and your field is $K=k[u]/(g)$, $K_w = k_v[u]/(g) = F[t]/(f)$