Showing that right quasi regular elements are invertible

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 $R$.)

• Are we 100% sure about this statement? In a ring where regular elements are rqr, then every rqr element is invertible? In such a ring, all regular elements are invertible? It's not implausible: is see that this is the case for domains... – rschwieb Sep 10 '15 at 16:14
• So my proof goes: Suppose $a\in R$ is invertible and $a\neq 0$. Since $aa^{-1}=1$ we have $aa^{-1}a=a$, so $a$ is regular. For the converse, suppose $a$ is regular, that is we have a $b\in R$ such that $aba=1$. Since the identity is unique in a ring, we have $ab=1=ba$, and hence $b=a^{-1}$. Apologies, I'm only just learning ring theory and am unaware what a domain is. Do you mean an integral domain? – Calvin Nesbitt Sep 11 '15 at 2:22
• Apologies, I have noticed that my proof for $(1 - a)$ being invertible only applies if $a$ is nilpotent, which isn't necessarily true. I've removed this. – Calvin Nesbitt Sep 11 '15 at 2:28
• lets gloss over my earlier comment and run with what you just said. There is no reason to think $aba=1$: you only have $aba=a$ for some b – rschwieb Sep 11 '15 at 2:28
• Agreed. I'm unsure why I wrote this at the time, my proof was done a few days ago. Regular iff invertible was a preamble to the above problem, so it should be true. I guess I need to find a proof for this too now. – Calvin Nesbitt Sep 11 '15 at 2:33

2 Answers

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:

1. The only idempotent elements in $R$ are $0$ and $1$;
2. Every non-zero regular element in $R$ is invertible;
3. 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.

Probably these results will serve to resolve your question

1. In an arbitrary ring $R$, if $rad(R)\neq R$ then $rad(R)=\{r\in R| \forall x,y\in R: \;xry\mbox{ is right quasi-regular}\}$

proof: Remember that $Rad(M)=\bigcap\{M|M\mbox{ maximal modular right ideal of }R\}$ and let $H=\{r\in R| \forall x,y\in R: \;xry\mbox{ is right quasi-regular}\}$. Note that if an element $a\in R$ is not a right quasi-regular then there is a modular maximal ideal $M$ such that $a\notin M$. in this way, if $a$ is not right quasi-regular, the right ideal $S=\{at-t| t\in R\}$ does not contain $a$.

In particular, applying the Zorn Lemma you can see that there exist a right ideal $M$ maximal in the set of all the right ideals which contains $S$ but not the element $a$. moreover, it is easy to verify that this ideal is a right maximal in $R$. On the other hand $M$ is modular since $ar-r\in M$ for each $r\in R$, so if $b\notin H$ there are elements $x$, $y$ in $R$ such that $xby$ is not right quasi-regular. then there is a maximal modular right ideal which does not contain $xby$ and hence, does not contain $b$, and $b\notin Rad(R)$.

Conversely, one has to prove first that if $M$ is a maximal modular right ideal in $R$ then $H\subseteq (R:M)=\{s\in R| \forall m\in M, sm\in R\}$ (the proof of this propiety you can found in T.Y.Lam, a first course of noncommutative ring or every book in ring theory). Since $rad(R)\neq R$ then there exists a maximal modular right ideal $M$ in $R$ and $H\subseteq \bigcap\{(R:M)| M\mbox{maximal modular right ideals in }R\}$. Hence $RH\subseteq M$ for each $M$ and $M$ being modular, we also have $H\subseteq M$. Finally, $H\subseteq rad(R)$.

1. In a ring with identity $R$, $rad(R)=\{r\in R|\forall s\in R\mbox{ the element }1-rs \mbox{ is right invertible}\}$.

proof: It's time to use your observation, you see that in a ring with identity $R$ an element $r\in R$ is right quasi-regular if and only if is right invertible ($r+s-rs=0\Leftrightarrow (1-r)(1-s)=1$). Now, we define

$M=\{r\in R|xry \mbox{ is right quasi-regular in }R,\forall x\in R\}$ and $N=\{r\in R|xr \mbox{ is right quasi-regular in }R,\forall x\in R\}$, because the ring having identity, then $M \subseteq N$. On the other hand if $xr+t-xrt=0$ then $xry+ty-xrty=(xr+t-xrt)y=0$ and so $N\subset M$, and you can use the previous result, so the identity is done.