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

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Projection from projective space over a variety is closed

Statement

For every classical variety $Y$ and $N \geq 0$ , the projection $p : Y \times P_{k}^{N} \to Y$ is a closed map.

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]

For a classical variety $Y$ and $N \geq 0$ , $Y \times P_{k}^{N}$ exists with its standard product charts. If $Y$ is affine with $A = k [Y]$ , its closed subsets are precisely the zero loci of finitely generated homogeneous ideals of $A [T_{0}, \dots, T_{N}]$ . For such an ideal $I$ and $y \in Y$ , the fibre is empty if and only if the specialized ideal $I (y) \subseteq k [T_{0}, \dots, T_{N}]$ contains every monomial of some positive degree $d$ . 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. (Projective space over a classical base and homogeneous closed loci).

[F2]

Assume the Axiom of Choice. Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ . (Assuming the Axiom of Choice, Nakayama's lemma).

Proof

1.1

Closedness can be tested on an affine open cover of $Y$ , so take $Y$ affine with coordinate ring $A$ and a closed subset defined by a homogeneous ideal $I \subseteq S = A [T_{0}, \dots, T_{N}]$ . Fix $y$ outside its image. The monomial criterion supplies $d \geq 1$ such that the fibre of the finite $A$ -module $M = (S / I)_{d}$ at $y$ is zero: $M / m_{y} M = 0$ .

2.1

Localize at $m_{y}$ . Nakayama applies to the finite module $M_{m_{y}}$ and the maximal ideal of the local ring, which is its Jacobson radical, giving $M_{m_{y}} = 0$ . Choose finitely many generators of $M$ ; each is annihilated by some $s_{j} \notin m_{y}$ . Their product $s$ annihilates $M$ , and $s (y) \neq 0$ . If $M = 0$ already, use $s = 1$ .

F2step 1.1

3.1

For every $z \in D (s)$ the degree- $d$ fibre quotient is zero, so all degree- $d$ monomials belong to $I (z)$ . The monomial criterion shows that every fibre there is empty. Hence every point outside the image has an open neighborhood outside it, so the image is closed. This proof works for empty closed subsets, reducible $Y$ and $N = 0$ ; if $Y$ is empty there is nothing to check.

F1step 2.1∎

Depends on

Used by

Projective fibre dimension is upper semicontinuous Theorem

Dependency tree · two levels

8 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

Milne Theorem 7.22, pp.164–165 (standard reference, not scraped)