Let $q$ be a prime number and define $\Phi_q = X^{q-1} + \cdots + X^2 + X + 1 \in \mathbb{Z}[X] $.
Let $p$ be a prime number and define $f_{q,p} = \Phi_q \bmod p \in \mathbb{F}_p[X]$.
Show that all irreducible factors of $f_{11, p} = X^{10} + \cdots + X + 1 \in \mathbb{F_p}[X]$ have the same degree.
I do not really now how to begin this problem. We can split the polynomial in 10 irreducible polynoms of degree 1 by looking at the zero points of $(X^{11} - 1)/(X-1)$ which are ten if $p = 11$, because for every element $a \in \mathbb{F}$ we get $a^{11} = 1$ such that $f_{11, p} =(a^{11}-1)/(a - 1) = 0$
But if we look at the more general case, that is $f_{11,p} = g_1 g_2... g_k$ we want to prove that for every $1 \le i \le k $ $\deg(g_i) = d$, with $d|10$ and $kd = 10$.
Does somebody have any hints on how to begin the proof?
Let $\zeta_q$ be a primitive $q$th root of unity over $\mathbb F_p$. We have $\zeta_q\in\mathbb F_{p^n}$ iff $\zeta_q^{p^n}=\zeta_q$, that is, $p^n\equiv1\bmod q$. Let $k$ be the least positive integer with the property $p^k\equiv1\bmod q$. Then $\zeta_q$ is in $\mathbb F_{p^k}$, but in no other proper subfield. This shows that the minimal polynomial of $\zeta_q$ over $\mathbb F_p$ has degree $k$, and this happens for any other primitive $q$th root of unity over $\mathbb F_p$.