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

1.1L1givenalgebra

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 $V (S) = V ((S))$ .

2.1step 1.1L1algebra

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 $V ((S)) = V (\sqrt{(S)})$ .

3.1step 1.1step 2.1∎

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

Depends on

An affine algebraic set in affine space

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 1 level

Sources