I just realized that finding the splitting field of a polynomial over finite fields is not as "straightforward" as in $\mathbb{Q}$
I am struggling with the following problem:
"Find the splitting field of $f(x)= x^{15}-2$ over $\mathbb{Z}_7=\Bbb F_7$, the finite field of $7$ elements."
By direct computation, $f(x)$ has no roots on $\mathbb{Z}_7$ ; however, I do not how to prove that $f$ is actually irreducible.
I just found this lecture http://hyperelliptic.org/tanja/teaching/CCI11/online-ff.pdf
Using lemma 67, I can conclude that my polynomial is irreducible (although the proof seems a little weird)
Therefore, I think that the splitting field is $F= \mathbb{Z}_7(\alpha,\zeta)$ where $\alpha^{15} = 2$ and $\zeta$ is the $15th-$root of unity.
I want to describe $F$ as $\mathbb{F}_{7^n}$ for a suitable $n$.
Best Answer
There is a typo in the statement of Lemma 67 in your source. The $n$th roots of unity are in $\Bbb{F}_p$ only if $n\mid p-1$ or, iff $p\equiv1\pmod n$ (not $n\equiv1\pmod p$ as is written there). Therefore that Lemma does not apply.
In fact, the polynomial $x^{15}-2$ is NOT irreducible in $\Bbb{F}_7[x]$. This follows trivially from the fact that $3^5=243\equiv-2\pmod 7$. Therefore $$ x^{15}-2=(x^3)^5+3^5=(x^3+3)(x^{12}-3x^9+3^2x^6-3^3x^3+3^4). $$
We immediately see that $x^3+3$ has no zeros in $\Bbb{F}_7$ (the cubes in that field are $0,\pm1$), so it is irreducible. Therefore the polynomial has a zero $\alpha$ in $\Bbb{F}_{7^3}$.
To get the splitting field of $x^{15}-2$ we need, as you observed, the primitive 15th roots of unity. We easily see that $$ 7^4=2401\equiv1\pmod{15}. $$ The multiplicative group of the field $\Bbb{F}_{7^4}$ is cyclic of order $7^4-1$, and thus it contains a primitive 15th root of unity $\zeta$.
A consequence of all this is that the splitting field of this polynomial is $$ \Bbb{F}_7[\alpha,\zeta]=\Bbb{F}_{7^{12}}. $$