TITLE: Idempotents in rings without unity
QUESTION [8 upvotes]: Suppose there are non-trivial idempotents in the ring without unity. Is it right that all of them are zero divisors? 

If we're given unitary ring with unity $e$ and $a$ is non-trivial idempotent then $e-a \neq 0$. But $a(e-a) = 0$ so $a$ is zero divisor.
But i'm not sure about the case when ring doesn't have unity.
EDIT
In the non-commutative case i wonder whether each non-zero idempotent must be SOME kind of zero divisors, i.e. left or right, or perhaps both.

REPLY [6 votes]: Suppose $e$ is idempotent and not a right or left zero divisor.
Since multiplication on either side by $e$ is injective, $(xe-x)e=0$ and $e(ex-x)=0$ for all x implies $e$ is the identity of the ring.
So you can conclude this:

In a ring without identity, a nonzero idempotent $e$ must be a zero divisor on one side or the other. If instead the ring has an identity other than $e$, $e$ is a zero divisor on both sides.