From: "Charles H. Giffen" <chg4k@virginia.edu>
Subject: Re: Rings with single sided inverses?
Date: Fri, 28 Jan 2000 10:55:24 -0500
Newsgroups: sci.math
Summary: [missing]

Michael Abbott wrote:
> 
> (Originally posted, perhaps in error, to alt.algebra.help)
> 
> Can anyone give an example of a ring with single sided inverses?
> 
> Ie, describe a (non-commutative) ring R with two elements a and b such that
> a.b = 1 but ¬(b.a = 1).  Are there many such rings?
> 
> Alternatively, can somebody show me how to prove that a.b=1 => b.a=1?
> 
> (A c.c. to michael@rcp.co.uk is welcome.)

There is a "universal example".  Form the free noncommutative ring
Z{x,y} and divide out by the 2-sided ideal generated by  (xy-1),
obtaining an example  L = Z{x,y}/(xy-1).

If you consider row- and column- finite infinite matrices of
integers

    M = ( m[i,j] in Z : i,j = 1,2,... ),

then these form a ring with unit.  Mapping  x -> s,  y -> t
defines a faithful representation (ie. imbedding) of  L  in
this ring of matrices, where  s = ( \delta[i,j+1] ) and
t = ( \delta[i+1,j] ),  \delta being the Kronecker delta
function.  In other words,  L  is isomorphic to the subring
of the infinite matrices generated by  s  and  t  (and  I, the
identity).

The ring  L \tensor A (where  A  is any ring with unit)
has interesting algebraic K-theoretic properties (which
is my main interest in this subject).

--Chuck Giffen
==============================================================================

From: behr@himalaya.math.niu.edu (Eric Behr)
Subject: Re: Semigroups (Not Rings) with single sided inverses?
Date: 1 Feb 2000 05:55:44 GMT
Newsgroups: sci.math

In article <389604B3.16B35469@math.missouri.edu>,
Stephen Montgomery-Smith  <stephen@math.missouri.edu> wrote:
>> A day ago, I (G. Paseman) wrote:
>> >>Can anyone give an example of a ring with single sided inverses?
>
>I don't think that there is a simpler example.  It can be shown
>that if R is a ring with an element a that has a left inverse,
>but no right inverse, then it must have infinitely many left 

It seems to be an easy exercise that in a left Noetherian ring a
left-invertible element x must be invertible. If I didn't make a late-
-night booboo:

suppose ax = 1; the ascending chain of left annihilators of x^i stops,
so if anything annihilates some x^n (e.g. x^n a^n - 1) then it also
annihilates x^{n-1}: (x^n a^n - 1)x^{n-1} = 0. Thus x^n a - x^{n-1} = 0,
so that x^{n-1} (xa - 1) = 0. Since x is left invertible, this means
xa - 1 = 0.

So yes, you do need an infinite-dimensional situation for an example.

-- 
Eric Behr         | NIU Mathematical Sciences      | (815) 753 6727
behr@math.niu.edu | http://www.math.niu.edu/~behr/ |  fax: 753 1112