Is $i \in \mathbb{Q}[\sqrt[4]{-2}]$? I have a homework question from Artin's Algebra that asks

Is $i \in \mathbb{Q}[\sqrt[4]{-2}]$?

I suspect that this is not true because $i \sqrt{2} \in \mathbb{Q}[\sqrt[4]{-2}]$ and $\sqrt{2}$ is of course not rational, but I am having a hard time proving it.  Perhaps I could consider $\mathbb{Q}[\sqrt[4]{-2}] = \mathbb{Q}[x] / (x^4 + 2)$.  Now if $i \in \mathbb{Q}[\sqrt[4]{-2}]$, then $\mathbb{Q}[i] = \mathbb{Q}[x] / (x^2 + 1) \leq \mathbb{Q}[x] / (x^4 + 2)$ and $x^2 + 1 \mid x^4 + 2$, a contradiction?  I have a feeling this is not right, but I'm stuck.  Also, we haven't covered any Galois theory.  Any thoughts would be appreciated!
 A: Notice that $\sqrt[4]{-2}$ is a root of $x^4+2$ and by Eisenstein's creteria it is irreducable over $\mathbb Q$.
So $[\mathbb Q(\sqrt[4]{-2}):\mathbb Q]=4$ and 
$$x_1=\sqrt[4]{2}\ \operatorname{cis}(\pi/4)$$
$$x_2=\sqrt[4]{2}\ \operatorname{cis}(3\pi/4)$$
$$x_3=\sqrt[4]{2}\ \operatorname{cis}(5\pi/4)$$
$$x_4=\sqrt[4]{2}\ \operatorname{cis}(7\pi/4)$$ are all roots. Notice that $x_{k+1}/x_k=i$ so if $i\in  \mathbb Q(\sqrt[4]{-2})$ then it must includes all  other roots as well and under this assuption you can say that $\mathbb{Q}(\sqrt[4]{-2})=\mathbb{Q}(\sqrt[4]{2},i)$. 
But clearly $[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}]=8$ which is a contradiction.
A: Hint 
Let $\omega = \sqrt[4]{-2}$, then $\mathbb{Q}(\omega)$ is degree 4 over $\mathbb{Q}$, and as a vector space it has basis $\{1,\omega,\omega^2,\omega^3\}$. To show that $i$ is not in $\mathbb{Q}(\omega)$, one only need to show that $\{1,\omega,\omega^2,\omega^3,i\}$ is linearly independent over $\mathbb{Q}$. Write them out explicitly, then it should be easy to see.
