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Proof of Plucker's formula
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Case X is smooth plane curve F(x,y,z) = 0 Then let p: X to P^1 be projection to [x:z]. By Hurwitz 'S formula, it suffices to compute deg R_p (ramification divisor). But R_p is equal to the intersection divisor div(dF/dy). Degree of the latter is equal to deg(X) deg(dF/dy) by Bezout. Case X has n nodes and no other singularities. Need to redefine intersection divisor and check that bezout still holds. In this case even though every ramification point is still a Zero of dF/dy, the converse is not true. If we choose coordinate so that no node point will be ramification point then div(dF/dy) = R_p + divisor D of points corresponding to nodes ( two points for each node, each has coefficient 1). So deg R_p = deg (div(dF/dy)) - deg D = d(d-1) - 2n, by Bezout
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