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?

The four distinct roots of $x^{4}+2$ are given by:-
$$\sqrt[4]{-2}=\sqrt[4]{2}\exp(\dfrac{(2k+1)i\pi}{4})\,,k=0,1,2,3$$.
For $k=0$ we have
$\sqrt[4]{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))=\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$
For $k=1$
$\sqrt[4]{2}(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))=\sqrt[4]{2}(\frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$
For $k=2$.
$\sqrt[4]{2}(\cos(\frac{5\pi}{4})+i\sin(\frac{5\pi}{4}))=\sqrt[4]{2}(\frac{-1}{\sqrt{2}}+\frac{-i}{\sqrt{2}})$
For $k=3$
$\sqrt[4]{2}(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}))=\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{-i}{\sqrt{2}})$
Say I pick $k=0$ and denote $\sqrt[4]{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))=\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})=\sqrt[4]{-2}$ for the rest of the answer.
Suppose $i\in \mathbb{Q(\sqrt[4]{-2})}$.
Then $\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}),i\in\mathbb{Q(\sqrt[4]{-2})}$
So taking their product we have
So $\sqrt[4]{2}(\frac{i}{\sqrt{2}}-\frac{1}{\sqrt{2}})\in\mathbb{Q(\sqrt[4]{-2})} $
Then their $\frac{\text{difference}}{2} =\sqrt[4]{2}\frac{1}{\sqrt{2}}\in \mathbb{Q(\sqrt[4]{-2})}\implies \frac{1}{\sqrt[4]{2}}\in\mathbb{Q(\sqrt[4]{-2})}$.
Thus $\mathbb{Q}(\sqrt[4]{2},i)\subset\mathbb{Q}(\sqrt[4]{-2})$ and it is evident from the expression $\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$ that $\mathbb{Q}(\sqrt[4]{-2})\subset \mathbb{Q}(\sqrt[4]{2},i)$.
So $\mathbb{Q}(\sqrt[4]{2},i)=\mathbb{Q}(\sqrt[4]{-2})$.
But this gives a contradiction since $[\mathbb{Q(\sqrt[4]{2},i):Q]}=8$ but
$[\mathbb{Q(\sqrt[4]{-2}):Q]}=4$ as the minimal polynomial for $\sqrt[4]{-2}$ is given by $x^{4}+2$. So $i$ does not belong to the $\Bbb{Q}(\sqrt[4]{-2})$
So in the proof you can clearly see that it does not depend on the $k$ chosen as you will end up picking a $-$ or a $+$ sign in the real or imaginary parts in the expression
$\sqrt[4]{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$ . The crux of the argument is that the moment $i$ lies in the field it becomes the splitting field of $x^{4}+2$ and also the splitting field of $x^{4}-2$ which should be a degree $8$ extension and not degree $4$.
