Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 10E / a

Suppose $I$ is prime. Ifthen $ab\in I$, so $a\in I$ or $b\in I$. Thus one factor is zero and $R/I$ is an integral domain.

Conversely, if $R/I$ is an integral domain and $ab\in I$, thenso $a+I=0+I$ or $b+I=0+I$. Therefore $a\in I$ or $b\in I$, and $I$ is prime. This proves the prime ideal quotient criterion.

In a Boolean ring, $(r+1{)}^2=r+1$. Using $r^2=r$ and $1^2=1$ givesso

If $R$ is also a nonzero integral domain, thenforces every $r$ to equal $0$ or $1$. Hence $R\cong {\mathbb{F}}_2$.

If $I$ is prime in a Boolean ring, then $R/I$ is a nonzero Boolean integral domain by part (i), and hence is ${\mathbb{F}}_2$. Since the quotient is a field,This is the prime ideals of a Boolean ring are maximal property.