LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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Every nonempty open of a classical affine variety is dense

Statement

Every nonempty open subset of an irreducible classical affine variety $X$ is dense. Every finite intersection of nonempty open subsets of $X$ is nonempty and dense; the intersection of the empty family is $X$ .

Facts & Assumptions

Given: An affine variety $X$ over algebraically closed $k$ , a nonempty open of $X$ , and a finite family of nonempty opens of $X$ .

[F1]

Nonempty opens of an irreducible space are dense and irreducible, and two such opens meet (Irreducibility is equivalent to the nonempty-open intersection criterion).

Proof

technique · direct

1.1F1given

The density assertion is F1 with the irreducible nonempty space $X$ . For a finite list $U_{1}, \dots, U_{r}$ , start with $W_{0} = X$ and set $W_{j} = W_{j - 1} \cap U_{j}$ . If $W_{j - 1}$ is nonempty open, F1 gives $W_{j} \neq \emptyset$ ; intersection preserves openness. Finite induction gives $W_{r}$ nonempty open.

2.1F1step 1.1∎

F1 now makes $W_{r}$ dense. This includes $r = 0$ , since $W_{0} = X$ is nonempty and its closure is itself.

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, §2h p. 45. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

Every nonempty open of a classical affine variety is dense

Statement

Facts & Assumptions

Proof

Sources

Depends on

Dependency tree · 2 levels

Sources