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.

✓ 15 results · all verified · 8 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 7 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Rings, Subrings, Integral Domains and Fields

1 · Prerequisites

2 · Summary

Objective. A group has one operation. Every number system this library has built has two, and every one of them re-proved the same handful of facts about how the two interact: that multiplying by zero gives zero, that a product of negatives is positive, that cancellation works away from zero. This page isolates that interaction into axioms. A ring is an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides. The first two facts are then proved once, for every structure satisfying the axioms. The third is not: cancellation away from zero is equivalent to the absence of zero divisors, not a consequence of the ring axioms, and that equivalence is what the definition of an integral domain is built on.

The convention, stated once and kept. Here a ring has a multiplicative identity, because axiom (R2) of Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides asks $(R, \cdot, 1)$ to be a monoid. A structure satisfying the other axioms without an identity is called a non-unital ring where it occurs and is not called a ring. That decision has consequences the page carries openly: a subring must contain the identity of the ambient ring, and the companion page's $2 Z$ is closed under addition, negation and multiplication and is not a subring of $Z$ , because it does not contain $1$ is the witness that the requirement is not automatic. Nothing here requires $1 \neq 0$ of a ring; the definitions that need it — integral domain, division ring, field — say so, and the companion page records the one ring where the difference bites.

This page does not define a field. Field is already in the library and is the definition; what this page adds is the translation between it and the ring vocabulary, and the translation is four numbered items, never a remark. Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring shows a field is a commutative ring with $1 \neq 0$ , an integral domain and a division ring; Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree shows every commutative division ring satisfies the published axioms, so the two words name the same structures; Every ordered field is an ordered ring, and its order is the one its positive cone induces does the same for the order; and A ring homomorphism between fields is a field homomorphism in the published sense, and every such map is injective shows a ring homomorphism between fields is exactly a field homomorphism in the published sense, and is injective. Getting these wrong would leave the library with two notions wearing one name, which is the defect the four items exist to prevent.

The field axioms, where their quantifiers are load bearing. Axiom (M) of Field asserts associativity, commutativity and $x \cdot 1 = x$ on all of $F$ , the element $0$ included; its Remarks record the two-element counterexample showing that quantifier cannot be restricted to $F ∖ {0}$ . Two steps of Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring spend exactly that clause: that $(F, \cdot, 1)$ is a monoid, which is a statement about all of $F$ including $0$ , and right distributivity, which follows from the left form only by commuting a product one of whose factors may be $0$ . The axiom was amended into this explicit form in the commit that published this level, after the strict reading of its earlier wording was shown to admit a model falsifying Multiplication by zero: $0 \cdot a = 0$ .

Elementary consequences, and integer multiples. In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ proves $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ , each on both sides because multiplication is not assumed commutative, and records that a ring with $1 = 0$ has exactly one element. 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 $a, b \in R$ develops the notation $m a$ for $m \in Z$ : it is not a product in the ring but the integer power of $a$ in the additive group (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ), so the exponent laws 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, and what is genuinely new is $(m a) b = m (a b) = a (m b)$ , together with the identity $m a = (m 1) a$ that the characteristic later depends on. The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring names the units and obtains the group $R^{\times}$ from The invertible elements of a monoid form a group under the restricted operation applied to the multiplicative monoid, adding only the ring-specific fact that $0$ is a unit exactly in the zero ring.

Domains, division rings, and the quaternions. Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors fixes one convention for a zero divisor and keeps it: $0$ is never one, and $1 \neq 0$ is a hypothesis rather than a consequence. Cancellation characterises domains: in a commutative ring with $1 \neq 0$ , the implication $a b = a c$ and $a \neq 0$ imply $b = c$ holds if and only if the ring has no zero divisors proves that in a commutative ring with $1 \neq 0$ the cancellation law and the absence of zero divisors are equivalent, so a commutative ring with $1 \neq 0$ is a domain exactly when cancellation works there. Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit asks instead that every nonzero element be invertible, and the two conditions are genuinely different in both directions: $Z$ is a domain and not a division ring, and $H$ is a division ring and not a domain, since a domain is required to be commutative. The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ builds $H$ as $R^{4}$ with an explicit product formula in the eight real coordinates, and $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ proves it is a division ring with $q^{- 1} = \hat{N (q)^{- 1}} \overset{ˉ}{q}$ , that $i j = k$ while $j i = - k$ , and therefore that it is not a field. Associativity is the whole cost of that proof, and it is paid without any notion of linearity: the product is put in coefficient form, both bracketings expand into the same shape of finite real sum, the sixty four basis triples reduce to twenty seven by the identity law and to nine by a cyclic symmetry that is checked rather than asserted.

Order. Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication asks for a total order compatible with addition whose positive elements are closed under multiplication, in the strict form that the published The integers form a totally ordered ring proves and the published Ordered field requires. The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order then shows the order presentation and the positive-cone presentation determine each other, which is what keeps one notion of "ordered" in the library rather than two, and Every ordered field is an ordered ring, and its order is the one its positive cone induces applies it to fields.

Substructures and maps. Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication asks for the ambient identity and closure under addition, additive inverses and multiplication, and shows the subset is then a ring with the same zero and the same identity; 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 closure conditions to $a - b \in S$ and $a b \in S$ and proves intersections of subrings are subrings. Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations adds closure under inverses of nonzero elements, discharges the claim that a subfield is a field by way of the two bridge lemmas, and records that a subfield of an ordered field inherits the order. Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ imposes $f (1) = 1$ rather than deriving it, for the reason Monoid homomorphism and group homomorphism gives: the additive part is a group homomorphism, where identity preservation is free, and the multiplicative part is only a monoid homomorphism, where it is not. A ring homomorphism satisfies $f (0) = 0$ , $f (- a) = - f (a)$ and $f (m a) = m f (a)$ for $m \in Z$ , carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms collects what does follow.

Two constructions and the characteristic. The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ gives $R \times S$ componentwise, computes its units as $R^{\times} \times S^{\times}$ , and records that a product of two rings with $1 \neq 0$ always has zero divisors. The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations gives $R^{X}$ pointwise, and records that it has zero divisors as soon as $X$ has two distinct points and $R$ is not the zero ring; it is minted here because the same set, with the same pointwise addition, carries a second structure elsewhere in the library's plan, whose second operation multiplies a function by a scalar rather than two functions together. Finally In a field, the additive multiple $n \cdot 1_{F}$ is the canonical natural $ι (n)$ : the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion $ι (0) = 0_{F}$ , $ι (σ (n)) = ι (n) + 1_{F}$ proves that in a field the multiple $n \cdot 1_{F}$ is the canonical natural $ι (n)$ of The canonical natural $ι (n) = n \cdot 1_{F}$ of a field — proved, because assuming it is exactly the two-notions defect this page is built to avoid — and The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise defines $char (R)$ as the least $n \geq 1$ with $n \cdot 1_{R} = 0$ , or $0$ when there is none. That value $0$ is the opposite convention to the $\infty$ of 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, deliberately, and 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 $1_{R}$ shows why: with it, "the characteristic divides $n$ " is a single statement covering both cases, and the characteristic is exactly the additive order of $1_{R}$ , shared by every nonzero element when the ring is a domain.

What is deliberately absent. Ideals, quotient rings and the isomorphism theorems belong to a later page and are used nowhere here; the injectivity of a ring homomorphism between fields is proved without them. Polynomial rings are not constructed. The further property of the characteristic of an integral domain that would say more than "it is $0$ or at least $2$ " requires the notion of a prime number, which no definition on this page or among the items it cites introduces, so it is not stated here. Twenty-eight items are proved on this page, thirteen of them marked as landmarks in the flowchart above. Every instance lives on the companion page except one: the quaternions are built here, because companion pages are leaves in the reading order and later pages need them.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-28Open item page →

Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides

Definition

A ring is a set $R$ carrying two binary operations (Binary operation on a set; associativity, commutativity, and a subset closed under the operation), addition $+$ and multiplication $\cdot$ , and two distinguished elements $0$ and $1$ of $R$ , such that

(R1) $(R, +, 0)$ is an abelian group (Group and abelian group): addition is associative and commutative, $0 + x = x = x + 0$ for every $x \in R$ , and every $x \in R$ has an additive inverse;
(R2) $(R, \cdot, 1)$ is a monoid (Semigroup and monoid): multiplication is associative and $1 \cdot x = x = x \cdot 1$ for every $x \in R$ ;
(R3) multiplication distributes over addition on both sides: for all $x, y, z \in R$ ,

$x \cdot (y + z) = x \cdot y + x \cdot z, (y + z) \cdot x = y \cdot x + z \cdot x .$

We write $x y$ for $x \cdot y$ , call $0$ the zero and $1$ the identity (or unity) of $R$ , write $- x$ for the additive inverse of $x$ , and abbreviate $x - y : = x + (- y)$ .

Why "the" zero, "the" identity and "the" additive inverse are legitimate. Each of $0$ and $1$ is a two-sided identity for its own operation, and a binary operation has at most one two-sided identity (A left identity and a right identity for the same binary operation are equal; hence there is at most one two-sided identity, Left identity, right identity, and two-sided identity for a binary operation), so each is determined by the operation and the notation is single-valued. An additive inverse of $x$ is an inverse in the monoid $(R, +, 0)$ (Left inverse, right inverse, and invertible element of a monoid), and in a monoid an element has at most one inverse (In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided); so $- x$ names one element, and a single equation $y + x = 0$ already forces $y = - x$ .

Convention: a ring has an identity. Axiom (R2) asks for a monoid, so the existence of $1$ is part of the definition and is not an extra hypothesis to be carried around. This is the convention used throughout this library. A structure satisfying (R1), (R3) and the associativity half of (R2), but not required to have a multiplicative identity, is called a non-unital ring (also rng); it is not called a ring here, and where such a structure occurs it is named as a non-unital ring in as many words. The distinction has content: the companion page exhibits a subset of $Z$ that is closed under addition, additive inverses and multiplication and is not a subring, precisely because it misses the identity.

Nothing above requires $1 \neq 0$ . A ring in which $1 = 0$ has exactly one element; the companion page records it as the zero ring. Definitions that need $1 \neq 0$ — integral domain, division ring, field — say so explicitly.

Remarks

Both distributive laws are stated, and neither is redundant here. Multiplication is not assumed commutative, so $(y + z) x = y x + z x$ does not follow from $x (y + z) = x y + x z$ . The published Field states only the left form, in a setting where multiplication is commutative; a ring is not that setting. The quaternions (The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ), constructed on this page, are a ring whose multiplication is not commutative, so for them the two laws are genuinely two statements and both are checked.
A ring packages two structures the library already has. Read (R1) alone and a ring is an abelian group; read (R2) alone and it is a monoid. Every theorem proved about groups and about monoids therefore applies to a ring without restatement, and this page uses that repeatedly rather than reproving cancellation, uniqueness of inverses, or the behaviour of finite products.
Notation for repeated addition. For $m \in Z$ and $a \in R$ the expression $m a$ means the integer power of $a$ in the additive group $(R, +, 0)$ , written additively as in Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ . It is not a product formed with the ring multiplication: $m$ is an external integer scalar, even when the underlying set of $R$ happens to contain integers. The arithmetic of these multiples 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 $a, b \in R$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-07-28Open item page →

Commutative ring

Definition

A ring $R$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) is commutative when its multiplication is commutative (Binary operation on a set; associativity, commutativity, and a subset closed under the operation):

$x y = y x for all x, y \in R .$

Addition is commutative in every ring, by axiom (R1), so the word refers to multiplication alone.

Remarks

One of the two distributive laws becomes redundant. In a commutative ring the right law $(y + z) x = y x + z x$ follows from the left one, since $(y + z) x = x (y + z) = x y + x z = y x + z x$ . Both are still stated in Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, because a ring is not assumed commutative there.
Every field is a commutative ring (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring), and $Z$ , $Q$ and $R$ are commutative rings; the companion page records each of those as an instance. The quaternions are a ring that is not commutative (The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ , $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ).

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

In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$

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 $a, b, c \in R$ . Then

$0 \cdot a = 0$ and $a \cdot 0 = 0$ ;
$(- a) b = - (a b)$ and $a (- b) = - (a b)$ ;
$(- a) (- b) = a b$ ;
$(- 1) a = - a$ and $a (- 1) = - a$ ;
$a (b - c) = a b - a c$ and $(b - c) a = b a - c a$ .

In particular, if $1 = 0$ in $R$ then $R = {0}$ .

No commutativity is assumed, which is why each claim is stated on both sides.

Facts & Assumptions

Given: A ring $R$ with zero $0$ , identity $1$ , addition $+$ and multiplication $\cdot$ , and elements $a, b, c \in R$ ; $x - y$ abbreviates $x + (- y)$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L1]

$(R, +, 0)$ is an abelian group: addition is associative and commutative, $0 + x = x = x + 0$ , and each $x$ has an additive inverse $- x$ with $x + (- x) = 0 = (- x) + x$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group, Left identity, right identity, and two-sided identity for a binary operation).

[L2]

Both distributive laws hold: $x (y + z) = x y + x z$ and $(y + z) x = y x + z x$ for all $x, y, z \in R$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L3]

$1 \cdot x = x = x \cdot 1$ for every $x \in R$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L4]

Cancellation in the additive group: $x + z = y + z$ implies $x = y$ , and $z + x = z + y$ implies $x = y$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution, Group and abelian group).

[L5]

In a group, $(g^{- 1})^{- 1} = g$ ; written additively, $- (- x) = x$ (In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential, Group and abelian group).

Proof

technique · direct

1.1

$0 \cdot a = (0 + 0) \cdot a = 0 \cdot a + 0 \cdot a$ , using $0 + 0 = 0$ and the right distributive law.

L1L2

1.2

$a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$ , using $0 + 0 = 0$ and the left distributive law.

L1L2

1.3

$(- a) b + a b = ((- a) + a) b = 0 \cdot b$ , by the right distributive law and $(- a) + a = 0$ .

L1L2

1.4

$a (- b) + a b = a ((- b) + b) = a \cdot 0$ , by the left distributive law and $(- b) + b = 0$ .

