We say that an associative ring $A$ is *von Neumann regular* if for every $a\in A$ there exists a $x\in A$ such that $axa = a$. That is a rather strange condition, isn’t it? But, you can think of $x$ as a pseudoinverse to $a$. This weakening of inverses has a homological counterpart: if every nonzero element of $A$ had an inverse, every $A$-module would be free. This weakening of the inverse property weakens the free condition on modules: a ring $A$ is von Neumann regular if and only if every left (and right!) $A$-module is flat. Every free module is flat, just like if $A$ were a ring whose nonzero elements had inverses, then every element $a$ would have a $x$ such that $axa = a$.

There is another interesting way to look at von Neumann regular rings: through annihilators, a perspective that was noted in Ming’s paper

Ming, Roger Yur Chi. On (von Neumann) regular rings. Proc. Edinburgh Math. Soc. (2) 19 (1974/75), 89—91

This perspective is based on the following observation: suppose $A$ is von Neumann regular. If $a\in A$, then by definition there exists a $x\in A$ such that $axa = a$. Rearranging gives $a(xa – 1) = 0$. On the other hand, if $t(xa – 1) = 0$ then $txa = t$ and so $t$ is in the left ideal $Aa$.

Recall that if $a\in A$, the left annihilator of $a$ is defined as

$$l(a) = \{ b\in A : ba = 0\}.$$ We denote similarly by $r(a)$ the right annihilator of $A$. What we have observed is that if $A$ is von Neumann regular, then every left ideal is the left annihilator of some element of $A$.

The converse is not quite true. To make it true, we need to introduce an extra hypothesis on $A$. Recall that we say a ring $A$ is reduced if $A$ contains no nonzero nilpotent elements. An interesting thing about reduced rings is that if $xy = 0$ in a reduced ring, then $yxyx = 0$ and so $yx = 0$. Therefore, $l(x) = r(x)$ for all $x$ in a reduced ring. Ming discovered that for reduced rings, there is the following equivalence:

**Theorem.**A

*reduced*ring $A$ is von Neumann regular if and only if every principal left ideal of $A$ is the left annihilator of some nonzero element of $A$.

*Proof.*We have already seen that if $A$ is von Neumann regular, then every principal left ideal of $A$ is the left annihilator of a nonzero element of $A$, and this is true even when $A$ is not reduced.

Now assume that $A$ is a ring in which every principal left ideal is the left annihilator of an element of $A$. Let $a\in A$. We want to show that there exists a $x\in A$ such that $axa = a$. We first assume that $a$ is a nonzerodivisor. In this case, let $Aa = l(b)$. Since $a$ is a nonzerodivisor, $b = 0$ and so $Aa = l(0) = A$. Hence, $a$ has a left inverse $x$. In particular, $axa = a$.

Now we assume $a$ is a zerodivisor. Let $Aa = l(b)$ with $b\not=0$. In particular, $ab = 0$. Notice that since $A$ is reduced, $0 = baba$ implies that $ba = 0$ too. Let $c = a + b$. We claim that $c$ is a nonzerodivisor. Assume for a moment that this is true. Then by what we have already proved in the previous paragraph, $c$ has a left inverse $x$. Then $ca = a^2$ implies that $a = xa^2$. A straightforward computation shows that $(a – axa)^2 = 0$ and so $axa = a$.

Therefore, to finish the entire proof, we need to prove that $c$ is a nonzerodivisor. To this end, let $y$ be an element such that $cy = 0 = (a + b)y$. In particulary, $ay = -by\in r(b)\cap r(a)$. Now, an arbitrary element $w\in r(b)\cap r(a)$ can be written $w = za$ because $r(b) = l(b) = Aa$. So $aza = aw = 0$. Therefore, $(za)^2 = 0$ and so $za = w = 0$, yet again using that $A$ is reduced. This means that $r(b)\cap r(a) = 0$, and in particular,

$$ay = -by = 0.$$

By definition, $y\in r(b)\cap r(a)$, and so also $y = 0$. This proves that $c = a + b$ is a nonzerodivisor, completing the proof.