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A unital ring $R$ is reversible iff $ab=0\implies ba=0.$ This condition implies the following one.

If $a\in R$ is a left-zero divisor, then $a$ is also a right-zero divisor. And the other way around.

Could you please tell me if such rings have a name? It is a strictly weaker condition because it holds in $M_2(\mathbb R)$, which isn't reversible. Indeed, $$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix},$$ but $$\begin{pmatrix}0&0\\1&1\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$

Also, is there any relation between this condition and Dedekind-finiteness? I know that reversibility implies Dedekind finiteness (the proof is here).

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I think this is actually quite a relaxed condition (unlike reversibility, which is pretty strong). I don't know a name for it, but I'd like to tell you about a slightly stronger condition I do know about. One equivalent definition of a right cohopfian ring is "nonunits are left zero divisors".

It's actually pretty hard to find a right-not-left cohopfian ring, because most conditions imply it on both sides. So, if you need examples, there are plenty of two-sided cohopfian rings, and in such rings, one-sided zero divisors will be two-sided zero divisors. For these purposes you have at your disposal any one-sided Artinian or perfect ring, and any pi-regular ring.

Lastly, "one-sided cohopfian" implies Dedekind finite, so this is consistent with your idea that your condition might imply it.

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Thanks. Do I understand correctly that two-sided cohopfian is the same as classical? – user23211 Apr 22 '12 at 10:41
No, not the same. According to that defintion, even a one-sided cohopfian ring is classical. A von Neumann regular ring which is not Dedekind finite is classical, but not cohopfian on either side. – rschwieb Apr 22 '12 at 12:32

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