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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Zero loci and vanishing ideals form a Galois connection

Statement

For $E \subseteq k^{n}$ and an ideal $J \subseteq R$ , $E \subseteq V (J)$ iff $J \subseteq I (E)$ . Both $I$ and $V$ reverse inclusion. Moreover $V (I (E)) = \overline{E}$ in the Zariski topology, and $V (I (V (J))) = V (J)$ .

Facts & Assumptions

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

[F1]

Zero loci use a universal condition on equations (Classical affine algebraic sets, including the empty boundaries).

[F2]

Vanishing ideals use a universal condition on points (The classical vanishing ideal).

[F3]

Every closed zero locus can be defined by an ideal (A classical zero locus depends only on the generated ideal and its radical).

[F4]

Zero loci are exactly the Zariski closed sets (Classical affine zero loci form the Zariski closed sets).

Proof

technique · direct

1.1

$E \subseteq V (J)$ says that for every $a \in E$ and every $f \in J$ , $f (a) = 0$ . Interchanging these two universal quantifiers says precisely $J \subseteq I (E)$ . Enlarging $E$ imposes more conditions on $I (E)$ ; enlarging $J$ imposes more equations on $V (J)$ . This proves both reversals.

F1F2given

2.1

Every point of $E$ lies in $V (I (E))$ . If a closed set $V (J)$ contains $E$ , step 1.1 gives $J \subseteq I (E)$ and therefore $V (I (E)) \subseteq V (J)$ . Thus $V (I (E))$ is the smallest closed set containing $E$ . Applying this to the already closed set $E = V (J)$ gives the last identity.

F3F4step 1.1∎

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, Proposition 2.10 pp. 38–39, Proposition 2.14 p. 41, and Remark 2.23 p. 44. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

Depends on

Used by

Classical affine algebraic sets correspond to radical ideals, and irreducible sets to prime ideals 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

J. S. Milne, Algebraic Geometry v6.10, Proposition 2.10 pp. 38–39, Proposition 2.14 p. 41, and Remark 2.23 p. 44 (standard reference, not scraped)