Characters on rings of residue classes modulo polynomials over finite fields First recall the following orthogonality relation on $\mathbb{Z}/n\mathbb{Z}$. 
Fix $n \in \mathbb{Z}$, $n \neq 0$. For $r \in \mathbb{Q}$, let $e(r) := e^{2 \pi i r}$. Let $x \in \mathbb{Z}$. Then one can show that
$$
\dfrac{1}{n} \sum_{h=0}^{n-1}e\left(h\dfrac{x}{n}\right) = \begin{cases}
                                      1 & \mbox{ if } n \mid x;\\
                                      0 & \mbox{ otherwise.}
                                      \end{cases}
$$
I wonder what is the analogue for polynomials over finite fields. How do the characters here look like? Note that $e(x)$ was defined on $\mathbb{Q}$ so that we have that orthogonality. So I'm guessing the characters on $F[x]/gF[x]$, for $F$ a finite field and $g \in F[x]\setminus\{0\}$,
should be defined on the function field $F(x)$. Thanks!
 A: I'm not sure that this exactly what you are looking for, but here comes anyway.

The additive group of $F[x]$, $F=\Bbb{F}_q$, $q=p^n$ is just an infinite dimensional vector space over the prime field $\Bbb{F}_p$. Therefore the values of all characters are $p$th complex roots of unity.
A popular (standard?) description the characters of the additive group $(F,+)$ is the following. Let $tr:F\to\Bbb{F}_p$ be the trace map
$$
tr(x)=x+x^p+x^{p^2}+\cdots+x^{p^{n-1}}.
$$
The trace is a non-trivial homomorphism from $(F,+)$ to $(\Bbb{F}_p,+)$, and we get a character of $(F,+)$ by composing it with the obvious character $e:\Bbb{F}_p\to\Bbb{C}^*$ given by $e(x)=e^{2\pi i x/p}$.
From all of the above it follows that for any $a\in F$ we get a character $\chi_a$ of $(F,+)$ from the recipe
$$
\chi_a(x):=e(tr(ax)).
$$
It is not hard to show that the characters $\chi_a$ are distinct for distinct choices of $a$, and that we get all the characters of $(F,+)$ in this way.
As a more or less immediate corollary of this we get all the characters of the additive group of $F[x]$ from the following recipe. Select an arbitrary sequence
$\vec{a}=(a_0,a_1,a_2,\ldots)$ of elements of $F$. A character $\chi_{\vec{a}}$ of $F[x]$ associated with this sequence is the mapping
$
\chi_{\vec{a}}:F[x]\to\Bbb{C}
$
defined by
$$
\chi_{\vec{a}}(c_0+c_1x+\cdots +c_mx^m):=e\left(\sum_{j=0}^{m}tr(a_jc_j)\right)
$$
for all polynomials $c(x)=\sum_{j=0}^mc_jx^j\in F[x]$, $m$ an arbitrary natural number.
When we consider the quotient ring $F[x]/\langle g(x)\rangle$ nothing much changes. We only need to make sure that we get a well-defined mapping. This means that we can only consider vectors $\vec{a}$ such that $\chi_{\vec{a}}(bx^jg(x))=1$ for all integers $j\ge0$ and all $b\in F$. What this means is that if $m=\deg g(x)$, then we can choose $a_0,a_1,a_2,\ldots,a_{m-1}$ any which way we want, but these then determine a unique choice of $a_m, a_{m+1},\ldots$. 
The additive group of the quotient ring $F[x]/\langle g(x)\rangle$ is, after all, isomorphic to that of the vector space $F^m$, so we expect the description of the characters to be given in terms of $m$ constants of $F$ as well. All in line with the fact that the character group of a finite abelian group is isomorphic to the group itself. The isomorphism is not natural in any way, and that is reflected in the somewhat arbitrary use of the trace function (any non-trivial homomorphism of additive groups could serve its role equally well).
