[proofplan] We verify the two ingredients in the definition of an [integral domain](/page/Integral%20Domain). The commutative ring structure with identity and the nonzero condition are part of the definition of a field. It remains only to prove that $F$ has no zero divisors, and this follows by multiplying an equation $ab=0_F$ by the inverse of a nonzero factor. [/proofplan]

[step:Use the field axioms to obtain the required ring structure] By definition, a field $F$ is a nonzero commutative ring with multiplicative identity $1_F$ such that every element of $F \setminus \{0_F\}$ has a multiplicative inverse. Therefore the ring-structure requirements in the definition of an integral domain are already satisfied for $F$. [/step]

[step:Multiply by an inverse to rule out zero divisors]Let $a,b \in F$ and suppose that $ab=0_F$. We prove that $a=0_F$ or $b=0_F$. If $a=0_F$, then the desired conclusion holds. Suppose instead that $a \ne 0_F$. Since $F$ is a field, there exists $a^{-1} \in F$ such that $a^{-1}a=1_F$. Multiplying the equation $ab=0_F$ on the left by $a^{-1}$ gives \begin{align*} a^{-1}(ab)=a^{-1}0_F. \end{align*} By associativity of multiplication in $F$, \begin{align*} a^{-1}(ab)=(a^{-1}a)b. \end{align*} Using $a^{-1}a=1_F$ and the identity property of $1_F$, we have \begin{align*} (a^{-1}a)b=1_F b=b. \end{align*} By the absorbing property of $0_F$ under multiplication, \begin{align*} a^{-1}0_F=0_F. \end{align*} Hence $b=0_F$. Thus $ab=0_F$ implies $a=0_F$ or $b=0_F$.[/step]

[guided]We need to prove the part of the integral domain condition that is not merely structural: there must be no zero divisors. So take arbitrary elements $a,b \in F$ and assume \begin{align*} ab=0_F. \end{align*} The goal is to force one of the two factors to be zero. There are two cases. If $a=0_F$, then the conclusion $a=0_F$ or $b=0_F$ is immediate. Now assume $a \ne 0_F$. This is exactly where the field axiom is used: every nonzero element of a field has a multiplicative inverse. Therefore there exists $a^{-1} \in F$ satisfying \begin{align*} a^{-1}a=1_F. \end{align*} Multiplying the equation $ab=0_F$ on the left by $a^{-1}$ gives \begin{align*} a^{-1}(ab)=a^{-1}0_F. \end{align*} Associativity of multiplication lets us regroup the left-hand side: \begin{align*} a^{-1}(ab)=(a^{-1}a)b. \end{align*} Using $a^{-1}a=1_F$ and the identity property of $1_F$, this becomes \begin{align*} (a^{-1}a)b=1_F b=b. \end{align*} On the right-hand side, multiplication by the zero element gives \begin{align*} a^{-1}0_F=0_F. \end{align*} Combining these equalities yields $b=0_F$. Therefore, whenever $ab=0_F$, either $a=0_F$ or $b=0_F$.[/guided]

[step:Conclude that the field satisfies the definition of an integral domain] The preceding step proves that $F$ has no zero divisors. Since $F$ is already a nonzero commutative ring with identity, $F$ satisfies all conditions in the definition of an integral domain. Therefore every field is an integral domain. [/step]