## If a is an integer greater than 1, then aᵈ-1 divides aⁿ-1 if and only if d divides n

Fix an integer $a > 1$. Prove that for all $d,n \in \mathbb{N}$, $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude that $\mathbb{F}_{p^d} \subseteq \mathbb{F}_{p^n}$ if and only if $d|n$.

If $d|n$, then by this previous exercise, $x^d-1$ divides $x^n-1$ in $\mathbb{Z}[x]$, and so $a^d-1$ divides $a^n-1$ in $\mathbb{Z}$.

Conversely, suppose $a^d-1$ divides $a^n-1$. Using the Division Algorithm, say $n = qd+r$. Now $a^n-1 = (a^d)^qa^r - a^r + a^r - 1$ $= a^r((a^d)^q-1) + (a^r-1)$. If $q = 1$, then we have $a^r-1 = 0$, so that $r = 0$ and $d|n$. If $q > 1$, then $a^n-1 = a^r(a^d-1)(\sum_{i=0}^{q-1} (a^d)^i) + (a^r-1)$, and again we have $r = 0$, so that $d|n$ as desired.

Recall that $\mathbb{F}_{p^k}$ is precisely the roots (in some splitting field) of $x^{p^k}-x$. Now $d|n$ if and only if $p^d-1$ divides $p^n-1$ by the above argument, if and only if $x^{p^d-1}-1$ divides $x^{p^n-1}-1$ by a previous exercise, if and only if $x^{p^d}-x$ divides $x^{p^n}-x$, if and only if $\mathbb{F}_{p^d} \subseteq \mathbb{F}_{p^n}$.