Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 4 2 Solution Created 2026-10-03 Updated 2026-10-07
Use on the target and on the source, so the map of coordinate rings sends to . A closed point of the real affine line corresponds to a maximal ideal generated either by , , or by a monic irreducible quadratic . The scheme-theoretic fibre over that point has coordinate algebraHere is the residue field, and in the real-point case. The fibre is a scheme, so its nilpotents must be retained.
For , put . The factors , , are pairwise coprime, with the last one irreducible over . The Chinese remainder theorem identifies with , giving three irreducible components.
For , put . The exact factorization isEach quadratic has discriminant , and they are distinct and coprime. Thus , giving two irreducible components.
For , . It has one prime ideal, generated by the class of , so one irreducible component. It is not the reduced point : its class of is a nonzero nilpotent element with nilpotency index four.
Finally write the irreducible quadratic as , with . Over , has four roots above and four above . They are all simple, because , and the two sets are disjoint. None is real. Conjugation pairs them into four distinct irreducible real quadratics. Hence and this fibre has four irreducible components.
| Target closed-point type | Real coordinate algebra of the fibre | Irreducible components |
|---|---|---|
| , | ||
| , | ||
| Irreducible quadratic |
All fibres in each row are isomorphic as affine schemes over . The rows are distinct: their numbers of irreducible components already distinguish them, and the zero fibre is additionally nonreduced. ThereforeThese are ordinary scheme components, not geometric components counted after complexifying the base. Counting only real roots would miss both the complex-residue factors and the nonreal closed points.