CounterexampleConstruction: Literature-sourcedVerification: 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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The total quotient ring of a nondomain need not be a field: $Q (Z / 6) ≅ Z / 6$

Statement refuted

The total quotient ring of every nonzero commutative ring is a field.

Facts & Assumptions

Given: The quotient ring $R = Z / 6 Z$ . By definition, a regular element $s$ has trivial annihilator, and the total quotient ring $Q (R)$ is the localisation at all regular elements.

[F1]

Units form a group, and a map already taking all denominators to units extends uniquely through localisation (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring, Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[F2]

The ring $Z / 6$ has congruence-class arithmetic (For every $n \in N$ , the congruence-class ring $Z / n$ is the quotient ring $Z / n Z$ ).

[F3]

In a field, every nonzero element is a unit (Field).

Counterexample

technique · direct

1.1

In $R = Z / 6 Z$ , the class $0$ is not regular because it annihilates $1$ ; the classes $2, 3, 4$ are nonzero zero divisors; and $1, 5$ are units and hence regular. Thus the regular elements are exactly $S = {1, 5} = R^{\times}$ .

givenF1F2algebra

2.1

Since the identity map of $R$ already sends every element of $S$ to a unit, [F1] gives an inverse to the localisation map, so $Q (R) = S^{- 1} R ≅ R$ .

F1step 1.1

3.1

The nonzero class of $2$ is not a unit because every product $2 a$ is even modulo $6$ and cannot equal $1$ . Hence $R$ , and therefore $Q (R)$ , is not a field by [F3].

F2F3step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

The Stacks Project, Section 10.25: Zerodivisors and total rings of fractions (standard reference, not scraped)

The total quotient ring of a nondomain need not be a field: Q(Z/6)≅Z/6Q(Z/6)≅Z/6