I am reviewing Galois Theory questions for an upcoming prelim, and I came across this question on a past prelim
"Find the Galois Group $\operatorname{Gal}(\mathbb{Q}(\sqrt{\sqrt{3}-1})$"
The minimal polynomial for $\sqrt{\sqrt{3}-1}$ can easily be shown to be $x^4+2x^2-2$. So the four roots that we need to be in our field are $\pm\sqrt{\sqrt{3}-1}$ and $\sqrt{\pm\sqrt{3}-1}$. However, $-\sqrt{3}-1 < 0$ so $\sqrt{-\sqrt{3}-1}$ will be imaginary. So this root will not be in $\mathbb{Q}(\sqrt{\sqrt{3}-1})$ because this extension is real. Thus $\operatorname{Gal}(\mathbb{Q}(\sqrt{\sqrt{3}-1})$ is not a Galois extension over $\mathbb{Q}$. So, the Galois closure of $\mathbb{Q}(\sqrt{\sqrt{3}-1})$ is actually $\mathbb{Q}(\sqrt{\sqrt{3}-1},\sqrt{-\sqrt{3}-1})$. Is this reasoning correct? If so, how can I find the Galois group of $\mathbb{Q}(\sqrt{\sqrt{3}-1},\sqrt{-\sqrt{3}-1})$?
Best Answer
Hint: Argue that the splitting field has degree $8$ and hence the galois group has order $8$. (You’re basically there already) and then use the fundamental theorem of Galois theory to say that the galois group must have a non normal subgroup forcing the Galois group to be $D_8$