Cyclotomic fields of finite fields There is a fact that if $K=\mathbb Q$, then the Galois group of the $n$th cyclotomic field over $\mathbb Q$ is isomorphic to $\mathbb Z_n^{*}$.
In the case were $K$ is an arbitrary field, we have that the Galois group of the $n$th cyclotomic field is isomorphic to a subgroup of $\mathbb Z_n^{*}$.
In particular if we have $K=\mathbb F_7$ and $f(x)=x^5-1$ we have that $Gal(L/K)\cong U\subset\mathbb Z_5^{*}$, where $L$ is the splitting field of $f$ over $\mathbb F_7$.
But what utilities do I have to determine the Galois group? Normally I completely know how the Galois group looks like, if $K=\mathbb Q$...
In our case now the possibilities are $\mathbb Z_4$, $\mathbb Z_2 \times \mathbb Z_2$, $\mathbb Z_2,\{e\}$.
 A: If you only want to know what the Galois group is, in case the base is a finite field, the answer is easy, when you remember that the groups always are cyclic. In the case $\Bbb F_7$ and $X^5-1$, all you need to do is ask what the first power of $7$ is such that $5|(7^n-1)$, that’s the order of the multiplicative group of $\Bbb F_{7^n}$. The answer here is $7^4=2401$, so the Galois group is cyclic of order $4$.
A: Indeed, the Galois group is isomorphic to $(\mathbb{Z}/5)^{\times}$. 
In general,  the splitting field of $x^n-1$ over $\mathbb{F}_q$  is $\mathbb{F}_{q^l}$ where $l$ smallest so that $n_0$ divides $q^l -1$. Indeed, $x^n-1 = (x^{n_0}-1)^{p^t}$ (over $\mathbb{F}_p$) , so you are only having $n_0$ distinct roots of $1$. Moreover, an $\mathbb{F}_Q$'s consists of $0$ and $(Q-1)^{\text{th}}$ roots of $1$, and you want the minimal extension containing your roots of $1$. 
The Galois group of $\mathbb{F}_{q^l}$ over $\mathbb{F}_q$ is cyclic of order $l$, generated by the map $x\mapsto x^q$, that is, all the maps of the form $x \mapsto x^{q^i}$, for $0 \le i \le l-1$. If $l$ is the one specified above - then you see this as a subgroup of $(\mathbb{Z}_{n_0})^{\times}$. 
For instance, $q=p=7$, $n=5$, $\ \ \ 7 = 2 \!\!\!\mod 5$ and $2$ is a generator of $(\mathbb{Z}_5)^{\times}$ so you got the full group. You notice that $x^5-1 = (x-1)(x^4 + x^3 + x^2 + x+1)$ does not factor any further over $\mathbb{F}_{7}$. 
Compare  with $q=p=11$, $n$ still $5$, $\ \ 11=1 \!\!\!\mod 5$, and so the extension $\mathbb{F}_{11}(\zeta_{5}) /\mathbb{F}_{11}$ is trivial. Indeed, $x^5-1 = (x-1)(x-3)(x-4)(x-5)(x-9)$ splits completely over $\mathbb{F}_{11}$.
