Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 113 3 i Solution Created 2026-10-03 Updated 2026-10-05
Let be the function field of the integral scheme and choose a nonzero rational section of a line bundle . Choose an open trivializing cover with generators for the invertible sheaf . On each nonempty , write with . If on an overlap, then , so is a regular unit. Consequently the local rational functions define a Cartier divisor .
The divisor line bundle restricts to . Define its local isomorphism from by . On an overlap,so the maps glue. HenceThe rational section maps to the rational function , which also fixes the sign convention for . Choosing a different nonzero rational section changes by a principal Cartier divisor, so its linear equivalence of Cartier divisors class is unchanged. No projectivity or Noetherian hypothesis is required.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 113 3 a Solution 2026-10-03
A standard sufficient hypothesis is that is both a Noetherian scheme and an integral scheme, and is regular in codimension one; in particular, a Noetherian normal scheme qualifies. A Weil divisor is then a finite sumover integral codimension-one closed subschemes . The local ring at the generic point of each is a discrete valuation ring, so every nonzero rational function has a principal divisorThe divisor class group is
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 113 2 c Solution Created 2026-09-28 Updated 2026-10-03
False. For , take the affine hypersurfaceIt is an integral scheme, and its only possible singular point is the origin, which has codimension two. Hence it is regular in codimension one; as a hypersurface it satisfies Serre's condition , so the Serre criterion for normality also makes it a normal scheme. The Divisor class group of an A-type surface singularity isgenerated by . Thus a closed affine subscheme satisfying can have nonzero torsion in its class group.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 113 1 c Solution 2026-09-28
For a point , let be the image of in its residue field . The scheme-theoretic fiber isIf , then . The quadratic polynomial is irreducible. Indeed, after setting , any hypothetical linear factors must restrict, up to nonzero scalars, to and ; comparing the and coefficients then forces both coefficients to vanish, contradicting the nonzero coefficient. Hence its homogeneous coordinate ring is an integral domain, so is an integral scheme.
At the origin , the fiber is , the union of the two distinct projective lines and , and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of .
Rational section of a line bundle 2026-10-05
On an integral scheme with generic point and function field , a rational section of a line bundle is an element of the one-dimensional -space . A nonzero rational section writes as in local trivializations, with . Ratios of these coefficients are regular units, and therefore define a Cartier divisor.