Field

A field is a set $F$ equipped with two binary operations $+$ (addition) and $\cdot$ (multiplication) and two distinguished elements $0\mathrm{\ne }1$ of $F$, satisfying:

• (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 $x+(-x)=0$.
• (M) Multiplication is associative and commutative on all of $F$, and $x\cdot 1=x$ for every $x\in F$; moreover $(F\setminus \{0\},\cdot )$ is an abelian group with identity $1$, so every $x\mathrm{\ne }0$ has a multiplicative inverse $x^{-1}$ with $x\cdot x^{-1}=1$.
• (D) Multiplication distributes over addition: $x\cdot (y+z)=x\cdot y+x\cdot z$.

We write $x-y:=x+(-y)$ and $x\mathrm{/}y:=x\cdot y^{-1}$ (for $y\mathrm{\ne }0$), and abbreviate $x\cdot y$ as $xy$.

• The rationals $\mathbb{Q}$ (The rationals form a field) and both constructions of the reals (The reals form a field, The Dedekind reals form a field) are fields. This definition states the field axioms once, abstractly, so that every field fact proved from them applies to all of these at once.
• The requirement $0\mathrm{\ne }1$ excludes the trivial one-element structure.
• Why (M) quantifies over all of $F$ explicitly (amended 2026-07-28, owner approved). An earlier wording headed (M) with the group $(F\setminus \{0\},\cdot )$ alone, and under that strict reading the axioms do not pin down the products $0\cdot y$: take ${\mathbb{F}}_2$ and redefine $0\cdot y:=y$, which leaves (A) untouched, leaves $(F\setminus \{0\},\cdot )$ the trivial group, and satisfies (D) in all eight instances, yet has $0\cdot 1=1$ — making Multiplication by zero: $0\cdot a=0$ false in a purported model of the axioms. Associativity, commutativity and $x\cdot 1=x$ are therefore asserted on all of $F$, the element $0$ included. Commutativity on all of $F$ is also what turns (D) into its right-hand form $(y+z)\cdot x=y\cdot x+z\cdot x$, which Multiplication by zero: $0\cdot a=0$ uses.
• The identities $0,1$ and the inverses $-x,x^{-1}$ are unique (Identities and inverses in a field are unique ↗), so writing them as single-valued notation, and the abbreviations $x-y:=x+(-y)$ and $x\mathrm{/}y:=x{y}^{-1}$, are legitimate.