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Background

Let $R$ be a ring or a semigroup. We say that $x\in R$ is a von Neumann regular element of $R$ if there exists $y\in R$ such that

$$xyx=x.$$

Any $y\in R$ satisfying the above equation is called a pseudoinverse of $x$. (It is usually not unique.) It is a useful notion which extends outside ring and semigroup theory. For example, it can be proven that for any $(m\times n)$-matrix $A$ over a field $K$ there exists (and can be effectively found) a $(n\times m)$-matrix $X$ over $K$ such that

$$AXA=A.$$

This pseudoinverse of $A$ can be used to solve the linear system

$$Ax=b.$$

This equation has solutions iff $Xb$ is a solution, which is easy to verify. It isn't much more difficult to check that if $Xb$ is a solution, then

$$x=Xb+(I-XA)y,$$

where $y$ is arbitrary, gives all solutions $x$ of the system.

Let's say that $y$ is a 2-pseudoinverse of $x$ if

$$yxy=y.$$

It is easy to see that if $x$ has a pseudoinverse $y$, then it has a 2-pseudoinverse too. Indeed, in this case, we have

$$(yxy)x(yxy)=yxyxy=yxy$$

so $yxy$ is a 2-pseudoinverse of $x.$

Question

I would like to know whether the existence of a 2-pseudoinverse implies the existence of a pseudoinverse. That is, if I have $y$ such that

$$yxy=y,$$

will I always find $y'$ such that

$$xy'x=x?$$

In yet other words, if $y$ is a regular element with a pseudoinverse $x,$ must $x$ be a regular element? (Is it true in semigroups? Is it true in rings?)

The problem is that nothing in the equation $yxy=y$ allows me to reduce some expression to $x,$ which is essentially what I would have to do in order to prove that the answer is "yes". On the other hand, to prove it is "no", I would have to find a counter-example and I have no idea where to look.

It is clear that I can't look for counter-examples in semigroups or rings in which all elements are regular. (Which are called regular semigroups and von Neumann regular rings respectively). Unfortunately, I know next to nothing about regular elements in semigroups which are not regular and in rings which are not von Neumann regular.

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1 Answer 1

up vote 1 down vote accepted

No:

Let $S$ be a non-regular semigroup with a zero element $0$. Then $0$ is regular, and every element of $S$ is a pseudoinverse of $0$ (i.e. $0x0 = 0$ for all $x\in S$). But not every element of $S$ is regular.

This works for rings for rings as well, because the multiplicative semigroup associated with a ring always has a zero.

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I assume this works for rings as well? –  Tara B Mar 10 '12 at 13:58
    
Thank you very much! Yes, of course it does. I'd ask what about semigroups without zero, but I'd better think first. –  user23211 Mar 10 '12 at 14:30
    
Oh yes, of course it does. I added that to my answer now. –  Tara B Mar 10 '12 at 15:06

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