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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The map $n \mapsto (n, 0)$ from $Z$ to $Z \times Z$ preserves addition and multiplication and does not preserve $1$ , so the clause $f (1) = 1$ is not redundant

Statement refuted

False claim: if $R$ and $S$ are rings and $f : R \to S$ satisfies $f (x + y) = f (x) + f (y)$ and $f (x y) = f (x) f (y)$ for all $x, y \in R$ , then $f$ is a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ); that is, clause (RH3), $f (1_{R}) = 1_{S}$ , is redundant.

The map

$f : Z ⟶ Z \times Z, f (n) : = (n, 0)$

refutes it, $Z \times Z$ being the product ring (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ). It satisfies both displayed conditions and sends $1$ to $(1, 0)$ , which is not the identity $(1, 1)$ of $Z \times Z$ .

Facts & Assumptions

Given: The commutative ring $Z$ , the product ring $Z \times Z$ with componentwise operations, zero $(0, 0)$ and identity $(1, 1)$ , and the map $f (n) = (n, 0)$ ( $Z$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ).

[L1]

$Z$ is a commutative ring ( $Z$ 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, The integers as equivalence classes of pairs of naturals, Arithmetic on the integers).

[L2]

$Z \times Z$ is a ring whose operations are componentwise, whose zero is $(0, 0)$ and whose identity is $(1, 1)$ ; 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 $R^{\times} \times S^{\times}$ ).

[L3]

$0 \cdot x = x \cdot 0 = 0$ 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 $a (b - c) = a b - a c$ ).

[L4]

$1 \neq 0$ in $Z$ , since $1 = ι (1)$ , $0 = ι (0)$ , $ι$ is injective, and $1 = σ (0) \neq 0$ in $N$ by Peano axiom (P1) (The naturals embed in the integers, Arithmetic on the integers, The von Neumann naturals form a Peano system).

[L5]

A ring homomorphism must satisfy (RH1) additivity, (RH2) multiplicativity and (RH3) $f (1_{R}) = 1_{S}$ ; (RH1) alone makes $f$ a homomorphism of the additive groups (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ , Monoid homomorphism and group homomorphism).

[L6]

The refuted claim: (RH1) and (RH2) imply (RH3).

Counterexample

technique · direct

1.1

$f$ is additive: $f (m + n) = (m + n, 0) = (m, 0) + (n, 0) = f (m) + f (n)$ , the middle equality being componentwise addition with $0 + 0 = 0$ . So $f$ satisfies (RH1) and is a homomorphism of the additive groups.

L1L2L5

1.2

$f$ is multiplicative: $f (m n) = (m n, 0)$ and $f (m) f (n) = (m, 0) (n, 0) = (m n, 0 \cdot 0) = (m n, 0)$ , using componentwise multiplication and $0 \cdot 0 = 0$ . So $f$ satisfies (RH2).

L2L3

1.3

$f (1) = (1, 0) \neq (1, 1)$ , since the second components differ: $0 \neq 1$ in $Z$ by [L4]. So (RH3) fails for $f$ .

L2L4

2.1

By steps 1.1, 1.2 and 1.3 the map $f$ satisfies (RH1) and (RH2) and fails (RH3), so it is not a ring homomorphism and the claim of [L6] is false: clause (RH3) of Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ is not redundant.

step 1.1step 1.2step 1.3L5L6∎

Remarks

The contrast with groups is the point. By (RH1) alone the map $f$ is a homomorphism of the additive groups, and for groups preservation of the identity is automatic (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed); indeed $f (0) = (0, 0)$ here. The multiplicative structures are only monoids, and for monoids the analogous statement is false, which is exactly why Monoid homomorphism and group homomorphism imposes $f (e) = e^{'}$ on monoid homomorphisms and Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ imposes (RH3).
The image is not a subring. $f (Z) = Z \times {0}$ is closed under subtraction and multiplication and does not contain $(1, 1) = 1_{Z \times Z}$ , so it fails clause (T1) of Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication — the same clause that $2 Z$ is closed under addition, negation and multiplication and is not a subring of $Z$ , because it does not contain $1$ fails, for the same reason, namely that closure alone never supplies the ambient identity. It follows that A ring homomorphism satisfies $f (0) = 0$ , $f (- a) = - f (a)$ and $f (m a) = m f (a)$ for $m \in Z$ , carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms, whose claim 4 puts a subring as the image of a ring homomorphism, really does use (RH3).
$(1, 0)$ is an identity for the image, but not the identity of the ambient ring. For $(m, 0)$ in the image, $(1, 0) (m, 0) = (m, 0)$ . So the failure is not that the image has no identity; it is that its identity is not $1_{Z \times Z}$ , and Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication asks for the ambient identity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

44 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

Ring homomorphism (Wikipedia) (standard reference, not scraped)
Product of rings (Wikipedia) (standard reference, not scraped)

The map n↦(n,0)n↦(n,0) from ZZ to Z×ZZ×Z preserves addition and multiplication and does not preserve 11, so the clause f(1)=1f(1)=1 is not redundant