“Algebraization" of $p$-adic fields

Question

Part 1: a single finite place. Let $K$ be a finite extension of $\mathbf{Q}_p$.

Does there exist a number field $F/\mathbf{Q}$ and a finite place $v$ lying over $p$, such that for the completion of $F$ at $v$ we have a topological isomorphism of topological field extensions of $\mathbf{Q}_p$:

$$F_v\simeq K ?$$

In imprecise words, does any $p$-adic field come as a completion of a number field?

Part 2: a family of $p$-adic fields. (ASKED AS A SEPARATE QUESTION, here) Let $(K_p)_p$ be a collection of $p$-adic fields. Assume the first part of the question has positive answer. Does there exist a condition on $(K_p)_p$ such that there is one number field $K$ recovering each $K_p$ as a completion?

Answer 1

The answer to part 1 is yes. Given $K/\mathbb{Q}_p$, let $\alpha\in K$ be a primitive element, with minimal monic polynomial $f(x)=x^n+\sum_{i=1}^n a_i x^{n-i}$, $a_i\in\mathbb{Q}_p$. So we have $K\cong\mathbb{Q}_p[x]/(f(x))$. Krasner's lemma implies there exists a positive integer $N$ such that if $b_i\in\mathbb{Q}_p$ satisfy $v_p(b_i-a_i)\geq N$, then the polynomial $g(x)=x^n+\sum_i b_i x^{n-i}$ satisfies $\mathbb{Q}_p[x]/(f(x))\cong\mathbb{Q}_p[x]/(g(x))$. In particular, since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, we can choose the $b_i$ to be rational. Then $F:=\mathbb{Q}[x]/(g(x))$ is a finite extension of $\mathbb{Q}$, and $F\otimes\mathbb{Q}_p\cong\mathbb{Q}_p[x]/(g(x))\cong K$. In general, tensoring a number field with $\mathbb{Q}_p$ gives the product of the completions of that number field at the primes lying over $p$, so in this case there is only one prime $v$ of $F$ over $p$, and $F_v\cong K$.

I do not know an answer to part 2, but I will mention some necessary conditions on the family $(K_p)$: only finitely many can be ramified, only finitely many degrees occur, and $\{p:K_p=\mathbb{Q}_p\}$ has positive lower density.