Is Every Ring Where a Squared Equals a Itself Commutative?

[Bashyboy]

Homework Statement

If $a^2 = a$ for all $a \in R$, then $R$ is commutative.

Homework Equations

The Attempt at a Solution

I have been working on this problem for a few hours without any success; I literally have nothing but a's and b's scrawled over a bunch of papers. I could use a hint.

 

[fresh_42]

Have you calculated $(a-b)^2$ on your papers? And $(1+1)^2$.

 

[Bashyboy]

I just did now, but I don't see how this is helpful. Here are some of my calculations:

$a-b = (a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2$ or $a-b = a - ab -ba + b$ or $2b = ab + ba$...

In fact, here is something strange I encountered: $(-1)^2 = -1$ becomes $1 = -1$. Isn't this a contradiction? If so, then Boolean rings cannot exist.

 

[fresh_42]

I just did now, but I don't see how this is helpful. Here are some of my calculations:

$a-b = (a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2$ or $a-b = a - ab -ba + b$ or $2b = ab + ba$...

In fact, here is something strange I encountered: $(-1)^2 = -1$ becomes $1 = -1$. Isn't this a contradiction? If so, then Boolean rings cannot exist.

No, it's no contradiction. Therefore you should compute $(1+1)^2=1^2+ 1 \cdot 1+ 1 \cdot 1 + 1^2=1+1$, hence $1+1=0$, i.e. the characteristic of this ring is two. This means, as you've noticed: $+1=-1$.
Now you know what $b+b$ and $-ab$ are.