Galois group and intermediate subfields of finite field extension Let $K$ be the splitting field of $f(x)$ over $F$. Determine $Gal(K/F)$ and find all the intermediate subfields of $K/F$. In the case, I will consider $F=\mathbb{F}_{5}$ and $f(x)=x^{4}-7$.
I know how to solve for $F=\mathbb{Q}$ and I already know that this extension will be cyclic(because it's finite field extension), but I can't figure out how to determine $[K:F]$ and all automorphism to determine all intermediate fields.

From Fermat's Little Theorem we have that $x^4 -7$ doesn't have a root in $\mathbb{F}_5$. Additionally you can prove that the polynomial isn't a product of two quadratic factors and hence you conclude that it's irreducible over $\mathbb{F}_5$.
Now let $\alpha$ be a root of it. Then as you've mentioned the group $\text{Gal}(K/F)$ will be cyclic. Moreover it will be generated by the Frobenius Automorphism, i.e $\sigma: \alpha \to \alpha^5$. So you can conclude that $K=F(\alpha)$. Indeed the other roots of the polynomial are $\alpha^5,\alpha^{25}$ and $\alpha^{125}$. Now you can see that the $\text{Gal}(K/F) = \{Id,\sigma,\sigma^2,\sigma^3\}$ and we have a single intermediate field, which corresponds to the subgroup $\{Id, \sigma^2\}$. It's not hard to conclude that it is the field $F(\alpha^2)$
