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Prove that $f(x)$ divides $x^{p^n} - x$ if and only if $d := \deg f(x)$ divides $n$.

I believe that I have the backward direction covered: Let $d \mid n$ say $n = dq$ for some $q$ in $\mathbb{F}_p[x]$. Consider the field $\mathbb{F}_p[x]/(f(x))$ which has $p^d$ elements. Take an element $x+I$ from the field (here $I = (f(x))$) so we have: $(x+I)^{p^n} = (x+I)^{p^{dq}}$. As long as you keep factoring out $(x+I)$ with the $p^d$ power you will get $(x+I)$ so $x^{p^n} - x \in (f(x))$.

I am having trouble getting to the other direction.

  • 2
    Is $f(x)$ supposed to be irreducible over $\mathbb{F}_p$? Otherwise, $\mathbb{F}_p[x]/(f(x))$ won't be a field.2016-02-25

2 Answers 2

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Hints:

(i) Show that the splitting field of the polynomial $\,p(x):=x^{p^n}-x\in\Bbb F_p[x]\,$ over the prime field $\,\Bbb F_p\,$ is the field $\,\Bbb F_{p^n}\,$

(ii) One way to go: show that the set of roots of the above polynomial $\,p(x)\,$ in some algebraic closure of $\,\Bbb F_p\,$ is a field...

(iii) Prove that $\,\Bbb F_{p^d}\,$ is a subfield of $\,\Bbb F_{p^n}\,$ iff $\,d\mid n\,$

Of course, take into account that $\,f(x)\mid p(x)\Longrightarrow\,$ all the roots of $\,f(x)\,$ are also roots of $\,p(x)\,$

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    About proving (iii): show that $(x^{p^d}-x)\mid(x^{p^n}-x)\Longleftrightarrow d\mid n$2012-10-22
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This question is very old, but there is a more direct solution worth noting.

Proof.

Suppose $f(x)$ divides $h(x):=x^{p^n} -x$. Then since $h(x)$ splits over $\mathbb{F}_{p^n}$, so does $f(x)$. Let $\alpha \in \mathbb{F}_{p^n}$ be a root of $f(x)$. Then $\mathbb{F}_{p}(\alpha)\subset \mathbb{F}_{p^n}$, and $[\mathbb{F}_p(\alpha): \mathbb{F}_p] = d$. Finally, we have that $n = [\mathbb{F}_{p^n}: \mathbb{F}_p]= [\mathbb{F}_{p^n}: \mathbb{F}_p(\alpha)][\mathbb{F}_p(\alpha): \mathbb{F}_p]$, completing the proof that $d | n$.