LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12

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Irreducibility is equivalent to every pair of nonempty open sets meeting

Statement

Let $k$ be an algebraically closed field, let $n \geq 0$ , and let $X \subseteq A_{k}^{n}$ be a nonempty subset with the subspace Zariski topology. Then the following are equivalent:

$X$ is irreducible.
Every pair of nonempty open subsets of $X$ has nonempty intersection.
Every nonempty open subset of $X$ is dense in $X$ .

Facts & Assumptions

Given: An algebraically closed field $k$ , an integer $n \geq 0$ , and a nonempty subset $X \subseteq A_{k}^{n}$ with the subspace Zariski topology.

Proof

technique · direct

1.1

Assume $X$ is irreducible. If $U, V \subseteq X$ are nonempty open subsets with $U \cap V = \emptyset$ , then $X = (X ∖ U) \cup (X ∖ V)$ is a union of two proper closed subsets, contrary to irreducibility. Therefore every two nonempty open subsets of $X$ meet.

given

1.2

Conversely, assume every two nonempty open subsets of $X$ meet. If $X = A \cup B$ with $A$ and $B$ proper closed subsets, then $X ∖ A$ and $X ∖ B$ are nonempty disjoint open subsets, a contradiction. Hence $X$ is irreducible.

givenalgebra

1.3

If every nonempty open subset of $X$ is dense and $U, V$ are nonempty open subsets, then $\overline{U} = X$ , so $U$ meets the nonempty open set $V$ . Thus every pair of nonempty open subsets meets.

givenalgebra

2.1

Assume every two nonempty open subsets of $X$ meet, and let $U \subseteq X$ be nonempty open. If $\overline{U} \neq X$ , then $X ∖ \overline{U}$ is a nonempty open set disjoint from $U$ , contradiction. So $U$ is dense.

step 1.2given

3.1

Steps 1.1 and 1.2 show that irreducibility is equivalent to pairwise intersection of nonempty opens, and steps 2.1 and 1.3 show that this is also equivalent to density of every nonempty open subset.

step 1.1step 1.2step 2.1step 1.3∎

Depends on

Used by

Every nonempty open subset of an affine variety is dense Lemma

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, §1.5 (standard reference, not scraped)