Let $p$ be a prime and $P(x)\in\mathbb Z_p[x]$ a monic irreducible polynomial with $\deg{P}=k$. Let $q=p^{mk}$ for any positive integer $m$ and define $e_q(x) = x^q-x$. I want to prove that $P$ divides $e_q$.
I know that the field $\mathbb Z_p$ is a subfield of $\mathbb Z_q$. Moreover, in $\mathbb Z_{q}[x]$ the polynomial $e_q$ splits completely because every element is a root. I have been trying to show that $P$ divides $e_q$ in $\mathbb Z_q$, and I've been thinking that there might be some reason why $P$ must also split completely over $\mathbb Z_q$. However, I can't see why $P$ would need to have any particular factorization, or equivalently, why it should have any particular roots in $\mathbb Z_q$.
Perhaps $\mathbb Z_q$ is the splitting field of $P$? What do I know about the splitting field? It's $\mathbb Z_{p}[x]/(P)$ which is a $\mathbb Z_p$ vector space and the dimension of this vector space is the degree of $P$, which is $k$. So the number of elements in this vector space is $p^k$. So the splitting field of $P$ is $\mathbb Z_{p^k}$ if I have all that right, and so this looks like a dead-end. Or at least I don't see how to carry this line of thought further.
I think this question is asking the same thing as mine: Irreducible polynomial f in polynomial ring of finite field order n divides a particular polynomial if and only if degree of f divides n
But I don't understand the algebraic closure well enough to know how to use the hints.