L1L2

2.1

$0 \cdot a = 0$ : step 1.1 gives $0 + 0 \cdot a = 0 \cdot a = 0 \cdot a + 0 \cdot a$ , and cancelling $0 \cdot a$ on the right gives $0 = 0 \cdot a$ .

step 1.1L1L4

2.2

$a \cdot 0 = 0$ : step 1.2 gives $0 + a \cdot 0 = a \cdot 0 = a \cdot 0 + a \cdot 0$ , and cancelling $a \cdot 0$ on the right gives $0 = a \cdot 0$ . This proves claim 1.

step 1.2L1L4

3.1

$(- a) b = - (a b)$ : by step 1.3 and step 2.1, $(- a) b + a b = 0 \cdot b = 0 = - (a b) + a b$ , and cancelling $a b$ on the right gives $(- a) b = - (a b)$ .

step 1.3step 2.1L1L4

3.2

$a (- b) = - (a b)$ : by step 1.4 and step 2.2, $a (- b) + a b = a \cdot 0 = 0 = - (a b) + a b$ , and cancelling $a b$ on the right gives $a (- b) = - (a b)$ . This proves claim 2.

step 1.4step 2.2L1L4

4.1

$(- a) (- b) = - (a (- b)) = - (- (a b)) = a b$ : the first equality is step 3.1 applied with $b$ replaced by $- b$ , the second is step 3.2, and the third is $- (- x) = x$ . This proves claim 3.

step 3.1step 3.2L5

4.2

$(- 1) a = - (1 \cdot a) = - a$ and $a (- 1) = - (a \cdot 1) = - a$ , by step 3.1 and step 3.2 applied with $a$ or $b$ equal to $1$ , together with the identity law. This proves claim 4.

step 3.1step 3.2L3

4.3

$a (b - c) = a (b + (- c)) = a b + a (- c) = a b + (- (a c)) = a b - a c$ , and $(b - c) a = (b + (- c)) a = b a + (- c) a = b a + (- (c a)) = b a - c a$ , using the two distributive laws and step 3.1 and step 3.2. This proves claim 5.

step 3.1step 3.2L2

5.1

If $1 = 0$ then for every $x \in R$ we have $x = 1 \cdot x = 0 \cdot x = 0$ by the identity law and step 2.1, so $R = {0}$ .

step 2.1L3∎

Remarks

These are the general statements; the field versions already in the library are instances of them. For a field $F$ , claim 1 is Multiplication by zero: $0 \cdot a = 0$ and claims 2 and 3 are Sign rules for products: $(- a) b = - (a b)$ and $(- a) (- b) = a b$ , both proved earlier from the field axioms alone. Once Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring identifies a field as a commutative ring, those two lemmas are exactly the present ones read in $F$ . The direction of generality matters for citation: a proof about an arbitrary ring must use this lemma, never the field lemmas, since a ring need not be a field.
Claim 1 is where the additive group is doing the work. The identity $0 \cdot a = 0$ is not an axiom and does not follow from the multiplicative structure; it follows from distributivity together with the fact that addition cancels, which is available because $(R, +, 0)$ is a group and not merely a monoid.
The last sentence of the statement is why the zero ring exists at all. A ring with $1 = 0$ collapses to one element, so requiring $1 \neq 0$ , as Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors and Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit do, is exactly the requirement that the ring is not that one-element ring.

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

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.

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

The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring

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). An element $u \in R$ is a unit of $R$ when it is invertible in the multiplicative monoid $(R, \cdot, 1)$ (Left inverse, right inverse, and invertible element of a monoid), that is, when there is $v \in R$ with $u v = 1 = v u$ . Write $R^{\times}$ for the set of units. Then:

a unit has exactly one inverse, written $u^{- 1}$ , and a single equation $v u = 1$ or $u v = 1$ with $u$ already known to be a unit forces $v = u^{- 1}$ ;
$R^{\times}$ contains $1$ , is closed under multiplication and under inversion, and $(R^{\times}, \cdot, 1)$ is a group (Group and abelian group), the group of units of $R$ ;
$0 \in R^{\times}$ if and only if $1 = 0$ , that is, if and only if $R = {0}$ .

Facts & Assumptions

Given: A ring $R$ with zero $0$ and identity $1$ , and $R^{\times} = {u \in R : u v = 1 = v u for some v \in R}$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Left inverse, right inverse, and invertible element of a monoid).

[L1]

$(R, \cdot, 1)$ is a monoid: multiplication is associative and $1$ is a two-sided identity for it (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Semigroup and monoid).

[L2]

In a monoid a left inverse and a right inverse of the same element are equal; so an invertible element has exactly one two-sided inverse, and one of the two equations already determines it (In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided).

[L3]

The invertible elements of a monoid $M$ contain the identity, are closed under the operation and under inversion, and form a group under the restricted operation (The invertible elements of a monoid form a group under the restricted operation, Group and abelian group).

[L4]

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

Proof

technique · direct

1.1

By [L1] the pair $(R, \cdot, 1)$ is a monoid, so "unit of $R$ " as defined above is exactly "invertible element of the monoid $(R, \cdot, 1)$ ", and $R^{\times}$ is the set of units of that monoid in the sense of Left inverse, right inverse, and invertible element of a monoid.

1.2

Claim 1 is [L2] applied to the monoid $(R, \cdot, 1)$ .

L1L2

1.3

Claim 2 is [L3] applied to the same monoid: $1 \in R^{\times}$ because $1 \cdot 1 = 1$ , the set is closed under multiplication and under inversion, and $(R^{\times}, \cdot, 1)$ is a group.

L1L3

1.4

Conversely, if $1 = 0$ then $0 \cdot 0 = 0 = 1$ , so $0$ is its own two-sided inverse and $0 \in R^{\times}$ ; and $R = {0}$ .

2.1

If $0 \in R^{\times}$ , choose $v \in R$ with $0 \cdot v = 1$ . But $0 \cdot v = 0$ , so $1 = 0$ , and then $R = {0}$ .

step 1.1L4

3.1

Steps 2.1 and 1.4 give claim 3: $0 \in R^{\times}$ exactly when $1 = 0$ , exactly when $R$ is the one-element ring.

step 2.1step 1.4L4∎

Remarks

Nothing is reproved here. The group structure comes from The invertible elements of a monoid form a group under the restricted operation, which was proved for an arbitrary monoid; all this item does is name the monoid ( $(R, \cdot, 1)$ , which exists by axiom (R2) of Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) and record the one ring-specific fact, claim 3, which needs $0 \cdot v = 0$ and so is not a monoid statement.
Claim 3 is the reason $1 \neq 0$ appears as a hypothesis elsewhere. In the one-element ring every element, $0$ included, is a unit, so "every nonzero element is a unit" is vacuously true there. Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit therefore requires $1 \neq 0$ separately, and so does Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors.
$R^{\times}$ is written multiplicatively and is in general not all of $R ∖ {0}$ : in $Z$ it is ${1, - 1}$ , as the companion page records. The rings in which it is all of $R ∖ {0}$ are exactly the division rings.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors

Definition

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). An element $a \in R$ is a zero divisor when

$a \neq 0 and a b = 0 or b a = 0 for some b \in R with b \neq 0.$

The ring $R$ has no zero divisors when no element of $R$ is a zero divisor; equivalently, when $a b = 0$ implies $a = 0$ or $b = 0$ , for all $a, b \in R$ .

An integral domain, or simply a domain, is a commutative ring (Commutative ring) $R$ such that

(D1) $1 \neq 0$ in $R$ ;
(D2) $R$ has no zero divisors.

In a commutative ring the two clauses $a b = 0$ and $b a = 0$ of the definition of a zero divisor say the same thing, so there the notion is one-sided without ambiguity.

Two conventions fixed here, and kept. First, $0$ is not a zero divisor, because the clause $a \neq 0$ is part of the definition. This matters: $0 \cdot b = 0$ holds for every $b$ in every 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 $a (b - c) = a b - a c$ ), so without that clause $0$ would be a zero divisor in every ring with more than one element and the phrase "no zero divisors" would name the empty class. Second, (D1) is a hypothesis and not a consequence of (D2). The one-element ring, in which $1 = 0$ , has no zero divisors at all — vacuously, since it has no nonzero element — so (D2) alone would admit it. It is excluded by (D1), and by nothing else; the companion page records that ring explicitly.

Remarks

Where the definition is used. Cancellation is what a domain really buys: in a commutative ring with $1 \neq 0$ , the absence of zero divisors is equivalent to the cancellation law $a b = a c$ and $a \neq 0$ imply $b = c$ , which is Cancellation characterises domains: in a commutative ring with $1 \neq 0$ , the implication $a b = a c$ and $a \neq 0$ imply $b = c$ holds if and only if the ring has no zero divisors.
Commutativity is part of the definition of a domain, and is not part of the definition of a zero divisor. A non-commutative ring may perfectly well have no zero divisors — the quaternions ( $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ) are such a ring, since every nonzero element there is invertible — and it is still not called an integral domain here.
Every field has no zero divisors (A field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$ ), so every field is an integral domain once it is known to be a commutative ring; that is part of Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring. The converse fails, and $Z$ is the standard witness, recorded on the companion page.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-28Open item page →

Cancellation characterises domains: in a commutative ring with $1 \neq 0$ , the implication $a b = a c$ and $a \neq 0$ imply $b = c$ holds if and only if the ring has no zero divisors

Statement

Let $R$ be a commutative ring (Commutative ring) with $1 \neq 0$ . Consider the two conditions

(C) for all $a, b, c \in R$ : if $a b = a c$ and $a \neq 0$ then $b = c$ ;
(Z) $R$ has no zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

Then (C) holds if and only if (Z) holds; that is, (C) holds exactly when $R$ is an integral domain.

Facts & Assumptions

Given: A commutative ring $R$ with zero $0$ and identity $1$ , and $1 \neq 0$ (Commutative ring, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L1]

$(R, +, 0)$ is an abelian group and multiplication is commutative and distributes over addition (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring, Group and abelian group).

[L2]

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

[L3]

Cancellation in the additive group: $x + z = y + z$ implies $x = y$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution, Group and abelian group).

[L4]

$a$ is a zero divisor when $a \neq 0$ and $a b = 0$ for some $b \neq 0$ ; $R$ has no zero divisors exactly when $a b = 0$ implies $a = 0$ or $b = 0$ ; and an integral domain is exactly a commutative ring with $1 \neq 0$ and no zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

Proof

technique · direct

1.1

Assume (Z), and let $a b = a c$ with $a \neq 0$ . Then $a (b - c) = a b - a c = 0$ .

L2given

1.2

Assume (C), and let $a b = 0$ with $a \neq 0$ . Then $a b = 0 = a \cdot 0$ .

L2given

2.1

From step 1.1, (Z) gives $a = 0$ or $b - c = 0$ ; since $a \neq 0$ , we get $b + (- c) = 0$ . As also $c + (- c) = 0$ , cancelling $- c$ gives $b = c$ . So (Z) implies (C).

step 1.1L1L3L4

2.2

From step 1.2, (C) applied with $a$ , $b$ and $0$ gives $b = 0$ . So whenever $a b = 0$ and $a \neq 0$ we have $b = 0$ , which says exactly that $a b = 0$ implies $a = 0$ or $b = 0$ ; hence (Z). So (C) implies (Z).

step 1.2L4

3.1

By steps 2.1 and 2.2 the two conditions are equivalent, and (Z) together with commutativity and $1 \neq 0$ is the definition of an integral domain.

step 2.1step 2.2L4∎

Remarks

The hypothesis $1 \neq 0$ is used nowhere in the equivalence itself. It is carried in the statement only so that "(C) holds exactly when $R$ is an integral domain" is literally true, since (D1) of Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors is part of being a domain. In the one-element ring both (C) and (Z) hold vacuously and the ring is still not a domain.
Cancellation is by $a$ , not by $0$ . The clause $a \neq 0$ cannot be dropped: $0 \cdot b = 0 \cdot c$ holds for all $b$ and $c$ in every 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 $a (b - c) = a b - a c$ ), so cancelling $0$ would collapse the ring.
Multiplicative cancellation does not make the nonzero elements a group. It makes $(R ∖ {0}, \cdot, 1)$ a cancellative commutative monoid, and $Z$ shows that is strictly weaker than being a group: $2$ cancels and is not invertible. The rings where the nonzero elements do form a group are the fields (Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree).

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit

Definition

A division ring (also skew field) is a ring $D$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) such that

(V1) $1 \neq 0$ in $D$ ;
(V2) every $x \in D$ with $x \neq 0$ is a unit, that is, has a two-sided multiplicative inverse (Left inverse, right inverse, and invertible element of a monoid, The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

Equivalently, $D^{\times} = D ∖ {0}$ : by (V2) every nonzero element is a unit, and by (V1) together with The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring the element $0$ is not a unit, since $0$ is a unit only when $1 = 0$ . Consequently $(D ∖ {0}, \cdot, 1)$ is a group (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring); in particular $D ∖ {0}$ is closed under multiplication, so a division ring has no zero divisors.

A commutative division ring is a division ring whose multiplication is commutative. Those are exactly the fields (Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree, Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring).

Remarks

(V1) is not implied by (V2). In the one-element ring, where $1 = 0$ , there is no nonzero element at all, so (V2) holds vacuously; (V1) is what excludes it, exactly as in Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors. Without (V1) the one-element ring would count as a division ring and the sentence "a division ring has no zero divisors" would still be true but useless.
The definition is not vacuous beyond the commutative case. The quaternions $H$ are a division ring that is not commutative, hence not a field ( $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ); they are constructed on this page for that reason.
"Has no zero divisors" is genuinely weaker. $Z$ has no zero divisors and is not a division ring, since $2$ is not a unit; the companion page records this. So (V2) is a real strengthening of the domain condition, and no argument on this page derives one from the other.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring

Statement

Let $F$ be a field (Field), with addition $+$ , multiplication $\cdot$ , and distinguished elements $0 \neq 1$ . Then

$(F, +, \cdot, 0, 1)$ is a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides), and it is commutative (Commutative ring), with $1 \neq 0$ ;
$F$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors);
$F$ is a division ring (Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit), and hence a commutative division ring.

The field structure is not changed by this: the ring operations are the field operations, and the ring's zero and identity are the field's $0$ and $1$ .

Facts & Assumptions

Given: A field $F$ with operations $+$ and $\cdot$ and distinguished elements $0 \neq 1$ , satisfying the axioms (A), (M) and (D) of Field.

[A1]

Axiom (M) of Field: multiplication is associative and commutative on all of $F$ , and $x \cdot 1 = x$ for every $x \in F$ , the element $0$ included; moreover $(F ∖ {0}, \cdot)$ is an abelian group with identity $1$ , so every $x \neq 0$ has a multiplicative inverse $x^{- 1}$ with $x \cdot x^{- 1} = 1$ .

[A2]

Axiom (A): $(F, +)$ is an abelian group with identity $0$ ; addition is associative and commutative, $x + 0 = x$ for all $x$ , and every $x$ has an additive inverse $- x$ with $x + (- x) = 0$ (Field, Group and abelian group).

[A3]

Axiom (D), left distributivity: $x (y + z) = x y + x z$ for all $x, y, z \in F$ (Field).

[A4]

$0 \neq 1$ (Field).

[L1]

A ring is an abelian group under addition, a monoid under multiplication, and satisfies both distributive laws; it is commutative when its multiplication is (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring, Semigroup and monoid).

