If $d \mid n$ then $x^{q^d} - x \mid x^{q^n} - x$.

Question

I've only been able to resolve this far $$ d \mid n \Rightarrow \exists l \in \mathbb{Z}\; \textrm{such that} \; n=dl \Rightarrow x^{q^n} - x = x^{q^{dl}} - x.$$

I can't move forward.

Answer 1

If $l=\frac{n}{d}$ is $1$, then the statement is obvious. We can proceed from here by simply inducting on which multiple of $d$ $n$ is. Suppose the statement has been proven when $n=d(l-1)$. We want to prove it when $n=dl$.

$$x^{q^n}-x=x^{q^{dl}}-x=x^{q^dq^{d(l-1)}}-x=\left(x^{q^d}\right)^{q^{d(l-1)}}-x\\=\left(\left(x^{q^d}-x\right)+x\right)^{q^{d(l-1)}}-x$$

By the binomial theorem, then $$x^{q^n}-x=\sum_{i=0}^{q^{d(l-1)}}\left(\binom{q^{d(l-1)}}{i}\left(x^{q^d}-x\right)^ix^{q^{d(l-1)}}\right)-x$$

Clearly, whenever the index $i$ is at least $1$, the term in the series is divisible by $x^{q^d}-x$. The only terms left aside from those are the $0$th index term and the $-x$ at the end.

The $0$th index term is $x^{q^{d(l-1)}}$. So the only terms remaining that are not obvious multiples of $x^{q^d}-x$ are $x^{q^{d(l-1)}}$ and $-x$. But by the induction hypothesis $(x^{q^d}-x)\mid (x^{q^{d(l-1)}}-x) $.

Thus, the statement is true for $n=dl$.

Answer 2

ETA: I used $b$ and $d$ instead of $d$ and $n$. I show that $b|d$ implies $(x^{q^b}-x)|(x^{q^d}-x)$.

If $x$ is a root of $x^{q^b}-x$, then $x^{q^b} = x$. Which implies $ (x^{q^b})^{q^b} = x^{(q^b)^2} = x^{q^b} = x$. Which implies (repeating this $\frac{d}{b}$ times) that $x^{(q^b)^{d/b}}=x$. Which implies that every root of $x^{q^b}-x$ is also a root of $x^{q^d}-x$.

EDIT due to an observation from cardioid_ass_22: We have shown that every root of $x^{q^b}-x$ is also a root of $x^{q^d}-x$; to finishing showing that $(x^{q^b}-x)|(x^{q^d}-x)$ it suffices to show that every root of $x^c-x$ has multiplicity 1. To see this, we note that $x$ is a root of a polynomial $p(x)$ with multiplicity at least two only if $x$ is a root of both $p(x)$ and the formal derivative of $p(x)$. Thus it suffices to check that the roots of the derivative $cx^{c-1}-1$ of $x^c-x$ indeed do not overlap the roots of $x^c-x$, for each integer $c \ge 2$. To this end we note that $0$ is not a root of $cx^{c-1}-1$, so now it suffices to observe that all nonzero roots of $cx^{c-1}-1$ are not roots of $x^c-x$ for all positive integers $c \ge 2$. However, if $x$ is a root of $cx^{c-1}-1$ then [mulitplying both sides by $x$ gives] $x^c = c^{-1}x$. However for $c \not = 1$ and $x$ nonzero $c^{-1}x \not = x$ so $x$ is not a root of $x^c-x$. So indeed the roots of the derivative $cx^{c-1}-1$ of $x^c-x$ do not overlap the roots of $x^c-x$. Thus the roots of $x^c-x$ all have multiplicity 1 for every $c \ge 2$ namely $c=q^b$. This, and that that every root of $x^{q^b}-x$ is also a root of $x^{q^d}-x$, imply together that indeed, $(x^{q^b}-x)|(x^{q^d}-x)$