calculating $[\mathbb{Q}(2^{1/4})(\sqrt{5} ) ] : \mathbb{Q}(\sqrt{5}]$

Question

calculating $[\mathbb{Q}(2^{1/4})(\sqrt{5}) : \mathbb{Q}(\sqrt{5}]$ by definition this is the degree of min. polynomial of $2^{1/4}$ over $\mathbb{Q}(\sqrt{5})$ but I am having troubles understanding what this even means, and how I would find such a polynomial.

Answer 1

Very roughly speaking, if the minimal polynomial of $\alpha$ over $\Bbb Q$ is $f(x)$, and if $K$ is a field that has nothing to do with $\Bbb Q(\alpha)$ (*), then $f$ will also be the minimal polynomial for $\alpha$ over $K$.

(*) “Nothing to do with”— I don’t want to make this explicit. But the moral of my sermon is that you should expect $x^4-2$ still to be the minimal polynomial over $\Bbb Q(\sqrt5\,)$. Now prove it.