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Edit: I am working over $\mathbb C$ here, but a similar answer work over an arbitrary algebraically closed field. See my comment below as well as Andrea Ferreti's. The degree of the divisor is equal to the degree of the image of $\varphi$, let's call it $C$, times the topological degree of the map $ \varphi : X \to C$. |
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Edit: I am working over $\mathbb C$. The degree of the divisor is equal to the degree of the image of $\varphi$, let's call it $C$, times the topological degree of the map $ \varphi : X \to C$. |
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The degree of the divisor is equal to the degree of the image of $\varphi$, let's call it $C$, times the topological degree of the map $ \varphi : X \to C$. |
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