Prime Ideal iff Quotient Ring is Integral Domain

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity.

Let $J$ be an ideal of $R$.

Then $J$ is a prime  ideal iff the quotient ring $R  / J$ is an integral domain.

Proof
Since $J \subset R$, it follows from Quotient Ring of Commutative Ring is Commutative and Quotient Ring of Ring with Unity is Ring with Unity that  $R / J$ is a commutative ring with unity.

Let $0_{R/J}$ be the additive identity of $R/J$.

First suppose that $J$ is prime.

We need to show that if $x+J,\ y+J\in\left({R / J, +, \circ}\right)$ such that $(x+J)\circ(y+J)=(x\circ y)+J=0_{R/J}$ then $x + J=0_{R/J}$ or $y+J=0_{R/J}$.

The additive identity of $R/J$ is $0_{R/J}=J$.

Therefore, $(x\circ y)+J=0_{R/J}$ implies that $x\circ y\in J$.

Because $J$ is prime, it follows that either $x\in J$ or $y\in J$. Without loss of generality we assume that $x\in J$.

But then $x+J=J=0_{R/J}$.

Conversely, suppose that $A/J$ is an integral domain, and $x,y\in R$ are such that $x\circ y\in J$. Then:


 * $0_{R/J}=J=x\circ y+J=(x+J)\circ (y+J)$.

Therefore, because $A/J$ is an integral domain it follows that $x+J=0_{R/I}$ or $y+J=0_{R/I}$. Without loss of generality assume that $x+J=0_{R/I}$. Then $x\in J$, and therefore $J$ is prime.