Commutativity of Boolean ring

Question

I could not prove that "Every Boolean ring is commutative", but I found it on the Internet. I am giving the idea of the proof that I "learned".

Let $R$ be the Boolean ring.

We can easily prove that $x+x = 0 \; \forall \; x \in R$. Then, we again use the same idea that $(x+y)^2 = (x+y), \; \forall \; x,y \in R$ to get that $xy=-yx$ and use the fact that $xy= -xy$ to get the result.

We had to find the new property for the Boolean ring that $x+x =0 $ to solve that it is commutative. Even with this solution, I cannot understand what makes a Boolean ring commutative. Is there any intuitive way to understand that Boolean ring is commutative? I am not looking for just a computational answer (if possible, there should be some idea behind the computation). I am hoping for any sort of intuition or idea (mathematical or in plain English) that tells us why Boolean ring is commutative.

P.S.: A Boolean ring $R$ is a ring in which $a^2 = a, \; \forall \; a \in R $.

Answer 1

$$x+y = (x+y)^2 = x^2 + xy + yx + y^2 = x + xy + yx + y$$

This shows $xy + yx = 0$. That means $xy + \underbrace{xy + yx}_{=0} = xy$. From your statement ($X + X = 0$) we can rearrange to see:

$$\underbrace{xy + xy}_{=0} + yx = xy \Rightarrow yx = xy$$

Answer 2

First a small claim: If $R$ is Boolean, then $-a=a$ for all $a \in R$.

$\textbf{Proof of claim:}$ We know that

$$-a=(-a)^2=(-1)^2a^2=a$$

and the proof of the claim is done.

Next let $a,b \in R$, we aim to show $ab=ba$. Consider the sum

\begin{align} a+b &=(a+b)^2 && \text{as $R$ is Boolean}\\ &=a^2+ab+ba+b^2 && \text{factoring}\\ &=a+ab+ba+b && \text{as $R$ is Boolean} \end{align}

This forces

$$ab+ba=0$$

replace $ab$ with $-ab$ and we conclude that

$$ba=ab$$

as needed.

Answer 3

$(a+b)^2 = a+b \implies ab+ba=0$.

Now multiply $ab+ba=0$ with $a$ from left side we have $ab+aba=0$ and multiply $ab+ba=0$ with $a$ from right side $aba+ba=0$.

Now equating $(-aba)$'s value we have $ab=ba$ for all $a,b$ in the ring.