[Math] Commutative ring addition where $a^2 = a$

Question

I'm trying to solve following question: If $a^2=a$ for all $a \in R$ where $R$ is a commutative ring, then $a+a=0$.

I have tried to solve this problem for a while now and I'm more or less stuck. I will show you what I have come up with, and perhaps I'm on the right track – otherwise please point me in the correct direction.

$(1)$ If $a^2 = a$ then $(a^2)^k = a^k = a$ for any integer $k$, e.g. $a^3 = a^2a = a^2 = a$ and so on. This made me think that $a$ is either 1 or 0.

$(2)$ $a^2 = a \implies a^2+a = 2a \implies a(a+1) = 2a$. This in turn implies that either $a=1$ (and $a+1 = 2a$ holds) or $a=2a$ (and $a+1 = 1$). In the second case: $2a=a \implies a+a = a \implies a = 0$.

Now for $a=0$ then clearly $a+a = 0$, but if $a=1$ then the statement is false.

Does this prove anything?

Answer 1

$a+a = (a+a)^2 = a^2 + a^2+a^2 + a^2 = a+a+a+a$, and so $a+a=0$.