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.
Best Answer
First, the fact that the roots of a quartic do not lie in a field is not enough to show that the polynomial is irreducible. For example, $(x^2+1)^2$ is reducible over $\mathbb{R}$ but has no roots in $\mathbb{R}$. Another detail is that you have to show it is indeed a splitting field (Which is fairly clear). Look at the tower lemma to understand why the basis given is indeed a basis.