If $1-ba$ is a regular element, then can we claim that $1-ab$ is regular?

Question

Let $R$ be a ring with unity and $a,b \in R$ and $1-ba$ is a regular element, then can we claim that $1-ab$ is regular?

$a\in R$ is regular if there exists $b \in R$ such that $a=aba.$

Answer 1

$1-ba$ is regular means that there's $x$ so that $(1-ba) = (1-ba)x(1-ba).$

The trick for these types of problems is to use a factorization that a priori makes no sense, but ultimately will produce something that does. Write, inspired by the geometric series formula,

$$(1-ab)^{-1} = 1 + ab + abab + \cdots = 1 + a(b + bab + \cdots) = 1 + a(1 + ba + baba + \cdots)b = 1 + a(1+ba)^{-1}b.$$

Now, if we are really abusive, the regularity relation means that $x = (1-ba)^{-1}.$ So let's just try and solve for some $y$ so that

$$y = 1 + axb.$$

Forgetting all that formal nonsense, the definition of $y$ makes sense. And now we compute. We have $$(1-ab)y(1-ab) = (y-aby)(1-ab) = y - yab - aby + (ab)y(ab).$$ A similar expansion reveals $$1-ba = (1-ba)x(1-ba) = x - xba - bax + (ba)x(ba).$$

So $$(ab)y(ab) = (ab)(1 + axb)(ab) = (ab)(ab + axbab) = abab + a((ba)x(ba))b.$$

The term $$yab = (1+axb)(ab) = ab + axbab = ab + a(xba)b$$ and $$aby = (ab)(1+axb) = ab + (ab)axb = ab + a(bax)b.$$

So, $$a(1-ba)b = (axb) - a(xba)b - a(bax)b + a(baxba)b = (-1 + y) - (yab - ab) - (aby - ab) + (abyab -abab) = (1-ab)y(1-ab) - 1 + ab + ab - abab.$$

But $a(1-ba)b = ab - abab.$ So, moving all the terms to one side, we get

$$(1-ab)y(1-ab) = 1-ab,$$ as desired!