TITLE: Showing that right quasi regular elements are invertible
QUESTION [6 upvotes]: I am trying to show that a right quasi regular element $(\neq 0$ or $1$) in a ring $R$, where all regular elements different from $1$ are right quasi regular, is invertible. 
I have managed to show that an element is regular iff invertible so was hoping to show that my element is regular, but can't see how to use this. Any hints? I'm also interested in what the meaning behind these properties are, they seem very arbitrary!
(By regular I mean an element a in a ring $R$ such that $aba=a$ for some $b$ in $R$. By right-quasi regular I mean an element a such that $a + x -ax=0$ for some $x$ in $R$.)

REPLY [5 votes]: I will give an elementary proof that every right quasi-regular element $a \in R\setminus \{0,1\}$ is invertible. To do so we prove:

The only idempotent elements in $R$ are $0$ and $1$;
Every non-zero regular element in $R$ is invertible;
Every right quasi-regular element $a \in R\setminus \{0,1\}$ is invertible.

Proof of 1:
Note that if $e$ is an idempotent element in $R$, then $e$ is regular and hence either $e=1$ or $e$ is right quasi-regular. If $e$ is right quasi-regular we have $e + x - ex=0$ for some $x$ and so $e = e^2 +ex - e^2x = e(e+x-ex)=e0=0$. It follows that every idempotent element in $R$ is either $0$ or $1$.
Proof of 2:
Suppose that $a$ is a regular element and $a \neq 0$, then by definition there exists $b$ such that $aba=a$ and hence $ab$ and $ba$ are idempotent (i.e. $(ab)(ab)=(aba)b=ab$ and $(ba)(ba)=b(aba)=ba$). Since $ab$ can't be $0$ because then $a=aba=0$ it follows from 1 that $ab=1$ and similarly that $ba=1$. This proves that every non-zero regular element is invertible.
Proof of 3: Suppose that $a$ is a right regular element such that $a \neq 0$ and $a \neq 1$. By assumption there exists $x$ such that $a + x -ax=0$. It follows that $(1-a)(1-x) = 1-x-a+ax=1$ and hence $(1-a)(1-x)(1-a) = 1-a$ so that $1-a$ is a non-zero regular element. By assumption there exists $y$ such that $1-a + y -(1-a)y=0$.  However this means that $1 -a + ay=0$ and hence $1 = a -ay= a(1-y)$. This means that $a(1-y)a=a$ and hence $a$ is invertible by 2.