unique von neumann inverse in rings

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December 15th 2011, 07:02 AM
ymar

unique 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?
December 15th 2011, 02:52 PM
NonCommAlg

Re: unique von neumann inverse in rings

Quote:

Originally Posted by ymar

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?

your questions are not very clear. for example, in your first question, you mean "the inverses, if they exist, are unique", right? i mean you're not assuming that your ring is von Neumann regular?
December 15th 2011, 03:01 PM
ymar

Re: unique von neumann inverse in rings

Quote:

Originally Posted by NonCommAlg

your question is not very clear. for example, in your first question, you mean "the inverses, if they exist, are unique", right? i mean you're not assuming that your ring is von Neumann regular?

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.
December 15th 2011, 03:13 PM
NonCommAlg

Re: unique von neumann inverse in rings

Quote:

Originally Posted by ymar

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.

the answer to the first question is no. any von Neumann regular domain is a division ring. because from $a = axa$ we get $a(1-xa)=0$ and since your ring is a domain and $a$ is non-zero, we have $xa=1$ .

your second question is more interesting and i need to think about it.
December 15th 2011, 05:06 PM
NonCommAlg

Re: unique von neumann inverse in rings

Quote:

Originally Posted by ymar

2) there are any von Neumann regular rings which are not domains, and whose non-zero elements have unique von Neumann inverses.

ok, i can't believe that i thought the answer was yes! the answer is actually no. to see this, let $R$ be a ring with this property that for every nonzero element $a \in R,$ the equation $a = axa$ has a unique solution in $R.$ then $R$ will have no non-trivial idempotent element because if $e^2=e,$ where $0 \neq e \in R,$ then the equation $e=exe$ will have (at least) two solutions $x = 1, e$ and so $e = 1.$ now let $0 \neq a \in R$ and let $x \in R$ be such that $axa=a.$ then $(ax)^2=ax$ and $ax \neq 0$ because otherwise $a = axa=0,$ which is false. thus $ax$ is a nonzero idempotent of $R$ and so $ax = 1.$ hence $R$ is a division ring.
December 16th 2011, 06:27 AM
ymar

Re: unique von neumann inverse in rings

Quote:

Originally Posted by NonCommAlg

ok, i can't believe that i thought the answer was yes! the answer is actually no. to see this, let $R$ be a ring with this property that for every nonzero element $a \in R,$ the equation $a = axa$ has a unique solution in $R.$ then $R$ will have no non-trivial idempotent element because if $e^2=e,$ where $0 \neq e \in R,$ then the equation $e=exe$ will have (at least) two solutions $x = 1, e$ and so $e = 1.$ now let $0 \neq a \in R$ and let $x \in R$ be such that $axa=a.$ then $(ax)^2=ax$ and $ax \neq 0$ because otherwise $a = axa=0,$ which is false. thus $ax$ is a nonzero idempotent of $R$ and so $ax = 1.$ hence $R$ is a division ring.

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 $a,$ as I understand it, is an $x$ such that

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

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.
December 16th 2011, 01:38 PM
NonCommAlg

Re: unique von neumann inverse in rings

Quote:

Originally Posted by ymar

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 $a,$ as I understand it, is an $x$ such that

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

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.

ok, with that definition, the answer is yes and an example is $R=F \oplus F.$ where $F$ is a field.
December 16th 2011, 05:28 PM
NonCommAlg

Re: unique von neumann inverse in rings

Quote:

Originally Posted by NonCommAlg

ok, with that definition, the answer is yes and an example is $R=F \oplus F.$ where $F$ is a field.

i just realized that the above example is just a very special case of the following nice fact:

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

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

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