Galois Group of $\mathbb{Q}(\sqrt{\sqrt{3}-1})$

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 $$\mathbb{Q}(\sqrt{\sqrt{3}-1},\sqrt{-\sqrt{3}-1})$$?

• You’re reasoning is correct – Dionel Jaime Aug 19 at 1:06

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 $$D_8$$
• I can say that $\big[\mathbb{Q}(\sqrt{\sqrt{3}-1},\sqrt{-\sqrt{3}-1}): \mathbb{Q}\big]=\big[\mathbb{Q}(\sqrt{\sqrt{3}-1})(\sqrt{-\sqrt{3}-1}): \mathbb{Q}(\sqrt{\sqrt{3}-1})\big]\cdot \big[\mathbb{Q}(\sqrt{\sqrt{3}-1}):Q\big]=2\cdot 4=8$. By the fundamental Theorem, the subgroup of our Galois group corresponding to the fixed subfield $\mathbb{Q}((\sqrt{\sqrt{3}-1})$ cannot be normal since otherwise the extension $\mathbb{Q}((\sqrt{\sqrt{3}-1})$ would be Galois. The only subgroup of order 8 with a non-normal subgroup of order 4 is $D_8$, so the Galois group is $D_8$. Is this reasoning correct? – MEG Aug 19 at 1:27
• Close. An extension of degree $4$ must correspond to a subgroup of order 2 by the Fundamental Theorem. So we want a non normal subgroup of order 2. In this case it doesn’t matter because the only group of order $8$ with a non normal subgroup is $D_8$ – Dionel Jaime Aug 19 at 1:40
• No problem. There are other ways to conclude its $D_8$ if you don’t know the fact I used about normal subgroups of groups of order $8$ but I thought this to be the fastest. – Dionel Jaime Aug 19 at 4:48
• @nguyenquangdo I’m not really sure what you’re talking about. We have that the galois group is of order 8. We know that the splitting field has a non normal subextension of $\mathbb{Q}$ (the one stated by the OP) and thus our galois group must have a normal subgroup. This leaves $D_8$ as our only option. By “it doesn’t matter” I mean you don’t even have to look at the orders of subgroups of $D_8$. All you have to do is know that $D_8$ is the only group of order $8$ with a non normal subgroup. – Dionel Jaime Aug 19 at 17:06