A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors.
That is, if $ab=0$ for $a, b \in R$, then either $a=0$ or $b=0$.

Proof.

We give two proofs.

Proof 1.

Let $r \in R$ be a nonzero element in $R$.
We show that $r$ is a unit.

Consider the map $f: R\to R$ sending $x\in R$ to $f(x)=rx$.
We claim that the map $f$ is injective.
Suppose that we have $f(x)=f(y)$ for $x, y \in R$. Then we have
\[rx=ry\]
or equivalently, we have
\[r(x-y)=0.\]

Since $R$ is an integral domain and $r\neq 0$, we must have $x-y=0$, and thus $x=y$.
Hence $f$ is injective. Since $R$ is a finite set, the map is also surjective.

Then it follows that there exists $s\in R$ such that $rs=f(s)=1$, and thus $r$ is a unit.
Since any nonzero element of a commutative ring $R$ is a unit, $R$ is a field.

Proof 2.

Let $r\in R$ be a nonzero element.
We show that the inverse element of $r$ exists in $R$ as follows.
Consider the powers of $r$:
\[r, r^2, r^3,\dots.\]
Since $R$ is a finite ring, not all of the powers cannot be distinct.
Thus, there exist positive integers $m > n$ such that
\[r^m=r^n.\]

Equivalently we have
\[r^n(r^{m-n}-1)=0.\]
Since $R$ is an integral domain, this yields either $r^n=0$ or $r^{m-n}-1=0$.
But the former gives $r=0$, and this is a contradiction since $r\neq 0$.

Hence we have $r^{m-n}=1$, and thus
\[r\cdot r^{m-n-1}=1.\]
Since $r-m-1 \geq 0$, we have $r^{m-n-1}\in R$ and it is the inverse element of $r$.

Therefore, any nonzero element of $R$ has the inverse element in $R$, hence $R$ is a field.

Related Question.

Problem. Let $R$ be a finite commutative ring with identity $1$. Prove that every prime ideal of $R$ is a maximal ideal of $R$.

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[…] Problem Finite Integral Domain is a Field, any finite integral domain is a field. This yield that $R/I$ is a field, and hence $I$ is a […]