Galois group of $x^4 - 25$.

Question

To find the Galois group of $x^4 - 25 = (x^2 - 5)(x^2 + 5)$, I first note that all the roots are in $\mathbb{Q}(i,\sqrt{5})$, which is a degree 4 extension of $\mathbb{Q}$. A root can only go to itself or its negative countetrpart, so then is it the Klein-4 group?

Answer 1

Yes. $F = \mathbb{Q}(i,\sqrt{5})$ is the splitting field of the polynomial $f(x) = x^4 - 25$ and $[F:\mathbb{Q}] = 4$, so $\mathrm{Gal}(F/\mathbb{Q})$ is the Klein-four group, since you can send $i \mapsto \pm i$ and similarly $\sqrt{5} \mapsto \pm \sqrt{5}$.

Answer 2

You are correct, and your result also generalizes. It is a theorem that, given $\alpha, \beta \in F$ such that $\alpha, \beta$, and $\alpha\beta$ are not perfect squares in $F$, then the biquadratic extension $F[\sqrt{\alpha}, \sqrt{\beta}]$ is a splitting field of the polynomial $f(x) = (x^2 - \alpha)(x^2 - \beta)$ with corresponding Galois group $\mathbb{Z}_2 \times \mathbb{Z}_2$.

This theorem is proven here.