Consistent Choice of Root of Unity in Characteristic $p > 0$

Question

Let $\zeta_n$ be a primitive root of $x^n - 1$ in $\bar{\mathbb{F}}_p$ where $p \nmid n$. I'm assuming there's no canonical choice for $\zeta_n$ like in characteristic $0$, where we can just take $e^{2\pi i /n}$. However, I'm wondering how to prove one can choose roots of unity consistently such that for $d \mid n$, we have $\zeta_n^{n/d} = \zeta_{d}$. I imagine one would have an inductive argument, where we assume we have chosen $\zeta_{d}$ for $d < n$. Then it suffices to show for $q$ prime such that $q \mid n$, we have $\zeta_n^q = \zeta_{n/q}$ since every proper divisor $d \mid n/q$ for some $q$, so $n/q = dk$ and thus $\zeta_n^{n/d} = \zeta_n^{qk} = \zeta_{n/q}^k = \zeta_{n/qk} = \zeta_d$ by the inductive hypothesis. My other idea was to first define $\zeta_q$ for all primes $q$ since they vacuously satisfy this condition, and then construct the rest of the roots but I'm not sure how this argument would work.

Answer 1

One simple way to do this is to first recursively choose $\zeta_{p^{n!}-1}$ (i.e., a generator of the multiplicative group of $\mathbb{F}_{p^{n!}}$) for each $n$. Having chosen $\zeta_{p^{n!}-1}$, choose $\zeta_{p^{(n+1)!}-1}$ to be a primitive $(p^{(n+1)!}-1)$st root of unity such that $\zeta_{p^{(n+1)!}-1}^{(p^{(n+1)!}-1)/(p^{n!}-1)}=\zeta_{p^{n!}-1}$. Then, for arbitrary $m$, define $\zeta_m$ to be $\zeta_{p^{n!}-1}^{(p^{n!}-1)/m}$ for any $n$ such that $m$ divides $p^{n!}-1$. This is independent of the choice of such an $n$ since if $n<N$ then $\zeta_{(p^{N!}-1)!}^{(p^{N!}-1)/(p^{n!}-1)}=\zeta_{p^{n!}-1}$ by construction so $\zeta_{p^{N!}-1}^{(p^{N!}-1)/m}=\zeta_{p^{n!}-1}^{(p^{n!}-1)/m}$. Similarly, if $d$ divides $m$, then for sufficiently large $n$, $\zeta_d=\zeta_{(p^{n!}-1)}^{(p^{n!}-1)/d}=(\zeta_{(p^{n!}-1)}^{(p^{n!}-1)/m})^{m/d}=\zeta_m^{m/d}$.

(The only actually relevant property of the numbers $a_n=p^{n!}-1$ used here is that they form a sequence $a_0\mid a_1\mid a_2\mid\dots$ such that any integer $m$ that is coprime to $p$ divides some $a_n$. The point is that this gives a sequence that is cofinal in the divisibility poset for your roots of unity, so you can just go up the sequence step by step instead of trying to make a bunch of non-linearly ordered roots of unity compatible with each other. Another approach that is useful in more general contexts where you don't necessarily have such a sequence would be to recursively choose $\zeta_{q^n}$ for each prime $q$ different from $p$, and then combine them using the Chinese remainder theorem to get $\zeta_m$ for arbitrary $m$.)