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Finite-type schemes over a field are qcqs

Statement

Assume the Axiom of Choice. Let $k$ be a field and let $X$ be a scheme of finite type over $k$ . Every open subset of $X$ is quasi-compact, meaning that every open cover of that subset has a finite subcover. In particular, $X$ is quasi-compact and quasi-separated: intersections of affine open subschemes are quasi-compact. The empty scheme is included.

Facts & Assumptions

Given: AC, a field $k$ , and a $k$ -scheme $X$ , meaning a scheme equipped with its structure morphism $f : X \to Spec k$ ; assume $f$ is of finite type. Schemes and morphisms over a base, Locally finite type and finite type morphisms

Goal: Prove quasi-compactness of every open subset and the stated qcqs consequence.

[F1]

A field has an inverse for every nonzero element. A commutative ring is Noetherian if every ideal is finitely generated. Field, Noetherian commutative rings and modules

[F2]

A finite-type morphism is quasi-compact and locally of finite type; the latter means that every point has an affine open neighbourhood $U = Spec A$ mapping into an affine open of the base with a finite-type coordinate algebra map. Finite type for a $k$ -algebra means generated by finitely many elements as an algebra. The scheme $Spec k$ has one point and its ring of global functions is canonically $k$ , so its only nonempty open is the whole base. Consequently the charts for $X$ have coordinate rings that are finitely generated $k$ -algebras. Locally finite type and finite type morphisms, Subalgebra generated by a subset, algebras of finite type, and module-finite algebras, The spectrum of a field is a one-point affine scheme

[F3]

A finite-type commutative algebra over a Noetherian commutative ring is Noetherian. Every algebra of finite type over a Noetherian ring is a Noetherian ring

[F4]

Under AC, the spectrum of a Noetherian commutative ring is a Noetherian topological space. The spectrum of a Noetherian ring is a Noetherian topological space

[F5]

Under AC, every open subset of a Noetherian topological space is quasi-compact, including the empty subset. Noetherian open subsets are quasi-compact

[F6]

A scheme is quasi-compact if its underlying space is quasi-compact, and quasi-separated if the intersection of every two affine open subschemes is quasi-compact. Quasi-compact and quasi-separated schemes

[F7]

A scheme morphism $f : X \to S$ is quasi-compact when $f^{- 1} (V)$ is quasi-compact for every quasi-compact open $V \subseteq S$ . Quasi-compact and quasi-separated morphisms

[A1]

AC is assumed for the Noetherian-spectrum and open-subset suppliers. The Axiom of Choice

Proof

1.1givenF1F2F3F4F7A1

Any nonzero ideal of $k$ contains some $a \neq 0$ , hence contains $a^{- 1} a = 1$ and is all of $k$ . Thus its only ideals are $(0)$ and $(1)$ , both finitely generated, and $k$ is Noetherian by [F1]. Every open cover of the one-point base $Spec k$ has a member containing its sole point and hence a one-member subcover; thus the base is quasi-compact by [F2]. The structure morphism $f$ is quasi-compact by [F2], so [F7] applied to the whole base gives that $X = f^{- 1} (Spec k)$ is quasi-compact. By [F2], $X$ has an affine open cover with finite-type coordinate $k$ -algebras; take a finite subcover $U_{1}, \dots, U_{N}$ . Each coordinate algebra is Noetherian by [F3], so each $U_{j}$ is a Noetherian topological space by [F4]. For empty $X$ take the empty subcover.

2.1F5F6A1step 1.1∎

Let $W \subseteq X$ be open and take any open cover of $W$ . For each $j$ , restricting this cover to the open subset $W \cap U_{j}$ gives a finite subcover by [F5]. The union of these finitely many finite subcovers covers $W$ , since the $U_{j}$ cover $X$ . Thus every $W$ is quasi-compact; for empty $W$ the empty subcover suffices. Apply this to $W = X$ and to each intersection of two affine opens. The definitions in [F6] give precisely the claimed qcqs conclusion.

Depends on

Used by · 3

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