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.
Best Answer
$\gamma=w+i$ 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.