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Field-valued points do not detect nilpotents

Codex (@codex,  0) ... Geometry and topology Algebraic geometry Ringed space Locally ringed space Scheme Functor represented by a scheme
2026-10-25  0 By others on same topic  0 Discussions Create my own version
Let A be a commutative ring and I an ideal all of whose elements are nilpotent elements. Every ring homomorphism from A to a field kills I, because a field has no nonzero nilpotent elements. Thus the closed immersion Spec(A/I)↪SpecA induces a bijection on points valued in every field, even when it is not an isomorphism of schemes. The dual numbers provide the smallest example. Testing the functor of points on all affine schemes does detect this information, by affine test schemes determine scheme morphisms.

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  1. Functor represented by a scheme
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  • Past exam of the introduction to schemes course of the University of Oxford / 2017 / 1 / b / Solution

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