LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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.

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 $a, b \in R$

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). For $m \in Z$ and $a \in R$ write $m a$ for the integer power of $a$ in the additive group $(R, +, 0)$ , read in additive notation as in Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ : thus $0 a = 0$ , $σ (n) a = n a + a$ for $n \in N$ , and $m a = - (k a)$ when $m < 0$ and $- m$ is the image of $k \in N$ under the embedding of The naturals embed in the integers. Throughout, a natural number written where an integer is expected means its image under that embedding.

Then for all $m, n \in Z$ and all $a, b \in R$ :

$(m + n) a = m a + n a$ ;
$m (a + b) = m a + m b$ ;
$(m a) b = m (a b) = a (m b)$ ;
$(m a) (n b) = (m n) (a b)$ ;
$m a = (m 1) a = a (m 1)$ , where $1$ is the identity of $R$ .

No commutativity of $R$ is assumed. The symbol $m a$ is not a product in $R$ : an integer is not an element of $R$ , and claim 5 is the precise statement that the multiple $m a$ is nevertheless a product in $R$ , namely the product of $a$ with the ring element $m 1$ .

Facts & Assumptions

Given: A ring $R$ with zero $0$ and identity $1$ , elements $a, b \in R$ , integers $m, n \in Z$ , and multiples $m a$ as described in the Statement (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L1]

$(R, +, 0)$ is an abelian group, $(R, \cdot, 1)$ is a monoid, and both distributive laws hold (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group).

[L2]

The defining recursion for multiples: $0 a = 0$ and $σ (n) a = n a + a$ for $n \in N$ ; and $m a = - (k a)$ when $m < 0$ and $- m = ι (k)$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L3]

Exponent laws in a group, read additively in $(R, +, 0)$ : $(x + y) a = x a + y a$ ; $(- x) a = - (x a)$ ; $y (x a) = (x y) a$ ; and $x (u + v) = x u + x v$ whenever $u$ and $v$ commute, all for $x, y \in Z$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L4]

Ring arithmetic: $0 \cdot x = x \cdot 0 = 0$ and $(- x) y = - (x y) = x (- y)$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ).

[L5]

Induction on $N$ (The principle of mathematical induction), and $N$ contains $0$ (The natural numbers $N$ (von Neumann)).

[L6]

$ι : N \to Z$ is injective, preserves addition, multiplication and order, and its image is exactly the nonnegative integers (The naturals embed in the integers, The integers as equivalence classes of pairs of naturals).

[L7]

$Z$ is a totally ordered commutative ring: multiplication on $Z$ is commutative, exactly one of $0 \leq x$ and $x < 0$ holds, and $x < 0$ implies $0 < - x$ (The integers form a commutative ring, The integers form a totally ordered ring, Order on the integers, Arithmetic on the integers).

Proof

technique · induction

1.1

Claim 1 is the first exponent law of [L3] read additively in the group $(R, +, 0)$ : $(m + n) a = m a + n a$ .

L1L3

1.2

Claim 2: addition in $R$ is commutative, so any two elements of $(R, +, 0)$ commute, and the last law of [L3] applies with $u = a$ , $v = b$ to give $m (a + b) = m a + m b$ .

L1L3

1.3

Base of claim 3 at the exponent $0$ : $(0 a) b = 0 \cdot b = 0$ , $0 (a b) = 0$ and $a (0 b) = a \cdot 0 = 0$ , so all three agree.

baseL2L4

1.4

Inductive hypothesis for claim 3: fix $n \in N$ and assume $(n a) b = n (a b)$ and $a (n b) = n (a b)$ for all $a, b \in R$ .

2.1

Successor step: $(σ (n) a) b = (n a + a) b = (n a) b + a b = n (a b) + a b = σ (n) (a b)$ , by the recursion, the right distributive law and the hypothesis; and $a (σ (n) b) = a (n b + b) = a (n b) + a b = n (a b) + a b = σ (n) (a b)$ , by the recursion, the left distributive law and the hypothesis.

step 1.4L1L2

3.1

By induction, $(k a) b = k (a b) = a (k b)$ for every $k \in N$ and all $a, b \in R$ .

step 1.3step 2.1L5

4.1

Negative exponents. Let $m < 0$ , so $0 < - m$ and $- m = ι (k)$ for a unique $k \in N$ . Then $(m a) b = (- (k a)) b = - ((k a) b) = - (k (a b)) = ((- k) (a b)) = m (a b)$ , and likewise $a (m b) = a (- (k b)) = - (a (k b)) = - (k (a b)) = m (a b)$ , using the second clause of the recursion, the sign rules of [L4], step 3.1, and $(- x) c = - (x c)$ from [L3].

step 3.1L2L3L4L6L7

5.1

Claim 3 in full: for $m \in Z$ either $0 \leq m$ , in which case $m = ι (k)$ for some $k \in N$ and step 3.1 applies, or $m < 0$ , in which case step 4.1 applies; exactly one of the two holds.

step 3.1step 4.1L6L7

6.1

Claim 4: applying claim 3 first with the pair $(a, n b)$ and then with the pair $(a, b)$ , $(m a) (n b) = m (a (n b)) = m (n (a b)) = (n m) (a b) = (m n) (a b)$ , the third equality being $y (x c) = (x y) c$ from [L3] with $x = n$ , $y = m$ , and the fourth commutativity of multiplication in $Z$ .

step 5.1L3L7

6.2

Claim 5: applying claim 3 with the pair $(1, a)$ gives $(m 1) a = m (1 \cdot a) = m a$ , and applying it with the pair $(a, 1)$ gives $a (m 1) = m (a \cdot 1) = m a$ , using the identity law of the multiplicative monoid.

step 5.1L1

7.1

Claims 1 to 5 are established: claim 1 in step 1.1, claim 2 in step 1.2, claim 3 in step 5.1, claim 4 in step 6.1 and claim 5 in step 6.2.

step 1.1step 1.2step 5.1step 6.1step 6.2discharge-induction∎

Remarks

Nothing here is a new recursion. The multiples $m a$ are the integer powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ applied to the additive group of $R$ and nothing else, so claims 1 and the law $n (m a) = (n m) a$ are quoted from Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute rather than reproved. What is genuinely new is claim 3, which is the only place the multiplicative structure enters, and it is exactly the point at which distributivity is used.
Why claim 5 is worth stating separately. It converts a statement about repeated addition into a statement about a single ring product, and that is what makes the characteristic of a ring (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) a statement about the element $m 1$ rather than about $a$ for every $a$ at once; 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 $1_{R}$ uses it in exactly that way.
The proof needs the order on $Z$ only to know that every integer is either the image of a natural number or the negative of one. That is the same case split Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ performs when it defines $g^{x}$ , and it is performed on the sign of the integer rather than on a representative of it.

Depends on

Used by

Dependency tree · two levels

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

Integer multiples in a ring: (m+n)a=ma+na(m+n)a=ma+na, m(a+b)=ma+mbm(a+b)=ma+mb, (ma)b=m(ab)=a(mb)(ma)b=m(ab)=a(mb) and (ma)(nb)=(mn)(ab)(ma)(nb)=(mn)(ab) for all m,n∈Zm,n∈Z and a,b∈Ra,b∈R