CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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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$2\mathbb{Z}$ is closed under addition, negation and multiplication and is not a subring of $\mathbb{Z}$ , 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 $\mathbb{Z}$ and

$2\mathbb{Z} \;:=\; \{\, x \in \mathbb{Z} \;:\; 2 \mid x \,\} \;=\; \{\, 2k : k \in \mathbb{Z} \,\},$

divisibility being the relation of Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ 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 $\mathbb{Z}$ .

Facts & Assumptions

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

[L1]

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

[L2]

$d \mid a$ means $a = dq$ for some $q \in \mathbb{Z}$ , and $d \mid 0$ for every $d$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L3]

Divisibility is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all $x, y \in \mathbb{Z}$ ; and $d \mid a$ implies $d \mid ac$ and $d \mid -a$ (Divisibility is reflexive and transitive on $\mathbb{Z}$ , and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$ ; also $d \mid a$ implies $d \mid ac$ , $-d \mid a$ and $d \mid -a$ ).

[L4]

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

[L5]

The order on $\mathbb{Z}$ is total and compatible with addition, and $\iota : \mathbb{N} \to \mathbb{Z}$ is injective and order preserving with $\iota(0) = 0$ , $\iota(1) = 1$ (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 $ab \in S$ (Subring criterion: $S \subseteq R$ is a subring if and only if $1_R \in S$ and $a - b \in S$ and $ab \in S$ 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

$0 \in 2\mathbb{Z}$ , since $2 \mid 0$ by [L2].

1.2

$2\mathbb{Z}$ is closed under multiplication: if $2 \mid a$ then $2 \mid ab$ for every $b \in \mathbb{Z}$ , by [L3].

1.3

$0 < 1 < 2$ 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 $1 < 2$ ; and adding $-1$ to $0 < 1$ gives $-1 < 0$ . Hence $2 \ne 1$ and $2 \ne -1$ .

2.1

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

step 1.1L1L3L7

2.2

$1 \notin 2\mathbb{Z}$ : if $2 \mid 1$ 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 $2\mathbb{Z}$ contains $0$ and is closed under addition, additive inverses and multiplication; by step 2.2 it does not contain $1 = 1_{\mathbb{Z}}$ , so clause (T1) of [L6] fails and $2\mathbb{Z}$ is not a subring of $\mathbb{Z}$ . The claim of [L8] is therefore false.

step 1.1step 2.1step 1.2step 2.2L6L8∎

Remarks

What $2\mathbb{Z}$ is, since it is not a subring. It is a subgroup of $(\mathbb{Z},+,0)$ 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 $ab \in S$ 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 $ab \in S$ " but leaves $1_R \in S$ standing on its own, and the present witness is the reason: no amount of closure implies it.
$2\mathbb{Z}$ has no identity at all, not merely a different one. If $e \in 2\mathbb{Z}$ satisfied $ex = x$ for every $x \in 2\mathbb{Z}$ , then taking $x = 2$ gives $2e = 2 = 2 \cdot 1$ , so $e = 1$ by multiplicative cancellation in $\mathbb{Z}$ (The integers have no zero divisors; multiplicative cancellation), which is not in $2\mathbb{Z}$ by the argument above.

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 71 results over 20 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

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)

2Z2\mathbb{Z}2Z is closed under addition, negation and multiplication and is not a subring of Z\mathbb{Z}Z, because it does not contain 111