Finding the fields between $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$ and $\mathbb{Q}$

Question

I would like to find the fields between $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$ and $\mathbb{Q}$. I know the degree of $\mathbb{Q}=(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})/\mathbb{Q}$ is $8$ and its Galois group is isomorphic to $D_4$ (sometimes called $D_8$). Hence, there are exactly $8$ fields between $\mathbb{Q}=(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$ and $\mathbb{Q}$. I've been able to find $6$ of them: $\mathbb{Q}(\sqrt{3+\sqrt{7}}),\mathbb{Q}(\sqrt{3-\sqrt{7}}),\mathbb{Q}(\sqrt{2},\sqrt{7}),\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{7}),\mathbb{Q}(\sqrt{14})$. Which ones am I missing? I believe the fields should be between $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$ but I can't find them. Thank you in advance.

Answer 1

Subgroups of the dihedral group... four reflections, each of which generates a subgroup of order $2$/index $4$. The rotation group, of order $4$/index $2$. The $180^\circ$ rotation, of order $2$/index $4$. Since that last one is the center, it combines with the reflections to make two more groups of order $4$/index $2$; each has two of the reflections and the two central elements.

The three order $4$ groups each correspond to fields of degree $2$ over $\mathbb{Q}$, generated by a single square root. You've found those three. The five groups of order $2$ correspond to fields of degree $4$, each generated by something with two square roots in it. You've got three of those, but there should be two more.

Now, another way to look at this - it's the splitting field of a fourth degree polynomial with those two roots and their negatives. Explicitly, that polynomial is $$\left(x^2-(3+\sqrt{7})\right)\left(x^2-(3-\sqrt{7})\right)=x^4-6x^2+2$$ How do the order-2 elements in the group act on the roots? A permutation of order $2$ is some combination of $1$-cycles and $2$-cycles; either we have two pairs of roots that swap with each other, or one pair that swaps and two elements that stay fixed. In the dihedral group, the $180^\circ$ rotation and two of the reflections are two-pair swaps, while the other two reflections each swap one of the pairs from the rotation.

So, which ones have you found? Fixing one of those roots $\sqrt{3+\sqrt{7}}$ or $\sqrt{3+\sqrt{7}}$ will also fix its negative. Those are the two corner reflections, which each swap a pair of elements and fix the other two. The $180^\circ$ reflection, in the center of the group, comes from intersecting any two order-$4$ subgroups - or from combining their fields. That's $\mathbb{Q}(\sqrt{2},\sqrt{7})=\mathbb{Q}(\sqrt{9+2\sqrt{14}})$ (An arbitrary choice for generator there).

That leaves two subgroups, the side reflections. Those will swap across pairs. The first interchanges $\sqrt{3+\sqrt{7}}$ with $\sqrt{3-\sqrt{7}}$ - so it'll fix their sum $$\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}} = \sqrt{3+\sqrt{7}+3-\sqrt{7}+2\sqrt{3^2-7}}=\sqrt{6+2\sqrt{2}}$$ The second interchanges $\sqrt{3+\sqrt{7}}$ with $-\sqrt{3-\sqrt{7}}$, so it'll fix their sum $$\sqrt{3+\sqrt{7}}-\sqrt{3-\sqrt{7}} = \sqrt{3+\sqrt{7}+3-\sqrt{7}-2\sqrt{3^2-7}}=\sqrt{6-2\sqrt{2}}$$ The final two fields are $\mathbb{Q}(\sqrt{6+2\sqrt{2}})$ and $\mathbb{Q}(\sqrt{6-2\sqrt{2}})$, corresponding to the edge reflections in the group.

Incidentally, we can see that those two fields each have $\sqrt{2}$ in them, so $\mathbb{Q}(\sqrt{2})$ corresponds to the order-$4$ subgroup of the center and the two edge reflections. The group of the center and the two corner reflections corresponds to $\mathbb{Q}(\sqrt{7})$, and the rotation subgroup to $\mathbb{Q}(\sqrt{14})$. There - not just all the fields, but a way to match them to the groups, aside from the arbitrary choice of which reflections take the plus signs inside those square roots.

I believe the fields should be between $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$

Indeed they are.