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let $X$ a smooth algebraic curve over a field , let $A$ be its jacobian and $n:A\rightarrow A$ the multiplication by $n$ map (can assume $n$ coprime with the char of the base field if this semplifies things ). Since the curve embedds in its Jacobian which is the corresponding map $X\rightarrow X$, or $X\rightarrow Y$ where this is a finite map and $Y$ is a smooth curve?

thanks

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    I'm afraid, I can't understand OP's last sentence...2011-06-15
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    A self-map of the curve will induce an endomorphism of the Jacobian. The converse however does not hold: why do you think there should be a map from $X$ to $X$ "corresponding" to multiplication by $n$ on the Jacobian? What "correspondence" do you have in mind?2011-06-15

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