multiplication of noncommutative idempotents

Question

Let $R$ be a ring (possibly without unity element) and let $I(R)$ be the set of all idempotents of the ring $R$, i.e. $I(R)=\{a\in R \; : a^2=a\}$. Moreover, let's assume that: $$\forall_{e,f\in I(R)}\; ef=0\iff fe=0.$$ Is the set $I(R)$ closed under multiplication? So, is the product $ef$ of two arbitrary idempotents is an idempotent again ($(ef)^2=ef$)?

Answer 1

There is a proof if the ring has identity. I'm currently working to see if the argument can be adapted or not to rings without identity. (Thanks for your patience.)

You can check that $e+ef-efe$ is another idempotent.

Now (if $R$ has identity) $1-e$ annihilates this on the left, and so it annihilates it on the right too. That yields $ef(1-e)=0$, so that $ef=efe$. Then $efef=eff=ef$.

Work on remainder of problem

I had hoped that unitization might just work, where the idempotents of the Dorroh extension $\mathbb Z\times R$ are things of the form $(0,e)$ and $(1, -e)$ for each idempotent $e\in R$. The problem is that the condition of symmetric annihilation seems hard to enforce for products like $(0,e)(1,-f)=(0, e-ef)$. This would amount to showing that $e-ef=0$ implies $e-fe=0$.

In the way of counterexamples, I was hoping a semigroup ring over the semigroup $\{a,b\}$ given by $a^2=a=ab$ and $b^2=b=ba$ would yield a counterexample. No counterexample emerged after trying a few small coefficient rings, though.