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 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 zero of $R / J$.

Sufficient Condition
Let $J$ be a prime ideal.

We need to show that if:
 * $x + J, \ y + J \in \left({R / J, +, \circ}\right)$

such that:
 * $\left({x + J}\right) \circ \left({y + J}\right) = \left({x \circ y}\right) + J = 0_{R / J}$

then:
 * $x + J = 0_{R/J}$ or $y + J = 0_{R/J}$.

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

Therefore:
 * $\left({x \circ y}\right) + J = 0_{R / J} \implies x \circ y \in J$

Because $J$ is prime, it follows that either $x \in J$ or $y \in J$.

we assume that $x \in J$.

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

Necessary Condition
Let $A / J$ be an integral domain.

Let $x, y \in R$ be such that $x \circ y \in J$.

Then:


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

Because $A / J$ is an integral domain it follows that:
 * $x + J = 0_{R / J}$ or $y + J = 0_{R / J}$

assume that $x + J = 0_{R / J}$.

Then $x \in J$, and therefore $J$ is prime.