Let $p$ be a prime number. I want to find a conceptual way of proving that $i = \sqrt{-1}$ is not an element of $F = \Bbb Q(\sqrt[4]{-p})$.
In this question, the solution involves too much computations for me.
Notice that it is easy to show that $i \notin \Bbb Q(i \sqrt[4]{p})$. But if $i \in F$, then so would be $\sqrt p = -i (\sqrt[4]{-p})^2$. Equality of degrees would imply that $ F= \Bbb Q(i, \sqrt p)$, but what to do next?
If $i\in F$, then $[F(i):\Bbb Q]=4$, and if $i\notin F$, then $[F(i):\Bbb Q]=8$.
Now consider $[F(i):\Bbb Q(i)]$. It is either $2$ or $4$ by the results above, and it's obtained adjoining a specific root of $x^4+p$ to $\Bbb Q(i)$. So we just have to show that no second degree factor of $x^4+p$ has coefficients in $\Bbb Q(i)$.
That's six different possibilities to check (or three of we only include the ones which contain $(x-\sqrt[4]{-p})$) but for each such second degree polynomial $f(x)$, we have that $-f(ix)$ is another one, and this way we only need to check two of them: one with diametrically opposite roots, and one with adjacent roots. Assuming $a=\sqrt[4]{-p}$ is the root with complex argument $\pi/4$, we have $$ (x-a)(x+a)=x^2-a^2\notin\Bbb Q(i)[x]\\ (x-a)(x-ia)=x^2-(a+ia)x+ia^2\notin\Bbb Q(i)[x] $$ which finishes the proof.