[L2]

In a field, $a b = 0$ implies $a = 0$ or $b = 0$ (A field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$ ).

[L3]

In a field the identities $0$ , $1$ and the inverses $- x$ , $x^{- 1}$ are unique, so the notation is single-valued (Identities and inverses in a field are unique, Left inverse, right inverse, and invertible element of a monoid).

Proof

technique · direct

1.1

$(F, +, 0)$ is an abelian group: this is axiom (A), and $0 + x = x + 0 = x$ follows from $x + 0 = x$ and commutativity of addition.

1.2

$(F, \cdot, 1)$ is a commutative monoid: multiplication is a binary operation on $F$ , it is associative and commutative on all of $F$ by axiom (M), and $x \cdot 1 = x$ for every $x \in F$ by the same axiom, whence $1 \cdot x = x \cdot 1 = x$ by commutativity.

A1L1

1.3

Right distributivity: for all $x, y, z \in F$ , $(y + z) x = x (y + z) = x y + x z = y x + z x$ , the first and third equalities being commutativity of multiplication at the pairs $(y + z, x)$ , $(x, y)$ and $(x, z)$ from axiom (M) as stated in [A1], and the middle one axiom (D).

A1A3

1.4

$F$ has no zero divisors: if $a b = 0$ then $a = 0$ or $b = 0$ by [L2], which is exactly the condition of Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors.

2.1

By steps 1.1, 1.2 and 1.3 together with axiom (D), $F$ satisfies (R1), (R2) and (R3) of Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, so $F$ is a ring; its multiplication is commutative by step 1.2, so it is a commutative ring; and $1 \neq 0$ by [A4]. This is claim 1.

step 1.1step 1.2step 1.3A3A4L1

3.1

Claim 2: by step 2.1 the ring $F$ is commutative with $1 \neq 0$ , and by step 1.4 it has no zero divisors, so it is an integral domain.

step 2.1step 1.4L2

3.2

Claim 3: $1 \neq 0$ by [A4]; and if $x \in F$ with $x \neq 0$ , axiom (M) supplies $x^{- 1} \in F$ with $x \cdot x^{- 1} = 1$ , and $x^{- 1} \cdot x = 1$ as well, by the commutativity of multiplication that (M) asserts. So $x$ is a unit of the ring $F$ , and $F$ is a division ring; it is commutative by step 2.1.

step 2.1A1A4L3

4.1

Claims 1, 2 and 3 are established in steps 2.1, 3.1 and 3.2.

step 2.1step 3.1step 3.2∎

Remarks

Where (M)'s unrestricted quantifier is load bearing. Two of the ring axioms are about all of $F$ : that $(F, \cdot, 1)$ is a monoid needs associativity and $x \cdot 1 = x$ at $x = 0$ as well, and right distributivity is obtained from the left form only by commuting a product one of whose factors may be $0$ . Axiom (M) of Field asserts associativity, commutativity and $x \cdot 1 = x$ on all of $F$ outright, which is what steps 1.2 and 1.3 spend; its Remarks record the two-element counterexample showing that the quantifier cannot be restricted to $F ∖ {0}$ .
Where else the same clause is in play. The published Multiplication by zero: $0 \cdot a = 0$ cites Field for the right distributive law $(y + z) x = y x + z x$ , which is licensed by (M)'s unrestricted commutativity together with axiom (D), exactly as step 1.3 above, and by nothing else in the axioms.
The converse direction is a separate item. Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree shows every commutative division ring satisfies the axioms of Field, so the two vocabularies name the same structures and this page never needs a second notion of field. Note that a commutative division ring satisfies the unrestricted clause of (M) outright, since its multiplication is commutative on all of $D$ and $1$ is a monoid identity by definition.
Nothing is reproved. The absence of zero divisors is quoted from A field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$ rather than rederived; the general ring statement In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ is not available as a substitute, because a ring may perfectly well have zero divisors.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-28 (claude-fable-5)Open item page →

Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree

Statement

Let $D$ be a commutative division ring (Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit, Commutative ring), with addition $+$ , multiplication $\cdot$ , zero $0$ and identity $1$ . Then $D$ , with the same operations and the same two distinguished elements, satisfies the axioms (A), (M) and (D) of Field; that is, $D$ is a field.

Together with Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring this says that "field" and "commutative division ring" name exactly the same structures, so the published definition of a field and the ring-theoretic description of one agree and no second notion of field is introduced on this page.

Facts & Assumptions

Given: A commutative division ring $D$ with zero $0$ , identity $1$ , $1 \neq 0$ , and $x^{- 1}$ the two-sided multiplicative inverse of each $x \neq 0$ (Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit, Commutative ring).

[L1]

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

[L2]

$1 \neq 0$ , and every $x \neq 0$ has a two-sided inverse $x^{- 1}$ ; equivalently $D^{\times} = D ∖ {0}$ (Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit).

[L3]

$D^{\times}$ contains $1$ , is closed under multiplication and under inversion, and is a group under the restricted multiplication; and $0 \in D^{\times}$ only when $1 = 0$ (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring, Left inverse, right inverse, and invertible element of a monoid, Group and abelian group).

[L4]

$x \cdot 0 = 0 = 0 \cdot x$ for every $x \in D$ (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]

The field axioms to be verified: (A) $(F, +)$ is an abelian group with identity $0$ ; (M) multiplication is associative and commutative on all of $F$ with $x \cdot 1 = x$ for every $x \in F$ , and $(F ∖ {0}, \cdot)$ is an abelian group with identity $1$ , each $x \neq 0$ having an inverse; (D) $x (y + z) = x y + x z$ ; and $0 \neq 1$ (Field).

Proof

technique · direct

1.1

Axiom (A) holds: $(D, +, 0)$ is an abelian group by [L1], which is precisely what (A) asserts.

L1L5

1.2

Axiom (D) holds: the left distributive law $x (y + z) = x y + x z$ is one of the two distributive laws of a ring.

L1L5

1.3

$0 \neq 1$ holds, by [L2].

L2L5

1.4

$D^{\times} = D ∖ {0}$ : every nonzero element is a unit by [L2]; and $0$ is not a unit, since $0 \cdot v = 0$ for every $v$ by [L4], so $0 \cdot v = 1$ would force $1 = 0$ , contradicting [L2].

L2L3L4

2.1

$D ∖ {0}$ is a group under the restricted multiplication, with identity $1$ : this is [L3] applied to $D^{\times}$ , which by step 1.4 is $D ∖ {0}$ . In particular $D ∖ {0}$ is closed under multiplication, so $D$ has no zero divisors.

step 1.4L3

2.2

That group is abelian, since multiplication is commutative on all of $D$ and therefore on the subset $D ∖ {0}$ .

step 1.4L1

3.1

Axiom (M) holds in both of its clauses: multiplication is associative and commutative on all of $D$ with $x \cdot 1 = x$ for every $x \in D$ , since $(D, \cdot, 1)$ is a commutative monoid by [L1]; and $(D ∖ {0}, \cdot)$ is an abelian group with identity $1$ by steps 2.1 and 2.2.

step 2.1step 2.2L1L5

4.1

By steps 1.1, 1.2, 1.3 and 3.1 the structure $(D, +, \cdot, 0, 1)$ satisfies (A), (M), (D) and $0 \neq 1$ , so it is a field.

step 1.1step 1.2step 1.3step 3.1L5∎

Remarks

Both directions are needed, and each is a numbered item. This lemma turns a commutative division ring into a field; Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring turns a field into a commutative division ring. Without the pair, the page would carry two unrelated words for one class of structures, which is exactly the defect the page exists to avoid.
Both clauses of axiom (M) are verified separately. (M) asserts associativity, commutativity and $x \cdot 1 = x$ on all of $F$ , and that $(F ∖ {0}, \cdot)$ is an abelian group. A commutative division ring supplies the first clause directly from its ring axioms, its multiplication being a commutative monoid operation on all of $D$ , and the second from steps 2.1 and 2.2; nothing about $0$ is left implicit.
Why step 1.4 is not a triviality. "Every nonzero element is a unit" does not by itself say that the units are exactly the nonzero elements: the extra content is that $0$ is not a unit, and that needs $0 \cdot v = 0$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ) together with $1 \neq 0$ . In the one-element ring, where $1 = 0$ , the units are all of the ring and axiom (M) of Field would fail for want of $0 \neq 1$ .

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$

Definition

Let $R$ be the field of real numbers (The real numbers, The reals form a field, Field) and let $4 = {0, 1, 2, 3}$ be the von Neumann natural number $4$ (The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ). The set of quaternions is the set of all functions $4 \to R$ ,

$H : = R^{4} = {x : 4 \to R},$

whose members are written as quadruples $x = (x_{0}, x_{1}, x_{2}, x_{3})$ with $x_{p} \in R$ for $p \in 4$ . Note that the coordinates are indexed from $0$ , because $4$ is the set ${0, 1, 2, 3}$ .

Addition is componentwise:

$x + y : = (x_{0} + y_{0}, x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3}) .$

Multiplication is given by the explicit formula

$x y : = (x_{0} y_{0} - x_{1} y_{1} - x_{2} y_{2} - x_{3} y_{3}, x_{0} y_{1} + x_{1} y_{0} + x_{2} y_{3} - x_{3} y_{2}, x_{0} y_{2} + x_{2} y_{0} + x_{3} y_{1} - x_{1} y_{3}, x_{0} y_{3} + x_{3} y_{0} + x_{1} y_{2} - x_{2} y_{1}) .$

The distinguished elements are

$0_{H} : = (0, 0, 0, 0), 1_{H} : = (1, 0, 0, 0) .$

Both rules are functions $H \times H \to H$ , since every coordinate on the right is a real number formed from real numbers by the field operations of $R$ ; so each is a binary operation on $H$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation), and no separate well-definedness question arises. That these data satisfy the ring axioms (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides), with $1_{H}$ a two-sided identity (Left identity, right identity, and two-sided identity for a binary operation), is proved in $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ and is not assumed here.

The basis quaternions and the table. Write

$e_{0} = 1 = (1, 0, 0, 0), e_{1} = i = (0, 1, 0, 0), e_{2} = j = (0, 0, 1, 0), e_{3} = k = (0, 0, 0, 1),$

and for a real $λ$ write $\hat{λ} : = (λ, 0, 0, 0)$ , so that $\hat{1} = 1$ and $\hat{0} = 0_{H}$ . Evaluating the multiplication formula on these sixteen pairs gives the table

$i^{2} = j^{2} = k^{2} = - 1, i j = k, j k = i, k i = j, j i = - k, k j = - i, i k = - j,$

together with $1 x = x 1 = x$ for $x \in {1, i, j, k}$ , where $- 1$ abbreviates $\hat{- 1} = (- 1, 0, 0, 0)$ and similarly for $- i$ , $- j$ , $- k$ . In the usual notation $x = x_{0} + x_{1} i + x_{2} j + x_{3} k$ , and the displayed product formula is precisely what the table forces once products are expanded and real coefficients are collected; but the formula, not the table, is the definition, so nothing is "extended by linearity" and no unproved notion is used.

Conjugate and norm. For $x \in H$ put

$\overset{ˉ}{x} : = (x_{0}, - x_{1}, - x_{2}, - x_{3}), N (x) : = x_{0}^{2} + x_{1}^{2} + x_{2}^{2} + x_{3}^{2} \in R .$

Both are defined by explicit real formulas; $\overset{ˉ}{x} \in H$ and $N (x) \in R$ , and $N$ is a function $H \to R$ , not a quaternion.

Remarks

Why $H$ is defined on this page rather than among the examples. It is a division ring that is not a field ( $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ), so it is the witness that Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit is strictly weaker than Field, and it is a ring whose multiplication is not commutative. Companion pages are leaves in the reading order, so no later page may cite an item homed on one; $H$ is placed here so that later pages can use it.
Indexing from $0$ is not a stylistic choice. A quaternion is a function on the von Neumann natural $4$ , which is the set ${0, 1, 2, 3}$ (On $N$ the order is membership: $m < n ⟺ m \in n$ ), so its coordinates are $x_{0}, x_{1}, x_{2}, x_{3}$ and the real part is $x_{0}$ . Every sum over basis indices below runs over $p \in 4$ , that is over $p = 0, 1, 2, 3$ .
The map $λ \mapsto \hat{λ}$ is how real scalars enter. There is no separate scalar multiplication in the definition: multiplying by a real number means multiplying by the quaternion $\hat{λ}$ , and the formula shows at once that $\hat{λ} x = x \hat{λ} = (λ x_{0}, λ x_{1}, λ x_{2}, λ x_{3})$ . This is used throughout the proof of $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ and is what keeps that proof free of any notion of linearity.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

$H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$

Statement

Let $H$ be the quaternions, with the addition, the multiplication, the elements $0_{H}$ and $1_{H}$ , the basis elements $e_{0} = 1, e_{1} = i, e_{2} = j, e_{3} = k$ , the real embedding $λ \mapsto \hat{λ}$ , the conjugate $\overset{ˉ}{x}$ and the norm $N (x)$ of The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ . Then:

$(H, +, \cdot, 0_{H}, 1_{H})$ is a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides);
it is not commutative (Commutative ring): $i j = k$ and $j i = - k$ , and $k \neq - k$ ;
$x \overset{ˉ}{x} = \overset{ˉ}{x} x = \hat{N (x)}$ for every $x \in H$ , and $N (x) > 0$ in $R$ whenever $x \neq 0_{H}$ ;
$H$ is a division ring (Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit): $1_{H} \neq 0_{H}$ and every $x \neq 0_{H}$ is a unit, with

$x^{- 1} = \hat{N (x)^{- 1}} \overset{ˉ}{x};$

consequently $H ∖ {0_{H}}$ is a group under multiplication;
$H$ is not a field (Field).

Facts & Assumptions

Given: The set $H$ of quadruples of real numbers with the operations, distinguished elements, basis elements $e_{p}$ for $p \in 4$ , real embedding $\hat{λ}$ , conjugate and norm of The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ; $[x]_{m}$ denotes the $m$ -th coordinate of $x \in H$ , for $m \in 4 = {0, 1, 2, 3}$ (The natural numbers $N$ (von Neumann)).

[L1]

$R$ is a field: $(R, +, 0)$ is an abelian group, multiplication is associative and commutative with identity $1$ , multiplication distributes over addition, $0 \neq 1$ , and every $t \neq 0$ has an inverse $t^{- 1}$ (The reals form a field, The real numbers, Field, Group and abelian group). Also $0 \cdot t = t \cdot 0 = 0$ for every real $t$ (Multiplication by zero: $0 \cdot a = 0$ ).

[L2]

$R$ is a totally ordered field with positive cone $P$ : exactly one of $t \in P$ , $t = 0$ , $- t \in P$ holds, and $P$ is closed under addition and multiplication (The reals form a totally ordered field, Ordered field).

