TITLE: find a degree and splitting field for $x^4-2$ over $\mathbb{Q}(i)$
QUESTION [6 upvotes]: let $K=\mathbb{Q}(i)$ and let $f=x^4-2$. Find the splitting field, its degree and the basis.
My solution
First I find roots of the polynomial $x_{1,2}=\pm\sqrt[4]{2},\hspace{2mm}x_{3,4}=\pm i \sqrt[4]{2}$ and I notice that the polynomial $f=x^4-2$ is irreducible over $\mathbb{Q}(i)$ since neither of the above roots are in $K=\mathbb{Q}(i)$ hence $x^4-2=min pol_{\mathbb{Q}(i)}\sqrt[4]{2}$
Can I deduce from the above that 
$[\mathbb{Q}(i,\sqrt[4]{2}):\mathbb{Q}(i)]=4$
What will be the $\mathbb{Q}(i)$ basis for $\mathbb{Q}(i,\sqrt[4]{2})$?
In my book the answer is :
-degree is 4
-basis:$\{1,\alpha,\alpha^2,\alpha^3\}$ for $\alpha=\sqrt[4]{2}$
Something here I really do not understand. Please explain what is wrong in my solution.

REPLY [2 votes]: Your solution is true, 
notice that if you are extending your field by adding $a$, then you also add all power of $a$.
But not all power of $a$ are linerley independent.Thus,Just take linery independent ones as a basis.
in your case,$a^k$ for $4\leq k$ can be written as linear combination of $1,a^1,a^2,a^3$.Actualy, we expect it to   happen since $Q(i,2^{1/4}):Q(i)]=4$ thus,it is 4 dimensional vector space over $Q(i)$.
Or equivalantly,if you know that degree of minimal polynomial of $a$ over $F$,let say it is $n$,then $F(a)$ has a base over $F$  with ${ 1,α,α^2,...α^{n-1} }$.