A question regarding a central idempotent in a ring $R$

Question

I am trying to solve the problem:

Question: In a ring $R$ with identity, if every idempotent is central, then prove that for $a, b \in R$, $$ab =1 \implies ba=1$$

I have done in the following manner: $$ab=1\\ \implies b(ab)=b\\ \implies (ba)b-b=0\\ \implies (ba-1)b=0$$

Case 1: If $R$ contains no divisor of Zero then as $b \ne 0$ ,we get $$ba-1=0 \implies ba=1$$

Case 2: If $R$ contains divisor of Zeros then we may have $$ba-1\ne0\\ \implies ba\ne1\\ \implies (ab)a\ne a$$ But $ab=1$ hence $a\ne a$, an absurd condition. So $ba-1=0$ or $ba=1$.

I think I have solved the problem but I haven't used the given conditions "Every idempotent is central". So I think there is something wrong which I have done but can't find out! Please rectify my mistake if I am wrong and provide any hint to solve it.

Answer 1

Here's kind of a slick solution (don't read further if you want to figure it out for yourself!)

$$ab = 1$$ $$\Longrightarrow (ba)^{2} = b(ab)a = ba$$

So $ba$ is an idempotent and therefore central. So we have:

$$b(ba) = (ba)b$$ $$\Rightarrow ab(ba) = a(ba)b$$ $$\Rightarrow (ab)(ba) = (ab)(ab)$$ $$\Rightarrow ba = 1$$

Answer 2

As mentioned in the comments, $ba\neq 1$ does not imply $aba\neq a$. In rings for which there exists $ab=1$ and $ba\neq 1$ (see here for examples) you have obviously that $aba=(ab)a=a$.

But clearly, if $ab=1$, $ba$ is at least idempotent, hence central by your hypotheses.

Then

$ba=(ba)ab =a(ba)b=abab =1$.