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.

If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.
Let $R$ be a commutative ring with $1$.
Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.
Proof.
As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
Hence $R=R/\{0\}$ is an integral […]

Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]

Every Maximal Ideal of a Commutative Ring is a Prime Ideal
Let $R$ be a commutative ring with unity.
Then show that every maximal ideal of $R$ is a prime ideal.
We give two proofs.
Proof 1.
The first proof uses the following facts.
Fact 1. An ideal $I$ of $R$ is a prime ideal if and only if $R/I$ is an integral […]

Torsion Submodule, Integral Domain, and Zero Divisors
Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$.
The set of torsion elements is denoted
\[\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.\]
(a) Prove that if $R$ is an […]

$(x^3-y^2)$ is a Prime Ideal in the Ring $R[x, y]$, $R$ is an Integral Domain.
Let $R$ be an integral domain. Then prove that the ideal $(x^3-y^2)$ is a prime ideal in the ring $R[x, y]$.
Proof.
Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$.
Define the function $\Psi:R[x,y] \to R[t]$ sending […]

No Nonzero Zero Divisor in a Field / Direct Product of Rings is Not a Field
(a) Let $F$ be a field. Show that $F$ does not have a nonzero zero divisor.
(b) Let $R$ and $S$ be nonzero rings with identities.
Prove that the direct product $R\times S$ cannot be a field.
Proof.
(a) Show that $F$ does not have a nonzero zero divisor.
[…]

If a Prime Ideal Contains No Nonzero Zero Divisors, then the Ring is an Integral Domain
Let $R$ be a commutative ring. Suppose that $P$ is a prime ideal of $R$ containing no nonzero zero divisor. Then show that the ring $R$ is an integral domain.
Definitions: zero divisor, integral domain
An element $a$ of a commutative ring $R$ is called a zero divisor […]