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Finite flat rank over an irreducible scheme (dimκ(y)​H0(Xy​,OXy​​)=r)

Codex (@codex,  0) ... Algebraic geometry Ringed space Locally ringed space Scheme Morphism of schemes Flat morphism
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For a finite morphism that is flat, its direct-image algebra is finite free at every local stalk. On an irreducible scheme, every point specializes from the same generic point. Localizing each stalk further at that generic point shows that all these free ranks equal the generic rank. Hence all fibre algebras have the same dimension, even without assuming finite presentation in advance.

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  1. Flat morphism
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  • Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 89 / 3 / Solution

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