In a general ring with 1, a right (left) zero divisor cannot have a right (left) inverse. In a matrix ring over a field, a stronger condition is satisfied: a (right or left) zero divisor cannot have a (right or left) inverse, with the proof I know invoking the rank plus nullity theorem (which makes sense, because infinite dimensional linear transformations can have right-inverses and be right-zero divisors). Is this more or less a peculiarity of matrices, or am I missing a more general property that they (and perhaps other spaces) possess?
[Math] In a matrix ring, no zero divisors may have an inverse
linear algebraring-theory
Best Answer
I'm glad to see that Dedekind finite rings have been raised as the right concept to look at, but I think a little more can be done to help you see where the condition lies with respect to other conditions you might know.
Here is a diagram reproduced from a portion of my notes that gives a view (at least a partial one):
There are a lot of strange names up there, but between Lam's two books (First course on noncommutative rings and Lectures on modules and rings) you would be able to reproduce this table. In addition to those books, there is an additional paper by Lam that talks about stable range and its immediate relatives: A crash course on...
Of course, matrix rings over division rings lie in the "semisimple" box. In addition to the Dedekind finite box, I think you should also know about "stably finite". A ring $R$ is said to be stably finite if all of the $n\times n$ matrix rings over $R$ are Dedekind finite.
Since Noetherian and perfect rings have already been mentioned, I'd like to contribute by pointing out two "new" examples given here. Since there are two-sided self-injective rings which aren't Noetherian and aren't semiperfect, and unit regular rings which aren't Noetherian and aren't semiperfect, nor injective on both sides, I think both of those classes of rings are "new" to this post.
Finally, there is one thing to add to this diagram. I don't normally keep a box for "commutative ring," but if it were in this picture, it would have an arrow pointing to "Stably finite."
There is one more condition which might interest you. A ring in which every element which isn't a left zero divisor is a unit is called a right cohopfian ring. This means essentially that if multiplication by an element on the left makes an injective map $R\to R$, then it is an isomorphism. This relates closely to an alternative definition of Dedekind finite: If multiplication on the left by $a$ is surjective then it is an isomorphism. This is saying that $R$ is "hopfian". It turns out that for a ring, this condition is left-right symmetric, but the cohopfian version isn't :)
I've got a bit of info on them in this solution: https://math.stackexchange.com/a/135051/29335 . One-sided cohofian rings are also Dedekind finite. I would have to double check which things on the graph point to "cohopfian" :) They're a bit of an oddball...