In my Analytic Number Theory course, we are going through Davenport's Multiplicative Number Theory (3rd ed.), and I am having some trouble working through a certain part of section 3 (Cyclotomy). I include this for context:
Let $g$ be a primitive root to the modulus $q$, and let $\nu(n)$ denote the index of $n$ relative to $g$. As $n$ assumes the values $1,2,\cdots,q-1$. its index assumes the same values in another order.
Now consider any factorization of $q-1$, say $q-1=ef$. The roots of unity $\zeta^i$, for $i=1,\cdots,q-1$ can be subdivided according to the residue class (mod $e$) to which $\nu(n)$ belongs. There will be $e$ such sets, each comprising of $f$ numbers. The sums of the various subsets are called the $\textbf{Gaussian periods}$ of $f$ terms, and are denoted $\eta_0, \cdots, \eta_e-1$. Thus $$\eta_j = \sum_{\substack{n = 1\\ \nu(n) \equiv j\pmod e}}^{q-1} \zeta^n$$
The book then presents the following argument:
Let $F(\zeta)$ be any polynomial in $\zeta$, say $$F(\zeta) = \sum_{r=1}^{q-1}A_r\zeta^r$$ And suppose $F(\zeta)$ has the property that $F(\zeta^m) = F(\zeta)$ whenever $\nu(n) \equiv 0 \pmod e$ Then, by the uniqueness of representation of a polynomial, we have $$A_r =A_s$$ whenever $r \equiv sm \pmod {q-1}. $$\textbf{This is my first question}$, this should be $(\operatorname{mod} q) $ , right? I hope it's a typo.
Then the book proceeds:
[...] Hence $A_r$ depends only on the residue class $(\operatorname{mod} e$) to which $\nu(r)$ belongs. Grouping together the terms in the same residue class, we obtain $$F(\zeta) = A_1\eta_1 + \cdots + A_n\eta_n$$
My second question is slightly subtle. Is it OK to assume that each $A_i$ is $\textbf{exactly}$ the coefficient of $\eta_i$ in this linear combination? It is not inmediately evident that, for instance, $A_1$ would correspond with $\zeta^1$. This would be false in cases where $\nu(1) \not\equiv 1 \pmod e$ (which I think is every case). Is it necessary to reorder the $A_i$'s in some way? I believe this reordering would depend directly on $g$, and obviously on $e$ and $f$.