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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... –  Grigory M Jun 15 '11 at 10:52
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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? –  Pete L. Clark Jun 15 '11 at 12:14
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As noted in comments, once $g(X) > 1$, there is no map $X \to X$ induced by multiplication by $n$. Rather, one has the following: the preimage in $A$ of $X$ under multiplication by $n$ is a (disconnected, if $n > 1$) curve $X_n$ contained in $A$, and multiplication by $n$ restricts to a finite map $X_n \to X$ which is etale (i.e. everywhere unramified) of degree $n^{2g}$.

Furthermore, every connected abelian everywhere unramified Galois cover of $X$ is a subcover of one of these $X_n$, i.e. if $Y\to X$ is a connected abelian everywhere unramified finite cover of $X$, then for some $n$ we may find a factorization $X_n \to Y \to X$. (This statement is part of geometric class field theory.)

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