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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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Have you tried anything? My first thought was: $f$ is irreducible iff $R$ is a field. –  Bey Oct 22 '12 at 0:18
    
I believe that's part of it but there's not much to go on by defining a linear map and saying that 1 must belong to the image –  Student Oct 22 '12 at 2:02
    
Dear Student, The only if direction is the easier of the two: you get to assume that $R$ is a field, and you have to prove some basic properties of the Frobenius endomorphism (or perhaps you have already covered them in class). Before moving onto the harder if direction, do you completely understand this easier direction? Regards, –  Matt E Oct 22 '12 at 2:19
    
What are the basic properties of a Frobenius endomorphism? The only property that I'm aware of is that $f(a) = a^p$. This is for a first semester course in algebra it's supposed to be not complicated. –  Student Oct 22 '12 at 2:22
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@Andrew: Dear Andrew, I don't really understand your hint. But my advice to student would be in a different direction: there is a temptation to think of elements of $R$ in terms of coset representatives (which are certain polynomials, and seems to be in the spirit of your hint). My experience is that this is normally not very helpful. It is usually better to think of $R$ as a ring in its own right, with certain properties that it inherits from properties of $f$ (e.g. it is a field iff $f$ is irreducible). Best wishes, –  Matt E Oct 22 '12 at 2:38

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