Subfields of $\mathbb{Q}(\sqrt[6]{5})$.

The problem is to find all subfields of $K=\mathbb{Q}(\sqrt[6]{5})$.

Big idea approach: Determine the Galois closure of $K$, and its Galois group. Then determine the subgroups of this Galois group. Apply the fundamental theorem to the Galois closure, and see which subfields are contained in $K$.

The Galois closure will be the splitting field of the minimal polynomial of $\sqrt[6]{5}$ (right?), which is $x^6-5$. The roots of this polynomial are $\pm\sqrt[6]{5},\frac{\sqrt[6]{5}\pm i\sqrt{3}\sqrt[6]{5}}{2},$ and $\frac{-\sqrt[6]{5}\pm i\sqrt{3}\sqrt[6]{5}}{2}$ (yes?).

OK, so this means that the splitting field of $x^6-5$ is $E=\mathbb{Q}(\sqrt[6]{5}, i\sqrt{3})$. Any automorphism of $E$ is determined by its action on generators, so we have 4 possibilities: $1$, $\sigma$, $\tau$, and $\sigma\tau$, with $$\sigma(\sqrt[6]{5})=\sqrt[6]{5},\;\sigma(i\sqrt{3})=-i\sqrt{3}$$ $$\tau(\sqrt[6]{5})=-\sqrt[6]{5},\;\tau(i\sqrt{3})=i\sqrt{3}.$$

THIS is where my confusion begins. $E/\mathbb{Q}$ is a Galois extension, so the number of automorphisms should equal the degree of the extension — which is 16 (?). But there are only 4 automorphisms? What is going wrong here?

EDIT: Going off of Alex Wertheim's comment: Let $r_1,r_2,\ldots,r_6$ be the six roots of $x^6-5$. Let $\sigma_j^{+}$ be the automorphism of $E$ which sends $\sqrt[6]{5}↦ r_j$, and $i\sqrt{3}↦ i\sqrt{3}$, and let $\sigma_j^{-}$ be the automorphism of $E$ which sends $\sqrt[6]{5}↦ r_j$, and $i\sqrt{3}↦ -i\sqrt{3}$; for $j∈\{1,2,\ldots,6\}$. Then $\text{Gal}(E/\mathbb{Q})=\{\sigma_1^+,\sigma_2^+,\ldots,\sigma_6^+,\sigma_1^-,\sigma_2^-,\ldots,\sigma_6^-\}$. From this point is it just a matter of working out the orders of each element//analyzing the relations and figuring out which group it is?

Let $F$ be a subfield of $K$. Let $Hom_F(K,\overline{K})$ be the set of $F$-algebra homomorphisms $K\to \overline{K}$. You'll have that $$F =\{ a\in K, \forall \sigma\in Hom_F(K,\overline{K}), a=\sigma(a)\}$$ In particular $F$ is the subfield fixed by a subset of $Hom_\Bbb{Q}(K,\overline{K})$.

Conveniently $Hom_\Bbb{Q}(K,\overline{K})$ is very easy to understand (as it is the restriction to $K$ of a cyclic group of automorphisms of the Galois closure), as well as the subfield fixed by any set of elements.

You'll find that for each $d| 6$ there is a subfield fixed by $\sigma_d : \sqrt[6]{5}\to e^{2i\pi d/6} \sqrt[6]{5}$ and no other subfield.