[L3]

In an ordered field the square of a nonzero element is positive (Squares of nonzero elements are positive).

[L4]

Finite sums in the commutative monoid $(R, +, 0)$ are defined by The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity read additively, and their value is unchanged by regrouping the summands or permuting them (Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either, Semigroup and monoid, Binary operation on a set; associativity, commutativity, and a subset closed under the operation).

[L5]

A ring is an abelian group under addition, a monoid under multiplication, and satisfies both distributive laws; a division ring is a ring with $1 \neq 0$ in which every nonzero element is a unit; the units of a ring form a group (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit, Left inverse, right inverse, and invertible element of a monoid, The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

[L6]

Every field is a commutative ring (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring).

Proof

technique · direct

1.1

The additive group. Addition on $H$ is defined coordinatewise from addition on $R$ , so it is associative and commutative, $0_{H}$ is a two-sided identity, and $(- x_{0}, - x_{1}, - x_{2}, - x_{3})$ is a two-sided additive inverse of $x$ . Hence $(H, +, 0_{H})$ is an abelian group and $- x = (- x_{0}, - x_{1}, - x_{2}, - x_{3})$ .

L1L5given

1.2

Coefficient form of the product. For $m, p, q \in 4$ let $ε_{m} (p, q)$ be the real coefficient of the monomial $x_{p} y_{q}$ in the $m$ -th coordinate of the defining product formula, and $ε_{m} (p, q) = 0$ when that monomial does not occur there; every such coefficient is $1$ , $- 1$ or $0$ . Reading the four coordinates of the formula off one at a time, $[x y]_{m} = \sum_{p < 4} \sum_{q < 4} ε_{m} (p, q) x_{p} y_{q}$ for all $x, y \in H$ and all $m \in 4$ , the right-hand side being a sum of sixteen real numbers.

L1L4given

1.3

The cyclic symmetry of the table. Let $γ : 4 \to 4$ fix $0$ and send $1 \mapsto 2 \mapsto 3 \mapsto 1$ , and let $Γ : H \to H$ be the coordinate permutation determined by $[Γ x]_{γ (p)} = x_{p}$ , that is $Γ (x_{0}, x_{1}, x_{2}, x_{3}) = (x_{0}, x_{3}, x_{1}, x_{2})$ . Then $Γ$ is a bijection, $Γ (e_{p}) = e_{γ (p)}$ , $Γ$ fixes each $\hat{λ}$ , and $Γ (λ x) = λ Γ (x)$ , all immediately from the definition of $Γ$ .

L1given

1.4

Claim 2: from the table, $i j = k = (0, 0, 0, 1)$ and $j i = - k = (0, 0, 0, - 1)$ . By [L1], $1 \neq 0$ , so [L3] gives $1 = 1^{2} > 0$ ; the ordered-field definition in [L2] then gives $- 1 < 0$ , hence $1 \neq - 1$ . Thus the two products differ, and multiplication on $H$ is not commutative.

L1L2L3given

1.5

Claim 3, the norm identity. Evaluating the product formula at $y = \overset{ˉ}{x}$ gives coordinates $x_{0} x_{0} - x_{1} (- x_{1}) - x_{2} (- x_{2}) - x_{3} (- x_{3}) = N (x)$ , then $x_{0} (- x_{1}) + x_{1} x_{0} + x_{2} (- x_{3}) - x_{3} (- x_{2}) = 0$ , then $x_{0} (- x_{2}) + x_{2} x_{0} + x_{3} (- x_{1}) - x_{1} (- x_{3}) = 0$ , then $x_{0} (- x_{3}) + x_{3} x_{0} + x_{1} (- x_{2}) - x_{2} (- x_{1}) = 0$ ; so $x \overset{ˉ}{x} = \hat{N (x)}$ . Evaluating it at $x : = \overset{ˉ}{x}$ , $y : = x$ gives $x_{0} x_{0} - (- x_{1}) x_{1} - (- x_{2}) x_{2} - (- x_{3}) x_{3} = N (x)$ and, in the same way, $0$ in each of the other three coordinates; so $\overset{ˉ}{x} x = \hat{N (x)}$ as well.

L1given

1.6

Claim 3, positivity. Let $x \neq 0_{H}$ ; then $x_{p} \neq 0$ for at least one $p \in 4$ . Each $x_{p^{'}}^{2}$ with $x_{p^{'}} \neq 0$ is positive, and each $x_{p^{'}}^{2}$ with $x_{p^{'}} = 0$ equals $0$ ; a sum in which at least one summand is positive and the rest are positive or $0$ is positive, since $P$ is closed under addition and $u + 0 = u$ . Hence $N (x) > 0$ , and in particular $N (x) \neq 0$ .

L1L2L3

2.1

Real scalars pass through the product. For $λ \in R$ the formula gives $\hat{λ} x = x \hat{λ} = (λ x_{0}, λ x_{1}, λ x_{2}, λ x_{3})$ , an element we abbreviate $λ x$ ; in particular $\hat{λ} \hat{μ} = \hat{λ μ}$ , $1_{H} x = x 1_{H} = x$ , and $(- 1) x = - x$ by step 1.1.

step 1.1L1given

2.2

The coefficients are the multiplication table: $ε_{m} (p, q) = [e_{p} e_{q}]_{m}$ for all $m, p, q \in 4$ . Fix $p$ and $q$ and substitute $x = e_{p}$ , $y = e_{q}$ into step 1.2: then $x_{p^{'}} = 0$ for $p^{'} \neq p$ and $y_{q^{'}} = 0$ for $q^{'} \neq q$ , and a real product with a factor $0$ is $0$ , so every one of the sixteen summands vanishes except the one indexed by $(p, q)$ , which equals $ε_{m} (p, q) \cdot 1 \cdot 1 = ε_{m} (p, q)$ .

step 1.2L1L4

2.3

Both distributive laws hold. By step 1.2 and distributivity in $R$ , $[(x + x^{'}) y]_{m} = \sum_{p, q} ε_{m} (p, q) (x_{p} + x_{p}^{'}) y_{q} = \sum_{p, q} (ε_{m} (p, q) x_{p} y_{q} + ε_{m} (p, q) x_{p}^{'} y_{q}) = [x y]_{m} + [x^{'} y]_{m}$ , the last equality being a regrouping of a finite sum of thirty-two real terms; the same computation in the second argument gives $x (y + y^{'}) = x y + x y^{'}$ .

step 1.2L1L4

2.4

The nine table checks that establish $e_{γ (p)} e_{γ (q)} = Γ (e_{p} e_{q})$ for all $p, q \in {1, 2, 3}$ , read off the table of The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ : $j j = - 1 = Γ (- 1) = Γ (i i)$ ; $j k = i = Γ (k) = Γ (i j)$ ; $j i = - k = Γ (- j) = Γ (i k)$ ; $k j = - i = Γ (- k) = Γ (j i)$ ; $k k = - 1 = Γ (- 1) = Γ (j j)$ ; $k i = j = Γ (i) = Γ (j k)$ ; $i j = k = Γ (j) = Γ (k i)$ ; $i k = - j = Γ (- i) = Γ (k j)$ ; $i i = - 1 = Γ (- 1) = Γ (k k)$ . The cases with $p = 0$ or $q = 0$ are immediate, both sides being $e_{γ (q)}$ or $e_{γ (p)}$ respectively.

step 1.3given

2.5

Claim 5: every field is a commutative ring by [L6], and the multiplication of $H$ is not commutative by step 1.4; so $H$ is not a field.

step 1.4L6

3.1

Real scalars pass through a triple product too: by step 1.2 and step 2.1, $[(λ x) y]_{m} = \sum_{p, q} ε_{m} (p, q) (λ x_{p}) y_{q} = λ [x y]_{m}$ and likewise $[x (λ y)]_{m} = λ [x y]_{m}$ , so $(λ x) y = λ (x y) = x (λ y)$ . Taking $λ = - 1$ gives $(- x) y = - (x y) = x (- y)$ .

step 1.2step 2.1L1L4

3.2

Reduction of associativity to the sixty-four basis triples. Applying step 1.2 twice and rearranging, $[(x y) z]_{m} = \sum_{s, r} ε_{m} (s, r) [x y]_{s} z_{r} = \sum_{p, q, r} (\sum_{s} ε_{m} (s, r) ε_{s} (p, q)) x_{p} y_{q} z_{r}$ , and applying step 2.2 twice, $[(e_{p} e_{q}) e_{r}]_{m} = \sum_{s} ε_{m} (s, r) [e_{p} e_{q}]_{s} = \sum_{s} ε_{m} (s, r) ε_{s} (p, q)$ ; hence $[(x y) z]_{m} = \sum_{p, q, r < 4} x_{p} y_{q} z_{r} [(e_{p} e_{q}) e_{r}]_{m}$ . The same computation with the other bracketing gives $[x (y z)]_{m} = \sum_{p, q, r < 4} x_{p} y_{q} z_{r} [e_{p} (e_{q} e_{r})]_{m}$ . Therefore, if $(e_{p} e_{q}) e_{r} = e_{p} (e_{q} e_{r})$ holds for all $p, q, r \in 4$ , then $(x y) z = x (y z)$ for all $x, y, z \in H$ .

step 1.2step 2.2L1L4

3.3

Basis triples containing the index $0$ . Since $e_{0} = 1_{H}$ is a two-sided identity by step 2.1, each of $(e_{0} e_{q}) e_{r} = e_{q} e_{r} = e_{0} (e_{q} e_{r})$ , $(e_{p} e_{0}) e_{r} = e_{p} e_{r} = e_{p} (e_{0} e_{r})$ and $(e_{p} e_{q}) e_{0} = e_{p} e_{q} = e_{p} (e_{q} e_{0})$ holds. So only the twenty-seven triples with $p, q, r \in {1, 2, 3}$ remain.

step 2.1

3.4

The multiplication of $H$ commutes with $Γ$ : $Γ (x) Γ (y) = Γ (x y)$ for all $x, y$ . By step 1.2 and $[Γ x]_{p} = x_{γ^{- 1} (p)}$ , $[Γ (x) Γ (y)]_{γ (m)} = \sum_{p, q} ε_{γ (m)} (γ (p), γ (q)) x_{p} y_{q}$ , while $[Γ (x y)]_{γ (m)} = [x y]_{m} = \sum_{p, q} ε_{m} (p, q) x_{p} y_{q}$ ; by step 2.2 the two families of coefficients are $[e_{γ (p)} e_{γ (q)}]_{γ (m)}$ and $[e_{p} e_{q}]_{m} = [Γ (e_{p} e_{q})]_{γ (m)}$ , which agree by step 2.4.

step 1.2step 2.2step 1.3step 2.4L1L4

4.1

Reduction of the twenty-seven triples to nine. Suppose $(e_{p} e_{q}) e_{r} = e_{p} (e_{q} e_{r})$ for a triple $(p, q, r)$ . Applying step 3.4, $(e_{γ (p)} e_{γ (q)}) e_{γ (r)} = Γ (e_{p} e_{q}) Γ (e_{r}) = Γ ((e_{p} e_{q}) e_{r})$ and $e_{γ (p)} (e_{γ (q)} e_{γ (r)}) = Γ (e_{p}) Γ (e_{q} e_{r}) = Γ (e_{p} (e_{q} e_{r}))$ , so the identity holds for $(γ (p), γ (q), γ (r))$ as well. Since $γ$ restricted to ${1, 2, 3}$ is a cycle of length three, for each $p \in {1, 2, 3}$ there is exactly one $t \in {0, 1, 2}$ with $γ^{t} (p) = 1$ ; hence every triple in ${1, 2, 3}^{3}$ is obtained by iterating $γ$ from a triple whose first entry is $1$ , and it suffices to check the nine triples $(1, q, r)$ with $q, r \in {1, 2, 3}$ .

step 2.4step 3.4

4.2

The nine remaining checks, using the table of The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ and the sign rule of step 3.1: $(i i) i = (- 1) i = - i$ and $i (i i) = i (- 1) = - i$ ; $(i i) j = - j$ and $i (i j) = i k = - j$ ; $(i i) k = - k$ and $i (i k) = i (- j) = - (i j) = - k$ ; $(i j) i = k i = j$ and $i (j i) = i (- k) = - (i k) = j$ ; $(i j) j = k j = - i$ and $i (j j) = i (- 1) = - i$ ; $(i j) k = k k = - 1$ and $i (j k) = i i = - 1$ ; $(i k) i = (- j) i = - (j i) = k$ and $i (k i) = i j = k$ ; $(i k) j = (- j) j = - (j j) = 1$ and $i (k j) = i (- i) = - (i i) = 1$ ; $(i k) k = (- j) k = - (j k) = - i$ and $i (k k) = i (- 1) = - i$ . All nine agree.

step 3.1step 2.1given

5.1

Multiplication on $H$ is associative: by steps 3.3, 4.1 and 4.2 the identity $(e_{p} e_{q}) e_{r} = e_{p} (e_{q} e_{r})$ holds for all sixty-four basis triples, and step 3.2 transfers it to all of $H$ .

step 3.2step 3.3step 4.1step 4.2

6.1

Claim 1: by step 1.1 the additive structure is an abelian group; by step 5.1 and step 2.1 multiplication is associative with two-sided identity $1_{H}$ , so $(H, \cdot, 1_{H})$ is a monoid; and both distributive laws hold by step 2.3. So $H$ is a ring.

step 1.1step 2.1step 2.3step 5.1L5

7.1

Claim 4. First $1_{H} \neq 0_{H}$ , because $1 \neq 0$ in $R$ . Let $x \neq 0_{H}$ and put $λ : = N (x)^{- 1}$ , which exists by step 1.6, and $y : = λ \overset{ˉ}{x} = \hat{N (x)^{- 1}} \overset{ˉ}{x}$ . Then $x y = x (λ \overset{ˉ}{x}) = λ (x \overset{ˉ}{x}) = λ \hat{N (x)} = \hat{λ N (x)} = \hat{1} = 1_{H}$ by step 3.1, step 1.5 and step 2.1, and $y x = (λ \overset{ˉ}{x}) x = λ (\overset{ˉ}{x} x) = 1_{H}$ in the same way. So $x$ is a unit of the ring $H$ with $x^{- 1} = \hat{N (x)^{- 1}} \overset{ˉ}{x}$ , and $H$ is a division ring; by [L5] its units form a group, and by the description of a division ring that group is $H ∖ {0_{H}}$ .

step 2.1step 3.1step 6.1step 1.5step 1.6L1L5

8.1

Claims 1 to 5 are established: claim 1 in step 6.1, claim 2 in step 1.4, claim 3 in step 1.5 together with step 1.6, claim 4 in step 7.1 and claim 5 in step 2.5.

step 6.1step 1.4step 1.5step 1.6step 7.1step 2.5∎

Remarks

No notion of linearity is used, and none is available here. The reduction of associativity to basis triples is carried out entirely inside $R$ : the product is a fixed real formula, its coefficients are named, and the two bracketings are expanded into the same shape of finite sum, whose coefficients are then recognised as the coordinates of the corresponding basis products. The only tools are the field arithmetic of $R$ and the regrouping law for finite sums (Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either).
How the count of cases falls. Sixty-four basis triples; those in which one of the three indices is $0$ collapse by the identity law, leaving twenty-seven; the cyclic symmetry $i \mapsto j \mapsto k \mapsto i$ is a bijection commuting with multiplication, and it acts on the twenty-seven triples with every orbit of size three, so nine representatives suffice. The symmetry is checked, not asserted: it rests on nine equations of the table.
$H$ separates three notions this page keeps apart. It is a ring that is not commutative; it is a division ring that is not a field; and it has no zero divisors without being an integral domain, since a domain is required to be commutative (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).
The inverse formula $x^{- 1} = \hat{N (x)^{- 1}} \overset{ˉ}{x}$ is the exact analogue of $\overset{ˉ}{z} / ∣ z ∣^{2}$ for complex numbers, and the proof is the same computation; the only quaternionic subtlety is that $x \overset{ˉ}{x}$ and $\overset{ˉ}{x} x$ have to be computed separately, which the norm-identity step above does.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication

Definition

An ordered ring is a ring $R$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) together with a total order $\leq$ on $R$ (Partial order and partially ordered set) such that, for all $a, b, c \in R$ :

(OR1) if $a \leq b$ then $a + c \leq b + c$ ;
(OR2) if $0 < a$ and $0 < b$ then $0 < a b$ .

