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

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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.

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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

1.1L1L3given

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.

1.2L1L2choosealgebra

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 $m = m_{a}$ .

2.1L1step 1.1algebra

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.

3.1step 2.1step 1.2∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources