# Is every non-invertible element of a commutative von Neumann regular ring a zero-divisor? (answered: yes!)

As I learned from a previous old question, every commutative von Neumann regular ring is a subdirect product of a family of fields. For a direct product of fields, it seems clear to me that every non-invertible element is a zero-divisor. But a subdirect product is more subtle than a direct product.

However, the question

Is every non-invertible element of a commutative von Neumann regular ring a zero-divisor?

is completely independent from subtleties of the subdirect product. A simple counterexample would be enough to answer it, if it should turn out to be false.

• Every non-invertible element of a vN regular ring (not necessarily commutative) is a zero divisor: given $x \neq 0$, there exists $y$ with $xyx = x$, so $e = xy$ is idempotent; if $xy \neq 1$, then $(1-e)x = 0$ implies $x$ is a left zero divisor. If $yx \neq 1$, then $x$ is similarly a right zero divisor. [And of course, there exist vN regular rings with elements $x$, $y$, such that $xy = 1 \neq yx$.] Nov 2, 2014 at 14:34
• @DavidHandelman Why not make this an answer, for extra visibility? Nov 2, 2014 at 18:20
• @Yemon Choi Because it's elementary. Nov 3, 2014 at 1:29
• Dear @DavidHandelman : you might consider that this policy of "if a question turns out to have an elementary solution, put it in the comments" diminishes the usefulness of the Unanswered Questions feature. It just doesn't seem right to deny the OP a posted answer just because you think the answer is too easy. If you are worried about appearances, you could always make it a CW answer. Regards Jan 1, 2015 at 18:48

Let $$H$$ be a (multiplicatively written) monoid. We say that $$H$$ is Dedekind-finite if $$xy = 1_H$$ implies $$yx = 1_H$$; and von Neumann regular (VNR) if, for every $$x \in H$$, there is an element $$y \in H$$ such that $$xyx = x$$. On the other hand, we call an element $$x \in H$$ singular if $$x$$ is neither left- nor right-cancellative. Of course, every commutative monoid is Dedekind-finite; and if $$H$$ is the multiplicative monoid of a unital ring $$R$$, then the singular elements of $$H$$ are precisely the (two-sided) zero divisors of $$R$$.
Claim. Every non-unit of a Dedekind-finite VNR monoid $$H$$ is singular.
Proof. Let $$x \in H$$ be a non-unit. Since $$H$$ is a VNR monoid, we have $$x = xyx$$ for some $$y \in H$$. So, if $$x$$ is left- or right-cancellative (i.e., if $$x$$ is not singular), then $$xy = 1_H$$ or $$yx = 1_H$$, which shows in turn that $$x$$ is a unit (absurd), by the hypothesis that $$H$$ is Dedekind-finite. ▢
When dropping the assumption that $$H$$ is Dedekind-finite, one can still show (by the same argument) that every non-unit of $$H$$ is either left- or right-singular (cf. David Handelman's first comment under the OP).