ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$Z$ is an integral domain of characteristic $0$ whose group of units is ${1, - 1}$ , so it is not a field: $2$ is nonzero and not invertible

Example

Let $Z$ be the integers with the commutative ring structure of $Z$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, and let $2 : = 1 + 1$ . Then:

$Z$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors);
$Z^{\times} = {1, - 1}$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , with the unit group structure from 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);
$2 \neq 0$ and $2 \notin Z^{\times}$ , so $Z$ is not a field (Field);
$char (Z) = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise).

So an integral domain need not be a field: claim 1 with claim 3 is the witness that the two notions differ.

Facts & Assumptions

Given: The integers $Z$ with the operations of Arithmetic on the integers and the order of Order on the integers, and the numeral $2 : = 1 + 1$ (The integers as equivalence classes of pairs of naturals).

[L1]

$(Z, +, \cdot, 0, 1)$ is a commutative ring ( $Z$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring).

[L2]

If $x, y \in Z$ are nonzero then $x y \neq 0$ (The integers have no zero divisors; multiplicative cancellation).

[L3]

$(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , Left inverse, right inverse, and invertible element of a monoid, 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).

[L4]

The order on $Z$ is total and compatible with addition, and $Z$ is a commutative ring (The integers form a totally ordered ring, The integers form a commutative ring, Order on the integers).

[L5]

$ι : N \to Z$ , $ι (n) = [(n, 0)]$ , is injective and preserves addition, multiplication and order (The naturals embed in the integers); the formulas for $0_{Z} = [(0, 0)]$ and $1_{Z} = [(1, 0)]$ give $ι (0) = 0$ and $ι (1) = 1$ (Arithmetic on the integers).

[L6]

Induction on $N$ , and $n + 1 = σ (n)$ on $N$ (The principle of mathematical induction, Addition of natural numbers, The natural numbers $N$ (von Neumann)).

[L7]

The additive multiples in a ring satisfy $0 \cdot 1 = 0$ and $σ (n) \cdot 1 = n \cdot 1 + 1$ for $n \in N$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L8]

$char (R)$ is the least $n \geq 1$ with $n \cdot 1_{R} = 0_{R}$ , or $0$ if there is none (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise); a field is a commutative ring in which every nonzero element is a unit (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, Field).

Verification

technique · direct

1.1

Claim 1: $Z$ is a commutative ring by [L1]; $1 \neq 0$ , since $1 = ι (1)$ , $0 = ι (0)$ and $ι$ is injective while $1 \neq 0$ in $N$ ; and $Z$ has no zero divisors by [L2]. So $Z$ is an integral domain.

L1L2L5

1.2

Claim 2 is [L3]: the multiplicative monoid of the ring $Z$ is $(Z, \cdot, 1)$ , and its group of units is ${1, - 1}$ .

L1L3

1.3

$0 < 1 < 2$ : $1 = ι (1)$ lies in the image of $ι$ , so $0 \leq 1$ , and $1 \neq 0$ by injectivity of $ι$ ; adding $1$ to $0 < 1$ gives $1 < 1 + 1 = 2$ .

L4L5

1.4

The map $n \mapsto n \cdot 1$ from $N$ to $Z$ is $ι$ . Both send $0$ to $0$ , since $0 \cdot 1 = 0$ and $ι (0) = 0$ ; and if $n \cdot 1 = ι (n)$ then $σ (n) \cdot 1 = n \cdot 1 + 1 = ι (n) + ι (1) = ι (n + 1) = ι (σ (n))$ , because $ι$ preserves addition. Induction on $N$ gives the claim.

L5L6L7

2.1

Claim 3: by step 1.3, $0 < 1 < 2$ , so $2 \neq 0$ and $2 \neq 1$ ; and $- 1 < 0 < 2$ by adding $- 1$ to $0 < 1$ and using transitivity, so $2 \neq - 1$ . Hence $2 \notin {1, - 1} = Z^{\times}$ by step 1.2. A field has every nonzero element a unit, so $Z$ is not a field.

step 1.2step 1.3L4L8

2.2

Claim 4: for $n \in N$ with $n \geq 1$ we have $n \cdot 1 = ι (n)$ by step 1.4, and $ι (n) \neq ι (0) = 0$ because $ι$ is injective and $n \neq 0$ . So no $n \geq 1$ satisfies $n \cdot 1 = 0$ , and $char (Z) = 0$ .

step 1.4L5L8

3.1

Claims 1 to 4 are established in steps 1.1, 1.2, 2.1 and 2.2.

step 1.1step 1.2step 2.1step 2.2∎

Remarks

This is the standard witness that "integral domain" is strictly weaker than "field". $Z$ satisfies every clause of Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors and fails the one extra clause a field asks for. The gap is exactly the failure of $2$ to be invertible, and $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ is what pins the units down.
Characteristic $0$ and infinite additive order are the same statement here. By The characteristic of a ring is the additive order of $1_{R}$ , with $0$ recording infinite order; $n \cdot 1_{R} = 0$ holds exactly when $char (R) ∣ n$ ; and in an integral domain every nonzero element has the same additive order as $1_{R}$ the characteristic is the additive order of $1$ , so claim 4 says that $1$ has infinite additive order in $(Z, +)$ ; and since $Z$ is a domain, the same lemma says every nonzero integer has infinite additive order too.
The units are read off $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , not off the order. That lemma proves $Z^{\times} = {1, - 1}$ from the divisibility relation and the bound on divisors; nothing on this page reproves it, and no group-theoretic example page is cited for it.

Depends on

Used by

A product of two rings with 1 ≠ 0 always has zero divisors: (1,0)(0,1) = (0,0) in ℤ × ℤ, so a product of integral domains is never an integral domain Example

Dependency tree · two levels

64 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

Integral domain (Wikipedia) (standard reference, not scraped)
Unit (ring theory) (Wikipedia) (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §16.4: Integral Domains and Fields (standard reference, not scraped)

ZZ is an integral domain of characteristic 00 whose group of units is {1,−1}{1,−1}, so it is not a field: 22 is nonzero and not invertible