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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 $y^p$.

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    Dear @MattE, thanks for pointing this out, I should have made my hint clearer. I meant for my element $y$ to be as in the second paragraph, with $y=y^p.$ Then, expressing $y$ as I said, the basic properties of the Frobenius and the fact that $\ker(F)=0$ lead to an easy proof that $a_d=\cdots=a_1=0.$ Indeed, the first thing to notice is that $R$ is a field iff $f$ is irreducible, as you say! Also, thank you for the general advice on how to approach this kind of problem. Regards2012-10-22

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