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

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The characteristic of a ring is the additive order of $1_{R}$ , with $0$ recording infinite order; $n \cdot 1_{R} = 0$ holds exactly when $char (R) ∣ n$ ; and in an integral domain every nonzero element has the same additive order as $1_{R}$

Statement

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) and let $ord (1_{R})$ be the order of $1_{R}$ in the abelian group $(R, +, 0_{R})$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity), with $ord (1_{R}) = \infty$ when no $n \geq 1$ satisfies $n \cdot 1_{R} = 0_{R}$ . Throughout, a natural number written where an integer is expected means its image under the embedding $ι_{Z} : N \to Z$ of The naturals embed in the integers. Then:

if $ord (1_{R})$ is finite then $char (R) = ord (1_{R})$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise), and if $ord (1_{R}) = \infty$ then $char (R) = 0$ ; so the characteristic is the additive order of $1_{R}$ , with the value $0$ recording infinite order;
for every $n \in Z$ , the equation $n \cdot 1_{R} = 0_{R}$ holds if and only if $char (R)$ divides $n$ in $Z$ , divisibility being the relation of Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ;
if $R$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors) then for every $a \in R$ with $a \neq 0_{R}$ and every $n \in Z$ , $n a = 0_{R}$ if and only if $n \cdot 1_{R} = 0_{R}$ ; consequently every nonzero element of $R$ has the same additive order as $1_{R}$ .

Facts & Assumptions

Given: A ring $R$ with zero $0_{R}$ and identity $1_{R}$ ; multiples $m a$ for $m \in Z$ , $a \in R$ , as in Integer multiples in a ring: $(m + n) a = m a + n a$ , $m (a + b) = m a + m b$ , $(m a) b = m (a b) = a (m b)$ and $(m a) (n b) = (m n) (a b)$ for all $m, n \in Z$ and $a, b \in R$ ; and $char (R)$ as in The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise.

[L1]

$char (R)$ is the least $n \in N$ with $n \geq 1$ and $n \cdot 1_{R} = 0_{R}$ when such an $n$ exists, and is $0$ otherwise (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise).

[L2]

$ord (g)$ , for $g$ in a group, is the least $k \in N$ with $k \geq 1$ and $g^{k} = e$ when such a $k$ exists, and is $\infty$ otherwise; read additively in $(R, +, 0_{R})$ this is the least $k \geq 1$ with $k \cdot 1_{R} = 0_{R}$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Group and abelian group).

[L3]

If $ord (g) = n$ with $n \geq 1$ , then for $k \in Z$ we have $g^{k} = e$ if and only if $n ∣ k$ ; and if $ord (g) = \infty$ then $g^{j} = g^{k}$ implies $j = k$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L4]

$(m a) b = m (a b) = a (m b)$ and $m a = (m 1_{R}) a$ for all $m \in Z$ , $a, b \in R$ (Integer multiples in a ring: $(m + n) a = m a + n a$ , $m (a + b) = m a + m b$ , $(m a) b = m (a b) = a (m b)$ and $(m a) (n b) = (m n) (a b)$ for all $m, n \in Z$ and $a, b \in R$ ).

[L5]

$0_{R} \cdot b = 0_{R} = b \cdot 0_{R}$ for every $b \in R$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ).

[L6]

In an integral domain, $x y = 0$ implies $x = 0$ or $y = 0$ (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors, Commutative ring).

[L7]

$Z$ is a commutative ring, so $0 \cdot q = 0$ for every integer $q$ ; and $d ∣ a$ means $a = d q$ for some $q \in Z$ (The integers form a commutative 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 $a (b - c) = a b - a c$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). The embedding $ι_{Z}$ is injective and preserves addition, multiplication and order, its image being the nonnegative integers (The naturals embed in the integers).

Proof

technique · direct

1.1

Claim 1. Read additively in the abelian group $(R, +, 0_{R})$ , the set $S_{1_{R}}$ of [L2] is ${k \in N : k \geq 1, k \cdot 1_{R} = 0_{R}}$ , which is the set $T_{R}$ of [L1]. So the two definitions take the minimum of the same set: if that set is nonempty both $ord (1_{R})$ and $char (R)$ equal its least element, and if it is empty then $ord (1_{R}) = \infty$ while $char (R) = 0$ .

