I am trying on this exercise and I have proved that $\sqrt{2}\in \Bbb Q(\alpha)$ where $\alpha$ is the root of $x^4-4x^2+2$ as follows:
Solve for roots of $x^4-4x^2+2$ we have the roots are $\alpha_1=\sqrt{2+\sqrt{2}}$, $\alpha_2=-\sqrt{2+\sqrt{2}}$, $\alpha_3=\sqrt{2-\sqrt{2}}$, $\alpha_4=-\sqrt{2+\sqrt{2}}$. And taking the square of each root and subtract $2$ gives $\sqrt{2}$.
But when I tried to find the minimal polynomial of $\alpha$ over $\Bbb Q(\sqrt{2})$, since I have not been told to find the minimal polynomial of which root among these four. So I think the answer is $x^2-2-\sqrt{2}$ OR $x^2-2+\sqrt{2}$.
Could someone please check if it is correct? Or if it is wrong, could someone give a solution?
Thanks.