Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 23 5 Solution Created 2026-10-03 Updated 2026-10-14
Use the finite map from the preceding solution. The inverse images of the two standard affine charts and of their intersection are affine, because is finite. The vanishing of quasi-coherent cohomology on an affine scheme therefore lets their two-term Čech cochain complex compute . It has terms only in degrees zero and one, provingThis actually applies to every quasi-coherent sheaf on .
For a smooth projective curve with , define its genus of a smooth projective curve byThe preceding Serre duality also gives . The condition on constants holds in particular for a geometrically connected smooth projective curve. It makes equal to the arithmetic genus . We state the theorem with this usual convention and then give the formula valid without that condition.
For a divisor on an algebraic curve and a canonical divisor with , the Riemann-Roch theorem isThe degree over an arbitrary field is weighted by residue degrees: when . Equivalently, for a line bundle ,We now prove it via Riemann-Roch via elementary modifications.
For any closed point and any line bundle , the exact sequencehas a skyscraper sheaf as its quotient, with -dimension . This follows locally from a uniformizer of the discrete valuation ring at : a local line-bundle generator modulo that uniformizer spans one copy of . The quotient has zero higher cohomology, as it is the direct image from a finite affine scheme. The long exact sequence in sheaf cohomology and the proved higher vanishing giveEvery line bundle has a nonzero rational section of a line bundle, whose local orders give a divisor on an algebraic curve with . Iterating the displayed identity, adding or subtracting closed points with their multiplicities, givesFor the constants convention above, . The dualizing sheaf on a smooth projective curve gives , which converts this Euler-characteristic identity into the stated Riemann-Roch theorem. This proves the formula over arbitrary , without assuming the closed points are -rational.
For vector bundles, the corresponding formula isTo see it, saturate a rational rank-one subspace to a line subbundle and induct on rank. Both Euler characteristic of a coherent sheaf and the degree of the determinant are additive in the resulting short exact sequence. The line-bundle formula just proved starts the induction.
Taking and using duality gives , , and hence . In particular a line bundle of degree greater than has zero first cohomology: its dual twisted by has negative degree, and a negative-degree line bundle has no nonzero section, since any section would have an effective divisor of zeros.
If the term “curve over ” is used without requiring , the unconditional formula we proved isWriting then replaces by . For example, viewed as a -curve for a finite separable extension has , so a literal constant term would be incorrect. Alternatively the arithmetic genus gives the formula with constant . The constants hypothesis for the usual genus is recorded in Stacks Project, genus of a curve.