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 domain.

Let $a$ be an arbitrary nonzero element in $R$.
We prove that $a$ is invertible.
Consider the ideal $(a^2)$ generated by the element $a^2$.

If $(a^2)=R$, then there exists $b\in R$ such that $1=a^2b$ as $1\in R=(a^2)$.
Hence we have $1=a(ab)$ and $a$ is invertible.

Next, if $(a^2)$ is a proper ideal, then $(a^2)$ is a prime ideal by assumption.
Since the product $a\cdot a=a^2$ is in the prime ideal $(a^2)$, it follows that $a\in (a^2)$.
Thus, there exists $b\in R$ such that $a=a^2b$.
Equivalently, we have $a(ab-1)=0$.

We have observed above that $R$ is an integral domain. As $a\neq 0$, we must have $ab-1=0$, and hence $ab=1$.
This implies that $a$ is invertible.

Therefore, every nonzero element of $R$ is invertible.
Hence $R$ is a field.

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 […]

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 […]

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 Prime Ideal is Maximal if $a^n=a$ for any Element $a$ in the Commutative Ring
Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.
Then prove that every prime ideal is a maximal ideal.
Hint.
Let $R$ be a commutative ring with $1$ and $I$ be an ideal […]

Is the Quotient Ring of an Integral Domain still an Integral Domain?
Let $R$ be an integral domain and let $I$ be an ideal of $R$.
Is the quotient ring $R/I$ an integral domain?
Definition (Integral Domain).
Let $R$ be a commutative ring.
An element $a$ in $R$ is called a zero divisor if there exists $b\neq 0$ in $R$ such that […]

$(x^3-y^2)$ 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.
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