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Irreducibility is equivalent to the nonempty-open intersection criterion
Statement
For a nonempty topological space , irreducibility is equivalent to the intersection of every two nonempty open subsets being nonempty. Every nonempty open subspace of an irreducible space is itself irreducible and dense. This applies to the classical Zariski spaces.
Facts & Assumptions
Given: A nonempty topological space . For the inheritance assertions assume irreducible and let be nonempty open.
Irreducibility excludes a union of two proper closed subsets (A classical affine variety).
Proof
Two disjoint nonempty opens give the proper closed cover . Conversely a proper closed cover gives disjoint nonempty opens . Taking complements proves both directions of the criterion.
If is nonempty open in irreducible , each nonempty open of meets by step 1.1. Thus no proper closed subset of contains , which says . If are nonempty opens of , they are opens of because is open, so they intersect. The criterion makes irreducible.
Sources
Source comparison: Milne, Algebraic Geometry, v6.10, §2h p. 45. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.
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Sources
- J. S. Milne, Algebraic Geometry v6.10, §2h p. 45 (standard reference, not scraped)