ExampleConstruction: AI-generatedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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 radical of a product and intersection computation

Example

In the polynomial ring $k [x, y]$ , one has $\sqrt{(x^{2}, x y) \cap (y^{3})} = (x y)$ .

Facts & Assumptions

Given: A field $k$ .

[L1]

The radical of an intersection is the intersection of the radicals (The radical of a finite intersection).

[L2]

The radical of $(x^{2}, x y)$ is $(x)$ (Computing sqrt((x^2,xy)) from its containing primes).

[L3]

An element lies in the radical of an ideal exactly when some positive power lies in that ideal (Radical membership via positive powers).

Verification

technique · direct

1.1

By [L1] and [L2], $\sqrt{(x^{2}, x y) \cap (y^{3})} = \sqrt{(x^{2}, x y)} \cap \sqrt{(y^{3})} = (x) \cap \sqrt{(y^{3})}$ . Since $y^{3} \in (y^{3})$ , [L3] gives $y \in \sqrt{(y^{3})}$ , so $(y) \subseteq \sqrt{(y^{3})}$ . Conversely, if $g \in \sqrt{(y^{3})}$ , choose $m \geq 1$ with $g^{m} \in (y^{3})$ and write $g = a (x) + y h (x, y)$ with $a (x) \in k [x]$ ; if $a (x) \neq 0$ , then $g^{m} = a (x)^{m} + y b (x, y)$ for some $b (x, y) \in k [x, y]$ , so $g^{m}$ has a nonzero $y^{0}$ term and cannot lie in $(y^{3})$ , a contradiction. Hence $a (x) = 0$ , so $g \in (y)$ . Thus $\sqrt{(y^{3})} = (y)$ .

L1L2L3givenchoosealgebra

2.1

Step 1.1 reduces the problem to $(x) \cap (y)$ . A polynomial lies in both ideals exactly when it is divisible by both $x$ and $y$ , hence by $x y$ . Therefore $(x) \cap (y) = (x y)$ .

step 1.1algebra

3.1

Therefore $\sqrt{(x^{2}, x y) \cap (y^{3})} = (x y)$ .

step 1.1step 2.1∎

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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §13 and §17 (standard reference, not scraped)
M. Hochster, Introduction to Commutative Algebra, Math 614 notes (2020) (standard reference, not scraped)