Field

Thus, under the convention used here that a ring need not have a multiplicative identity, the unital condition is part of the definition of a field.

Written out in terms of the operations, a field is a set $F$ with binary operations $+$ and $\cdot$ and elements $0\neq 1$ such that for all $a,b,c\in F$:

• (Additive associativity) $(a+b)+c=a+(b+c)$.
• (Additive commutativity) $a+b=b+a$.
• (Additive identity) $a+0=a$.
• (Additive inverse) there exists $-a\in F$ such that $a+(-a)=0$.
• (Multiplicative associativity) $(ab)c=a(bc)$.
• (Multiplicative commutativity) $ab=ba$.
• (Multiplicative identity) $a1=a$.
• (Multiplicative inverse) if $a\neq 0$ then there exists $a^{-1}\in F$ such that $aa^{-1}=1$.
• (Distributivity) $a(b+c)=ab+ac$.

These are the field axioms. Together with the order axioms and the completeness axiom, they characterize the real numbers up to isomorphism.

For a commutative unital ring $R$ with $1\neq 0$, the following are equivalent: $R$ is a field; its only ideals are $(0)$ and $R$; and $(0)$ is maximal. The hypotheses matter: a nonunital ring can have only two ideals without being a field.

• $\mathbb{Q}$ and $\mathbb{R}$ are fields.
• For a prime $p$, $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$ is a field.
• $\mathbb{Z}$ is not a field since $2$ has no inverse in $\mathbb{Z}$.