When does $x^{q'}-x$ divide $x^q-x$ in $\mathbb{Z}[x]$? 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.
 A: 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 $k|r$.
A: 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)... 
A: Expanding on Mathemanic's answer for the forward direction,
Suppose $x^{q'}-x \mid x^q - x$ in $\mathbb{Z}[x]$. Then, reducing both polynomials mod $p$ preserves this property, so we have that $x^{q'}-x \mid x^q - x$ in $\mathbb{F}_p[x]$.
Now, any root of $x^{q'}-x$ must also be a root of $x^q - x$ by this divisibility relation. But, the roots of $x^{q'} - x$ are exactly the elements of $\mathbb{F}_{q'}[x]$, so we have that $\mathbb{F}_{q'}[x] \subset \mathbb{F}_q[x]$. Thus, $k \mid r$.
