CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.
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$\mathbb{Z}/1$ has one element and satisfies $[0]_1=[1]_1$ , so it is not a field

Statement refuted

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

Facts & Assumptions

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

[L1]

The quotient $\mathbb Z/1$ has exactly one class (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[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 $n$ , $(\mathbb{Z}/n,+)$ 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 $\mathbb Z/1$ is not a field.

step 1.1L2L3∎

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: 51 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

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

Z/1\mathbb{Z}/1Z/1 has one element and satisfies [0]1=[1]1[0]_1=[1]_1[0]1​=[1]1​, so it is not a field