When does $x^{q'}-x$ divide $x^q-x$ in $\mathbb{Z}[x]$?

Question

Let $p$ be a prime integer, and let $q=p^r$ and $q'=p^k$. For which values of $r$ and $k$ does $x^{q'}-x$ divide $x^q-x$ in $\mathbb{Z}[x]$?

From Artin's Algebra, Chapter 15, problem 7.12 from the section on finite fields.

I believe it's those $k$ and $r$ such that $k|r$, but I'm having a hard time rationalizing it. I believe I need to show $\mathbb{F}_{q'}\subset\mathbb{F}_q$, but not sure how to show this either.

Answer 1

For the forward direction: $x^{q'}-x|x^q-x$ in $\mathbb{Z}[x]$ $\implies x^{q'}-x|x^q-x$ in $\mathbb{F}_p[x]$. Therefore because the roots of $x^{q'}-x$ are all roots of $x^q-x$, and the roots of this are the elements in $\mathbb{F}_{q'}[x]$, then $\mathbb{F}_{q'}[x]\subset \mathbb{F}_q[x]$, which implies that $k|r$.

For the backward direction, $k|r \implies r=kt, t\in\mathbb{Z}$. So $p^k-1|p^{kt}-1$, which implies $x^{p^k-1}-1|x^{p^{kt}-1}-1$, and multiplying by $x$ gives $x^{p^k}-x|x^{p^{tk}}-x$. Therefore we have $x^{q'}-x|x^q-x$.

Just need some reasoning for some of the steps (especially in the backward direction)...

Answer 2

Consider a field $\Bbb F$ containing both $\Bbb F_{q,q'}$. Let $F$ the Froebenius morphism $x\in F \to x^p\in F$.

Hence $\Bbb F_q = \{x\in F: x^{p^r}=x\} = \ker F^r, \Bbb F_{q'} = \ker F^k$. Let $r = mk + s$, with $s<k$.

$$ (X^{p^k}-X) | (X^{p^r}-X) \iff \ker F^k\subset \ker F^r $$ If $s=0$, RHS is true.

If $\ker F^k\subset \ker F^r$: let $x\in F, F^k x = x$, such as $k = \min\{l>0:F^lx=x\}$. $$ x = F^rx = F^s(F^k)^mx = F^sx $$hence $s=0$ and $k|r$.