ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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.

Every element of k[X] is integral over k[X^2], and k[X] has basis 1, X over k[X^2]

Example

Let $k$ be a field. Then every polynomial $f (X) \in k [X]$ is integral over the subring $k [X^{2}]$ , and $k [X]$ is a free $k [X^{2}]$ -module with basis $1, X$ .

Facts & Assumptions

Given: A field $k$ and the polynomial ring $k [X]$ .

[L1]

Over a nonzero commutative ring $A$ , an element $b$ is integral over $A$ if and only if there exists a faithful $A [b]$ -module finitely generated over $A$ (Integrality and finite-module characterizations for one element).

Verification

technique · direct

1.1

Every polynomial $f (X) \in k [X]$ can be written uniquely as $a (X^{2}) + X b (X^{2})$ with $a (T), b (T) \in k [T]$ , by separating the even and odd powers of $X$ . Therefore $k [X] = k [X^{2}] \oplus X k [X^{2}]$ , so $1$ and $X$ form a basis of $k [X]$ over $k [X^{2}]$ .

givenalgebra

2.1

Fix $f (X) \in k [X]$ . Because $k [X]$ is a faithful module over the subring $k [X^{2}] [f (X)]$ and step 1.1 shows that it is finitely generated over $k [X^{2}]$ , [L1] implies that $f (X)$ is integral over $k [X^{2}]$ .

L1step 1.1

3.1

Thus $k [X]$ is finite free of rank $2$ over $k [X^{2}]$ , and every element of $k [X]$ is integral over that subring.

step 1.1step 2.1∎

Depends on

Integrality and finite-module characterizations for one element

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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., Exercise (10.24) (standard reference, not scraped)