Exercise 3.11

A ‘Boolean ring’ is a ring $R$ such that $\forall a\in R,a^2=a$.

• Prove that for all $a$ in a Boolean ring, $2a=0$.
• Prove that Boolean rings are commutative, that is, $ab=ba$ for all $a,b$. Hint: Consider $(a+a{)}^2,(a+b{)}^2$

Answer

Let $R$ be a Boolean ring.

a. Let $a\in R$. Since $R$ is a Boolean ring, $2a=a+a=(a+a{)}^2=a^2+a^2+a^2+a^2=4a^2=4a$. Thus, $2a=0$.

b. Let $a,b\in R$. Since $R$ is a Boolean ring, $a+b=(a+b{)}^2=a^2+ba+ab+b^2=a+ba+ab+b$. Thus, $ba+ab=0$. Thus, $ba=-ab$. From part a, we have $a=-a$ for all $a\in R$ . Therefore, $ba=-ab=ab$. $■$