Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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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.

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Localising at a zero divisor need not be injective: inverting 3 in Z/6 kills 2

Statement refuted

Every localisation map R→S−1R is injective.

Facts & Assumptions

Given: The quotient ring R=Z/6Z and the subset S={1,3}.

[F1]

The kernel of a localisation map consists of the elements annihilated by some denominator (Equality, vanishing, and the kernel of the localisation map).

[F2]

The ring Z/6 is Z/6Z with congruence-class arithmetic (For every n∈N, the congruence-class ring Z/n is the quotient ring Z/nZ).

Counterexample

technique · direct
1.1

In R=Z/6Z, let S={1,3}, the multiplicative set generated by the class of 3 because 32=3. The class of 2 is nonzero, while 3⋅2=0.

F2algebra
2.1

Since the denominator 3 annihilates 2, [F1] gives 2/1=0 in S−1R. Thus the localisation map kills a nonzero element and is not injective.

F1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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