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 $\mathbb Z_4$, $\mathbb Z_2 \times \mathbb Z_2$, $\mathbb Z_2,\{e\}$.

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 $4$.
