In von Neumann regular rings every element has a von Neumann inverse. Are there many rings in which these inverses are unique for non-zero elements? Are there any such domains that are not skew fields? Are there any such non-domains?

Printable View

- Dec 15th 2011, 07:02 AMymarunique von neumann inverse in rings
In von Neumann regular rings every element has a von Neumann inverse. Are there many rings in which these inverses are unique for non-zero elements? Are there any such domains that are not skew fields? Are there any such non-domains?

- Dec 15th 2011, 02:52 PMNonCommAlgRe: unique von neumann inverse in rings
- Dec 15th 2011, 03:01 PMymarRe: unique von neumann inverse in rings
Oh, you're right, sorry. I want to assume that. I would like to know if

1) there are any von Neumann regular domains which are not skew fields, and whose non-zero elements have unique von Neumann inverses;

2) there are any von Neumann regular rings which are not domains, and whose non-zero elements have unique von Neumann inverses. - Dec 15th 2011, 03:13 PMNonCommAlgRe: unique von neumann inverse in rings
the answer to the first question is no. any von Neumann regular domain is a division ring. because from $\displaystyle a = axa$ we get $\displaystyle a(1-xa)=0$ and since your ring is a domain and $\displaystyle a$ is non-zero, we have $\displaystyle xa=1$.

your second question is more interesting and i need to think about it. - Dec 15th 2011, 05:06 PMNonCommAlgRe: unique von neumann inverse in rings
ok, i can't believe that i thought the answer was yes! the answer is actually no. to see this, let $\displaystyle R$ be a ring with this property that for every nonzero element $\displaystyle a \in R,$ the equation $\displaystyle a = axa$ has a unique solution in $\displaystyle R.$ then $\displaystyle R$ will have no non-trivial idempotent element because if $\displaystyle e^2=e,$ where $\displaystyle 0 \neq e \in R,$ then the equation $\displaystyle e=exe$ will have (at least) two solutions $\displaystyle x = 1, e$ and so $\displaystyle e = 1.$ now let $\displaystyle 0 \neq a \in R$ and let $\displaystyle x \in R$ be such that $\displaystyle axa=a.$ then $\displaystyle (ax)^2=ax$ and $\displaystyle ax \neq 0$ because otherwise $\displaystyle a = axa=0,$ which is false. thus $\displaystyle ax$ is a nonzero idempotent of $\displaystyle R$ and so $\displaystyle ax = 1.$ hence $\displaystyle R$ is a division ring.

- Dec 16th 2011, 06:27 AMymarRe: unique von neumann inverse in rings
I think I must have used "von Neumann inverse" incorrectly. I really should be more careful when I ask questions here. I'm very sorry.

A von Neumann inverse of $\displaystyle a,$ as I understand it, is an $\displaystyle x$ such that

$\displaystyle axa=a\mbox{ and }xax=x$

So a unique inverse would be the unique solution of the system of equations. The first equation alone could have more solutions.

Maybe I will tell you why I'm asking this question. In semigroup theory there are regular semigroups and inverse semigroups. Regular semigroups are those which satisfy the von Neumann regularity condition. (It's well-defined for semigroups, since the condition doesn't employ addition.) Inverse semigroups are those in which every element has a unique weak inverse (in the sense defined above -- the two conditions). It's a very rich theory (of which I know barely anything), and I was wondering if it had any bearing on ring theory. - Dec 16th 2011, 01:38 PMNonCommAlgRe: unique von neumann inverse in rings
- Dec 16th 2011, 05:28 PMNonCommAlgRe: unique von neumann inverse in rings
i just realized that the above example is just a very special case of the following nice fact:

let $\displaystyle R$ be a commutative von Neumann regular ring and let $\displaystyle a \in R.$ there exists a unique $\displaystyle x \in R$ such that $\displaystyle a = xa^2$ and $\displaystyle x = ax^2.$

proof. since $\displaystyle R$ is a commutative von Neumann regular ring, there exists some $\displaystyle b \in R$ such that $\displaystyle a =ba^2.$ now let $\displaystyle x = ab^2$ and see that $\displaystyle a = xa^2$ and $\displaystyle x = ax^2.$ so we have proved the existence. to prove the uniqueness, suppose that $\displaystyle a = ya^2$ and $\displaystyle y = ay^2$ for some $\displaystyle y \in R.$ then

$\displaystyle x = ab^2 = ya^2b^2 = yab = y^2a^2b = y^2a = y.$