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Every nonempty open subset of an affine variety is dense
Statement
Let be a classical affine variety. Every nonempty open subset of is dense in .
Facts & Assumptions
Given: A classical affine variety and a nonempty open subset .
In a classical affine variety, irreducibility is equivalent to every pair of nonempty open subsets meeting, and also to every nonempty open subset being dense (Irreducibility is equivalent to every pair of nonempty open sets meeting).
Proof
Because is a classical affine variety, it is irreducible. By [L1], an irreducible space is one in which every nonempty open subset is dense.
Applying step 1.1 to the specific nonempty open subset shows that is dense in .
Depends on
Used by
- Rational maps between irreducible classical affine varieties Definition
- Dominant maps pull back function fields functorially Lemma
- Morphisms from an irreducible affine variety to an affine variety are determined by a dense open subset Lemma
- The rational-map equivalence relation is transitive Lemma
- Regular functions on a principal open are the principal localization of the coordinate ring Theorem
Dependency tree · two levels
4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- J. S. Milne, Algebraic Geometry, Chapter 2g (standard reference, not scraped)
- Donu Arapura, Notes on Basic Algebraic Geometry, irreducibility discussion in §1.5 (standard reference, not scraped)