Degrees of irreducible factors of a polynomial over finite field

Question

I'm working on an exercise in Janusz's Algebraic Number Fields. (I simplified it.)

Let $\Phi(x)$ be the minimal polynomial of the primitive $p$-th root of unity. ($p$ is an odd prime.) Let $q\ne p$ be a prime and consider the reduced polynomial $\bar \Phi(x)$ modulo $q$. Show that the splitting field of $\bar \Phi(x)$ over $GF(q)$ is the field $GF(q^m)$, where $m$ is the order of $q$ in the multiplicative group $\mathbb{F}_p^\times$. Conclude that every prime factor of $\bar \Phi(x)$ over $GF(q)$ has degree $m$.

I proved that the splitting field is $GF(q^m)$ for such $m$, but I don't know how to conclude the final statement. I expect that $\bar \Phi(x)$ splits as a product of linear polynomials over $GF(q^m)$ and $m$ linear polynomials make a irreducible factor of $\bar \Phi(x)$ over $GF(q)$ in some sense. I also thought that if we can show first that every irreducible factor of $\bar \Phi(x)$ over $GF(q)$ has the same degree then it should be $m$, since the degree of the splitting field over $GF(q)$ is equal to the lcm of all degrees of irreducible factors. But I cannot finish both approaches.

Answer 1

Note that if you add one single primitive $p$-th root of unity to any field, you add ALL $p$-th roots of unity, since you can obtain them just by taking powers of that one single primitive $p$-th root of unity.

Hence for any irreducible factor $f$ of $\Phi$, the field $\operatorname{GF}(q)[x]/(f)$ is a splitting field of $\Phi$, hence equal to $\operatorname{GF}(q^m)$. Thus $\deg f = m$.