= Solution
A <prime ideal> is a proper <ideal> $I\neq R$ such that $ab\in I$ implies $a\in I$ or $b\in I$, for every $a,b\in R$.
Suppose first that $I$ is prime. Since $I$ is proper, $1+I\neq0+I$ in the <quotient ring>. If $(a+I)(b+I)=0+I$, then $ab\in I$, so primeness gives $a+I=0+I$ or $b+I=0+I$. Thus the commutative <quotient ring> has a nonzero identity and no nonzero <zero divisors>, hence is an <integral domain>.
Conversely, if $R/I$ is an <integral domain>, its identity is nonzero, so $I\neq R$. If $ab\in I$, then $(a+I)(b+I)=0$ in $R/I$. Absence of nonzero <zero divisors> implies $a+I=0$ or $b+I=0$, namely $a\in I$ or $b\in I$. Therefore
$$
\boxed{I\text{ is prime}\ \Longleftrightarrow\ R/I\text{ is an integral domain}.}
$$
The properness condition matters: taking $I=R$ would produce the <zero ring>, which is not an <integral domain>.
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