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

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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.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$2 Z$ is closed under addition, negation and multiplication and is not a subring of $Z$ , because it does not contain $1$

Statement refuted

False claim: if $S$ is a subset of a ring $R$ that contains $0_{R}$ and is closed under addition, under additive inverses and under multiplication, then $S$ is a subring of $R$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

The even integers refute it. Let $2 : = 1 + 1$ in $Z$ and

$2 Z : = {x \in Z : 2 ∣ x} = {2 k : k \in Z},$

divisibility being the relation of Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ . This set contains $0$ , is closed under addition, additive inverses and multiplication, and does not contain $1$ ; so it fails clause (T1) of Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication and is not a subring of $Z$ .

Facts & Assumptions

Given: The commutative ring $Z$ , the numeral $2 = 1 + 1$ , and the set $2 Z = {x \in Z : 2 ∣ x}$ ( $Z$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

$Z$ is a commutative ring with the operations of Arithmetic on the integers ( $Z$ 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).

[L2]

$d ∣ a$ means $a = d q$ for some $q \in Z$ , and $d ∣ 0$ for every $d$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L3]

Divisibility is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all $x, y \in Z$ ; and $d ∣ a$ implies $d ∣ a c$ and $d ∣ - a$ (Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ ).

[L4]

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

[L5]

The order on $Z$ is total and compatible with addition, and $ι : N \to Z$ is injective and order preserving with $ι (0) = 0$ , $ι (1) = 1$ (The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers, The integers as equivalence classes of pairs of naturals, Arithmetic on the integers).

[L6]

A subring must satisfy (T1) $1_{R} \in S$ , (T2) closure under addition, (T3) closure under additive inverses and (T4) closure under multiplication (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication); equivalently $1_{R} \in S$ together with $a - b \in S$ and $a b \in S$ (Subring criterion: $S \subseteq R$ is a subring if and only if $1_{R} \in S$ and $a - b \in S$ and $a b \in S$ for all $a, b \in S$ ; and an intersection of subrings is a subring).

[L7]

A subgroup of an abelian group is a subset containing the identity and closed under the operation and under inverses (Subgroup).

[L8]

The refuted claim: a subset of a ring containing $0$ and closed under addition, additive inverses and multiplication is a subring.

Counterexample

technique · direct

1.1

$0 \in 2 Z$ , since $2 ∣ 0$ by [L2].

1.2

$2 Z$ is closed under multiplication: if $2 ∣ a$ then $2 ∣ a b$ for every $b \in Z$ , by [L3].

1.3

$0 < 1 < 2$ and $- 1 < 0$ in $Z$ : $1 = ι (1)$ is nonnegative and differs from $0 = ι (0)$ because $ι$ is injective, so $0 < 1$ ; adding $1$ gives $1 < 2$ ; and adding $- 1$ to $0 < 1$ gives $- 1 < 0$ . Hence $2 \neq 1$ and $2 \neq - 1$ .

2.1

$2 Z$ is closed under addition and under additive inverses: if $2 ∣ a$ and $2 ∣ b$ then $2 ∣ a \cdot 1 + b \cdot 1 = a + b$ by the linearity of [L3], and $2 ∣ - a$ by [L3]. So $2 Z$ is a subgroup of $(Z, +, 0)$ in the sense of [L7].

step 1.1L1L3L7

2.2

$1 \notin 2 Z$ : if $2 ∣ 1$ then $2 \in {1, - 1}$ by [L4], contradicting step 1.3.

step 1.3L2L4

3.1

By steps 1.1, 2.1 and 1.2 the set $2 Z$ contains $0$ and is closed under addition, additive inverses and multiplication; by step 2.2 it does not contain $1 = 1_{Z}$ , so clause (T1) of [L6] fails and $2 Z$ is not a subring of $Z$ . The claim of [L8] is therefore false.

step 1.1step 2.1step 1.2step 2.2L6L8∎

Remarks

What $2 Z$ is, since it is not a subring. It is a subgroup of $(Z, +, 0)$ closed under multiplication, and with the restricted operations it satisfies every clause of Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides except the existence of a multiplicative identity. Such a structure is called a non-unital ring in this library, and it is not called a ring, by the convention fixed in Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides.
This is why the subring criterion tests $1_{R} \in S$ separately. Subring criterion: $S \subseteq R$ is a subring if and only if $1_{R} \in S$ and $a - b \in S$ and $a b \in S$ for all $a, b \in S$ ; and an intersection of subrings is a subring compresses the three additive and multiplicative closure conditions into " $a - b \in S$ and $a b \in S$ " but leaves $1_{R} \in S$ standing on its own, and the present witness is the reason: no amount of closure implies it.
$2 Z$ has no identity at all, not merely a different one. If $e \in 2 Z$ satisfied $e x = x$ for every $x \in 2 Z$ , then taking $x = 2$ gives $2 e = 2 = 2 \cdot 1$ , so $e = 1$ by multiplicative cancellation in $Z$ (The integers have no zero divisors; multiplicative cancellation), which is not in $2 Z$ by the argument above.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

49 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

Subring (Wikipedia) (standard reference, not scraped)
Rng (algebra) (Wikipedia) (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §16.3: Rings (standard reference, not scraped)

2Z2Z is closed under addition, negation and multiplication and is not a subring of ZZ, because it does not contain 11