TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Projective intersection dimension and nonemptiness

Statement

Let $X, Y \subseteq P_{k}^{n}$ be irreducible closed subvarieties. Every nonempty irreducible component $Z$ of $X \cap Y$ satisfies $\dim Z \geq \dim X + \dim Y - n$ . If $\dim X + \dim Y \geq n$ , then $X \cap Y \neq \emptyset$ .

Work over a fixed algebraically closed field $k$ , with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

If $X \subseteq P_{k}^{N}$ is a nonempty projective algebraic set, then $\dim C (X) = \dim X + 1$ . Over $X_{i} = X \cap D_{+} (T_{i})$ , the locus $C (X) \cap D (T_{i})$ is isomorphic to $X_{i} \times G_{m}$ . If $X$ is irreducible, so is $C (X)$ . Work over a fixed algebraically closed field $k$ , with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (A nonempty projective cone raises dimension by one).

[F2]

For irreducible closed $X, Y \subseteq A_{k}^{n}$ , every nonempty irreducible component $Z$ of $X \cap Y$ satisfies $\dim Z \geq \dim X + \dim Y - n$ . Work over a fixed algebraically closed field $k$ , with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (Affine intersection bound via the diagonal).

[F3]

If $U$ is a nonempty open of an irreducible classical variety $X$ , then $\dim U = \dim X$ . Every proper closed subvariety $Z ⊊ X$ has $\dim Z < \dim X$ . Work over a fixed algebraically closed field $k$ , with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (Nonempty opens preserve irreducible dimension).

Proof

1.1

The irreducible affine cones have dimensions $\dim X + 1$ and $\dim Y + 1$ in $A^{n + 1}$ . Their intersection $H$ contains the vertex. The affine intersection bound gives every component of $H$ dimension at least $\dim X + \dim Y - n + 1$ . If $\dim X + \dim Y \geq n$ , this is at least one, so there is a nonzero point of $H$ , projecting to $X \cap Y$ .

F1F2

2.1

For any nonempty projective component $Z$ , choose a standard chart meeting it away from the other projective components. The corresponding portion of $H$ is the product of that intersection with $G_{m}$ . Its component corresponding to $Z$ is an open of a component of $H$ , so the cone-chart dimension comparison gives $\dim Z + 1 \geq \dim X + \dim Y - n + 1$ . This proves the bound, without asserting that a vertex-only $H$ is the cone of an empty projective set. For $n = 0$ both factors are the point.

F1step 1.1F3∎

Depends on

Used by

Dependency tree · two levels

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Sources

Milne Corollary 6.47 (standard reference, not scraped)