Let $E$ be the splitting field for $x^4-2$ over $\mathbb{Q}$ in $\mathbb{C}$. I want to show that $E=\mathbb{Q}[i + \sqrt[4]{2}]$.
I have a hint: Find at least five different elements in the orbit of $i + \sqrt[4]{2}$ under $\text{Gal}(E/\mathbb{Q})$.
First, by Eisenstein criterion $x^4-2$ is irreducible and its roots in $E$ are $\sqrt[4]{2}, -\sqrt[4]{2}, i\sqrt[4]{2}, -i\sqrt[4]{2}$. $f$ has distinct roots, so $f$ is separable over $E$. I have the following theorem: if $E$ is a splitting field over $F$ for some separable polynomial over $F$, then $E$ is a Galois extension of $F$. By this theorem, $E$ is Galois over $\mathbb{Q}$. Now I have the following sequence $$\mathbb{Q} \subset \mathbb{Q}[\sqrt[4]{2}] \subset \mathbb{Q}[\sqrt[4]{2},i]=E $$ and $|\mathbb{Q}[\sqrt[4]{2}] : \mathbb{Q}|=4$ , $|\mathbb{Q}[\sqrt[4]{2},i] : \mathbb{Q}[\sqrt[4]{2}]|=2$ , hence $|E:\mathbb{Q}|=8$. By Fundamental theorem of Galois Theory, $|\text{Gal}(E/\mathbb{Q})|=8$.
My question is that how can i show that $\mathbb{Q}[\sqrt[4]{2},i] = \mathbb{Q}[i + \sqrt[4]{2}]$ by using hint above.