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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.

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    $\begingroup$ 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 $xy = 1 \neq yx$.] $\endgroup$
    – David Handelman
    Commented Nov 2, 2014 at 14:34
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    $\begingroup$ @DavidHandelman Why not make this an answer, for extra visibility? $\endgroup$
    – Yemon Choi
    Commented Nov 2, 2014 at 18:20
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    $\begingroup$ @Yemon Choi Because it's elementary. $\endgroup$
    – David Handelman
    Commented Nov 3, 2014 at 1:29
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    $\begingroup$ 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 $\endgroup$
    – rschwieb
    Commented Jan 1, 2015 at 18:48

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[Sorry for necromancy, I just stumbled on this question while looking for something else.]

This carries over to monoids and has nothing to do with the additive structure of a ring (and little to do with commutativity).

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).

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