As always $a < b$ means $a \leq b$ and $a \neq b$ . The positive cone of an ordered ring is

$P : = {x \in R : 0 < x},$

and its members are the positive elements. An element $x$ with $x < 0$ is negative. An ordered ring whose multiplication is commutative (Commutative ring) is an ordered commutative ring; nothing in the definition requires commutativity.

Convention, and the one place it differs from another in use. (OR2) is stated in the strict form: a product of two positive elements is positive. Some texts, and the Wikipedia article named below, instead require $0 \leq a$ and $0 \leq b$ to imply $0 \leq a b$ . Given (OR1) and a total order the two are not equivalent: the strict form is the non-strict one together with the extra requirement that a product of two positive elements is nonzero. The strict form is the one adopted here, for a reason internal to this library: it is verbatim what the published The integers form a totally ordered ring proves of $Z$ (" $0 < x$ and $0 < y$ imply $0 < x y$ ") and verbatim what axiom (O2) of the published Ordered field requires of a positive cone, so with this convention both of those become instances of the present definition with nothing to adjust.

Remarks

Two presentations, one notion. An ordered ring may be presented by its order, as above, or by its positive cone, as Ordered field presents an ordered field. The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order proves that the two presentations determine each other, so no second notion is introduced by the difference in style, and Every ordered field is an ordered ring, and its order is the one its positive cone induces is the bridge for fields.
Both published order structures on this library's number systems are instances. $Z$ with the order of Order on the integers is one, by The integers form a totally ordered ring; every ordered field is another, by Every ordered field is an ordered ring, and its order is the one its positive cone induces. The companion page records $Z$ , $Q$ and $R$ explicitly.
The order is data. Writing " $R$ is an ordered ring" always means that a particular $\leq$ has been fixed alongside the ring operations, exactly as " $F$ is an ordered field" fixes a particular positive cone in Ordered field. Nothing below depends on there being only one such order, and no uniqueness is claimed.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order

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). Call a subset $P \subseteq R$ a cone when

(C1) trichotomy: for each $x \in R$ exactly one of $x \in P$ , $x = 0$ , $- x \in P$ holds;
(C2) closure: if $x, y \in P$ then $x + y \in P$ and $x y \in P$ .

Then:

If $\leq$ makes $R$ an ordered ring (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication), then $P : = {x \in R : 0 < x}$ is a cone, and for all $a, b \in R$ , $a < b$ if and only if $b - a \in P$ .
Conversely, let $P \subseteq R$ be a cone and define $a \leq_{P} b$ to mean $b - a \in P$ or $a = b$ . Then $\leq_{P}$ is a total order making $R$ an ordered ring, and its positive cone ${x \in R : 0 <_{P} x}$ is $P$ .
The two constructions are mutually inverse: starting from an ordered ring $(R, \leq)$ , the order $\leq_{P}$ built from its cone $P$ is $\leq$ itself; and starting from a cone $P$ , the cone of $\leq_{P}$ is $P$ .

Facts & Assumptions

Given: A ring $R$ with zero $0$ ; $x - y$ abbreviates $x + (- y)$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L1]

$(R, +, 0)$ is an abelian group: addition is associative and commutative, $x + 0 = x$ , and $x + (- x) = 0$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group).

[L2]

Identities of the abelian group $(R, +, 0)$ , where $x - y : = x + (- y)$ : $x - 0 = x$ ; $x - x = 0$ ; $- (x - y) = y - x$ , since $- (u + v) = (- v) + (- u)$ and $- (- y) = y$ ; $(y - x) + (z - y) = z - x$ , by associativity and commutativity; and $x - y = 0$ exactly when $x = y$ , by cancellation after adding $y$ (Group and abelian group, In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential, Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution).

[L3]

A total order is a reflexive, antisymmetric, transitive relation in which any two elements are comparable, and $a < b$ means $a \leq b$ with $a \neq b$ (Partial order and partially ordered set).

[L4]

An ordered ring is a ring with a total order satisfying (OR1) $a \leq b$ implies $a + c \leq b + c$ , and (OR2) $0 < a$ and $0 < b$ imply $0 < a b$ (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication).

Proof

technique · direct

1.1

Assume $\leq$ makes $R$ an ordered ring, and put $P = {x : 0 < x}$ . For $a, b \in R$ : $a < b$ if and only if $0 < b - a$ . Indeed, adding $- a$ to $a \leq b$ gives $a - a \leq b - a$ , that is $0 \leq b - a$ , by (OR1); adding $a$ to $0 \leq b - a$ gives $a \leq b$ likewise; and $a = b$ exactly when $b - a = 0$ . So $a < b$ exactly when $b - a \in P$ .

L1L2L4

1.2

Now let $P$ be any cone and define $\leq_{P}$ as in the statement. Reflexivity holds by the clause $a = a$ . Antisymmetry: if $a \neq b$ , $b - a \in P$ and $a - b \in P$ , then $- (b - a) = a - b \in P$ as well, contradicting trichotomy applied to $x = b - a$ ; so $a \leq_{P} b$ and $b \leq_{P} a$ force $a = b$ . Transitivity: if $a \leq_{P} b$ and $b \leq_{P} c$ and the two are not equalities, then $(c - b) + (b - a) = c - a \in P$ by closure; the cases where one of them is an equality are immediate. Comparability: given $a, b$ , trichotomy applied to $b - a$ gives $b - a \in P$ , or $b - a = 0$ and then $a = b$ , or $a - b = - (b - a) \in P$ . So $\leq_{P}$ is a total order.

L1L2L3

2.1

Trichotomy for the positive cone in claim 1. Return here to the ordered-ring order and its set $P = {x : 0 < x}$ from step 1.1. Let $x \in R$ . By totality and antisymmetry exactly one of $0 < x$ , $0 = x$ , $x < 0$ holds. By step 1.1 applied with $a = x$ , $b = 0$ , the last is equivalent to $0 - x = - x \in P$ . So exactly one of $x \in P$ , $x = 0$ , $- x \in P$ holds.

step 1.1L2L3

2.2

$\leq_{P}$ makes $R$ an ordered ring. (OR1): $(b + c) - (a + c) = b - a$ , so $a \leq_{P} b$ implies $a + c \leq_{P} b + c$ . (OR2): $0 <_{P} a$ means $a - 0 = a \in P$ , so if $0 <_{P} a$ and $0 <_{P} b$ then $a b \in P$ by closure, that is $0 <_{P} a b$ .

step 1.2L1L2L4

3.1

Closure for $P$ . Let $x, y \in P$ . Then $0 < x$ and, adding $x$ to $0 \leq y$ , $x \leq x + y$ ; with $0 < x$ and transitivity this gives $0 < x + y$ , so $x + y \in P$ . And $x y \in P$ is (OR2) verbatim. So $P$ is a cone, which with step 2.1 and step 1.1 proves claim 1.

step 1.1step 2.1L1L3L4

3.2

The cone of $\leq_{P}$ is $P$ : $0 <_{P} x$ means $x - 0 = x \in P$ and $x \neq 0$ , and $0 \notin P$ by trichotomy, so ${x : 0 <_{P} x} = P$ . With steps 1.2 and 2.2 this proves claim 2.

step 1.2step 2.2L2

4.1

Claim 3. Starting from an ordered ring $(R, \leq)$ with cone $P$ , step 1.1 says $a < b$ exactly when $b - a \in P$ , hence $a \leq b$ exactly when $b - a \in P$ or $a = b$ , which is $a \leq_{P} b$ ; so $\leq_{P}$ and $\leq$ are the same relation. Starting from a cone $P$ , step 3.2 says the cone of $\leq_{P}$ is $P$ .

step 1.1step 2.2step 3.2∎

Remarks

This is what keeps one notion of "ordered" in the library rather than two. The published The integers form a totally ordered ring presents the order on $Z$ as a relation; the published Ordered field presents the order on a field by its positive cone. Without this lemma the two would be different-looking hypotheses and every later statement would have to choose one. With it, Every ordered field is an ordered ring, and its order is the one its positive cone induces is a two-line consequence.
Trichotomy is doing the work in both directions. In claim 1 it comes from totality plus antisymmetry of the order; in claim 2 it is what supplies comparability and antisymmetry. Closure under addition, by contrast, is a strict statement in one direction and needs transitivity to recover in the other, which is why step 3.1 argues through $x \leq x + y$ rather than quoting (OR1) directly.
Nothing here uses multiplication except (OR2) and (C2), which correspond to each other verbatim. That is why the lemma holds for rings that are not commutative as well.

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

Every ordered field is an ordered ring, and its order is the one its positive cone induces

Statement

Let $F$ be an ordered field with positive cone $P$ (Ordered field), and let $\leq$ be the relation $a \leq b : ⟺ (b - a \in P$ or $a = b)$ that Ordered field defines from $P$ . Then:

$F$ with the operations of Field is a commutative ring (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring), and $P$ is a cone in the sense of The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order;
$\leq$ is a total order making $F$ an ordered ring (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication), whose positive cone ${x \in F : 0 < x}$ is exactly $P$ ;
$1 \in P$ , that is $0 < 1$ .

So an ordered field is an ordered ring, and its order and its positive cone determine each other exactly as they do in any ordered ring.

Facts & Assumptions

Given: An ordered field $F$ with positive cone $P$ , and $\leq$ defined from $P$ by $a \leq b : ⟺ (b - a \in P$ or $a = b)$ (Ordered field).

[A1]

Axiom (O1): for each $x \in F$ exactly one of $x \in P$ , $x = 0$ , $- x \in P$ holds (Ordered field).

[A2]

Axiom (O2): if $x, y \in P$ then $x + y \in P$ and $x y \in P$ (Ordered field).

[A3]

The order of an ordered field is defined by $a < b : ⟺ b - a \in P$ , and $a \leq b$ means $a < b$ or $a = b$ (Ordered field).

[L1]

Every field is a commutative ring with $1 \neq 0$ (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, Field).

[L2]

In a ring, a subset satisfying trichotomy and closure under addition and multiplication is a cone, and the relation it induces is a total order making the ring an ordered ring whose positive cone is that subset (The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order, Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication).

[L3]

In an ordered field, $a \neq 0$ implies $a^{2} > 0$ , that is $a \cdot a \in P$ (Squares of nonzero elements are positive).

Proof

technique · direct

1.1

$F$ is a commutative ring under its own addition and multiplication, with the same $0$ and $1$ .

2.1

$P$ is a cone in the ring $F$ : trichotomy is axiom (O1) verbatim, and closure under addition and under multiplication is axiom (O2) verbatim. This proves claim 1.

step 1.1A1A2L2

3.1

By [L2] applied to the ring $F$ and the cone $P$ , the relation $a \leq_{P} b : ⟺ (b - a \in P$ or $a = b)$ is a total order making $F$ an ordered ring, and its positive cone is $P$ .

step 2.1L2

4.1

That relation is the order of the ordered field: [A3] defines $a < b$ as $b - a \in P$ and $a \leq b$ as $a < b$ or $a = b$ , which is the definition of $\leq_{P}$ word for word. So $\leq$ and $\leq_{P}$ are the same relation, and claim 2 follows.

step 3.1A3

5.1

Claim 3: $1 \neq 0$ by [L1], so $1 \cdot 1 \in P$ by [L3]; and $1 \cdot 1 = 1$ because $1$ is the multiplicative identity. Hence $1 \in P$ , that is $0 < 1$ .

step 4.1L1L3∎

Remarks

This is one of the four bridges the page carries. The published Ordered field and the Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication of this page are two presentations of compatible orders, and without this item a later result about ordered rings could not be applied to $Q$ or $R$ without redoing the translation each time.
The converse is false as stated, and no converse is claimed. $Z$ is an ordered ring and is not a field at all, as the companion page records. What is true, and is claim 2, is that on a field the two presentations of an order agree.
Where the strict form of (OR2) is used. Axiom (O2) of Ordered field is already strict, so claim 1 is a verbatim match with the cone conditions of The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order and no strengthening is needed; that is exactly why Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication states its multiplicative axiom in the strict form.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication

Definition

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) with zero $0_{R}$ and identity $1_{R}$ . A subset $S \subseteq R$ is a subring of $R$ when

(T1) $1_{R} \in S$ ;
(T2) $x, y \in S$ implies $x + y \in S$ ;
(T3) $x \in S$ implies $- x \in S$ ;
(T4) $x, y \in S$ implies $x y \in S$ .

Why $S$ is then a ring, with the same zero and the same identity. From (T1) and (T3), $- 1_{R} \in S$ , and from (T2), $0_{R} = 1_{R} + (- 1_{R}) \in S$ . So $S$ contains $0_{R}$ , is closed under addition and closed under additive inverses, which are exactly conditions (S1), (S2) and (S3) of Subgroup for the abelian group $(R, +, 0_{R})$ ; hence $S$ is a subgroup of $(R, +, 0_{R})$ , and Subgroup states that a subgroup with the restricted operation is itself a group, whose identity and whose inverses are those of the ambient group. So $(S, +, 0_{R})$ is an abelian group, commutativity being inherited.

