TITLE: splitting field of a polynomial over a finite field
QUESTION [9 upvotes]: 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 $n$.

REPLY [3 votes]: Here is a way to find the splitting field without having to factor the polynomial.  Observe that $2$ is a primitive cube root of $1$ in $\mathbb{F}_7$, so $x^{15}-2$ splits completely in $\mathbb{F}_{7^n}$ iff there is a primitive $45$th root of $1$ in $\mathbb{F}_{7^n}$.  It follows that the splitting field is $\mathbb{F}_{7^n}$ for the least $n$ such that $7^n-1$ is divisible by $45$.  Doing some arithmetic mod $45$, it is not hard to compute that this $n$ is $12$.