CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.
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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 $R \to S^{- 1} R$ is injective.

Facts & Assumptions

Given: The quotient ring $R = Z / 6 Z$ 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 $Z / 6 Z$ with congruence-class arithmetic (For every $n \in N$ , the congruence-class ring $Z / n$ is the quotient ring $Z / n Z$ ).

Counterexample

technique · direct

1.1

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

F2algebra

2.1

Since the denominator $3$ annihilates $2$ , [F1] gives $2 / 1 = 0$ in $S^{- 1} R$ . 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 · 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

The Stacks Project, Section 10.9: Localization (standard reference, not scraped)

Localising at a zero divisor need not be injective: inverting 33 in Z/6Z/6 kills 22