By (T4) multiplication restricts to a binary operation on $S$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation), and that restriction is associative because it is associative on $R$ . By (T1) the element $1_{R}$ lies in $S$ and satisfies $1_{R} x = x = x 1_{R}$ there, so it is a two-sided identity for the restricted multiplication (Left identity, right identity, and two-sided identity for a binary operation); thus $(S, \cdot, 1_{R})$ is a monoid. Both distributive laws hold in $S$ because they hold in $R$ and are equations between elements of $S$ .

Therefore $(S, +, \cdot, 0_{R}, 1_{R})$ is a ring, and

$0_{S} = 0_{R}, 1_{S} = 1_{R}, -_{S} x = -_{R} x (x \in S) .$

This last line is not decoration: it is what makes a statement proved in $S$ transportable to $R$ without translation, and it is used when a subfield of an ordered field is shown to inherit the order and when scalars are restricted along a subfield inclusion.

Remarks

(T1) is not redundant, and it is forced by the ring convention. Because Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides builds the identity into a ring, a subring must carry the same identity, and a subset closed under (T2), (T3) and (T4) need not contain it. The companion page exhibits the standard witness: the even integers are closed under addition, negation and multiplication and do not contain $1$ . Such a subset is a non-unital ring, not a subring.
A subring need not be an ideal and an ideal need not be a subring. Nothing on this page defines ideals; the point of the remark is only that closure under multiplication by arbitrary elements of $R$ is not asked for here.
The economical test. Conditions (T2) and (T3) can be replaced by the single condition $x - y \in S$ , exactly as One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ replaces the three subgroup conditions by one; that is 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, which also shows that an intersection of subrings is a subring.

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

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

Statement

$S$ is a subring of $R$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication) if and only if $1_{R} \in S$ , and $a - b \in S$ and $a b \in S$ for all $a, b \in S$ ;
if $S$ is a nonempty set of subrings of $R$ , then $K = ⋂_{S \in S} S$ is a subring of $R$ . In particular the intersection of two subrings is a subring.

Facts & Assumptions

Given: A ring $R$ with zero $0_{R}$ and identity $1_{R}$ , and a subset $S \subseteq R$ ; $a - b$ abbreviates $a + (- b)$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L1]

A subring is a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication; it is then a ring with the same zero, identity and additive inverses as $R$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

[L2]

$(R, +, 0_{R})$ is an abelian group (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group).

[L3]

One-step subgroup test, written additively: a nonempty $T \subseteq R$ with $a - b \in T$ for all $a, b \in T$ is a subgroup of $(R, +, 0_{R})$ ; and a subgroup contains $0_{R}$ and is closed under addition and under additive inverses (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[L4]

The intersection of a nonempty set of subgroups of a group is a subgroup (The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ ).

Proof

technique · direct

1.1

Suppose $S$ is a subring. Then $1_{R} \in S$ by (T1); for $a, b \in S$ we have $- b \in S$ by (T3) and hence $a - b = a + (- b) \in S$ by (T2); and $a b \in S$ by (T4).

1.2

Conversely, suppose $1_{R} \in S$ and that $a - b \in S$ and $a b \in S$ for all $a, b \in S$ . Then $S$ is nonempty, so by the one-step test it is a subgroup of $(R, +, 0_{R})$ ; hence $0_{R} \in S$ , $S$ is closed under addition and $- x \in S$ for every $x \in S$ . Together with $1_{R} \in S$ and closure under multiplication, that is exactly (T1) to (T4), so $S$ is a subring.

L1L2L3

2.1

Steps 1.1 and 1.2 prove claim 1.

step 1.1step 1.2L1

2.2

Claim 2. Each $S \in S$ is a subgroup of $(R, +, 0_{R})$ by [L1] and [L3], so $K$ is a subgroup of $(R, +, 0_{R})$ by [L4]; in particular $K$ is closed under addition and under additive inverses. Also $1_{R} \in S$ for every $S \in S$ , so $1_{R} \in K$ ; and if $a, b \in K$ then $a b \in S$ for every $S \in S$ , so $a b \in K$ . Hence $K$ satisfies (T1) to (T4) and is a subring.

step 1.1L1L3L4

3.1

Claims 1 and 2 are established in steps 2.1 and 2.2.

step 2.1step 2.2∎

Remarks

Why the identity clause survives the economy. The one-step test compresses the three additive conditions into one, and nothing compresses (T1): a subset can satisfy $a - b \in S$ and $a b \in S$ and still miss $1_{R}$ altogether. The companion page's even integers are exactly that, which is why $1_{R} \in S$ is stated separately in claim 1 rather than folded in.
Nonemptiness comes for free here. One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ requires the subset to be nonempty; in claim 1 that is supplied by $1_{R} \in S$ , so the criterion has no separate nonemptiness hypothesis.
Claim 2 is what would make "the subring generated by a set" meaningful. Intersections of subrings being subrings, the intersection of all subrings containing a given subset is the smallest one containing it. No such construction is used on this page; the claim is recorded because it costs one step and because The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ already supplies the additive half.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations

Definition

Let $F$ be a field (Field), regarded as a commutative ring by Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring. A subset $K \subseteq F$ is a subfield of $F$ when

(K1) $K$ is a subring of $F$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication);
(K2) $x^{- 1} \in K$ for every $x \in K$ with $x \neq 0_{F}$ .

Equivalently, by 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, $K$ is a subfield exactly when $1_{F} \in K$ , $a - b \in K$ and $a b \in K$ for all $a, b \in K$ , and $x^{- 1} \in K$ for every nonzero $x \in K$ .

Why $K$ is then a field, and with the same $0$ and $1$ . By (K1) and Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication, $K$ with the restricted operations is a ring whose zero is $0_{F}$ and whose identity is $1_{F}$ ; its multiplication is commutative, being the restriction of a commutative one (Commutative ring). Since $1_{F} \neq 0_{F}$ in $F$ and both lie in $K$ , we have $1_{K} \neq 0_{K}$ . Let $x \in K$ with $x \neq 0_{K}$ ; then $x \neq 0_{F}$ , so $x^{- 1} \in F$ exists and lies in $K$ by (K2), and $x x^{- 1} = 1_{F} = 1_{K} = x^{- 1} x$ . So every nonzero element of $K$ is a unit of the ring $K$ , and $K$ is a commutative division ring (Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit); by Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree it is a field. Moreover the inverse of $x$ computed in $K$ is its inverse computed in $F$ , since $x^{- 1}$ already satisfies the defining equation inside $K$ .

In particular

$0_{K} = 0_{F}, 1_{K} = 1_{F}, (- x)_{K} = (- x)_{F}, (x^{- 1})_{K} = (x^{- 1})_{F} (x \in K, x \neq 0_{F}) .$

A subfield of an ordered field inherits the order. Let $(F, P)$ be an ordered field (Ordered field) and $K$ a subfield. Put $P_{K} : = P \cap K$ . Then (O1) holds in $K$ : for $x \in K$ we have $- x \in K$ by (K1), and exactly one of $x \in P$ , $x = 0_{F}$ , $- x \in P$ holds in $F$ , so exactly one of $x \in P_{K}$ , $x = 0_{K}$ , $- x \in P_{K}$ holds. And (O2) holds: if $x, y \in P_{K}$ then $x + y$ and $x y$ lie in $P$ by (O2) in $F$ and in $K$ by (K1), hence in $P_{K}$ . So $(K, P_{K})$ is an ordered field, and its order is the restriction of the order of $F$ , because $a < b$ means $b - a \in P$ on both sides and $b - a$ is the same element in $K$ as in $F$ .

Remarks

The inverse-closure clause is not implied by (K1). The integers sit inside the rationals as a subring that is not a subfield, since $2$ is nonzero there and $2^{- 1}$ is not an integer; the companion page records that witness. So (K2) is doing work.
The agreement of the two zeros and the two identities is the load-bearing part. A later page restricts the scalars of a vector space along a subfield inclusion, and every axiom checked there uses that $1_{K}$ acts as $1_{F}$ does. Nothing would go through if a "subfield" were merely a subset that happens to be a field under some operations of its own.
Why the definition goes through subrings rather than restating the field axioms. All of (A), (M) and (D) except the existence of inverses are already guaranteed by Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication, and the two bridge lemmas of this page convert the resulting commutative division ring back into a field. Restating the axioms would create a second definition of a field on this page, which is exactly what Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring and Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree exist to prevent.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-28Open item page →

Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$

Definition

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). A ring homomorphism from $R$ to $S$ is a function $f : R \to S$ such that, for all $x, y \in R$ ,

(RH1) $f (x + y) = f (x) + f (y)$ ;
(RH2) $f (x y) = f (x) f (y)$ ;
(RH3) $f (1_{R}) = 1_{S}$ .

A ring homomorphism from a ring to itself is a ring endomorphism; a bijective ring homomorphism is a ring isomorphism.

(RH3) is imposed, not derived, and the reason is exactly the reason Monoid homomorphism and group homomorphism treats monoids and groups differently. Condition (RH1) says that $f$ is a homomorphism of the additive groups $(R, +, 0_{R}) \to (S, +, 0_{S})$ , and for groups preservation of the identity is free: it follows from (RH1) by cancellation (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed). Condition (RH2) says only that $f$ is a homomorphism of the multiplicative semigroups, and $(R, \cdot, 1_{R})$ is a monoid that need not be a group (Semigroup and monoid); for monoids preservation of the identity does not follow and must be assumed, which is precisely why Monoid homomorphism and group homomorphism puts the clause $f (e) = e^{'}$ into the definition of a monoid homomorphism and leaves it out of the definition of a group homomorphism (Left identity, right identity, and two-sided identity for a binary operation).

So a ring homomorphism is: a homomorphism of additive groups that is also a homomorphism of multiplicative monoids. The clause is not redundant: the companion page exhibits a map $Z \to Z \times Z$ satisfying (RH1) and (RH2) and failing (RH3).

Remarks

What follows automatically, and what does not. $f (0_{R}) = 0_{S}$ , $f (- x) = - f (x)$ and $f (m x) = m f (x)$ for every integer $m$ all follow from (RH1) alone, and units are carried to units once (RH3) is available; these are A ring homomorphism satisfies $f (0) = 0$ , $f (- a) = - f (a)$ and $f (m a) = m f (a)$ for $m \in Z$ , carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms. What does not follow from (RH1) and (RH2) is (RH3) itself.
Between fields there is no difference from the published notion. A ring homomorphism between fields is exactly a field homomorphism in the sense of Field homomorphism and embedding, and every such map is injective; that is A ring homomorphism between fields is a field homomorphism in the published sense, and every such map is injective.
Kernels, quotient rings and the isomorphism theorems are not defined on this page and nothing here uses them.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

A ring homomorphism satisfies $f (0) = 0$ , $f (- a) = - f (a)$ and $f (m a) = m f (a)$ for $m \in Z$ , carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms

Statement

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) and $f : R \to S$ a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ). Then:

$f (0_{R}) = 0_{S}$ and $f (- a) = - f (a)$ for every $a \in R$ ; consequently $f (a - b) = f (a) - f (b)$ ;
$f (m a) = m f (a)$ for every $a \in R$ and every $m \in Z$ , the multiples being those of 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 $a, b \in R$ ;
if $u \in R^{\times}$ then $f (u) \in S^{\times}$ and $f (u)^{- 1} = f (u^{- 1})$ ;
the image $f (R) = {f (a) : a \in R}$ is a subring of $S$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication);
if $g : S \to T$ is a ring homomorphism then so is $g \circ f : R \to T$ , and the identity map of $R$ is a ring homomorphism.

Facts & Assumptions

Given: Rings $R$ , $S$ with zeros $0_{R}$ , $0_{S}$ and identities $1_{R}$ , $1_{S}$ , and a function $f : R \to S$ with $f (x + y) = f (x) + f (y)$ , $f (x y) = f (x) f (y)$ and $f (1_{R}) = 1_{S}$ . For claim 5, let $T$ be a ring and let $g : S \to T$ be a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[A1]

$f (x + y) = f (x) + f (y)$ for all $x, y \in R$ .

[A2]

$f (x y) = f (x) f (y)$ for all $x, y \in R$ .

[A3]

$f (1_{R}) = 1_{S}$ .

[A4]

$g (y + z) = g (y) + g (z)$ for all $y, z \in S$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[A5]

$g (y z) = g (y) g (z)$ for all $y, z \in S$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[A6]

$g (1_{S}) = 1_{T}$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[L1]

$(R, +, 0_{R})$ and $(S, +, 0_{S})$ are abelian groups, and by [A1] the map $f$ is a homomorphism of these groups in the sense of Monoid homomorphism and group homomorphism (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L2]

A group homomorphism $h$ satisfies $h (e) = e^{'}$ , $h (g^{- 1}) = h (g)^{- 1}$ and $h (g^{n}) = h (g)^{n}$ for every $n \in Z$ ; read additively, $h (0) = 0$ , $h (- a) = - h (a)$ and $h (n a) = n h (a)$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L3]

The integer multiple $m a$ in a ring is the integer power of $a$ in its additive group, read additively (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 $a, b \in R$ , Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L4]

A unit of a ring is an invertible element of its multiplicative monoid, its inverse is unique, and a single equation $v u = 1$ or $u v = 1$ determines it (Left inverse, right inverse, and invertible element of a monoid, In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided, The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

[L5]

Subring criterion: $T \subseteq S$ is a subring exactly when $1_{S} \in T$ and $a - b \in T$ and $a b \in T$ for all $a, b \in T$ (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, Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

Proof

technique · direct

1.1

By [L1] the map $f$ is a homomorphism from the additive group of $R$ to the additive group of $S$ , so [L2] applies to it.

A1L1

1.2

Claim 5: for $x, y \in R$ , $(g \circ f) (x + y) = g (f (x) + f (y)) = g (f (x)) + g (f (y))$ and $(g \circ f) (x y) = g (f (x) f (y)) = g (f (x)) g (f (y))$ , while $(g \circ f) (1_{R}) = g (1_{S}) = 1_{T}$ . The identity map satisfies the three conditions trivially.

A1A2A3A4A5A6

1.3

Claim 3: let $u \in R^{\times}$ with inverse $u^{- 1}$ , so $u u^{- 1} = 1_{R} = u^{- 1} u$ . Applying $f$ and using [A2] and [A3], $f (u) f (u^{- 1}) = f (1_{R}) = 1_{S} = f (u^{- 1}) f (u)$ . So $f (u)$ has the two-sided inverse $f (u^{- 1})$ in $S$ , hence $f (u) \in S^{\times}$ , and $f (u)^{- 1} = f (u^{- 1})$ because inverses in a monoid are unique.

