ExampleConstruction: AI-generatedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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.

Localising Z/12Z kills exactly the torsion seen by the denominator set

Example

Let $M = Z / 12 Z$ . If $S_{2} = {2^{n} : n \in N}$ and $S_{3} = {3^{n} : n \in N}$ , then

\ker (M \to S_{2}^{- 1} M) = {[0], [3], [6], [9]}, \ker (M \to S_{3}^{- 1} M) = {[0], [4], [8]} .

So localisation kills exactly the torsion detected by the chosen denominator set.

Facts & Assumptions

Given: The module $M = Z / 12 Z$ and the multiplicative sets $S_{2} = {2^{n} : n \in N}$ and $S_{3} = {3^{n} : n \in N}$ .

[L1]

A fraction in a localised module is zero exactly when one denominator kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L2]

Elements of $S^{- 1} M$ are fractions $m / s$ with $s \in S$ (Localisation of a module at a multiplicative subset).

Verification

technique · direct

1.1

By [L1], $[a] \in \ker (M \to S_{2}^{- 1} M)$ exactly when $2^{r} [a] = 0$ in $Z / 12 Z$ for some $r \geq 0$ . This happens for $[0], [3], [6], [9]$ , and for no other class, because $12 ∣ 2^{r} a$ is possible exactly when the odd part of $a$ is divisible by $3$ .

L1algebra

1.2

Likewise, $[a] \in \ker (M \to S_{3}^{- 1} M)$ exactly when $3^{r} [a] = 0$ for some $r \geq 0$ . This happens for $[0], [4], [8]$ , and for no other class, because $12 ∣ 3^{r} a$ is possible exactly when the $2$ -primary part of $a$ is divisible by $4$ .

L1algebra

2.1

Steps 1.1 and 1.2 give the two kernels, and [L2] interprets them as the elements killed by the respective localisation maps.

step 1.1step 1.2L2∎

Depends on

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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., (12.2) (standard reference, not scraped)