ExampleConstruction: AI-generatedVerification: AI-generatedprecheck 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.

I(V(x^2, xy)) keeps only the radical information

Example

Let $k$ be algebraically closed and let $I = (x^{2}, x y) \subseteq k [x, y]$ . Then $I (V (I)) = (x) = \sqrt{I}$ , so the zero locus forgets the nilpotent multiplicity in $x^{2}$ .

Facts & Assumptions

Given: An algebraically closed field $k$ and the ideal $I = (x^{2}, x y) \subseteq k [x, y]$ .

[L1]

An ideal and its radical have the same zero locus (An ideal and its radical have the same zero locus).

Verification

technique · direct

1.1

A point $(a, b) \in k^{2}$ lies in $V (I)$ exactly when $a^{2} = 0$ and $a b = 0$ . Since $k$ is a field, this means $a = 0$ and $b$ is arbitrary. Hence $V (I) = {(0, b) : b \in k}$ .

given

2.1

The vanishing ideal of the $y$ -axis is $(x)$ . Also $\sqrt{I} = (x)$ because $x^{2} \in I$ and every element of $I$ is divisible by $x$ . Therefore $I (V (I)) = (x) = \sqrt{I}$ , in agreement with [L1].

L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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)