show the splitting field of polynomial 
Show that $\mathbb{Q}(\sqrt[4]{2}, i)$ is also the splitting field of $x^4 + 2$ over $\mathbb{Q}$.

I solve it as $x^4+2=0$
then $x^4=-2$
$\implies \mathbb Q(\sqrt[4]{-2}, \sqrt{-2})$
 A: Form of roots: $z^4 = -2 \Rightarrow z_k = 2^{\frac{1}{4}}e^{(\frac{pi+2pk}{4})i}$ for $k=0,1,2,3$.
More clearly: Let $\alpha = 2^{\frac{1}{4}} \omega$, where $ \omega = \frac{-\sqrt{2}+i\sqrt{2}}{2}$. And one can read off the roots by plugging in values of $k$ above.
Explanation: It suffices to have $\alpha,i$ since if we have $\alpha \Rightarrow$ we have $\alpha^2 = \sqrt{2}, \alpha^4 = 2, \frac{\alpha^2}{\alpha^4}=\frac{\sqrt{2}}{2}$. Now if we have $i$ then we are can get all of the roots by simply multiplying and taking powers. 
Hence, the splitting field of $f(x)=x^4-2$ is : $\mathbb{Q}(2^{\frac{1}{4}},i)$.
A: Note that the fourth roots of $-2$ are $\alpha=\sqrt[4]{2}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$, and $-\alpha$, $i\alpha$, and $-i\alpha$. 
These roots are all in $\mathbb{Q}(\sqrt[4]{2},i)$, so the splitting field $F$ is a subfield of $\mathbb{Q}(\sqrt[4]{2},i)$. 
To show that $F$ is all of $\mathbb{Q}(\sqrt[4]{2},i)$, it is enough to show that $\sqrt[4]{2}$ and $i$ are both in $F$.
To show $i$ is in $F$ is easy, for $i=\frac{\alpha i}{\alpha}$. 
To show that $\sqrt[4]{2}$ is in $F$, note that $\alpha^2=\sqrt{2}(-i)$. Since $i\in F$, it follows that $\sqrt{2}\in F$. But since $\alpha$, $i$, and $\sqrt{2}$ are in $F$, it follows that $\sqrt[4]{2}$ is in $F$. 
