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Every field is a commutative ring with $1 \ne 0$ ; 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 $0 \ne 1$ . 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 $1 \ne 0$ ;
$F$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors);
$F$ is a division ring (Division ring: a ring with $1 \ne 0$ 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 $0 \ne 1$ , 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 $(F \setminus \{0\}, \cdot)$ is an abelian group with identity $1$ , so every $x \ne 0$ 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: $x(y+z) = xy + xz$ for all $x, y, z \in F$ (Field).

[A4]

$0 \ne 1$ (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, $ab = 0$ implies $a = 0$ or $b = 0$ (A field has no zero divisors: $ab = 0 \Rightarrow a = 0$ 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 $x, y, z \in F$ , $(y+z)x = x(y+z) = xy + xz = yx + zx$ , 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 $ab = 0$ then $a = 0$ or $b = 0$ by [L2], which is exactly the condition of Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ 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 $1 \ne 0$ 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 $1 \ne 0$ , 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: $1 \ne 0$ by [A4]; and if $x \in F$ with $x \ne 0$ , 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 $F \setminus \{0\}$ .
Where else the same clause is in play. The published Multiplication by zero: $0 \cdot a = 0$ cites Field for the right distributive law $(y+z)x = yx + zx$ , 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: $ab = 0 \Rightarrow a = 0$ or $b = 0$ rather than rederived; the general ring statement In any ring $0 \cdot a = a \cdot 0 = 0$ , $(-a)b = a(-b) = -(ab)$ , $(-a)(-b) = ab$ , $(-1)a = -a$ and $a(b - c) = ab - ac$ is not available as a substitute, because a ring may perfectly well have zero divisors.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 \ne 01=0; it is an integral domain, and it is a commutative division ring