Loosely put, my question is:
What happens if we swap determinant by the trace in an Artin $L$-function?
This question is not very precise and can be a little misleading, so I explain the specific context of the number field case. (I am also interested in more geometric analogs).
Let $L \supseteq K \supseteq \mathbb{Q}$ be a Galois extension of algebraic number fields, with Galois group $G$. Given a complex representation $\rho\colon G \to \mathrm{GL}_m(\mathbb{C})$ with character $\chi := \mathrm{tr}(\rho)$ and a maximal ideal $P \in \mathrm{Spec} \, \mathcal{O}_K$, define \begin{align} \tag{1} \chi(P) := \begin{cases} \chi((L/K;P)), & \text{if $P$ is unramified}, \\ 0, & \text{otherwise.} \end{cases} \end{align}
As usual, $(L/K;P)$ denotes the Frobenius $G$-conjugacy class associated to the unramified prime $P$. Define \begin{align} \tag{2} M(s,\rho) := M(s,\chi) := \prod_{P} \left(1 - \dfrac{\chi(P)}{N(P)^s}\right)^{-1}, \end{align} where the product ranges over all maximal ideals in $\mathcal{O}_K$. $M(s,\rho)$ hit me in the face during my research, so I am not sure if the definition I give is what I really want. For instance, I realize I am ignoring ramified primes altogether (unlike Artin).
I am curious to know if Artin considered (2) back then. Both references and observations are very welcome. Thanks for taking the time to read this!
