Normal basis theorem splitting field

Question

Consider the field $F$ as the splitting field of $x^4-2$ over $\mathbb{Q}$. I know by the normal basis theorem that there exists $\alpha \in F$ such that $\{ g \alpha : g \in Gal(F/\mathbb{Q}) \}$ is a basis for $F$ over $\mathbb{Q}$. I know that a basis for $F$ is $\{1, 2^{1/4}, 2^{1/2}, 2^{3/4}, i, 2^{1/4}i, 2^{1/2}i, 2^{3/4}i\}$. I was wondering if any of the following could be valid choices for $\alpha$, $i, \sqrt[4]{2}, \sqrt[4]{2} + 1$ or the sum of the basis elements as previously given.

Answer 1

Write $\sigma$ and $\tau$ for the automorphisms with $\sigma(i)=i$ and $\sigma(\sqrt[4]2)=i\sqrt[4]2$ and $\tau(i)=-i$ and $\tau(\sqrt[4]2)=\sqrt[4]2$.

First of all, a normal basis cannot be in any proper subextension. That immediately rules out $i$, $\sqrt[4]2$ and $\sqrt[4]2+1$.

How about $\alpha=1+2^{1/4}+2^{1/2}+2^{3/4}+i+i2^{1/4}+i2^{1/2}+i2^{3/4}$? Well, $$\beta=\frac12(1+\tau)\cdot\alpha=\alpha=1+2^{1/4}+2^{1/2}+2^{3/4}$$ and by applying powers of $\sigma$ to $\beta$ and taking linear combinations we can isolate each of the powers of $\sqrt[4]2$. Likewise $$\gamma=\frac12(1-\tau)\cdot\alpha=\alpha=i(1+2^{1/4}+2^{1/2}+2^{3/4})$$ and we can get each $i2^{k/4}$ too. Therefore $\alpha$ is a normal basis.