LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-30

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 vanishing ideal is always radical

Statement

Let $k$ be a field and let $X \subseteq k^{n}$ . Then the vanishing ideal $I (X) = {f \in k [x_{1}, \dots, x_{n}] : f (a) = 0 for every a \in X}$ is a radical ideal.

Facts & Assumptions

Given: A field $k$ and a subset $X \subseteq k^{n}$ .

[L1]

Polynomial evaluation at a point is well defined (Evaluation and roots of a polynomial in a commutative target ring).

Proof

technique · direct

1.1L1given

Let $f \in k [x_{1}, \dots, x_{n}]$ and assume $f^{N} \in I (X)$ for some $N \geq 1$ . Then for every $a \in X$ , $f (a)^{N} = f^{N} (a) = 0$ by [L1]. Since $k$ is a field, $f (a) = 0$ for every $a \in X$ .

2.1step 1.1∎

Thus $f \in I (X)$ . By the definition of radical ideal, this proves that $I (X)$ is radical.

Depends on

Evaluation and roots of a polynomial in a commutative target ring

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem (15.7) (standard reference, not scraped)