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

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Classical affine points are maximal ideals

Statement

Assume the Axiom of Choice, inherited from the Nullstellensatz route. For every affine algebraic set $X$ and $A = k [X]$ , the map $x \mapsto m_{x} = \ker ({ev}_{x} : A \to k)$ is a bijection from $X$ to the maximal ideals of $A$ . Its residue-field map $A / m_{x} \to k$ is the canonical $k$ -isomorphism given by evaluation. Both sets are empty when $X$ is empty.

Facts & Assumptions

Given: AC, an algebraically closed field $k$ , and an affine algebraic set $X$ with coordinate ring $A$ .

[F1]

$A = R / I (X)$ (The coordinate ring of a classical affine algebraic set).

[F2]

Evaluation in the polynomial ring has maximal kernel $(x_{i} - a_{i})_{i}$ (Evaluation at a point has kernel (x_1-a_1, ..., x_n-a_n)).

[F3]

Ideals of the quotient correspond to ideals upstairs containing $I (X)$ (Correspondence theorem: ideals of $R / I$ correspond to ideals of $R$ containing $I$ ).

[F4]

Maximal ideals upstairs are uniquely the coordinate-point ideals (Over an algebraically closed field, every maximal ideal is an evaluation ideal).

[F5]

Closed algebraic sets satisfy $V (I (X)) = X$ (Classical affine algebraic sets correspond to radical ideals, and irreducible sets to prime ideals).

Proof

technique · direct

1.1

At $x \in X$ , evaluation on $R$ kills $I (X)$ and hence defines evaluation on $A$ ; constants make it surjective. Its kernel is maximal by the same argument as F2, or by correspondence with the maximal evaluation ideal upstairs. Two points with equal kernels have equal inverse images upstairs, and F4 makes the points equal.

F1F2F3F4given

2.1

If $M$ is maximal in $A$ , its inverse image in $R$ is maximal by F3. F4 identifies it with the evaluation ideal at a unique $x$ . Since it contains $I (X)$ , $x \in V (I (X)) = X$ . Thus $M = m_{x}$ . Evaluation induces a bijection $A / m_{x} \to k$ : equality of values is exactly equality modulo the kernel, and every constant is attained. It preserves all operations. For $X = \emptyset$ , $A = 0$ has no proper, hence no maximal, ideals.

F1F3F4step 1.1F5∎

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, Example 2.13, 2.20, and §3e, pp. 41, 43, 65. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

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Sources

J. S. Milne, Algebraic Geometry v6.10, Example 2.13, 2.20, and §3e, pp. 41, 43, 65 (standard reference, not scraped)