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When looking through some old exercises, I have found one which stumped me a bit, and which has been bugging me for quite a few days. The question is how the polynomial $X^{p^n}-X-c$ factorizes over $\mathbb{F}_{p^n}$ (with $c \in \mathbb{F}_{p^n}$). The case when $n=1$ is easy, we find that the polynomial is irreducible. However in the more general case, it gets quite hard! I can't (I think) easily generalize the argument I used for when $n=1$, there I just used that if $r$ is a root of the polynomial in question we also find that $r \notin \mathbb{F}_{p}$ and that $r+i$ is a root for all $i \in \mathbb{F}_p$. The rest follows by looking at the explicit factorization over $\overline{\mathbb{F}_{p}}$

However when $n \neq 1$, I can't easily use the fact that $r \notin \mathbb{F}_{p^n}$. So my earlier arguments stops working, there's a hint that I could look at how the Frobenius automorphism acts on the roots of our polynomial, so i think I could use something about our polynomial being separable, however I didn't really see how. Any hints or answers would be greatly appreciated!

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  • $\begingroup$ Does this help? Or possibly this? $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 4, 2017 at 17:14
  • $\begingroup$ Not really, I'm actually specifically looking for a solution using Frobenius acting on the roots. We also haven't talked about the trace in class so I have a feeling it should be possible to proof it without this. But thanks :) $\endgroup$
    – whatisaring
    Commented Mar 4, 2017 at 19:20
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    $\begingroup$ I see. May be this can be adapted? The idea would be that if $\alpha$ is a zero of your polynomial from some extension field of $\Bbb{F}_{p^n}$, then we get $$F(\alpha)=\alpha^{p^n}=\alpha+c$$ for the Frobenius conjugate $F(\alpha)$. Given that $c^{p^n}=c$ repeating the dose reveals that that $p$th iterate $F^p(\alpha)=\alpha$, Therefore $\alpha\in\Bbb{F}_{p^{pn}}$ et cetera. $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 4, 2017 at 19:39
  • $\begingroup$ @JyrkiLahtonen Maybe it's a dumb question but I don't really see why $\alpha^{p^n}$ would equal $\alpha+c$ at the moment. I do however get at the rest. We could use this using $\alpha$'s minimum polynomial right? If $m$ is said minimum polynomial we find that $deg(m) | pn$ and also $deg(m)|p^n$, and then we could do something else weird with $m$. I'm sorry if this doesn't make sense though, I'm having quite a hard time with this question and class. So still thanks :) $\endgroup$
    – whatisaring
    Commented Mar 4, 2017 at 21:14
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    $\begingroup$ Remember that $\alpha$ satisfies the equation $$\alpha^{p^n}-\alpha-c=0.$$ This implies that $\alpha^{p^n}=\alpha+c$. $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 4, 2017 at 22:04

1 Answer 1

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How about this approach? I hope that @JyrkiLahtonen will examine it closely, since I’m not completely convinced that all is well.

Call $p^n=q$, $k=\Bbb F_q$, $F(x)=x^q-x-c\in k[x]$, and $\varphi$ the $q$-Frobenius, sending $\xi\in\bar k$ to $\xi\,^q$. Let $\alpha$ be a root of $F$, i.e. of a $k$-irreducible factor $g$ of $F$. The $k$-conjugates of $\alpha$ are the finitely many $\varphi^m\alpha$, and the number of these is the degree of $g$. Now, $\varphi\alpha=\alpha^q=\alpha+c$, $\varphi^2\alpha=\varphi(\alpha+c)=\varphi\alpha+c=\alpha+2c$, and indeed $\varphi^n=\alpha+nc$ for any $n\in\Bbb Z$. Thus, if $c\ne0$, the number of $k$-conjugates of $\alpha$ is precisely $p$, and the irreducible factor $g$ of $F$ is of degree $p$.

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  • $\begingroup$ Yup, that's exactly what happens. $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 6, 2017 at 5:55
  • $\begingroup$ @JyrkiLahtonen Thanks, that's also the way how I solved the question (after long hours of thinking hahaha). Sorry I responded so late! I forgot about it. This argument is solid though right? One could use that the polynomial ring over the algebraic closure is a unique factorization ring, to make the argument that all irreducible factors should be of degree $p$ more rigorous I thought $\endgroup$
    – whatisaring
    Commented Mar 8, 2017 at 22:01
  • $\begingroup$ what is $\bar k$? $\endgroup$
    – Emptymind
    Commented Mar 26 at 0:13
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    $\begingroup$ Should be a given, fixed, algebraic closure of $k$, @Emptymind . $\endgroup$
    – Lubin
    Commented Mar 26 at 21:42

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