Question in finding a new $\mathbb{Q}$-basis for $F/\mathbb{Q}$. Let $F$ be the splitting field of $x^4 - 2$ over $\mathbb{Q}$. Let $G$ be its Galois group. When viewed as a $\mathbb{Q}$- vectorspace, $F$ has the following basis:
$$\mathcal{B}=\{1,2^{1/4},2^{1/2},2^{3/4},i,2^{1/4}i,2^{1/2}i,2^{3/4}i\}$$
According to the Normal Basis Theorem, there exists some $x\in F$ such that the set $\{g(x)|g\in G\}$ is a $\mathbb{Q}$-basis of $F$.
My questions is how we can find such $x\in F$? It seemed natural to me that we start doing something to the basis we already have. But, for example, $x$ definitely cannot be $i$ or $2^{1/4}$, since for any $g\in G$, we must have $g(i)=\pm i$ and $g(2^{1/4})=\pm 2^{1/4}$. For the same reason, if you take some linear combination, for example, $i+ 2^{1/4}$, then $g(i+2^{1/4}) = \pm i \pm 2^{1/4}$. So how should we find such $x$? 
 A: Here’s a sketch of a construction for a normal basis in this case, but I’m afraid it’s too late at night for me to go through all the details.
Let’s use the chain of field inclusions $\Bbb Q\subset k=\Bbb Q(i)\subset K=\Bbb Q(i,\lambda)$, where $\lambda^4=2$. Now if we’re just looking for a basis of $K$ over $\Bbb Q$, we know that we can take a (two-element) basis of $k$ over $\Bbb Q$ and a (four-element) basis of $K$ over $k$, and multiply them together element by element to get eight things that are $\Bbb Q$-linearly independent.
I propose to do the same thing with a normal basis of $K$ over $k$ and the well-known normal basis $\{1+i,1-i\}$ of $k$ over $\Bbb Q$. Roughy speaking, almost any random linear combination of all the basis elements of $K$ over $k$ should do the trick for the top layer of the chain, and I calculated the $4$-by-$4$ determinant for the four $k$-conjugates of $\mu=1+\lambda+2\lambda^2+4\lambda^3$, and found that this $\mu$ gave a normal basis. Remember that the Galois group is $\{e,\sigma,\sigma^2,\sigma^3\}$, where $\sigma(\lambda)=\lambda^\sigma=i\lambda$. That is, $\lambda^{\sigma^m}=i^m\lambda$.
Now I say that since $1+i$ gives a normal $\Bbb Q$-basis of $k$, and $\mu$ gives a normal $k$-basis of $K$, it’s fairly easy to see that $(1+i)\mu$ and its seven other conjugates under the total Galois group form a $\Bbb Q$-basis of $K$. It’s the details of this that I’m now too groggy to fill you in on.
