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.

Vanishing ideals and zero loci form a Galois connection

Statement

Let $k$ be an algebraically closed field, let $X \subseteq k^{n}$ , and let $J \subseteq k [x_{1}, \dots, x_{n}]$ be an ideal. Then $X \subseteq V (J) ⟺ J \subseteq I (X) .$ Consequently, $X \subseteq V (I (X)) and J \subseteq I (V (J)) .$

Facts & Assumptions

Given: An algebraically closed field $k$ , a subset $X \subseteq k^{n}$ , and an ideal $J \subseteq k [x_{1}, \dots, x_{n}]$ .

[L1]

$V (J)$ is the set of points where every polynomial in $J$ vanishes (An affine algebraic set in affine space).

[L2]

$I (X)$ is the set of polynomials vanishing at every point of $X$ (The vanishing ideal of a subset of affine space).

Proof

technique · direct

1.1

If $X \subseteq V (J)$ and $f \in J$ , then every point of $X$ lies in $V (J)$ , so [L1] says $f$ vanishes at every point of $X$ . Therefore $f \in I (X)$ by [L2], and hence $J \subseteq I (X)$ .

L1L2given

1.2

Conversely, if $J \subseteq I (X)$ and $x \in X$ , then every polynomial in $J$ lies in $I (X)$ , so [L2] says every polynomial in $J$ vanishes at $x$ . By [L1], this means $x \in V (J)$ . Hence $X \subseteq V (J)$ .

L1L2given

2.1

Applying steps 1.1 and 1.2 with $J = I (X)$ yields $X \subseteq V (I (X))$ , and applying them with $X = V (J)$ yields $J \subseteq I (V (J))$ .

step 1.1step 1.2∎

Depends on

Used by

Affine algebraic sets correspond to radical ideals, and irreducible ones to prime ideals Theorem

Dependency tree · one level

2 results within one dependency step 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, paragraph 2.23 (standard reference, not scraped)
Donu Arapura, Notes on Basic Algebraic Geometry, Proposition 1.3.1 and Lemma 1.6.1 (standard reference, not scraped)