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.

An ideal and its radical have the same zero locus

Statement

Let $k$ be a field and let $I \subseteq k [x_{1}, \dots, x_{n}]$ be an ideal. Then

V (I) = V (\sqrt{I}) .

Facts & Assumptions

Given: A field $k$ and an ideal $I \subseteq k [x_{1}, \dots, x_{n}]$ .

[L1]

The radical $\sqrt{I}$ consists of elements whose some positive power lies in $I$ (The radical of an ideal).

[L2]

Polynomial evaluation is multiplicative (Evaluation and roots of a polynomial in a commutative target ring).

Proof

technique · direct

1.1

Since $I \subseteq \sqrt{I}$ by [L1], every common zero of $\sqrt{I}$ is a common zero of $I$ . Thus $V (\sqrt{I}) \subseteq V (I)$ .

L1given

1.2

Let $a \in V (I)$ and let $g \in \sqrt{I}$ . By [L1], some power $g^{N}$ lies in $I$ , so $0 = g^{N} (a) = g (a)^{N}$ by [L2]. Because $k$ is a field, $g (a) = 0$ . Hence $a \in V (\sqrt{I})$ and $V (I) \subseteq V (\sqrt{I})$ .

L1L2given

2.1

The two inclusions show that $V (I) = V (\sqrt{I})$ .

step 1.1step 1.2∎

Depends on

Used by

I(V(x², xy)) keeps only the radical information Example

Dependency tree · two levels

6 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, A Primer of Commutative Algebra, v4.03, Theorem 13.10 (standard reference, not scraped)