L1L2

1.2

Claim 3. Let $R$ be an integral domain, $a \neq 0_{R}$ and $n \in Z$ . By [L4], $n a = (n 1_{R}) a$ . If $n \cdot 1_{R} = 0_{R}$ then $n a = 0_{R} \cdot a = 0_{R}$ by [L5]. Conversely if $n a = 0_{R}$ then $(n 1_{R}) a = 0_{R}$ , so by [L6] either $n \cdot 1_{R} = 0_{R}$ or $a = 0_{R}$ ; the second is excluded, so $n \cdot 1_{R} = 0_{R}$ .

L4L5L6

2.1

Claim 2, the case $char (R) = c$ with $c \geq 1$ . By step 1.1, $ord (1_{R}) = c$ , so [L3] read additively in $(R, +, 0_{R})$ gives, for every $n \in Z$ : $n \cdot 1_{R} = 0_{R}$ if and only if $c ∣ n$ .

step 1.1L3L7

2.2

Claim 2, the case $char (R) = 0$ . By step 1.1, $ord (1_{R}) = \infty$ , so [L3] gives that $n \cdot 1_{R} = m \cdot 1_{R}$ forces $n = m$ ; taking $m = 0$ and using the additive reading $0 \cdot 1_{R} = 0_{R}$ from [L2], we get $n \cdot 1_{R} = 0_{R}$ if and only if $n = 0$ . On the other side, $0 ∣ n$ means $n = 0 \cdot q$ for some integer $q$ , and $0 \cdot q = 0$ in $Z$ , so $0 ∣ n$ holds exactly when $n = 0$ . The two conditions therefore agree.

step 1.1L2L3L7

3.1

Claim 2 follows from steps 2.1 and 2.2, since $char (R)$ is either $0$ or at least $1$ .

step 2.1step 2.2L1

4.1

Consequently, for $a \neq 0_{R}$ in an integral domain the set ${k \in N : k \geq 1, k a = 0_{R}}$ equals the set $T_{R}$ of step 1.1, so the two have the same least element when nonempty and are empty together: $ord (a) = ord (1_{R})$ , finite or infinite alike. With steps 1.1, 3.1 and 1.2 all three claims are established.

step 1.1step 3.1step 1.2L2∎

Remarks

The convention $char = 0$ is what makes claim 2 a single statement. With $\infty$ in place of $0$ the right-hand side would have to be split into two cases, since $\infty$ is not an integer and " $\infty ∣ n$ " has no meaning. With $0$ , the divisibility relation of Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ does the work in both cases, because $0 ∣ n$ holds exactly for $n = 0$ .
Claim 3 is where the ring structure enters. Claims 1 and 2 are statements about the additive group alone, and would be true in any abelian group with a distinguished element. Claim 3 uses $n a = (n 1_{R}) a$ , which is Integer multiples in a ring: $(m + n) a = m a + n a$ , $m (a + b) = m a + m b$ , $(m a) b = m (a b) = a (m b)$ and $(m a) (n b) = (m n) (a b)$ for all $m, n \in Z$ and $a, b \in R$ , and then the absence of zero divisors. Without the domain hypothesis the argument breaks at a named place: the step deducing $n \cdot 1_{R} = 0_{R}$ from $(n 1_{R}) a = 0_{R}$ is an appeal to the absence of zero divisors, and nothing on this page replaces it for a ring that has them.
$char (R)$ is a natural number and $n$ is an integer, so the divisibility in claim 2 is a statement in $Z$ about $ι_{Z} (char (R))$ and $n$ , as the Statement says. The two are not the same kind of object, and reading the claim without the embedding would be a category error rather than an abbreviation.

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Sources

Characteristic (algebra) (Wikipedia) (standard reference, not scraped)
Order (group theory) (Wikipedia) (standard reference, not scraped)
Integral domain (Wikipedia) (standard reference, not scraped)

The characteristic of a ring is the additive order of 1R1R​, with 00 recording infinite order; n⋅1R=0n⋅1R​=0 holds exactly when char⁡(R)∣nchar(R)∣n; and in an integral domain every nonzero element has the same additive order as 1R1R​