Quartic unramified extension of $\mathbb{Q}_p$ and existence of generators of the form $\sqrt[4]{a}$ for an integer $a$

Question

Let $K=\mathbb{Q}_p$ where $p$ is a prime number such that $p-1$ is divisible by $4$. Furthermore, let $F/K$ be the unramified extension of degree $4$.

Question: Does $F/K$ always have a generator of the form $\sqrt[4]{a}$ (equivalently, $F = K(\sqrt[4]{a})$ for some $a \in \mathbb{Z}$?

I noticed that for $p=5,13$, it is $F = K(\sqrt[4]{2})$. However, this observation is false for $p=17$ since $2$ is not a quadratic residue modulo $17$. I am interested in this question because the claim is true if we assume that $F/K$ has degree $2$ instead of $4$ and I would like to know to what extent this can be generalized.

Answer 1

Alternatively let $g$ an integer having order $p-1$ modulo $p$.

A root of $x^4-g\bmod p$ will have order $4(p-1)$.

And $4(p-1)$ divides $p^4-1=(p^2+1)(p+1)(p-1)$ but not $p^2-1=(p-1)(p+1)$.

Whence $x^4-g\bmod p$ is irreducible ie. $F=\Bbb{Q}_p[x]/(x^4-g)$ is your unramified extension of degree $4$.

Answer 2

There is always such an $a$. The assumption on $p$ guarantees that the base field contains 4th roots of unity. It then follows from Kummer theory that any extension of degree $4$ is generated by taking a 4th root.