LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Points of an affine algebraic set correspond to maximal ideals of its coordinate ring

Statement

Let $k$ be an algebraically closed field and let $X \subseteq A_{k}^{n}$ be an affine algebraic set. Then the map $x ⟼ m_{x} : = {\overline{f} \in k [X] : \overline{f} (x) = 0}$ is a bijection from $X$ onto the set of maximal ideals of $k [X]$ .

Facts & Assumptions

Given: An algebraically closed field $k$ and an affine algebraic set $X \subseteq A_{k}^{n}$ .

[L1]

The coordinate ring is $k [X] = k [x_{1}, \dots, x_{n}] / I (X)$ (The coordinate ring of an affine algebraic set).

[L2]

Every maximal ideal of $k [x_{1}, \dots, x_{n}]$ has the form $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ for a unique point $a \in k^{n}$ (Over an algebraically closed field, every maximal ideal is an evaluation ideal).

[L3]

For a commutative ring $R$ , an ideal $M$ is maximal exactly when $R / M$ is a field ( $R / M$ is a field if and only if $M$ is a maximal ideal).

Proof

technique · direct

1.1

Fix $x \in X$ . Evaluation at $x$ gives a surjective homomorphism ${ev}_{x} : k [X] \to k$ because every constant is attained. Its kernel is exactly $m_{x}$ , and the quotient $k [X] / m_{x}$ is isomorphic to the field $k$ . Hence [L3] shows that $m_{x}$ is maximal.

L1L3given

1.2

Let $m$ be a maximal ideal of $k [X]$ . Under the quotient map $π : k [x_{1}, \dots, x_{n}] ↠ k [X]$ , the inverse image $M : = π^{- 1} (m)$ is a maximal ideal of the polynomial ring. By [L2], $M = (x_{1} - a_{1}, \dots, x_{n} - a_{n})$ for a unique point $a \in k^{n}$ . Because $I (X) \subseteq M$ , every polynomial vanishing on $X$ vanishes at $a$ , so $a \in V (I (X)) = X$ . Then $m = m_{a}$ .

L1L2choosealgebra

2.1

If $m_{x} = m_{y}$ , then the images of all coordinate functions agree at $x$ and $y$ , so $x_{i} = y_{i}$ for every coordinate. Thus $x = y$ , and the map $x \mapsto m_{x}$ is injective.

L1step 1.1algebra

3.1

Steps 2.1 and 1.2 prove that $x \mapsto m_{x}$ is a bijection from $X$ to the maximal ideals of $k [X]$ .

step 2.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

13 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

J. S. Milne, Algebraic Geometry, Nullstellensatz and affine variety correspondence (standard reference, not scraped)
Donu Arapura, Notes on Basic Algebraic Geometry, §1.2 and §1.5 (standard reference, not scraped)