Let $p$ be a prime number with $p \not |\ n.$ Let $q=p^e.$ Let $\Bbb F_q^{(n)}$ be the splitting field of $X^n-1$ over $\Bbb F_q.$ Let $\Phi_n$ denote the $n$-th cyclotomic polynomial over $\Bbb F_q.$ Then how many prime factors will $\Phi_n$ have in it's prime decomposition over $\Bbb F_q$?
Let $\mu_n \left (\Bbb F_q^{(n)} \right )$ denote the zero set of $X^n-1$ over $\Bbb F_q^{(n)}.$ Then I know that $$\Phi_n = \prod_{\substack {\zeta \in \mu_n \left (\Bbb F_q^{(n)} \right ) \\ \text {Ord}\ \zeta = n}} \left (X-\zeta \right )\ \text{in}\ \ \Bbb F_q^{(n)}[X].$$
I am studying it from NPTEL lecture series on Finite Fields and Galois Theory. There I found that $\Phi_n$ splits into $\frac {\varphi (n)} {\text {Ord}_n\ q}$ many distinct prime factors in $\Bbb F_q[X]$ each of degree $\text {Ord}_n\ q$ where $\varphi$ is the Euler's totient function and $\text {Ord}_n\ q = \text {Ord}_{\Bbb Z_n^{\times}}\ [q]_n.$ But why is it the case? It is very easy to see that $\deg \Phi_n = \varphi(n).$ Let $\zeta_n$ be a primitive $n$-th root of unity. Let $\mu_{\zeta_n,\Bbb F_q}$ denote the minimal polynomial of $\zeta_n$ over $\Bbb F_q.$ Then clearly $\mu_{\zeta_n,\Bbb F_q}\ \big |\ \Phi_n$ in $\Bbb F_q[X].$ Therefore $\Phi_n = \mu_{\zeta_n,\Bbb F_q}.g$ in $\Bbb F_q[X].$ Also from this I asked yesterday it is clear that $\deg \mu_{\zeta_n,\Bbb F_q} = \text {Ord}_n\ q.$ So we get one of the prime factors of $\Phi_n$ of degree $\text {Ord}_n\ q.$ How do I find other prime factors of $\Phi_n$ of degree $\text {Ord}_n\ q$? Why do they all have the same degree? Would anybody please explain it to me?
Thanks for reading.
Source $:$ https://youtu.be/A_andcVo1uU?list=PLOzRYVm0a65dsCb_gMYe3R-ZGs53jjw02&t=1176
Best Answer
Let $\pi$ be any prime factor of $\Phi_n$ in $\Bbb F_q[X].$ WLOG we may assume that $\pi$ is monic. Let $\zeta$ be a zero of $\pi$ in $\Bbb F_q^{(n)}.$ Then $\pi$ is the minimal polynomial of $\zeta$ over $\Bbb F_q.$ Since $\zeta$ is a primitive $n$-th root of unity it follows that $\Bbb F_q^{(n)} = \Bbb F_q[\zeta].$ So $\deg \pi = [\Bbb F_q^{(n)} : \Bbb F_q] = \text {Ord}_n\ q$ (by using this).