Finding intermediate fields of the extension $\mathbb{Q}=(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})/\mathbb{Q}(\sqrt{14})$

Question

I want to find the intermediate of the extension $\mathbb{Q}=(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})/\mathbb{Q}(\sqrt{14})$. So far, I've been able to show that there are exactly two of them since the Galois group of $\mathbb{Q}=(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})/\mathbb{Q}$ is isomorphic to $D_4$ and I've been able to build the subfield lattice except for these two missing fields. I don't know how to proceed. Any help would be appreciated.

Answer 1

Let $k=\mathbb Q(\sqrt 7)$. In the following, I'll refer to https://math.stackexchange.com/a/3066678/300700 (see also my comment on your other question), where it is shown that $k(\sqrt {3\pm\sqrt 7})$ are two distinct quadratic extensions of $k$, hence $L=\mathbb Q(\sqrt {3+\sqrt 7},\sqrt {3-\sqrt 7})$ is a biquadratic extension of $k$. The third quadratic subextension of $L/k$ is $k(\sqrt 2)=\mathbb Q(\sqrt 7,\sqrt 2)$ because $(3 +\sqrt 7)(3-\sqrt 7)=2$. It follows that $k(\sqrt 2)/\mathbb Q$ is a biquadratic extension, whose three quadratic subextensions are $k, \mathbb Q(\sqrt 2), \mathbb Q(\sqrt {14})$ (draw a Galois diagram).

We must now look at the nature of the extension $L/\mathbb Q$. Since the conjugation $\sigma: \sqrt 7 \to -\sqrt 7$ of $k/Q$ permutes the two elements $3\pm\sqrt 7$, $L/\mathbb Q$ is normal of degree 8, say with Galois group $G$. All groups $G$ of order 8 are known up to isomorphism: either $G$ is abelian, or $G\cong D_8$ (dihedral) or $H_8$ (quaternionic). Here $G$ is not abelian, as can be seen (a bit painfully) by lifting the conjugation $\sigma$ of $\mathbb Q(\sqrt 7)/\mathbb Q$ and the conjugation $\tau$ of $\mathbb Q(\sqrt 2)/\mathbb Q$, and computing their commutator (note that $\mathrm{Gal}(L/k(\sqrt 2))$ lies in the center of $G$. So $G$ is isomorphic to $D_8$ or $H_8$. A convenient way to distinguish between these two cases is: $G$ is quaternionic (resp. dihedral) iff its $3$ subgroups of order $4$ are (resp. one and only one is) cyclic. Here $G$ has $2$ subgroups of type $(2,2)$, which are $\mathrm{Gal}(L/k)$ and $\mathrm{Gal}(L/\mathbb Q(\sqrt 2))$, hence $G\cong D_8$ and its $3$-rd subgroup of order $4$, namely $\mathrm{Gal}(L/\mathbb Q(\sqrt 14))$, must be cyclic.

Conclusion: the only strict subextension of $L/\mathbb Q(\sqrt 14)$ is $k(\sqrt 2)=\mathbb Q(\sqrt 2, \sqrt 7)$ ./.