Let $p$ be a prime with $p\not=2,3$. Prove that the degree of the splitting field of $x^{12}-1$ over $\mathbb{F}_p$ is either $1$ or $2$. Give a rule to determine when the degree is $1$ and when the degree is $2$.
So clearly $x^{12}-1$ is a cyclotomic polynomial, and normally has splitting field $\mathbb{Q}(\zeta_{12})$ over $\mathbb{Q}$ where $\zeta_{12}$ is some primitive $12^\text{th}$ root of unity. The degree here is then $\varphi(12)=4$, where $\varphi$ denotes Euler's totient function. How can we figure this out over $\mathbb{F}_p$? Since this is a finite field, we know any extension is of the form $\mathbb{F}_{p^n}$, but I am not sure how to proceed. Any advice?
Addendum: does this have something to do with reducing mod $p$ not dividing the discriminant? If so, I imagine the only primes dividing the discriminant here would be $2$ or $3$, and maybe we can check for this by determining if $x^{12}-1$ is still separable mod $p\not=2,3$?
Let $p\not=2,3$ be prime, let $L$ denote the splitting field of $f(x)=x^{12}-1$ over $\mathbb{F}_p$, and let $k=[L:\mathbb{F}_p]$. Then $|L|=p^k$, so $|L^\times|=p^k-1$. Note that the roots of $f(x)$ form a cyclic group of order $12$. It is a well known fact that any cyclic group $C_r$ contains a cyclic subgroup $C_s$ if and only if $s\mid r$. It is also known that the multiplicative group of any finite field is always cyclic. Thus $L^\times$ is cyclic, and contains the group of roots of $f(x)$ if and only if $12\mid p^k-1$. Since $L$ is the splitting field, $k$ is the smallest positive integer such that $12\mid p^k-1$. Such an integer $k$ exists since $\gcd(p,12)=1$. To find $k$, note that if $p$ is one more than a multiple of $12$ then $k=1$. Otherwise, observe $p^2-1=(p+1)(p-1)$. Since $p$ is odd, these are two even factors, so their product is divisible by $4$. Furthermore, one of $p-1,p,p+1$ has to be divisible by $3$, but since $p$ is prime with $p\not=3$, it must be one of $p-1$ or $p+1$. Since $\gcd(3,4)=1$, we must have $3\cdot4=12\mid p^2-1$, so $k=2$ and we're done.
EDIT: I think I found another approach using Galois theory. It is known that if a polynomial $g(x)\in\mathbb{Q}[x]$ is separable modulo $p$, then the Galois group of the reduction $\overline{g}(x)=g(x)\;(\text{mod}\;p)$ over $\mathbb{F}_p$ is isomorphic to a cyclic subgroup of the Galois group of $g(x)$ over $\mathbb{Q}$ (see Dummit and Foote section 14.8). Let's try and apply this here.
Note that $\mathbb{Z}/12\mathbb{Z}^\times\cong C_2^2$ is the Galois group of $g(x)=x^{12}-1$ over $\mathbb{Q}$. Reducing modulo $p$, we have $\overline{g}(x)=x^{12}-1\in\mathbb{F}_p$. Taking the derivative, we see $\overline{g}\thinspace^\prime(x)=12x^{11}$, whose only root is $0$ (with multiplicity $11$) since $\text{char}(\mathbb{F}_p)=p\nmid12$. Therefore $g(x)$ is separable modulo any prime $p\not=2,3$. Using the result from above, the Galois group of $g(x)$ over $\mathbb{F}_p$ is isomorphic to a cyclic subgroup of $C_2^2$. The only cyclic subgroups of $C_2^2$ are $\{0\}$ and $C_2$. Hence the degree of the splitting field for $x^{12}-1$ over $\mathbb{F}_p$ is either $1$ or $2$.
(I suppose giving a "rule" for when it is $1$ or $2$ would be harder in this approach.)