Additive characters modulo reducible polynomial in $\mathbb{F}_q[T]$ Suppose we have a prime power $q$ and a polynomial $F \in \mathbb{F}_q[T]$. I am wanting to know what the additive characters look like modulo $F$. If $F$ is irreducible then $\mathbb{F}_q[T]/(F(T)) \cong \mathbb{F}_{q^r}$ where $r = \deg F$ so the additive characters are all described by
$$\psi(x) = e_p(\text{Tr}(bx)) $$
for some $b \in \mathbb{F}_q[T]/(F(T))$, where $p$ is the characteristic of $\mathbb{F}_q$ and $\text{Tr}$ is the absolute trace $\mathbb{F}_{q^r} \to \mathbb{F}_p$.
I am wondering about when $F$ is not reducible. I can see that the additive structure modulo $F$ isn't super different from the irreducible case, but I am still unsure of what exactly the characters look like. Any help here would be greatly appreciated.
 A: You can use the following description of all the additive characters irrespective of whether $F(T)$ is irreducible or not. After all, the additive group of the quotient ring is always isomorphic to $\Bbb{F}_q^r$.
Let $b=(b_0,b_1,\ldots,b_{r-1})$ be any vector from $\Bbb{F}_q^n$. Define the character $\psi_b$ as follows. For any low degree polynomial
$a(T)=a_0+a_1T+\cdots+a_{r-1}T^{r-1}$ declare the value of $\psi_b$ at the coset $\overline{a(T)}=a(T)+\langle F(T)\rangle$ to be
$$
\psi_b(\overline{a(T)})=e_p(Tr(a_0b_0+a_1b_1+\cdots a_nb_n)).
$$
Here $Tr$ stands for the absolute trace $Tr:\Bbb{F}_q\to \Bbb{F}_p$.
As we clearly have $\psi_b\cdot\overline{\psi_{b'}}=\psi_{b-b'}$ it suffices to check that $\psi_b$ is non-trivial for any $b\neq0$. After all, we have the correct number of (candidate) characters, so we only need to check they are distinct. This is straight forward. If $b_i\neq0$ for some $i, 0\le i<r$, we can find an element $a_i\in \Bbb{F}_q$ such that $Tr(b_ia_i)\neq0$. Hence $\psi_b(\overline{a_iT^i)\neq1$, and we are done.
