Determine the minimal polynomial of $\alpha$ over $\Bbb Q(\sqrt{-2})$

Question

Let $\alpha$ be a root of the polynomial $x^4-4x^2+2.$

How should I determine the minimal polynomial of $\alpha$ over $\Bbb Q(\sqrt{-2})$?

Answer 1

Apply the quadratic formula to $x^4-4x^2+2$ to conclude $x^2 = \frac{4\pm \sqrt{8}}{2}=2\pm\sqrt{2}$. Since both roots are positive we see that $x^4-4x^2+2$ has $4$ real roots. Hence the minimum polynomial of $\alpha$ over $\Bbb{Q}[\sqrt{-2}]$ is $x^4-4x^2+2$

Edit: Note that $f(x)=x^4-4x^2+2$ is irreducible over $\Bbb{Q}$ by Eisenstein's criterion. Suppose it factored as 2 quadratics, say $g(x)$ and $h(x)$, over $\Bbb{Q}[\sqrt{-2}]$ Complex conjugation fixes all the roots of $f$, so $g$ and $h$ are fixed by complex conjugation. Hence $g$ and $h$ have real coefficients, so $f$ factors over the real subfield of $\Bbb{Q}[\sqrt{-2}]$. But that implies that $f$ factors over $\Bbb{Q}$ which is a contradiction.

Answer 2

Let $\alpha$ be a root of the polynomial $x^4-4x^2+2$

$$x^4-4x^2+2$$ $$\left(x^2 - \sqrt{2} -2\right) \left( x^2 + \sqrt{2} -2\right)$$ $$\left(\sqrt{2 - \sqrt2} - x\right) \left(\sqrt{2 + \sqrt2} - x\right) \left(\sqrt{2 - \sqrt2} + x\right) \left(\sqrt{2 + \sqrt2} + x\right)$$ Then $$\alpha = \left(\sqrt{2 \pm \sqrt2} \pm x\right) $$

Where the polynomial $\alpha$ has no solutions in $\Bbb{Q}(i\sqrt2)$