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Projective space over a classical base and homogeneous closed loci

Statement

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.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

Products of nonempty classical varieties exist in the category of classical varieties, and $\dim (X \times_{k} Y) = \dim X + \dim Y$ . If both factors are irreducible, their product is irreducible. 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. (Dimensions add under products).

[F2]

Let $R$ be a Noetherian commutative ring. Then the polynomial ring $R [x]$ is a Noetherian commutative ring. No hypothesis beyond Noetherianity is placed on $R$ : it may have zero divisors, and it may be the zero ring. (Hilbert basis theorem: if $R$ is Noetherian then $R [x]$ is Noetherian).

[F3]

Assume the Axiom of Choice. Let $k$ be an algebraically closed field. 1. The assignments $X ⟼ I (X), J ⟼ V (J)$ induce mutually inverse inclusion-reversing correspondences between affine algebraic sets $X \subseteq A_{k}^{n}$ and radical ideals $J \subseteq k [x_{1}, \dots, x_{n}]$ . 2. Under this correspondence, nonempty irreducible affine algebraic sets correspond exactly to prime ideals. (Affine algebraic sets correspond to radical ideals, and irreducible ones to prime ideals).

[F4]

For every $i$ , normalization of the $i$ th coordinate identifies $D_{+} (x_{i})$ with $A_{k}^{n}$ . In particular $[a_{0} : \dots : a_{n}] \mapsto (a_{1} / a_{0}, \dots, a_{n} / a_{0})$ identifies $D_{+} (x_{0})$ with $A_{k}^{n}$ . (standard projective opens are affine spaces).

Proof

1.1

Product existence follows from the product construction when $Y$ is nonempty; for empty $Y$ glue the empty charts. Standard projective opens give the affine charts $Y \times A^{N}$ when $Y$ is affine. A homogeneous equation dehomogenizes on each chart and defines a closed subset there, so its global projective zero set is closed.

F1F4

1.2

If the fibre is empty, the affine zero set of its specialized homogeneous ideal is contained in the origin. The Nullstellensatz gives $T_{i} \in \sqrt{I (y)}$ for each $i$ , so choose $e_{i} \geq 1$ with $T_{i}^{e_{i}} \in I (y)$ . Every monomial of degree $d = 1 + \sum_{i} (e_{i} - 1)$ is divisible by one of these powers, hence belongs to $I (y)$ . This also holds for the unit ideal.

2.1

Conversely let $Z$ be closed in this product with affine $Y$ . On each chart $T_{i} \neq 0$ , choose polynomial equations $h$ over $A$ for $Z$ there. Homogenize each $h$ in the $T$ variables to $H$ with respect to $T_{i}$ , and multiply by $T_{i}$ . This homogeneous polynomial vanishes on all of $Z$ : on that chart $h$ vanishes, and off the chart $T_{i}$ vanishes. If a point is outside $Z$ , some chart containing it has an equation $h$ nonzero there, so the associated $T_{i} H$ is also nonzero. These global homogeneous equations cut out exactly $Z$ . Hilbert basis makes their ideal finitely generated by homogeneous elements, since $A$ is a quotient of a finite polynomial ring over $k$ .

F2step 1.1

3.1

Conversely if all degree- $d$ monomials belong to $I (y)$ , any nonzero vector has some $T_{i} \neq 0$ , so $T_{i}^{d}$ cannot vanish on it. Thus there is no projective zero. For $N = 0$ the only variable is $T_{0}$ and the same power test applies.

step 1.2∎

Depends on

Used by

Projective classical morphisms Definition
Projection from projective space over a variety is closed Theorem

Dependency tree · two levels

15 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 §6q Lemma 6.51(a,b), p.158 (standard reference, not scraped)