I am having trouble trying to show this:
Let $f \in \mathbb{F}_p[x]$ be a non-constant polynomial and let $F$ denote the Frobenius map $F: R \rightarrow R$ where $R = \mathbb F_p[x]/(f)$. Prove that $f$ is irreducible iff $\ker(F)=0$ and $\ker(F-I)=\mathbb F_p$, where $I$ is the identity map $R \rightarrow R$.
Here is an idea so far: When you take an element $y \in \ker(F-I)$ then $y^p = y$ in $R$ so we have the Frobenius map that sends $a^p$ to $a$. If the $\ker(F) \neq 0$ then there is a nonconstant polynomial in $\mathbb F_p[x]$ that belongs to the same equivalence class as $y^p$ and divides $y^p$.