Search the corpus

    The degree of a divisor on a projective curve

    PR-Y5S7V

    On a smooth projective curve, zeros and poles balance: every principal divisor has degree zero.Suggest a slogan

    For \(D = \sum n_i P_i\) on a curve, \(\deg D = \sum n_i [k(P_i) : k]\), which is \(\sum n_i\) when \(k = \bar{k}\).

    On a smooth projective curve, \(\deg \div(f) = 0\) for every \(f \in K(X)^*\). Degree therefore descends to a homomorphism \(\Pic(X) \to \ZZ\), and \(\Pic^0(X)\) is its kernel.

    Projectivity enters because a nonconstant \(f\) defines a finite morphism \(X\to\PP^1\), and \(\div(f)=f^*(0)-f^*(\infty)\), where both fibres have degree \(\deg f\) counted with multiplicity. Zeros and poles balance because the map has a well-defined degree, and the map has one because \(X\) is projective.

    On \(\AA^1\) the statement is false — \(\div(x) = [0]\) has degree one — and this is the cleanest illustration that divisor theory is a theory about complete curves.

    What is the degree of a divisor?
    Why does a principal divisor have degree zero?

    Dependencies

    • D-5PQ5WWeil and Cartier divisors, and when they agreeuses

    Backlinks

    • D-CRVDEGREESDegrees of points, finite morphisms, line bundles and coherent sheaves on curvesrelated-to
    • PR-DIVAMPCURVEAmpleness of a divisor on a curveuses
    • FE-DIVP1EDegree-zero divisors on the line and on an elliptic curveuses
    • T-MWDVLRiemann--Roch and Serre duality on a curveuses
    • T-U5QSYThe Jacobian, Abel's theorem, and the Abel--Jacobi mapuses

    What links to this