Past exam of the algebra 2 course of the University of Oxford / 2014 / 1 / a

A prime ideal is a proper ideal $I\ne R$ 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, $1+I\ne 0+I$ 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 $I\ne R$. 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 $b\in I$. ThereforeThe properness condition matters: taking $I=R$ would produce the zero ring, which is not an integral domain.