Field

A field is a set $F$ equipped with two binary operations $+$ (addition) and $\cdot$ (multiplication) and two distinguished elements $0 \ne 1$ 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 $x \ne 0$ 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 $x/y := x \cdot y^{-1}$ (for $y \ne 0$), 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 $0 \ne 1$ 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 $x/y := x y^{-1}$, are legitimate.