Let E be the splitting field of $x^4 - 2 $ over $\mathbb{Q}$, and $\alpha = i + 2^{\frac{1}{4}} \in E$
Prove that $E = \mathbb{Q} (\alpha)$
I need to show that $i, 2^{\frac{1}{4}} \in \mathbb{Q}(\alpha)$. But I have tried taking the square of $\alpha$ and computing its inverse, but I'm unable to isolate either $i$ or $2^{\frac{1}{4}}$.
Can someone give me a hint?
$(i+2^{1/4})^2=\alpha^2\in E$
$2*2^{1/4}i+2^{1/2}\in E$
$-4*2^{1/2}+4*2^{3/4}i\in E$
$-2^{1/2}+2^{3/4}i\in E$
the sum of $-2^{1/2}+2^{3/4}i$ and $2*2^{1/4}i+2^{1/2}$ is $\in E$
$2^{5/4}i+2^{3/4}i\in E$
$(2^{5/4}i+2^{3/4}i)^2\in E$
$2^{5/2}+2^{3/2}\in E$
$2^{1/2}\in E$
the difference of $2^{1/2}$ and $2*2^{1/4}i+2^{1/2}$ is $\in E$
$2^{1/4}i\in E$
$i\alpha^3=i(i+2^{1/4})^3=1-3*2^{1/2}+i*2^{1/4}(2^{1/2}-3)\in E$
the quotient of $i\alpha^3$ and $\alpha^3$ is $\in E$
$i\in E$
$2^{1/4}\in E$