Let $L$ be the splitting field of $f(x) = x^4+10x^2+20$ over $\mathbb{Q}$.
The roots of $f$ are
$\alpha=\sqrt{-5+\sqrt{5}},$
$-\alpha = -\sqrt{-5+\sqrt{5}},$
$\beta = \sqrt{-5-\sqrt{5}},$
$-\beta=\sqrt{-5-\sqrt{5}}$.
So $L = \mathbb{Q}(\alpha,\beta).$
But $\sqrt{5} \in \mathbb{Q}(\alpha)$ and $(\sqrt{-5+\sqrt{5}}) (\sqrt{-5-\sqrt{5}})=2\sqrt{5}$,
so $L = \mathbb{Q}(\alpha)$.
As $f$ is irreducible over $\mathbb{Q}$, it follows that
$[L:\mathbb{Q}]=4$.
$L$ is a Galois extension, so $|\mathcal{G}(L/\mathbb{Q})|= [L:\mathbb{Q}]=4$.
I also know that each $\mathbb{Q}$-automorphism takes $\alpha$ to one distinct root.
For example, let $\sigma \in \mathcal{G}(L/\mathbb{Q})$ with $\sigma(\alpha) = \beta$. Then it follows that $\sigma(-\alpha) = -\beta$,
but I don't know how to compute $\sigma(\beta)$. How do I determine where does $\sigma$ send $\beta$?
Best Answer
$\beta=\frac{2(\alpha^2+5)}\alpha$ hence $\sigma(\beta)=\frac{2(\beta^2+5)}\beta=-\alpha.$