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let $F$ be a field with char$(F)=p>0$ where $p$ is a prime.
given $a\in F^\times $ ($a\not=0$) denote \begin{equation*}f(x)=x^p-x-a\end{equation*} I'm trying to prove that the degree of every irreducible polynomial in $F[x]$ that divides $f$ is the same.
I noticed that $f$ does not have roots of multiplicity $>1$ in a splitting field and if the statement is correct then every irreducible polynomial that divides $f$ must be of degree $1$ (or $p$ if $f$ is irreducible). I don't really know how to progress further.

Nathan Sikora
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  • Compare with this question. Warning: my answer, while highly voted, may NOT be very helpful to you (I guess it can be tweaked to give what you want). I recommend that you look Greg Martin's or zyx's answer for this variant of the question. – Jyrki Lahtonen May 29 '15 at 16:04
  • (x-b)|f(x) implies f(b) = 0 but f(b) = $b^p$ - b - a = -a – Piquito May 29 '15 at 16:46

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I think I reached the answer. It is easy to check that for every $i\in\mathbb F_p$ \begin{equation*}f(x)=f(x+i)\end{equation*} Let $f(x)=u\cdot g_1(x)\cdot...\cdot g_k(x)$ where $u\in F^\times $ be the unique factorization of $f$ to irreducibles. $\mathbb F_p$ acts on the set $A=g_1(x),...,g_n(x)$ by mapping (up to multiplying by unit elements) \begin{equation*}g_j(x)\mapsto g_j(x+1)\in A\end{equation*} The action can be viewed as a permutation $\sigma\in S_n$. it follows that $\sigma ^p=id$ which applies that $p\mid n$ so $n$ must be a power of $p$. if $n>1$ then it means that $f$ is factored to product of more then $\deg f$ polynomials of at least degree 1, which can't be.

Nathan Sikora
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