Norm map on function fields of curves For a curve $C$, $f\in \bar{K}(C)^*$ and a divisor $D = \sum n_P(P) \in \text{Div}(C)$ such that $D$ and $\text{div}(f)$ have disjoint supports (support of a divisor is the set of points with non-zero coefficients in its sum), we define :
$$f(D) = \prod_{P\in C}f(P)^{n_P}$$
Let $\phi:C_1\rightarrow C_2$ be a non-constant map of smooth curves. Then show that $f(\phi^*D) = (\phi_*f)(D)$ for all $f\in \bar{K}(C_1)^*$ and all $D\in \text{Div}(C_2)$ whenever both sides are defined. Here, $\phi^*:\bar{K}(C_2)\rightarrow \bar{K}(C_1)$ is the induced map on function fields and $\phi_*:\bar{K}(C_1)\rightarrow\bar{K}(C_2)$ is the map defined as $$\phi_* = (\phi^*)^{-1}\circ N_{\bar{K}(C_1)/\phi^*\bar{K}(C_2)}$$
where $N$ denotes the usual norm of field extensions. This is question 2.10(a) from Silverman's Arithmetic of Elliptic Curves. So far, I have managed to reduce it to the case where $\phi$ is separable, but I don't understand how to proceed further, since I don't really have a good understanding of what the norm map does in this context. I would prefer if the solution uses only the algebraic geometry discussed so far in Silverman.
This question has been asked before at exercise 2.10 in silverman AEC, but it received no responses.
 A: It suffices to prove this for the case $D=(Q)$ where $Q$ is a point of $C_2$. Let $A$ be the image of the local ring of $C_2$ at $Q$ under the map $\phi^*$ (i.e. it is a discrete valuation ring of $\phi^*K(C_2)$). Let $B$ be the integral closure of $A$ in $K(C_1)$. Write $m_Q$ for the maximal ideal of $A$. Then $m_Q$ splits in $B$ into the product $\prod_{\phi(P)=Q}m_P^{e_\phi(P)}$ where we similarly write $m_P$ for the maximal ideal of $B$ corresponding to a point $P$ lying over $Q$.
We can identify the value $\phi_*f(D)$ with the image of $N_{B/A}(f)$ in $A/m_Q=\overline K$ (note that $f\in B$ as the supports of $\operatorname{div}f$ and $\phi^*D$ are disjoint by assumption)
Now we have the following commutative diagram:
$$\require{AMScd}\begin{CD}
B @>>> B/m_QB;\\
@V{N_{B/A}}VV @V{N_{(B/m_QB)/(A/m_Q)}}VV \\
A @>>> A/m_Q
\end{CD}$$
Denote by $\bar{f}$ the image of $f$ in $B/m_QB$. Since $B/m_QB$ splits as $\prod_{\phi(P)=Q} B/m_P^{e_\phi(P)}$, hence $$N_{(B/m_QB)/(A/m_Q)}(\bar{f})=\prod_{\phi(P)=Q}N_{(B/m_P^{e_\phi(P)})/(A/m_Q)}(\bar f).$$
Writing $f=f(P)+ g$ where $g(P)=0$ we see that the map given by multiplication by $\bar{f}$ in $B/m_P^{e_\phi(P)}$ is $f(P)\operatorname{id}+N$ where $N$ is nilpotent. It is easy to see that then $\det(f(P)\operatorname{id}+N)=\det(f(P)\operatorname{id})$ (e.g. by writing $N=M^{-1}\tilde{N}M$ where $N$ is upper triangular with zeros on the diagonal). Hence $N_{(B/m_P^{e_\phi(P)})/(A/m_Q)}(\bar f)=f(P)^{e_{\phi}(P)}$ and we get \begin{align*}
f(\phi^*D):=\prod_{\phi(P)=Q}f(P)^{e_{\phi}(P)}&=\prod_{\phi(P)=Q}N_{(B/m_P^{e_\phi(P)})/(A/m_Q)}(\bar f)\\&=N_{(B/m_QB)/(A/m_Q)}(\bar{f})
=\overline{N_{B/A}(f)}=(\phi_*f)(Q)
\end{align*}
