a. It is the splitting field of $x^4 - 5$, so it is Galois. (I assume I can give for granted that the splitting field of a polynomial is always Galois)
The roots of this polynomial are $\sqrt[4]{5}, -\sqrt[4]{5}, i\sqrt[4]{5}, -i\sqrt[4]{5}$ so its splitting field must contain $\sqrt[4]{5}$ and $\frac{i\sqrt[4]{5}}{\sqrt[4]{5}} = i$, so it must contain $F$; on the other hand, it's immediate that $F$ contains all the roots of $x^4 - 5$, so $F$ is Galois.
b. Let $\sigma$ be the automorphism such that $\sigma(\sqrt[4]{5}) = i\sqrt[4]{5}$ and $\sigma(i) = i$, and $\tau$ be the complex conjugation. Both live in $Gal(F/\mathbb{Q})$ and they form a dihedral group.
Let's show the relations explicitly. We have that $\sigma^4(\sqrt[4]{5}) = i^4 \sqrt[4]{5} = \sqrt[4]{5}$ and $\sigma^4(i) = i$, so $\sigma^4$ is the identity. $\tau^2$ is obviously the identity since the complex conjugation has order $2$.
Finally, $\sigma \tau(\sqrt[4]{5}) = \sigma (\sqrt[4]{5}) = i \sqrt[4]{5}$, and $\tau \sigma^{-1} (\sqrt[4]{5}) = \tau(-i \sqrt[4]{5}) = i \sqrt[4]{5}$; moreover $\sigma \tau(i) = \sigma(-i) = -i$, and $\tau \sigma^{-1} (i) = \tau(i) = -i$; it follows that $\sigma \tau = \tau \sigma^{-1}$.
We need to show that every automorphism of $F$ over $\mathbb{Q}$ belongs to this group. Let $\eta$ be any such automorphism. Clearly $\eta(i) \in \{i, -i\}$ and $\eta(\sqrt[4]{5}) \in \{\sqrt[4]{5}, -\sqrt[4]{5}, i\sqrt[4]{5}, -i\sqrt[4]{5}\}$.
Suppose $\eta(\sqrt[4]{5}) = i^a \sqrt[4]{5}$ for some $0 \leq i \leq 4$. If $\eta(i) = i$, then it is immediate that $\eta = \sigma^a$ (on the generators, and hence on the whole field). If instead $\eta(i) = -i$, then $\eta = \sigma^a \tau$. In either case, $\eta$ belongs to the group.
Bonus remark, ignore if you don't need it: for an intuitive view of why this happens, the roots of $x^4 - 5$ form a square centered in $0$ in the complex plane, and the automorphisms of $F$ over $\mathbb{Q}$ correspond to the automorphisms of this square, which indeed are given by a dihedral group.
c. By the fundamental theorem, subextensions of degree $4$ correspond to subgroups of order $2$. We have five such subgroups, generated by $\tau, \sigma \tau, \sigma^2 \tau, \sigma^3 \tau, \sigma^2$. The subextensions corresponds to the fields given by the fixed points of these elements.
The subfield corresponding to $\tau$ is the real subfield $\mathbb{Q}(\sqrt[4]{5})$. The minimum polynomial of this generator has degree $4$, so of course it generates the subfield.
The subfield corresponding to $\sigma^2 \tau$ is the subfield $\mathbb{Q}(i\sqrt[4]{5})$. Indeed, $\sigma^2 \tau (i\sqrt[4]{5}) = \sigma^2 (-i\sqrt[4]{5}) = i\sqrt[4]{5}$. The minimum polynomial of this generator has degree $4$, so it generates the whole subfield.
The subfield corresponding to $\sigma^2$ must contain $i$ and $\sqrt{5}$. In fact, $\sqrt{5} = (\sqrt[4]{5})^2$, so $\sigma^2(\sqrt{5}) = \sigma^2(\sqrt[4]{5})^2 = (-\sqrt[4]{5})^2 = \sqrt{5}$. These are algebraically independent, so $\mathbb{Q}(i, \sqrt{5})$ has degree $4$ and this is the wanted subfield.
The subfield corresponding to $\sigma \tau$ is $\mathbb{Q}((1+i)\sqrt[4]{5})$. Indeed, \[ \sigma \tau ((1+i)\sqrt[4]{5}) = \sigma ((1-i)\sqrt[4]{5}) = (1+i)\sqrt[4]{5} \] so this elements belongs to the subfield. Moreover, $((1+i)\sqrt[4]{5})^4 = -4 \cdot 5 = -20$ so it has degree at most $4$, and it is clear that the degree cannot be $2$, so it has to be exactly $4$.
For the same reason, the subfield corresponding to $\sigma^3 \tau$ is $\mathbb{Q}((1-i)\sqrt[4]{5})$. Indeed, \[ \sigma^3 \tau ((1-i)\sqrt[4]{5}) = \sigma^3 ((1+i)\sqrt[4]{5}) = (1-i)\sqrt[4]{5} \] and by the same argument as before this generates the subfield.
Alessandro Iraci
