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Closed immersions are proper

Statement

Assume the Axiom of Choice. Every closed immersion of schemes is finite, hence proper; the empty closed immersion is included, no Noetherian, reducedness or nonemptiness hypothesis is used, and the Axiom of Choice enters through the affine quotient structure of closed immersions and the properness of finite morphisms.

Facts & Assumptions

Given: The Axiom of Choice and a closed immersion $i : Z \to Y$ .

[F1]

A morphism is a closed immersion when its underlying map is a homeomorphism onto a closed subset of its target and the structure-sheaf map is surjective. (Closed immersions of schemes)

[F2]

Assume AC. For a closed immersion $i : Z \to Y$ and every affine open $U = Spec A \subseteq Y$ there is a unique ideal $I \subseteq A$ with $i^{- 1} (U) ≅ Spec (A / I)$ over $U$ ; conversely every quotient map $A \to A / I$ induces a closed immersion, every base change of a closed immersion is a closed immersion, and the empty subscheme of $Spec A$ corresponds to $I = A$ . (Closed immersions are affine quotients and survive base change)

[F3]

A morphism $f : X \to S$ is finite when for every affine open $U = Spec A \subseteq S$ the inverse image is affine, $f^{- 1} (U) = Spec B$ , and $B$ is module-finite over $A$ ; the zero ring is allowed. (Finite morphisms of schemes)

[F4]

A module is cyclic when it is generated by one element and finitely generated when it is generated by a finite subset; the zero module is generated by the empty family. (Generated submodule, cyclic and finitely generated modules, module basis and free module)

[F5]

An $R$ -algebra $A$ is module-finite over $R$ when $A$ is finitely generated as an $R$ -module. (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras)

[F6]

Assume AC. Every finite morphism of schemes is proper. (Finite morphisms are proper)

[F7]

The Axiom of Choice states that every family of nonempty sets has a choice function. (The Axiom of Choice)

Proof

technique · direct: on each affine target the closed immersion is a quotient map, whose coordinate ring is a cyclic module, hence module-finite; the morphism is therefore finite, and finite morphisms are proper

1.1F1F2

Let $U = Spec A$ be an affine open of $Y$ . By [F1] and [F2] the restriction of $i$ over $U$ is a closed immersion with $i^{- 1} (U) ≅ Spec (A / I)$ over $U$ for the unique ideal $I \subseteq A$ ; in particular the inverse image is affine.

1.2F4F5

The quotient $A / I$ is cyclic as an $A$ -module: the class $1 + I$ generates it, because every $a + I = a \cdot (1 + I)$ lies in the submodule generated by $1 + I$ , while that submodule always lies in $A / I$ ; if $I = A$ then $A / I = 0$ is generated by the empty family. Hence $A / I$ is finitely generated, equivalently module-finite, over $A$ by [F4] and [F5].

2.1F2F3step 1.1step 1.2

Steps 1.1 and 1.2 verify the condition of [F3] on every affine open $U$ of $Y$ : the inverse image is affine, $i^{- 1} (U) = Spec (A / I)$ , and its coordinate algebra is module-finite over the coordinate algebra of $U$ . Therefore the closed immersion $i$ is finite.

3.1F2F3F6F7step 2.1∎

By [F6] the finite morphism $i$ is proper. The empty closed immersion is included: over each affine $U = Spec A$ it corresponds by [F2] to the ideal $I = A$ , so its inverse image is $Spec 0$ , an affine chart whose coordinate ring is the zero ring and is module-finite by [F3], and the same argument applies to it. The Axiom of Choice [F7] is assumed and is used only through [F2] and [F6], the two AC-qualified suppliers; no further selection occurs, and no Noetherian, reducedness or nonemptiness hypothesis is used.

Closed immersions are proper

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources