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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.2012-10-22
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    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 image2012-10-22
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    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,2012-10-22
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    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.2012-10-22
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    Dear student, What you have written is the definition of Frobenius, and you will need to know (or discover) more than that. Have you tried reading [the wikipedia page](http://en.wikipedia.org/wiki/Frobenius_endomorphism)? Regards,2012-10-22
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    Hint: For the first implication, write $y=a_dx^d+\cdots+a_1x+a_0$ with $d=\deg(f),$ and use the "freshman rule" for the Frobenius.2012-10-22
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    Dear Student, For the *only if* direction, the properties of Frobenius that you can read about on wikipedia will suffice. For the converse direction, you will want to consider the factorization of $f$ into irreducibles, and apply the Chinese Remainder Theorem. Regards,2012-10-22
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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,2012-10-22
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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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