A2A3L4

2.1

Claim 1: [L2] read additively gives $f (0_{R}) = 0_{S}$ and $f (- a) = - f (a)$ ; hence $f (a - b) = f (a + (- b)) = f (a) + f (- b) = f (a) - f (b)$ .

step 1.1A1L2

2.2

Claim 2: by [L3] the multiple $m a$ is the integer power of $a$ in $(R, +, 0_{R})$ , and $m f (a)$ is the integer power of $f (a)$ in $(S, +, 0_{S})$ ; so the claim is the third part of [L2] read additively.

step 1.1L2L3

3.1

Claim 4: $1_{S} = f (1_{R}) \in f (R)$ by [A3]; for $f (a), f (b) \in f (R)$ we have $f (a) - f (b) = f (a - b) \in f (R)$ by step 2.1 and $f (a) f (b) = f (a b) \in f (R)$ by [A2]. So $f (R)$ satisfies the subring criterion.

step 2.1A2A3L5

4.1

Claims 1 to 5 are established in steps 2.1, 2.2, 1.3, 3.1 and 1.2.

step 1.2step 2.1step 2.2step 1.3step 3.1∎

Remarks

Claim 1 is free, claim 3 is not. $f (0_{R}) = 0_{S}$ comes from additivity alone, because the additive structure is a group and cancellation is available there (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed). The multiplicative analogue would be $f (1_{R}) = 1_{S}$ , and it is not free: it is axiom (RH3) of Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ , and the step proving claim 3 above uses it twice.
The image is a subring, not merely a non-unital one, and that too rests on (RH3): without it the image would still be closed under subtraction and multiplication but might miss $1_{S}$ , and the companion page's map $n \mapsto (n, 0)$ has exactly that image.
Claim 2 is a dictionary entry, not new arithmetic. The multiples on both sides are the additive powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , so the statement is the power law of A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed with additive notation; nothing is proved twice.

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

A ring homomorphism between fields is a field homomorphism in the published sense, and every such map is injective

Statement

Let $F$ and $G$ be fields (Field), regarded as rings by Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, and let $f : F \to G$ be a function. Then:

$f$ is a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ) if and only if $f$ is a field homomorphism (Field homomorphism and embedding); the two definitions impose the same three conditions;
every such $f$ is injective (Injection, surjection, bijection).

So "ring homomorphism between fields" and "field homomorphism" name the same maps, and no second notion of homomorphism of fields is introduced.

Facts & Assumptions

Given: Fields $F$ and $G$ , with zeros $0_{F}, 0_{G}$ and identities $1_{F}, 1_{G}$ , and a function $f : F \to G$ (Field).

[L1]

$F$ and $G$ are commutative rings with $1 \neq 0$ , their ring operations, zeros and identities being the field ones (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L2]

A ring homomorphism is a map satisfying (RH1) $f (x + y) = f (x) + f (y)$ , (RH2) $f (x y) = f (x) f (y)$ and (RH3) $f (1_{F}) = 1_{G}$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[L3]

A field homomorphism is a map satisfying $φ (x + y) = φ (x) + φ (y)$ , $φ (x y) = φ (x) φ (y)$ and $φ (1_{F}) = 1_{G}$ (Field homomorphism and embedding).

[L4]

A ring homomorphism satisfies $f (a - b) = f (a) - f (b)$ (A ring homomorphism satisfies $f (0) = 0$ , $f (- a) = - f (a)$ and $f (m a) = m f (a)$ for $m \in Z$ , carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms).

[L5]

In a ring, $0 \cdot x = x \cdot 0 = 0$ (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 a field, every $x \neq 0$ has an inverse $x^{- 1}$ with $x^{- 1} x = 1$ , and $1_{G} \neq 0_{G}$ (Field).

[L7]

A map is injective when $f (a) = f (b)$ forces $a = b$ (Injection, surjection, bijection).

Proof

technique · direct

1.1

By [L1] the ring structures on $F$ and $G$ are the field structures, so the expressions $f (x) + f (y)$ , $f (x) f (y)$ , $1_{F}$ and $1_{G}$ mean the same thing in [L2] and in [L3].

2.1

Claim 1: the three conditions of [L2] and the three conditions of [L3] are the same three equations, so $f$ satisfies one triple exactly when it satisfies the other.

step 1.1L2L3

3.1

Let $f$ be such a map and suppose $f (a) = f (b)$ with $a \neq b$ . Then $a - b \neq 0_{F}$ , and $f (a - b) = f (a) - f (b) = 0_{G}$ by [L4].

step 2.1L4

4.1

Put $c : = (a - b)^{- 1}$ , which exists by [L6]. Then $1_{G} = f (1_{F}) = f (c (a - b)) = f (c) f (a - b) = f (c) \cdot 0_{G} = 0_{G}$ , using (RH3), (RH2) and [L5]. This contradicts $1_{G} \neq 0_{G}$ .

step 3.1L2L5L6

5.1

Hence $f (a) = f (b)$ forces $a = b$ , so $f$ is injective; with step 2.1 this proves both claims.

step 2.1step 4.1L7∎

Remarks

This is one of the page's bridges to the published vocabulary. Together with Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring and Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree, it means that everything proved on this page about rings, subrings and ring homomorphisms applies to the library's fields with no translation, and that a reader meeting Field homomorphism and embedding and Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ is meeting one notion twice, not two notions.
The injectivity proof here uses no ideals. The Remarks of Field homomorphism and embedding sketch the standard kernel-is-an-ideal argument; ideals are not defined on this page, and the argument above needs only an inverse and $0 \cdot x = 0$ .
Injectivity really does need $1_{G} \neq 0_{G}$ , that is, it needs the target to be a field rather than an arbitrary ring. The zero map from a field to the one-element ring satisfies (RH1), (RH2) and (RH3) there, since $1 = 0$ in that ring, and it is not injective.

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$

Definition

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). The product ring $R \times S$ is the cartesian product of the underlying sets with the componentwise operations

$(a, b) + (a^{'}, b^{'}) : = (a + a^{'}, b + b^{'}), (a, b) \cdot (a^{'}, b^{'}) : = (a a^{'}, b b^{'}),$

and the distinguished elements

$0_{R \times S} : = (0_{R}, 0_{S}), 1_{R \times S} : = (1_{R}, 1_{S}) .$

These data make $R \times S$ a ring. Both rules take a pair of elements of $R \times S$ to an element of $R \times S$ , so each is a binary operation. Every ring axiom is an equation between elements of $R \times S$ , and two such elements are equal exactly when their components are; so each axiom holds componentwise from the corresponding axiom in $R$ and in $S$ . Explicitly: $(R \times S, +, (0_{R}, 0_{S}))$ is an abelian group with $- (a, b) = (- a, - b)$ ; multiplication is associative with two-sided identity $(1_{R}, 1_{S})$ ; and both distributive laws hold.

Units. An element $(u, v)$ is a unit of $R \times S$ if and only if $u$ is a unit of $R$ and $v$ is a unit of $S$ , in which case $(u, v)^{- 1} = (u^{- 1}, v^{- 1})$ . Indeed, if $u u^{'} = 1_{R} = u^{'} u$ and $v v^{'} = 1_{S} = v^{'} v$ then $(u, v) (u^{'}, v^{'}) = (1_{R}, 1_{S}) = (u^{'}, v^{'}) (u, v)$ ; conversely, if $(u, v) (u^{'}, v^{'}) = (1_{R}, 1_{S}) = (u^{'}, v^{'}) (u, v)$ then reading each component gives $u u^{'} = 1_{R} = u^{'} u$ and $v v^{'} = 1_{S} = v^{'} v$ . So

$(R \times S)^{\times} = R^{\times} \times S^{\times},$

and by The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring that set is a group under multiplication (Left inverse, right inverse, and invertible element of a monoid).

Commutativity. $R \times S$ is commutative (Commutative ring) if and only if both $R$ and $S$ are. If both are, the componentwise product is commutative. Conversely, if $R \times S$ is commutative then for $a, a^{'} \in R$ , $(a a^{'}, 1_{S}) = (a, 1_{S}) (a^{'}, 1_{S}) = (a^{'}, 1_{S}) (a, 1_{S}) = (a^{'} a, 1_{S})$ , so $a a^{'} = a^{'} a$ ; the same argument in the second component settles $S$ .

Zero divisors. If $1_{R} \neq 0_{R}$ and $1_{S} \neq 0_{S}$ , then $R \times S$ has zero divisors and is therefore never an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors), whatever $R$ and $S$ are. Indeed $(1_{R}, 0_{S})$ and $(0_{R}, 1_{S})$ are both nonzero, and

$(1_{R}, 0_{S}) \cdot (0_{R}, 1_{S}) = (1_{R} 0_{R}, 0_{S} 1_{S}) = (0_{R}, 0_{S}),$

using $x \cdot 0 = 0 = 0 \cdot x$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ).

Remarks

The product of two domains is never a domain, by the zero-divisor computation above, since a domain has $1 \neq 0$ by definition. This is the cleanest source of zero divisors available at this point, and the companion page records the instance $Z \times Z$ .
The two projections are ring homomorphisms; the standard injections need not be, and fail when the other factor is nonzero. The maps $(a, b) \mapsto a$ and $(a, b) \mapsto b$ satisfy (RH1), (RH2) and (RH3) of Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ . The map $a \mapsto (a, 0_{S})$ satisfies (RH1) and (RH2) but sends $1_{R}$ to $(1_{R}, 0_{S})$ , which is not the identity of $R \times S$ when $1_{S} \neq 0_{S}$ ; the companion page uses exactly that map to show (RH3) is not redundant.
Only the product of two rings is defined here, which is all this page and its companion use. Nothing below needs an indexed family.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations

Definition

Let $X$ be a set and $R$ a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). Write

$R^{X} : = {f : f is a function X \to R}$

and define, for $f, g \in R^{X}$ , the pointwise operations

$(f + g) (x) : = f (x) + g (x), (f \cdot g) (x) : = f (x) g (x) (x \in X),$

with distinguished elements the constant functions

$0_{R^{X}} (x) : = 0_{R}, 1_{R^{X}} (x) : = 1_{R} (x \in X) .$

These data make $R^{X}$ a ring. Each rule assigns to a pair of functions $X \to R$ another function $X \to R$ , so each is a binary operation on $R^{X}$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation). Two elements of $R^{X}$ are equal exactly when they agree at every $x \in X$ , so every ring axiom, being an equation between elements of $R^{X}$ , holds as soon as it holds at each point; and at each point it is the corresponding axiom of $R$ . In particular $- f$ is the function $x \mapsto - f (x)$ , the additive group of $R^{X}$ is abelian, multiplication is associative with two-sided identity $1_{R^{X}}$ , and both distributive laws hold.

Commutativity. If $R$ is commutative (Commutative ring) then so is $R^{X}$ , pointwise. If $X \neq \emptyset$ the converse holds: fixing $x_{0} \in X$ and taking constant functions $f \equiv a$ , $g \equiv b$ gives $a b = b a$ from $(f g) (x_{0}) = (g f) (x_{0})$ . If $X = \emptyset$ then $R^{X}$ has exactly one element, the empty function, so $1_{R^{X}} = 0_{R^{X}}$ and $R^{X}$ is commutative whatever $R$ is; the converse therefore needs the hypothesis $X \neq \emptyset$ and is stated with it.

Zero divisors. Suppose $R$ is not the one-element ring, that is $1_{R} \neq 0_{R}$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ), and suppose $X$ has two distinct elements $x_{1} \neq x_{2}$ . Define $f, g \in R^{X}$ by

$f (x) : = {\begin{cases} 1_{R} & x = x_{1} \\ 0_{R} & x \neq x_{1} \end{cases} g (x) : = {\begin{cases} 0_{R} & x = x_{1} \\ 1_{R} & x \neq x_{1} . \end{cases}$

Then $f \neq 0_{R^{X}}$ because $f (x_{1}) = 1_{R} \neq 0_{R}$ , and $g \neq 0_{R^{X}}$ because $g (x_{2}) = 1_{R} \neq 0_{R}$ ; and $(f g) (x) = 0_{R}$ for every $x$ , since one of the two factors is $0_{R}$ at each point and $0_{R} y = 0_{R} = y 0_{R}$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ). So $f$ and $g$ are zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors) and $R^{X}$ is not an integral domain.

Remarks

The same underlying set carries other structures. $R^{X}$ is the set of all functions $X \to R$ , with no continuity, boundedness or finiteness condition. When $R$ is a field, this same set with this same addition carries a second structure of interest, obtained by replacing the pointwise product with a scalar multiplication. The addition of that second structure is this same pointwise rule $(f + g) (x) = f (x) + g (x)$ ; what differs is the second operation, which multiplies a function by a scalar rather than two functions together, so the two are not special cases of one another.
This is where non-domains become plentiful. Products of two nonzero rings (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ) and function rings on a set with at least two points are the two standard sources of zero divisors, and neither needs any arithmetic beyond $0 \cdot y = 0$ .
The empty index set is a genuine case, not an edge case to be waved away. $R^{\emptyset}$ has exactly one element and is the one-element ring, in which $1 = 0$ ; that is why the commutativity statement above carries the hypothesis $X \neq \emptyset$ in the direction where it is needed, and why the zero-divisor statement asks for two distinct points.

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

In a field, the additive multiple $n \cdot 1_{F}$ is the canonical natural $ι (n)$ : the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion $ι (0) = 0_{F}$ , $ι (σ (n)) = ι (n) + 1_{F}$

Statement

Let $F$ be a field (Field), which is a ring by Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring. Two functions $N \to F$ are in play:

$ι$ , the canonical natural of The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, defined by $ι (0) = 0_{F}$ and $ι (n + 1) = ι (n) + 1_{F}$ ;
$n \mapsto n \cdot 1_{F}$ , the additive natural power of the element $1_{F}$ in the abelian group $(F, +, 0_{F})$ , defined by Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ read additively: $0 \cdot 1_{F} = 0_{F}$ and $σ (n) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ .

These are the same function: $ι (n) = n \cdot 1_{F}$ for every $n \in N$ . In particular the notation $n \cdot 1_{F}$ used by The canonical natural $ι (n) = n \cdot 1_{F}$ of a field and the notation $n \cdot 1_{F}$ used by Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ denote the same element of $F$ , and no second notion is in play.

Facts & Assumptions

Given: A field $F$ with $0_{F}$ and $1_{F}$ , the map $ι : N \to F$ of The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, and the additive natural powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ in the group $(F, +, 0_{F})$ .

[L1]

$F$ is a ring, so $(F, +, 0_{F})$ is an abelian group (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group).

