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.

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

Statement

Let $k$ be an algebraically closed field and let $S \subseteq k [x_{1}, \dots, x_{n}]$ . If $(S)$ denotes the ideal generated by $S$ , then $V (S) = V ((S)) = V (\sqrt{(S)}) .$

Facts & Assumptions

Given: An algebraically closed field $k$ , a natural number $n$ , and a set $S \subseteq k [x_{1}, \dots, x_{n}]$ .

[L1]

For any subset $T \subseteq k [x_{1}, \dots, x_{n}]$ , $V (T) = {a \in k^{n} : f (a) = 0 for every f \in T}$ (An affine algebraic set in affine space).

Proof

technique · direct

1.1

Because $S \subseteq (S)$ , every point annihilating $(S)$ certainly annihilates $S$ , so $V ((S)) \subseteq V (S)$ . Conversely, if $a \in V (S)$ and $g = \sum_{i = 1}^{r} h_{i} f_{i} \in (S)$ with $f_{i} \in S$ , then $g (a) = \sum_{i = 1}^{r} h_{i} (a) f_{i} (a) = 0$ . Hence $a \in V ((S))$ . Therefore $V (S) = V ((S))$ .

L1givenalgebra

2.1

Since $(S) \subseteq \sqrt{(S)}$ , we have $V (\sqrt{(S)}) \subseteq V ((S))$ . Conversely, if $a \in V ((S))$ and $g \in \sqrt{(S)}$ , choose $m \geq 1$ with $g^{m} \in (S)$ . Then $g (a)^{m} = g^{m} (a) = 0$ , and a field has no nonzero nilpotent elements, so $g (a) = 0$ . Thus $a \in V (\sqrt{(S)})$ , proving $V ((S)) = V (\sqrt{(S)})$ .

step 1.1L1algebra

3.1

Combining steps 1.1 and 2.1 gives $V (S) = V ((S)) = V (\sqrt{(S)})$ .

step 1.1step 2.1∎

Depends on

An affine algebraic set in affine space

Used by

Dependency tree · one level

1 result 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, Chapter 2c (standard reference, not scraped)
Donu Arapura, Notes on Basic Algebraic Geometry, Corollary 1.3.2 (standard reference, not scraped)