TITLE: Proving a Ring is commutative if every element is idempotent
QUESTION [9 upvotes]: I had this question on a final exam.  I was wondering if anyone knows a proof for it.
Let R be a ring not necessarily with unity ($1 \neq 0$ and $x \in R$ so that $x*1=x=1*x$) and let R have the property that every element of R is idempotent that is $\forall x \in R\quad x*x=x$.  Prove that $R$ is commutative.
Not to put my whole answer up because I know I was wrong.  I will put the part where I had trouble going to the next step. 
My argument went like this:  let $a,b \in R$ then $ab\in R$ cause $R$ is a ring.  Therefore $abab=ab$.   
Now $aabb=ab$ as well, so $$abab=aabb\ \Rightarrow abab-aabb=0 \Rightarrow\ a(ba-ab)b=0\ \because\text{ distributive property of $R$}.$$ 
Now because R is not a division ring then the element $a$ could be a divisor of zero or $b$ could be as well so you cannot just assume $ba-ab=0$.  This is where I am stuck.

REPLY [16 votes]: Note that $(x+y)=(x+y)^2=x^2+xy+yx+y^2$. Now, using the property that $x^2=x$ and $y^2=y$, we have $xy+yx=0$ so that $xy=-(yx)$. Also, since $x=x^2=(-x)^2=-x$, each element is its own additive inverse, i.e., $xy=yx$ so the ring is commutative.