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Let $p$ and $d$ be prime numbers and $f\in\mathbb{F}_p[X]$ with $\deg f=d$ and no roots over $\mathbb{F}_p$. I want to show $$X^{p^d}\equiv X\text{ mod }f\;\;\;\Rightarrow\;\;\;f\text{ is irreducible over }\mathbb{F}_p$$ As Martin Brandenburg mentioned for $q=p^d$ and $n\in\mathbb{N}$, it holds $$(*)\;\;\;X^{q^n}-X=\prod_{p\in\mathbb{F}_q[X]\;:\;p \text{ is monic and irreducible with }\deg p\;\mid\; n}p$$ That means $$X^q-X=\prod_{(X+\alpha)\in\mathbb{F}_q[X]}(X+\alpha)=X^q\cdot o(X^q)$$ What can we conclude from that?

user26857
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0xbadf00d
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    Hmm. I believe that what Martin is saying is $$X^{p^d}-X=\prod_{g\in\Bbb{F}_p[x]: g\ \text{monic irreducible}, \deg g\mid d}g.$$ Given that $d$ is a prime, what does this say about the degrees of divisors of $x^{p^d}-x$? – Jyrki Lahtonen Jul 01 '14 at 14:35
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    The polynomial $X^{p^d}-X$ has exactly $p$ linear monic factors, and $(p^d-p)/d$ irreducible factors of $d$. This has been covered earlier, for example here. – Jyrki Lahtonen Jul 01 '14 at 15:11
  • @JyrkiLahtonen [You're right. I've almost forgotten that we can easily count the number of monic irreducible polynomials] But I think the conclusion now becomes very easy: If $f=gh$ for some polynomials $g$ and $h$, the previous result yields $\deg g=1$ and $\deg h=d$, or vice versa. But that means $f$ is irreducible.

    So, we know that $X^q-X$ is the product of monic irreducible factors of degree $1$ and $d$. Now we assume $f=gh$ for a monic irreducible polynomial $g$ - how does the previous result help to proceed?

     – 0xbadf00d Jul 01 '14 at 15:27
  • @JyrkiLahtonen Yes, I meant $\deg g=0$. The part "Now we assume ..." should not be there. I've updated my comment and have forgotten to delete this passage. – 0xbadf00d Jul 01 '14 at 18:36
  • If $g\mid f$, $g$ irreducible, then also $g\mid X^{p^d}-x$. By the previous result $g$ is of degree $d$ or of degree $1$. In the former case $f=g$, in the latter... Assuming that we agree what previous result we refer to :-) – Jyrki Lahtonen Jul 01 '14 at 18:42
  • @JyrkiLahtonen In the latter case $f$ should have a root over $\mathbb{F}_p$ - contradicting our assumption – 0xbadf00d Jul 01 '14 at 18:52
  • Correct. Well done! – Jyrki Lahtonen Jul 01 '14 at 18:53

1 Answers1

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Hint. $X^{p^d}-X$ is the product of irreducible monic polynomials whose degree divides $d$. In fact, these are the minimal polynomials of the elements of $\mathbb{F}_{p^d} = \{a \in \overline{\mathbb{F}_p} : a^{p^d}=a\}$.

2nd Hint. $d$ is a prime number.

3rd Hint. Assume that $g$ is a monic irreducible factor of $f$. Show that $g$ has degree $d$. Conclude $g=f$.

Martin Brandenburg
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