LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06

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Classical k-points give closed points over an algebraically closed field

Statement

Let $k$ be algebraically closed and let $A$ be a reduced finite-type $k$ -algebra. The closed points of $Spec A$ are exactly the kernels of the $k$ -algebra maps $A \to k$ . They need not exhaust all points of $Spec A$ .

Facts & Assumptions

Given: An algebraically closed field $k$ and a reduced finite-type $k$ -algebra $A$ .

[F1]

Every maximal ideal of an affine $k$ -algebra is the kernel of a $k$ -algebra map to $k$ (Over an algebraically closed field, maximal ideals of an affine algebra are kernels of points).

[F2]

Closed points of an affine scheme are its maximal ideals (Closed points of an affine scheme).

Proof

technique · direct

1.1F1F2

A closed point is maximal by [F2], and [F1] identifies every such ideal with a kernel $A \to k$ .

1.2F2algebra

Every unital $k$ -algebra map $A \to k$ is surjective, so its kernel is maximal and hence closed by [F2].

2.1step 1.1step 1.2∎

Thus classical $k$ -points and closed points agree, but this makes no assertion that every prime is maximal.

Depends on

Used by

Minimal tangent dimension and homogeneous regularity Corollary
Good and geometric quotients for group actions Definition
Irreducible classical varieties and integral separated finite-type schemes Theorem

Dependency tree · two levels

11 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

James S. Milne, Algebraic Geometry, 10.24 (standard reference, not scraped)