Let $C$ be a non-singular curve over $\mathbb{F}_q$. Denote by $d$ the degree map from the group of divisors to $\mathbb{Z}$ and denote by $P$ the set of prime divisors w.r.t. to the function field.
Let $C_n$ be the curve obtained from $C$ by extending $\mathbb{F}_q$ to $\mathbb{F}_q^n$. Is there a bijection between $\{P \text{ prime divisor of } C_n: d(P)=1\}$ and $\{P \text{ prime divisor of } C: d(P)|n\}$
Following is the context (see also C. Moreno, curves over finite fields):
Let $C$ be a non-singular curve over $\mathbb{F}_q$. Denote by $d$ the degree map from the group of divisors to $\mathbb{Z}$ and denote by $P$ the set of prime divisors w.r.t. to the function field.
The zeta function of $C$ is defined by $Z(t,C):= \sum_{D} {t^{d(D)}} $, where the sum goes through all positive divisors.
One can show that
Theorem: $Z(t,C)=exp(\sum_{m=1}^{\infty}N_m \frac{t^m}{m})$, where $N_m:=\sum_{P,d(P)|m}{d(P)}$ and $N_1$ is equal the number of prime divisors of degree $1$.
Theorem: $N_n=N_1(C_n)$, where $N_1(C_n)$ denotes corresponding $N_1$ w.r.t. to $Z(t,C_n)$ and $N_n$ w.r.t. $Z(t,C)$ as defined above.
Now obviously, we have that $N_1(C_n)$ is equal to the number of prime divisors $P$ of $C_n$ with $d(P)=1$. But now it's immediately deduced that this is equal to number of prime divisors of $C$ with $d(P)|n$, but I don't unterstand this deduction. How this works?
I think it's better to write the zeta function using points, i.e.
$$Z(t, C) = \exp\left(\sum_{n=1}^\infty \#\overline{C}(\mathbb F_{q^n}) \frac{t^n}{n}\right),$$
where $\overline{C}$ denotes the base change of $C$ to $\overline{\mathbb F_q}$.
A prime divisor $P$ on $C_n$ of degree $1$ is precisely the same thing as an $\mathbb F_{q^n}$-rational point of $C_n$. The points in $C_n(\mathbb F_{q^n})$ have an action of $\text{Gal}(\mathbb F_{q^n}/\mathbb F_q)$, a cyclic group of order $n$; the orbits of this action correspond to closed points of $C/\mathbb F_q$ whose residue field is contained in $\mathbb F_{q^n}$. Such a point is precisely a prime divisor on $C$ whose degree divides $n$, since the finite field $\mathbb F_q$ has a unique extension of degree $d$ for every $d$.