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.

Z[square-root of 2, square-root of 3] is finite over Z and contains the sum and product of its generators

Example

In the ring $Z [\sqrt{2}, \sqrt{3}]$ , every element is a $Z$ -linear combination of $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ . In particular the ring is finite over $Z$ , and the sum $\sqrt{2} + \sqrt{3}$ and product $\sqrt{6}$ of the two quadratic generators are integral over $Z$ .

Facts & Assumptions

Given: The subring $R : = Z [\sqrt{2}, \sqrt{3}] \subseteq R$ .

[L1]

A subalgebra generated by finitely many integral elements is module-finite (A subalgebra generated by finitely many integral elements is module-finite).

[L2]

Over a nonzero base ring, integral elements form a subring (Integral elements over a nonzero base ring form a subring).

Verification

technique · direct

1.1

The elements $\sqrt{2}$ and $\sqrt{3}$ satisfy the monic equations $T^{2} - 2 = 0$ and $T^{2} - 3 = 0$ over $Z$ , so they are integral over $Z$ . Therefore [L1] implies that $R$ is a finitely generated $Z$ -module.

L1given

2.1

Every element of $R$ is a polynomial in $\sqrt{2}$ and $\sqrt{3}$ . Replacing $(\sqrt{2})^{2}$ by $2$ and $(\sqrt{3})^{2}$ by $3$ reduces every monomial to a $Z$ -linear combination of $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ , so these four elements span $R$ over $Z$ .

step 1.1givenalgebra

3.1

Since $\sqrt{2}$ and $\sqrt{3}$ are integral over $Z$ , [L2] implies that their sum and product are integral over $Z$ . Here the product is exactly $\sqrt{6}$ , and the spanning result of step 2.1 places both elements inside one explicit finite $Z$ -module.

L2step 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Theorem 6.5 (standard reference, not scraped)
Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem (10.28) (standard reference, not scraped)