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

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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.

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A classical zero locus depends only on the generated ideal and its radical

Statement

For $S \subseteq R = k [x_{1}, \dots, x_{n}]$ , one has $V (S) = V ((S)) = V (\sqrt{(S)})$ .

Facts & Assumptions

Given: An algebraically closed field $k$ , a nonnegative integer $n$ , and a subset $S \subseteq R = k [x_{1}, \dots, x_{n}]$ .

[F1]

Zero loci mean simultaneous vanishing (Classical affine algebraic sets, including the empty boundaries).

[F2]

Elements of $(S)$ are finite sums $\sum r_{i} s_{i}$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

[F3]

Membership in the radical means a positive power lies in the ideal (The radical of an ideal).

Proof

technique · direct

1.1

If $a \in V (S)$ , then every $q = \sum_{i} r_{i} s_{i} \in (S)$ satisfies $q (a) = \sum_{i} r_{i} (a) s_{i} (a) = 0$ . Conversely $S \subseteq (S)$ , so vanishing on $(S)$ implies vanishing on $S$ . Thus $V (S) = V ((S))$ .

F1F2givenalgebra

2.1

Write $J = (S)$ . If $a \in V (J)$ and $q^{m} \in J$ for $m \geq 1$ , then $q (a)^{m} = 0$ , hence $q (a) = 0$ in the field $k$ . Thus $V (J) \subseteq V (\sqrt{J})$ . The reverse inclusion follows from $J \subseteq \sqrt{J}$ .

F1F3step 1.1algebra∎

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, §2a p. 36; Theorem 2.16 preamble p. 42. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

Depends on

Used by

Dependency tree · two levels

10 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, §2a p. 36; Theorem 2.16 preamble p. 42 (standard reference, not scraped)