Analogue of Dirichlet $L$-function for $\mathbb{F}_q[T]$, does $L_c(s, \chi)$ necessarily equal $1$? Consider an analogue of Dirichlet $L$-function for $\mathbb{F}_q[T]$. Let $g \in \mathbb{F}_q[T]$, $g \neq 0$, let $\chi: (\mathbb{F}_q[T]/(g))^\times \to \mathbb{C}^\times$ be a homomorphism, let $c \in \mathbb{C}^\times$, and consider$$L_c(s, \chi) = \sum_f \chi(f) \cdot c^{\text{deg}(f)} \cdot \#(\mathbb{F}_q[T]/(f))^{-s}\tag*{(*)}$$where $f$ ranges over all monic polynomials $\mathbb{F}_q[T]$ which are coprime to $g$, and deg means the degree. In the above (*), consider the case $g = T$ and $\chi$ is not a trivial homomorphism. My question is, does $L_c(s, \chi)$ necessarily have to equal $1$?
 A: The function you wrote down is $\sum_{f} \chi(f) (c/q^s)^{\deg f}$. Hence it equals $1$ for any non-zero $c$ iff $L(u,\chi) := \sum_{f}  \chi(f) u^{\deg f}$ is a constant polynomial, specifically $1$ (plug $u=0$). From now on we'll adopt that convention.
The $L$-function of a character divides the $\zeta$-function of some "cyclotomic" function field (certain extension of $\mathbb{F}_q(t)$) and one can apply GRH for function fields, and find that $L(u,\chi)$ is a polynomial of degree $\le \deg g-1$ with roots of absolute value $\sqrt{q}$ (and one possible "trivial zero" $u=1$). This settles $\deg g=1$, and in some cases also $\deg g=2$ (I think that if $\chi$ is not constant on $\mathbb{F}_{q}^{\times}$ then $L(u,\chi)=1$ too).
Of course, as KConrad pointed out, the fact that the $\zeta$-function is a polynomial is completely elementary, but the theory of GRH and cyclotomic fields gives the higher explanation for what's happening - the $\zeta$-function your $L$-function divides is that of a function field isomorphic to the rational function field (such as $F_q(\sqrt{T})$), and such $\zeta$-function has no roots, so its factors can't have roots either, i.e. are constants.
