LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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.

Depends on

Used by

Dependency tree · two levels

53 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

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​