# Prime Ideal iff Quotient Ring is Integral Domain

## Theorem

Let $\struct {R, +, \circ}$ be a commutative ring with unity.

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

Then $J$ is a prime ideal of $R$ if and only if 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:

$x + J, \ y + J \in \struct {R / J, +, \circ}$

such that:

$\paren {x + J} \circ \paren {y + J} = \paren {x \circ y} + J = 0_{R / J}$

then:

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

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

Therefore:

$\paren {x \circ y} + J = 0_{R / J} \implies x \circ y \in J$

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

Without loss of generality we assume that $x \in J$.

But then:

$x + J = J = 0_{R / J}$

$\Box$

### Necessary Condition

Let $R / J$ be an integral domain.

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

Then:

$0_{R / J} = J = x \circ y + J = \paren {x + J} \circ \paren {y + J}$

Because $R / J$ is an integral domain it follows that:

$x + J = 0_{R / J}$

or:

$y + J = 0_{R / J}$

If $x + J = 0_{R / J}$, then $x \in J$.

If $y + J = 0_{R / J}$, then $y \in J$.

Thus either $x \in J$ or $y \in J$, and so by definition $J$ is a prime ideal of $R$.

$\blacksquare$