CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31

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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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$Z / 1$ has one element and satisfies $[0]_{1} = [1]_{1}$ , so it is not a field

Statement refuted

The modular operations do not make $Z / n$ a field for every positive modulus $n$ : the boundary modulus $n = 1$ is a counterexample.

Facts & Assumptions

Given: The quotient $Z / 1$ with its modular operations.

[L1]

The quotient $Z / 1$ has exactly one class (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L2]

Its modular operations satisfy the abelian-group, commutative-monoid and distributive identities, with additive identity $[0]_{1}$ and multiplicative identity $[1]_{1}$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L3]

A field must have distinct additive and multiplicative identities (Field).

Counterexample

technique · direct

1.1

By [L1], all integers determine the same class modulo $1$ , so in particular $[0]_{1} = [1]_{1}$ .

2.1

Although the other algebraic identities hold by [L2], the equality in step 1.1 violates the distinct-identities clause [L3]. Hence $Z / 1$ is not a field.

step 1.1L2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

K. Conrad, Modular Arithmetic (standard reference, not scraped)
CSUSM, Polynomials over residue rings (standard reference, not scraped)

Z/1Z/1 has one element and satisfies [0]1=[1]1[0]1​=[1]1​, so it is not a field