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 Commutative Quotient Ring and Quotient 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.