Find the splitting field of polynomial $f(X)=X^{15}-2\in \mathbb{F}_ 7[X]$.
Splitting field of this polynomial should be of the form $\mathbb{F}_{7^n}$ for some suitable $n$.
For example, I know how to solve the much easier problem if we consider the polynomial $g(X)=X^{15}-1\in \mathbb{F}_7[X]$. In this case we have to take minimal $n$ such that $15 \mid7^n-1$.
Can anyone give the detailed answer how to do this in the case $X^{15}-2$?
I have seen the similar post in MSE but all of them is very bried and without details.
Would be very grateful if anyone can give a detailed answer!
Note that $15=3\times 5$ and that $5$ does not divide $p-1=6$. Because $4^5\equiv2\pmod{7}$ we see that $$X^{15}-2=(X^3)^5-4^5=(X^3-4)(X^{12}+4X^9+2X^6+X^3+4).$$ Hence the splitting field is the compositum of the splitting field of these two factors.
The factor $X^3-4$ has no roots in $\Bbb{F}_7$, so it is irreducible and its splitting field is isomorphic to $\Bbb{F}_{7^3}$.
The splitting field of $X^{12}+4X^9+2X^6+X^3+4$ is obtained by adjoining cube roots of the roots of $$P:=X^4+4X^3+2X^2+X+4,$$ to $\Bbb{F}_7$. This polynomial is irreducible because $$2P(4X)=X^4+X^3+X^2+X+1=\Phi_5,$$ which is the fifth cyclotomic polynomial, and $\Bbb{F}_7$ contains no primitive fifth roots of unity. So the splitting field of $P$ is isomorphic to $\Bbb{F}_{7^4}$. Hence the splitting field of the second factor is either $\Bbb{F}_{7^4}$ or $\Bbb{F}_{7^{12}}$, either way the splitting field of the original polynomial is $\Bbb{F}_{7^{12}}$.