Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 14 1 c Solution Created 2026-10-03 Updated 2026-10-14
The genus-degree formula gives arithmetic genus three for a plane quartic. Each of its three double points has multiplicity of a plane curve at a point two, so the arithmetic genus drop under a point blowup is one at each point. The resulting integral projective curve hasFor its finite normalization of an algebraic curve , the arithmetic genus identity isBoth terms are nonnegative, so . Over the algebraically closed field, the rationality of a smooth projective genus-zero curve identifies with the projective line: a point exists, and Riemann-Roch theorem gives a degree-one map from the two-dimensional space . Consequently is birational to . This proves the rationality of a plane quartic with three double points even when a double point is not an ordinary node.
Three distinct points of multiplicity two use the entire arithmetic genus three of an integral projective curve of degree four. One point blowup at each gives arithmetic genus zero. The normalization of an algebraic curve then has genus zero and is the projective line over an algebraically closed field, by rationality of a smooth projective genus-zero curve. This allows double points with one branch as well as ordinary nodes.