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the exercise 2.10(a) in silverman AEC says $(\phi_*f)(D)=f(\phi^*D)$.(any $f\in \bar K \ (C_1),D\in Div(C_2) ) ( \phi:C_1 \rightarrow C_2 is \ a \ nonconstant \ map \ of\ smooth\ curve $ )

(Here we need the support of divisor D and the support of div(f) disjoint)

I want to prove the simple case. When $D$ is $Q(Q \;\text{is a point in} \ C_2$).

Then we need prove $$N_{K(C_1)/\phi^* K(C_2))}(f)(\phi^{-1}(Q))=\prod_{P\in\phi^{-1}(Q)}f(P)^{e_{\phi}(P)}$$ For $$N_{K(C_1)/\phi^* K(C_2))}(f)=\prod_{\sigma\in Gal({K(C_1)/\phi^* K(C_2))}}\sigma(f)$$,
then i want to prove $\sigma(f)(P)=f(P_1) (some\ P_1\ is \ in \ \phi^{-1}(Q))$

and it is true for all Q in $C_2$.

Then i guess $Gal({K(C_1)/\phi^* K(C_2)})$ may have action on $\phi^{-1}(Q)$,
$Gal({K(C_1)/\phi^* K(C_2)}) \times\phi^{-1}(Q)\rightarrow\phi^{-1} Q)$,$(\sigma,P)\rightarrow \sigma (P)$
which keeps $\sigma(f)(P)=f(\sigma (P)) $.

Then we can prove $N_{K(C_1)/\phi^* K(C_2))}(f)(\phi^{-1}(Q))=\prod_{P\in\phi^{-1}(Q)}f(P)^{e_{\phi}(P)}$.

But i don't know whether it is true. Thank you very much for any help.

zheng
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  • It might help your cause to explain what $D$, $f$, $\phi_$, and $\phi^$ mean. True, many of the people who can help you will have access to Silverman's book, but not all. Or, their copy may be in their office, but now it's weekend. I can guess that $D$ may be a divisor, $\phi$ a map between two curves, but what is $f$? Edit the question body, please. The goal is to make the question as self-contained as possible. – Jyrki Lahtonen Apr 14 '18 at 04:52
  • Sorry, I edit the question and give the mean of D,f,$\phi$. – zheng Apr 14 '18 at 12:01
  • I have written a solution here – leoli1 May 15 '22 at 19:14

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