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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.

Depends on

Used by

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Field (mathematics) (Wikipedia) (standard reference, not scraped)
Integral domain (Wikipedia) (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §16.4: Integral Domains and Fields (standard reference, not scraped)

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