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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.

Devang Agarwal
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  • For $P \in C_1,Q=\phi(P)\in C_2, f(X) \in k(C_1), g(X) = N_{k(C_1)/\phi^* k(C_2)}(f(X))$ you want to show $g(P) = \prod_{R\in \phi^{-1}(Q)} f(R)^{e(R)}$ where $e(R)$ is the ramification index defined by $e(R)=ord_R h(\phi(X))$ for $h(Y) \in k(C_2)$ with a simple zero at $Q$. If $k(C_1)/\phi^* k(C_2)$ is a normal extension this is easy because $N_{k(C_1)/\phi^* k(C_2)}(f(X)) = (\prod_{j=1}^n f(\psi_j(X)))^d$ where $f(X) \mapsto f(\psi_j(X))$ are the automorphisms $\in Aut(k(C_1)/\phi^* k(C_2))$ and $n = [k(C_1):\phi^* k(C_2)]_s$ and $d =[k(C_1):\phi^k(C_2)]_i=\frac{[k(C_1):\phi^ k(C_2)]}{n}$. – reuns Mar 23 '19 at 11:17
  • The $\psi_j$ are some birational maps $C_1 \to C_1$ such that $\phi( \psi_j(X))=\phi(X)$, equality of rational maps so there are a few problematic points at the poles of the $\psi_j$. Now if $k(C_1)/\phi^* k(C_2)$ is not a normal extension (for example in the elliptic curve/isogeny case) then the $\psi_j$ are replaced by the embeddings from $k(C_1)$ to $L$ the normal closure of $k(C_1)/\phi^* k(C_2)$ and I'm not sure how to deal with it. – reuns Mar 23 '19 at 11:26
  • Even in the case when $k(C_1)/\phi^*k(C_2)$ is a normal extension, I don't see why the ramification indices should appear in the product expression for the norm. – Devang Agarwal Mar 23 '19 at 21:08
  • Let $u(X)$ with a unique zero of order $l$ at $P$ then $N_{k(C_1)/\phi^* k(C_2)}(u(X)) = (\prod_j u(\psi_j(X))^d=v(X)=w(\phi(X))$ where the zeros of $v(X)$ are at the $R \in \phi^{-1}(\phi(P)$ so $w(Y)$ has a unique zero of order $m$ at $\phi(P)$. If necessary replace $u(X)$ by $1/(1/u(X)+c)$ so there is no cancellation in the zeros/poles of the $u(\psi_j(X))$ so that $l = m$ obtaining $e(R) /d$ is the number of $\psi_j$ sending $P$ to $R$ – reuns Mar 24 '19 at 13:09
  • Thanks! I don't really see what to do in case of a non-normal extension either. Assuming a result from Hartshorne, that given any finitely generated field of trancendence degree one over $k$ is a function field of some curve, using that on the normal closure of the extension $k(C_1)/\phi^k(C_2)$ and that $N_{K/M} = N_{L/M}\circ N_{K/L}$, I found that $f(\phi^D)^d = ((\phi_f)(D))^d$, where $d = [K:k(C_1)]_s$ and $K$ is the normal closure of $k(C_1)/\phi^k(C_2)$, but I am not sure how to get rid of the exponent. – Devang Agarwal Mar 24 '19 at 19:29

1 Answers1

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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*}

leoli1
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