ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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.

$Q \otimes_{Z} Z / n = 0$ for every positive $n$

Example

For every positive integer $n$ ,

$Q \otimes_{Z} Z / n = 0.$

The positivity hypothesis matters: $Z / 0 ≅ Z$ , so the tensor product at $n = 0$ is $Q$ , not zero.

Facts & Assumptions

Given: A positive integer $n$ .

[L1]

For a right module $M$ and an ideal $I$ of a commutative ring $R$ , $M \otimes_{R} R / I ≅ M / I M$ ( $M \otimes_{R} R / I ≅ M / I M$ naturally).

[L2]

$Q$ is a field, so the nonzero integer $n$ has an inverse in $Q$ (The rationals form a field).

Verification

technique · direct

1.1

By [L1], $Q \otimes_{Z} Z / n ≅ Q / n Q$ .

givenL1

1.2

For every $q \in Q$ , the element $q / n$ belongs to $Q$ by [L2] and $q = n (q / n)$ , so $n Q = Q$ .

givenL2

2.1

Hence $Q / n Q = 0$ , and step 1.1 proves the claim.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Wenqi Li, Commutative Algebra, Lecture 10 (standard reference, not scraped)

Q⊗ZZ/n=0Q⊗Z​Z/n=0 for every positive nn