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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.

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    $\begingroup$ I think it would be cyclic of order 4. I assumed you checked irreducibility. Suppose you have $r^4=7$ as a root $F_5(r)$. Now over this ring it suffices to solve $x^4-1=0$ as you can solve $(\frac{x}{r})^4-\frac{7}{r^4}=0$ instead. However, $x^4-1=0$ splits completely over $F_5$ as $F_5^\star$ is exactly the cyclic group $x^4=1$ solution. So once adjoining $r$, you see the equation splits completely. Thus $F(r)$ is the splitting field of $f(x)$ over $F$. $\endgroup$ – user45765 Jul 2 '18 at 18:49
  • $\begingroup$ It looks like @Stefan4024 has given a sufficient answer, but I thought I'd add this: The Galois group for a polynomial of the form $x^n-a$ over a field whose characteristic is prime to $n$ and which contains all $n$th roots of $1$ will always be cyclic. In this case, $1,2,3,4$ are all $4$th roots of $1$ in $\Bbb F_5$. $\endgroup$ – Nicholas Camacho Jul 3 '18 at 0:01
  • $\begingroup$ I fully endorse Stefan's answer. Just adding an alternative argument tproving that $f(x)$ is irreducible (my standard way, so just a comment). The order of $7=2$ in $F^*$ is four. It follows that any root $\alpha$ of $f(x)$ has order sixteen in $K^*$ (this is not trivial but not difficult either, two is the only prime involved). By Lagrange, $16\mid |K^*|$. Because $16\nmid(5^2-1)$ this rules out a quadratic factor. $\endgroup$ – Jyrki Lahtonen Jul 3 '18 at 8:05
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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 $F(\alpha^2)$

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