[L2]

$ι (0) = 0_{F}$ and $ι (n + 1) = ι (n) + 1_{F}$ for every $n \in N$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L3]

The additive natural powers satisfy $0 \cdot a = 0_{F}$ and $σ (n) \cdot a = n \cdot a + a$ for every $n \in N$ and every $a \in F$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L4]

On $N$ : $m + 0 = m$ and $m + σ (n) = σ (m + n)$ , so $n + 1 = σ (n)$ (Addition of natural numbers, The natural numbers $N$ (von Neumann)).

[L5]

The recursion theorem: for a set $A$ , an element $a \in A$ and a function $u : A \to A$ there is exactly one $g : N \to A$ with $g (0) = a$ and $g (σ (n)) = u (g (n))$ (The recursion theorem).

Proof

technique · direct

1.1

Let $u : F \to F$ be the function $u (t) = t + 1_{F}$ , which is a function from $F$ to $F$ because addition is a binary operation on $F$ . By [L5] applied with $A = F$ , $a = 0_{F}$ and this $u$ , there is exactly one function $g : N \to F$ satisfying $g (0) = 0_{F}$ and $g (σ (n)) = g (n) + 1_{F}$ for every $n \in N$ .

L1L5

1.2

The map $ι$ satisfies those two equations: $ι (0) = 0_{F}$ by [L2], and $ι (σ (n)) = ι (n + 1) = ι (n) + 1_{F}$ by [L2] together with $n + 1 = σ (n)$ .

L2L4

1.3

The map $n \mapsto n \cdot 1_{F}$ satisfies them too: $0 \cdot 1_{F} = 0_{F}$ and $σ (n) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ , both by [L3] with $a = 1_{F}$ .

2.1

By the uniqueness clause of step 1.1, the two functions of steps 1.2 and 1.3 are equal, so $ι (n) = n \cdot 1_{F}$ for every $n \in N$ .

step 1.1step 1.2step 1.3L5∎

Remarks

This is the item the characteristic rests on. The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise is stated for an arbitrary ring using the additive powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ; for a field a reader may already know The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, and without the present lemma the page would carry two notations for one element and invite the reader to assume they agree. That assumption is exactly the defect this item removes, and it is removed by a proof rather than by a remark.
Uniqueness, not a computation, is what does the work. Both functions are characterised by the same recursion, and The recursion theorem says a recursion of that shape has exactly one solution. An induction on $N$ would prove the same thing directly; nothing is gained by writing it out, since the uniqueness clause of The recursion theorem is that induction.
$N$ contains $0$ , and $ι (0) = 0_{F}$ , not $1_{F}$ . So $ι$ is not the map $n \mapsto$ " $1_{F}$ added to itself $n$ times" for $n \geq 1$ only: the value at $0$ is a genuine value of the recursion. The canonical natural $ι (n) = n \cdot 1_{F}$ of a field records the same point, and the published Canonical naturals are positive and strictly increasing states its own recursion from $n = 1$ , which agrees because $ι (1) = ι (0) + 1_{F} = 1_{F}$ .

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise

Definition

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) with identity $1_{R}$ and zero $0_{R}$ . For $n \in N$ let $n \cdot 1_{R}$ be the additive natural power of $1_{R}$ in the abelian group $(R, +, 0_{R})$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Group and abelian group): thus $0 \cdot 1_{R} = 0_{R}$ and $σ (n) \cdot 1_{R} = n \cdot 1_{R} + 1_{R}$ . Put

$T_{R} : = {n \in N : n \geq 1 and n \cdot 1_{R} = 0_{R}} \subseteq N .$

The characteristic of $R$ is

$char (R) : = {\begin{cases} \min T_{R} & if T_{R} \neq \emptyset, \\ 0 & if T_{R} = \emptyset . \end{cases}$

Why the least element exists. $T_{R}$ is a subset of $N$ , so when it is nonempty it has a least element by the well-ordering principle (The well-ordering principle), and that element is unique: two least members are below one another and hence equal by antisymmetry of the natural-number order ( $\leq$ is a linear order on $N$ , Order on the natural numbers). So $char (R)$ is a well-defined natural number in both cases.

The clause $n \geq 1$ is not decoration. $N$ contains $0$ (The natural numbers $N$ (von Neumann)) and $0 \cdot 1_{R} = 0_{R}$ holds in every ring, so without that clause $T_{R}$ would always contain $0$ and the definition would say nothing.

Convention: the value in the empty case is $0$ , and this is the OPPOSITE of the convention for the order of a group element. 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 writes $ord (g) = \infty$ when no positive power of $g$ is the identity. Here the value in the corresponding case is the natural number $0$ , not a symbol $\infty$ . A reader coming straight from 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 should notice the difference: it is deliberate, it is the standard convention for the characteristic, and it is what makes the divisibility statement of 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 $1_{R}$ uniform across the two cases.

$char (R)$ is a natural number, hence a set, not an element of $R$ . A natural number in this library is a von Neumann natural (The natural numbers $N$ (von Neumann)), so $char (R) \notin R$ in general and the expression $n \cdot 1_{R}$ is not a product in $R$ but the additive multiple just described.

Dictionary for fields. When $R$ is a field $F$ (Field), the element $n \cdot 1_{F}$ is exactly the canonical natural $ι (n)$ of The canonical natural $ι (n) = n \cdot 1_{F}$ of a field; this is proved, not assumed, in In a field, the additive multiple $n \cdot 1_{F}$ is the canonical natural $ι (n)$ : the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion $ι (0) = 0_{F}$ , $ι (σ (n)) = ι (n) + 1_{F}$ . So for a field $T_{F} = {n \geq 1 : ι (n) = 0_{F}}$ and the characteristic is the least $n \geq 1$ with $ι (n) = 0_{F}$ , or $0$ if there is none.

Remarks

Characteristic $1$ occurs, exactly once. $1 \cdot 1_{R} = 0 \cdot 1_{R} + 1_{R} = 1_{R}$ , so $1 \in T_{R}$ exactly when $1_{R} = 0_{R}$ , that is exactly when $R$ is the one-element ring; and then $char (R) = 1$ , since $1$ is the least member of $N$ with $n \geq 1$ . The companion page records that ring. Every other ring has characteristic $0$ or a value at least $2$ .
Characteristic $0$ means "no such $n$ ", not " $n = 0$ works". The value $0$ is a flag, and the flag is chosen so that " $n \cdot 1_{R} = 0_{R}$ exactly when $char (R)$ divides $n$ " is true in both cases: when the characteristic is $0$ the right-hand side says $n = 0$ , which is exactly right. That statement is 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 $1_{R}$ .
A further theorem about the possible characteristics of an integral domain is not stated here. The current corpus does not yet state or prove that result. A future algebra page may add it from 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 $1_{R}$ once the needed prime-number interface is brought into scope.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

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.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

A finite sum in a commutative monoid indexed by an arbitrary finite set

Definition

Let $(M, +, 0)$ be a commutative monoid (Semigroup and monoid), let $S$ be a finite set with $n : = ∣ S ∣$ (The cardinality $∣ A ∣$ of a finite set), and let $a : S \to M$ . Choose a bijection $ϕ : n \to S$ and define

$\sum_{s \in S} a_{s} : = a_{ϕ (0)} + a_{ϕ (1)} + \dots + a_{ϕ (n - 1)},$

where the right side is the finite monoid product of The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity written additively. In particular, the sum over the empty set is $0$ .

This value is independent of the enumeration. If $ψ : n \to S$ is another bijection, then $ϕ^{- 1} \circ ψ$ is a permutation of $n$ (Injection, surjection, bijection), and generalised commutativity gives

$\sum_{k < n} a_{ψ (k)} = \sum_{k < n} a_{ϕ (k)}$

(Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either). Thus the displayed notation names one element of $M$ and does not select a preferred enumeration.

When $M$ is the additive monoid of $R$ or of $N$ , this definition agrees with The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form: both enumerate $S$ , apply the same finite recursion with identity $0$ , and are independent of the enumeration.

Remarks

The coefficient object is an arbitrary commutative monoid. This is stronger than the published real- and natural-valued definition and is the form needed for sums in a commutative ring.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31Open item page →

Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule

Statement

Let $(M, +, 0)$ be a commutative monoid and let all index sets below be finite.

If $h : T \to S$ is a bijection and $a : S \to M$ , then $\sum_{t \in T} a_{h (t)} = \sum_{s \in S} a_{s}$ .
If $S$ and $T$ are disjoint and $a : S \cup T \to M$ , then $\sum_{u \in S \cup T} a_{u} = \sum_{s \in S} a_{s} + \sum_{t \in T} a_{t}$ .
If $a : S \times T \to M$ , then $\sum_{(s, t) \in S \times T} a_{s, t} = \sum_{s \in S} (\sum_{t \in T} a_{s, t}) = \sum_{t \in T} (\sum_{s \in S} a_{s, t}) .$

Facts & Assumptions

Given: A commutative monoid $(M, +, 0)$ , finite sets $S, T$ , and functions and a bijection as in the Statement.

[F1]

A finite commutative-monoid sum is obtained from any enumeration of its finite index set, and its value is independent of that enumeration (A finite sum in a commutative monoid indexed by an arbitrary finite set).

[L1]

A finite monoid product splits at a cut, may be regrouped into consecutive blocks, and in a commutative monoid is invariant under permutations (Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either).

[L2]

A finite family of pairwise disjoint finite sets has finite union, and its cardinality is the sum of the cardinalities; in particular, $∣ S \cup T ∣ = ∣ S ∣ + ∣ T ∣$ when $S$ and $T$ are disjoint (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

[L3]

A Cartesian product of finite sets is finite with $∣ S \times T ∣ = ∣ S ∣ ∣ T ∣$ (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

[F2]

Composites and inverses of bijections are bijections (Injection, surjection, bijection).

Proof

technique · direct

1.1

For clause 1, choose an enumeration $ϕ : ∣ T ∣ \to T$ . Then $h \circ ϕ$ enumerates $S$ , and [F1] gives $\sum_{t \in T} a_{h (t)} = \sum_{k < ∣ T ∣} a_{h (ϕ (k))} = \sum_{s \in S} a_{s}$ .

F1F2

1.2

For clause 2, choose enumerations of $S$ and $T$ and concatenate them. By [L2] this gives an enumeration of $S \cup T$ of length $∣ S ∣ + ∣ T ∣$ , and the splitting law in [L1] turns the resulting finite sum into the sum over $S$ followed by the sum over $T$ .

F1L1L2

1.3

For clause 3, choose enumerations $ϕ : ∣ S ∣ \to S$ and $ψ : ∣ T ∣ \to T$ . The disjoint slices ${ϕ (i)} \times T$ cover $S \times T$ ; concatenate their $ψ$ -enumerations in the order of $i$ . By [L2] and [L3] this is an enumeration of $S \times T$ , and regrouping it into the $∣ S ∣$ consecutive slices gives $\sum_{(s, t) \in S \times T} a_{s, t} = \sum_{s \in S} (\sum_{t \in T} a_{s, t})$ .

F1L1L2L3

2.1

The column-major list also enumerates $S \times T$ , and permutation invariance followed by regrouping into columns gives $\sum_{(s, t) \in S \times T} a_{s, t} = \sum_{t \in T} (\sum_{s \in S} a_{s, t})$ .

step 1.3F1L1L3

3.1

Steps 1.1, 1.2, 1.3 and 2.1 prove reindexing, disjoint splitting and both finite Fubini equalities.

step 1.1step 1.2step 1.3step 2.1∎

5 · Examples, counterexamples and false statements

None yet.

Rings, Subrings, Integral Domains and Fields

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides

Definition

Remarks

Commutative ring

Definition

Remarks

In any ring 0⋅a=a⋅0=00⋅a=a⋅0=0, (−a)b=a(−b)=−(ab)(−a)b=a(−b)=−(ab), (−a)(−b)=ab(−a)(−b)=ab, (−1)a=−a(−1)a=−a and a(b−c)=ab−aca(b−c)=ab−ac

Statement

Facts & Assumptions

Proof

Remarks

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

Statement

Facts & Assumptions

Proof

Remarks

The units of a ring are the invertible elements of its multiplicative monoid, and R×R× is a group under multiplication; 0∈R×0∈R× only in the zero ring

Statement

Facts & Assumptions

Proof

Remarks

Zero divisor, and integral domain: a commutative ring with 1≠01=0 and no zero divisors

Definition

Remarks

Cancellation characterises domains: in a commutative ring with 1≠01=0, the implication ab=acab=ac and a≠0a=0 imply b=cb=c holds if and only if the ring has no zero divisors

Statement

Facts & Assumptions

Proof

Remarks

Division ring: a ring with 1≠01=0 in which every nonzero element is a unit

Definition

Remarks

Every field is a commutative ring with 1≠01=0; it is an integral domain, and it is a commutative division ring

Statement

Facts & Assumptions

Proof

Remarks

Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree

Statement

Facts & Assumptions

Proof

Remarks

The quaternions HH: real quadruples with componentwise addition and an explicit multiplication formula matching the table on 1,i,j,k1,i,j,k

Definition

Remarks

HH is a division ring that is not commutative, hence not a field: q−1=qˉ/N(q)q−1=qˉ​/N(q) for q≠0q=0, while ij=kij=k and ji=−kji=−k

Statement

Facts & Assumptions

Proof

Remarks

Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication

Definition

Remarks

The order presentation and the positive-cone presentation of an ordered ring determine each other: P={ x:0<x }P={x:0<x} satisfies trichotomy and closure, and a<b: ⟺ b−a∈Pa<b:⟺b−a∈P recovers the order

Statement

Facts & Assumptions

Proof

Remarks

Every ordered field is an ordered ring, and its order is the one its positive cone induces

Statement

Facts & Assumptions

Proof

Remarks

Subring: a subset containing 1R1R​ and closed under addition, additive inverses and multiplication

Definition

Remarks

Subring criterion: S⊆RS⊆R is a subring if and only if 1R∈S1R​∈S and a−b∈Sa−b∈S and ab∈Sab∈S for all a,b∈Sa,b∈S; and an intersection of subrings is a subring

Statement

Facts & Assumptions

Proof

Remarks

Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations

Definition

Remarks

Ring homomorphism: additive, multiplicative, and required to send 11 to 11

In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$

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 $a, b \in R$

The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring

Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors

Cancellation characterises domains: in a commutative ring with $1 \neq 0$ , the implication $a b = a c$ and $a \neq 0$ imply $b = c$ holds if and only if the ring has no zero divisors

Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit

Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring

The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$

$H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$

The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order

Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication

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

Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$

A ring homomorphism satisfies $f (0) = 0$ , $f (- a) = - f (a)$ and $f (m a) = m f (a)$ for $m \in Z$ , carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms

The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$

The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations

In a field, the additive multiple $n \cdot 1_{F}$ is the canonical natural $ι (n)$ : the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion $ι (0) = 0_{F}$ , $ι (σ (n)) = ι (n) + 1_{F}$

The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise

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 $1_{R}$