Ambiguity regarding $\mathbb Q(\sqrt[4]{-2})$ I am studying field theory.I found a question in Artin's algebra which asks if $i\in\mathbb Q(\sqrt[4]{-2})$.Now I am confused  with the meaning of $\sqrt[4]{-2}$ here because there are $4$ of them.It is not specified which one I should take.So,I am having problem.Can someone please help me?

This answer has many similarities to that of Oscar Lanzi, but I thought a bit more detail might be helpful.

It is true that for each root of $\mu^4+2=0$, $\mathbb{Q}[\mu]$ is a different ring, but all four of these rings are isomorphic.
Suppose $\mu^4+2=0$ and $i\in\mathbb{Q}[\mu]$. That is, we have $a,b,c,d\in\mathbb{Q}$ so that
$$
\begin{align}
0
&=\left(a\mu^3+b\mu^2+c\mu+d\right)^2+1\\
&=\underbrace{2(ad{+}bc)\vphantom{\left(a^2\right)}}_0\,\mu^3-\underbrace{\left(2a^2{-}2bd{-}c^2\right)}_0\,\mu^2-\underbrace{2(2ab{-}cd)\vphantom{\left(a^2\right)}}_0\,\mu+\underbrace{\left(d^2{-}2b^2{-}4ac{+}1\right)}_0\tag1
\end{align}
$$
Since $x^4+2$ is irreducible over $\mathbb{Q}$ (via Eisenstein), $\left\{1,\mu,\mu^2,\mu^3\right\}$ are independent over $\mathbb{Q}$. Thus, each of the coefficients in $(1)$ must be $0$.
Since $ad+bc=0$ and $2ab-cd=0$ we get
$$
\begin{align}
0
&=2(ad+bc)^2+(2ab-cd)^2\\
&=\left(2a^2+c^2\right)\left(2b^2+d^2\right)\tag2
\end{align}
$$
That is, $a=c=0$ or $b=d=0$.
If $a=c=0$, then $2a^2-2bd-c^2=0\implies b=0$ or $d=0$.
$\quad$if $b=0$, then $d^2-2b^2-4ac+1=0\implies d^2+1=0\quad\Rightarrow\Leftarrow\tag3$
$\quad$if $d=0$, then $d^2-2b^2-4ac+1=0\implies2b^2=1\quad\Rightarrow\Leftarrow\tag4$
If $b=d=0$, then $2a^2-2bd-c^2=0\implies2a^2=c^2\quad\Rightarrow\Leftarrow\tag5$
Cases $(4)$ and $(5)$ rely on the irrationality of $\sqrt2$.
Cases $(3)$-$(5)$ are exhaustive; thus, $i\not\in\mathbb{Q}[\mu]$.
Since we only used that $\mu^4+2=0$, this works for any root: $\mu=\frac{\pm1\pm i}{\sqrt[4]{2}}$.
