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Working towards the complete classification of finite fields in our algebra class, some final book-keeping involved proving the above (for a prime $p$, $n \geq 1$).

I've tried comparing the factorised forms:

$$x^{p^n} - x = \prod_{a \in \mathbb{F_{p^n}}} (x-a)$$

but nothing is jumping out at me - I'm not even convinced the statement is true myself, yet. Any hints or intuition on why this is true or how to proceed would be appreciated.

Thanks

jpmacmanus
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    @Servaes I'm not sure this is really a duplicate, although I agree it's related. You can use the result on polynomials to reduce it to $p^m-1|p^n-1$ and I see that if $m|n,$ the result on polynomials implies $p^m-1|p^n-1.$ But there's still a little work left in the other direction. (I guess the same argument will work, though.) – saulspatz Apr 01 '19 at 11:49
  • @saulspatz I agree that it is not an exact duplicate, but the result can indeed by applied twice to answer this question. Note that the linked question is not explicitly about polynomials, and the answer works for polynomials and integers identically; in fact it works for all $x\in\Bbb{Z}[X]$ that are not roots of unity, i.e. not $\pm1$. I've written up an answer just to clarify. – Servaes Apr 01 '19 at 12:04
  • @Servaes Thanks for the clarification. I see now. – saulspatz Apr 01 '19 at 12:16

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Both are divisible by $x$ so $$x^{p^m}-x\mid x^{p^n}-x\qquad\iff\qquad x^{p^m-1}-1\mid x^{p^n-1}-1.$$ By this question we have $x^a-1\mid x^b-1$ if and only if $a\mid b$, so applying this twice shows that $$x^{p^m-1}-1\mid x^{p^n-1}-1 \qquad\iff\qquad p^m-1\mid p^n-1 \qquad\iff\qquad m\mid n.$$

Servaes
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    Is the last equivalence truly an equivalence? I do see how $m \mid n$ implies that $X^m -1 \mid X^n -1$, then applied at $X=p$ give the result. However, how do you go from $X^m -1 \nmid X^n -1$ to $p^m -1 \nmid p^n -1$? – Cryme Feb 24 '20 at 17:13
  • The linked question contains a direct proof; the identity $$p^{qm+r}-1=p^r(p^{qm}-1)+p^r-1,$$ shows that if $n=qm+r$ then $p^m-1$ divides $p^n-1$ if and only if it divides $p^r-1$. I wouldn't recommend the detour through $X^m-1\mid X^n-1$. – Servaes Feb 25 '20 at 10:41