[proofplan] We verify the defining properties of an [integral domain](/page/Integral%20Domain). A field is already a commutative ring with identity, so the only substantive point is to prove that it has no zero divisors. Given $a,b \in k$ with $ab=0$, if $a \neq 0$, the field inverse $a^{-1}$ exists, and multiplying by it forces $b=0$. [/proofplan]

[step:Use the field axioms to obtain the ring structure required for an integral domain] By definition, a field $k$ is a commutative ring with multiplicative identity $1_k$ and with $0_k \neq 1_k$. These are exactly the ambient ring-theoretic structural conditions required in the definition of an integral domain. It remains only to prove that $k$ has no zero divisors. [/step]

[step:Cancel a nonzero factor by multiplying by its field inverse]Let $a,b \in k$ satisfy $ab=0_k$. Suppose first that $a \neq 0_k$. Since $k$ is a field, there exists an element $a^{-1} \in k$ such that $a^{-1}a=1_k$. Multiplying the equality $ab=0_k$ on the left by $a^{-1}$ gives \begin{align*} a^{-1}(ab)=a^{-1}0_k. \end{align*} By associativity of multiplication in the ring $k$, the left-hand side is \begin{align*} a^{-1}(ab)=(a^{-1}a)b=1_k b=b. \end{align*} By the absorbing property of $0_k$ under multiplication, the right-hand side is \begin{align*} a^{-1}0_k=0_k. \end{align*} Therefore $b=0_k$.[/step]

[guided]We must show that a product can be zero only when at least one factor is zero. Let $a,b \in k$ be elements satisfying $ab=0_k$. The useful extra structure in a field is that every nonzero element has a multiplicative inverse, so we split according to whether the first factor is zero. Assume $a \neq 0_k$. By the field axiom for inverses, there exists an element $a^{-1} \in k$ satisfying $a^{-1}a=1_k$. We now multiply the equation $ab=0_k$ on the left by $a^{-1}$. This is valid because multiplication in $k$ is a well-defined ring operation, and it gives \begin{align*} a^{-1}(ab)=a^{-1}0_k. \end{align*} Associativity allows us to regroup the left-hand side: \begin{align*} a^{-1}(ab)=(a^{-1}a)b. \end{align*} Using the inverse identity $a^{-1}a=1_k$ and then the identity property of $1_k$, we get \begin{align*} (a^{-1}a)b=1_kb=b. \end{align*} On the right-hand side, the ring absorbing law gives \begin{align*} a^{-1}0_k=0_k. \end{align*} Combining these equalities yields $b=0_k$. Thus, whenever $ab=0_k$ and $a$ is nonzero, the other factor must be zero.[/guided]

[step:Conclude that the field has no zero divisors] From the previous step, for all $a,b \in k$, if $ab=0_k$ and $a \neq 0_k$, then $b=0_k$. Equivalently, for all $a,b \in k$, \begin{align*} ab=0_k \implies a=0_k \text{ or } b=0_k. \end{align*} Thus $k$ has no zero divisors. Since $k$ is a commutative ring with identity and $0_k \neq 1_k$, it satisfies the definition of an integral domain. [/step]