Ways of finding primitive element of separable extension $\Bbb{Q}(\sqrt[4]{2},i)$ over $\Bbb{Q}$. Consider the field extension $L=\mathbb Q (\sqrt[4] 2 ,i)$  over $\mathbb Q$. This extension is separable as we know over a field of characterstic $0$. Now according to the primitive element theorem there exist an element $\gamma$ such that $L=\mathbb Q(\gamma)$. 
What are the possible ways of finding $\sqrt[4]2, i$ in terms of $\gamma$. 
(I think we can take $\gamma$ simply as a sum.)
Note that $L=\mathbb Q (\sqrt[4] 2 ,i)$ is a splitting field of $x^4-2$. 
Thanks. 
 A: $\gamma=w+i$ with $w=\sqrt[4]2$ is a good guess (and something similar, i.e. in rare cases maybe with an additional factor, always works, as per the proof of primitive element theorem).
If you write down powers of $\gamma$ and simplify, you will note that they all have the form 
$$\gamma^k=a+bw+cw^2+dw^3+ei+fwi+gw^2i+hw^3i$$
with the eight coefficients $a,\ldots,h$ depeding on $k$.
For example
$$ \gamma^5=w^5+5w^4i-10w^3-10w^2i+5w+i=7w-10w^3+11i-10w^2i.$$
If you thus compute $1,\gamma,\gamma^2,\ldots, \gamma^7$, you get $8$ vectors in $\mathbb Q^8$ that - hopefully - span the full $\mathbb Q^8$ (i.e. are linearly independant).
Especially, some rational linear combination of these vectors (i.e. a suitable rational polynomial in $\gamma$) equals $w$ and another equals $i$, showing the notrivial direction $\mathbb Q[\sqrt[4]2,i]\subseteq \mathbb Q[\gamma]$.
Where I said "hopefully" above, is the place where we may have to change the choice for $\gamma$ and retry with something like $\gamma=\sqrt[4]2+2i$ or $\gamma=\sqrt[4]2+42i$, perhaps. But in real life, the simple sum usually works.
A: If the Galois group is readily available, as it is in this case, it makes sense to use the fact that a polynomial with distinct roots is irreducible if and only if the Galois group is transitive on its roots.
So in this case you have just to check that the orbit of the obvious candidate $\alpha = \sqrt[4]{2} + i$ under the Galois group has length $8$. (Because it will tell you that the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has degree $8$.) This check is easy enough and then you're done. (If the first attempt is not successful, then follow the advice of @HagenvonEitzen, just try again with a different element.)
