False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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FALSE: In every commutative ring, each nonzero element is either a unit or a zero divisor

Statement

False claim: in every commutative ring $R$ (Commutative ring), every $a \in R$ with $a \ne 0$ is either a unit of $R$ (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) or a zero divisor (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors).

The integers refute it. With $2 := 1 + 1$ in $\mathbb{Z}$ , the element $2$ is nonzero, is not a unit, and is not a zero divisor.

Facts & Assumptions

Given: The commutative ring $\mathbb{Z}$ with the operations of Arithmetic on the integers and the order of Order on the integers, and the numeral $2 = 1 + 1$ ( $\mathbb{Z}$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, The integers as equivalence classes of pairs of naturals).

[L1]

$\mathbb{Z}$ is a commutative ring ( $\mathbb{Z}$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, The integers form a commutative ring, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring).

[L2]

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

[L3]

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

[L4]

The order on $\mathbb{Z}$ is total and compatible with addition; $\iota : \mathbb{N} \to \mathbb{Z}$ is injective and order preserving with $\iota(0) = 0$ and $\iota(1) = 1$ . Since $0+1=1$ in $\mathbb N$ , one has $0\le1$ there and hence $\iota(0)\le\iota(1)$ (The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers, Order on the natural numbers, Addition of natural numbers).

[L5]

$a$ is a zero divisor when $a \ne 0$ and $ab = 0$ or $ba = 0$ for some $b \ne 0$ (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors).

[L6]

The refuted claim: in every commutative ring, every nonzero element is a unit or a zero divisor.

Refutation

technique · direct

1.1

$0 < 1 < 2$ and $-1 < 0$ in $\mathbb{Z}$ : $1 = \iota(1)$ is nonnegative and differs from $0 = \iota(0)$ because $\iota$ is injective, so $0 < 1$ ; adding $1$ gives $1 < 2$ ; adding $-1$ to $0 < 1$ gives $-1 < 0$ . In particular $2 \ne 0$ , $2 \ne 1$ and $2 \ne -1$ .

2.1

$2$ is not a unit of $\mathbb{Z}$ : by [L2] the units are $1$ and $-1$ , and $2$ is neither, by step 1.1.

step 1.1L2

2.2

$2$ is not a zero divisor of $\mathbb{Z}$ : if $2b = 0$ with $b \ne 0$ , then $2 \ne 0$ by step 1.1 and $b \ne 0$ give $2b \ne 0$ by [L3], a contradiction; and $b \cdot 2 = 2b$ by commutativity. So no such $b$ exists.

step 1.1L1L3L5

3.1

$\mathbb{Z}$ is a commutative ring by [L1], and by step 1.1 the element $2$ is nonzero, while by steps 2.1 and 2.2 it is neither a unit nor a zero divisor. So the claim of [L6] is false.

step 1.1step 2.1step 2.2L1L6∎

Remarks

What the claim is confusing it with. In an integral domain every nonzero element is a non-zero-divisor, and the claim would follow if every nonzero non-zero-divisor were a unit. That last implication is what $\mathbb{Z}$ refutes: $2$ cancels, by The integers have no zero divisors; multiplicative cancellation, and is still not invertible.
The claim becomes true under a finiteness hypothesis, which it does not make. If $R$ is a commutative ring and $a \ne 0$ , consider $x \mapsto ax$ . If it is injective and $R$ is finite, it is surjective by A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$ , and equality holds if and only if $B = A$ , so $ax = 1$ for some $x$ and $a$ is a unit; if it is not injective, then $ax = ay$ with $x \ne y$ gives $a(x-y) = 0$ with $x - y \ne 0$ , so $a$ is a zero divisor. The hypothesis that $R$ is finite is exactly what $\mathbb{Z}$ fails, and the statement above assumes nothing of the kind. This paragraph applies the cited finite-set theorem as an observation about the claim; it is not part of the refutation.
The witness is $2$ , not $0$ . Under the convention of Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors the element $0$ is not a zero divisor, so a claim quantified over nonzero elements is not vacuously repaired by looking at $0$ ; the refutation has to exhibit a genuine nonzero element, and it does.

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 78 results over 21 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

Zero divisor (Wikipedia) (standard reference, not scraped)
Unit (ring theory) (Wikipedia) (standard reference, not scraped)
Integral domain (Wikipedia) (standard reference, not scraped)