How to prove that every subring and quotient ring of any boolean ring is also boolean?

Question

I have that a ring R is boolean if $\ \forall r\in R$ we have that $r^2=r$. I have proved that in any boolean ring $r=-r$ and R is conmutative. My problem is that I'm a little confused on how to see the elements of the quotient ring $R/I$, with $I$ a bilateral ideal; and also how to use the boolean property on subrings.

Answer 1

Clearly, the theorem is true for subrings, since if $S\subset R$ is a subring of $R$, then if $s\in S$, we have $s\in R$ and $s^2=s$.

Now consider the theorem for quotient rings. Let $I\subset R$ be an ideal of $R$, and consider the element $Ix\in R/I$, so that $x\in R$ and $I+x$ is a coset of $I$ (because $R$ is commutative as you stated in the question, it does not matter whether we consider left cosets $x+I$ or right cosets $I+x$). By using coset multiplication, we have that $(I+x)^2=(I+x)(I+x)=I+xx=I+x^2=I+x$ and so $(I+x)^2=I+x$, making $R/I$ boolean.