ExampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-28

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A product of two rings with $1 \neq 0$ always has zero divisors: $(1, 0) (0, 1) = (0, 0)$ in $Z \times Z$ , so a product of integral domains is never an integral domain

Example

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) with $1_{R} \neq 0_{R}$ and $1_{S} \neq 0_{S}$ , and let $R \times S$ be the product ring (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ). Then:

$(1_{R}, 0_{S})$ and $(0_{R}, 1_{S})$ are nonzero elements of $R \times S$ whose product is $(0_{R}, 0_{S})$ ; each is therefore a zero divisor (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors);
consequently $R \times S$ is never an integral domain, whatever $R$ and $S$ are; in particular a product of two integral domains is not one;
the concrete instance is $Z \times Z$ , where $(1, 0) (0, 1) = (0, 0)$ while $Z$ itself is an integral domain ( $Z$ 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).

Facts & Assumptions

Given: Rings $R$ , $S$ with $1_{R} \neq 0_{R}$ and $1_{S} \neq 0_{S}$ , and the product ring $R \times S$ with componentwise operations, zero $(0_{R}, 0_{S})$ and identity $(1_{R}, 1_{S})$ (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ).

[L1]

$R \times S$ is a ring; its operations are componentwise, and two of its elements are equal exactly when both components agree (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ , Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L2]

$x \cdot 0 = 0 = 0 \cdot x$ in any ring (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ).

[L3]

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

[L4]

$Z$ is a commutative ring with $1 \neq 0$ and is an integral domain ( $Z$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, $Z$ 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, The integers form a commutative ring, The integers as equivalence classes of pairs of naturals, Arithmetic on the integers).

Verification

technique · direct

1.1

$(1_{R}, 0_{S}) \neq (0_{R}, 0_{S})$ , since the first components differ, $1_{R} \neq 0_{R}$ ; and $(0_{R}, 1_{S}) \neq (0_{R}, 0_{S})$ , since the second components differ, $1_{S} \neq 0_{S}$ .

L1given

1.2

$(1_{R}, 0_{S}) \cdot (0_{R}, 1_{S}) = (1_{R} 0_{R}, 0_{S} 1_{S}) = (0_{R}, 0_{S})$ , the components being computed by [L2].

L1L2

2.1

Claim 1: by steps 1.1 and 1.2 the element $(1_{R}, 0_{S})$ is nonzero, the element $(0_{R}, 1_{S})$ is nonzero, and their product is the zero of $R \times S$ ; so each is a zero divisor.

step 1.1step 1.2L3

3.1

Claim 2: an integral domain has no zero divisors, and $R \times S$ has one by step 2.1; so $R \times S$ is not an integral domain. Every integral domain satisfies $1 \neq 0$ , so a product of two integral domains falls under the hypothesis and is never one.

step 2.1L3

4.1

Claim 3: $Z$ is a ring with $1 \neq 0$ by [L4], so steps 2.1 and 3.1 apply with $R = S = Z$ , giving $(1, 0) (0, 1) = (0, 0)$ with both factors nonzero; and $Z$ is an integral domain by [L4], so $Z \times Z$ is a product of two integral domains that is not one.

step 2.1step 3.1L4∎

Remarks

This is the cheapest source of zero divisors available. No arithmetic beyond $x \cdot 0 = 0$ is used, and the two witnesses are written down rather than found. The other standard source is a ring of functions from a set with at least two points to a coefficient ring with $1 \neq 0$ (The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations), and the two constructions are the same phenomenon: a point where one factor vanishes and the other does not.
The hypothesis is on both factors. If $S$ is the one-element ring then $(0_{R}, 1_{S}) = (0_{R}, 0_{S})$ is the zero of $R \times S$ and the argument collapses; indeed $R \times S$ is then essentially $R$ again, and may well be a domain. That is why both $1_{R} \neq 0_{R}$ and $1_{S} \neq 0_{S}$ are assumed.
The failure is not about commutativity. $R$ and $S$ need not be commutative for claim 1; commutativity enters only in claim 2, because Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors requires a domain to be commutative and the product of two commutative rings is commutative (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ).

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Sources

Product of rings (Wikipedia) (standard reference, not scraped)
Zero divisor (Wikipedia) (standard reference, not scraped)

A product of two rings with 1≠01=0 always has zero divisors: (1,0)(0,1)=(0,0)(1,0)(0,1)=(0,0) in Z×ZZ×Z, so a product of integral domains is never an integral domain