R/P is an integral domain iff P is prime

Statement

Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Then the quotient ring $R/P$ is an integral domain if and only if $P$ is a prime ideal.

Remarks

The condition $1 + P \neq 0 + P$ in $R/P$ is the condition $P \neq R$. The condition that $(a + P)(b + P) = 0$ implies $a + P = 0$ or $b + P = 0$ is the condition that $a b \in P$ implies $a \in P$ or $b \in P$. For example, $\mathbb{Z}/n\mathbb{Z}$ is an integral domain exactly when $n$ is prime or $n = 0$.

Explanations

L1 human

The theorem translates a property of the ideal into a property of the quotient ring. An element is zero in $R/P$ exactly when it lies in $P$. So "a product in $P$ has a factor in $P$" becomes "a zero product has a zero factor".