Why in a von Neumann regular ring do we have that $ax(1+x)=1+x?$

Why is it true that in a commutative von Neumann regular ring, we have that $$ax(1+x)=1+x?$$

Definition: We say that a unital ring $$R$$ is von Neumann whenever for every $$a \in R,$$ there exist an $$x \in R$$ such that $$a = axa.$$

After simplifying the LHS, I get that $$ax+axx = ax+xax = ax+x,$$ but I'm not getting from this how I can obtain $$1+x.$$

• Unless $R$ is a domain, I don't see why this is true. But if we assume that $R$ is a domain, the proof is as follows. We will assume that $a$ and $x$ satisfy $x = xax.$ We have therefore that $ax(1 + x) = ax + axx = ax + xax = ax + x.$ By multiplying both sides by $x,$ we find that $ax(1 + x)x = (ax + x)x = axx + xx = xax + xx = x + xx = (1 + x)x.$ Cancellation holds in $R$ (as it is a domain), so we obtain the desired $ax(1 + x) = 1 + x.$ – Carlo Jun 1 at 3:02
• @Carlo: If you had a domain, then from $axa=a$ you conclude immediately that $ax=xa=1$, in which case $ax(1+x)=1+x$ is trivial. – Arturo Magidin Jun 1 at 3:33
• Very true. In fact, a commutative von Neumann domain is a field. But the point is that I was building off the computation performed by OP. – Carlo Jun 1 at 4:08

Well, here we go, let's test that against the smallest nonfield von Neumann regular ring $$R=F_2\times F_2$$ where $$F_2$$ is the field of two elements.

Let $$x=a=(1,0)$$. You get

$$ax(1+x)=(1,0)[(1,1)+(1,0)]=(0,0)$$

and

$$1+x=(0,1)$$ which are not equal.

From what you've written it really looks like you've unclearly stated your hypotheses, but in this example $$axa=a$$ and $$xax=x$$, so it covers pretty much every interpretation of what you wrote.

So, something is wrong with the statement.

If, for example, you meant that $$a$$ and $$x$$ have to be related by $$a=axa$$, then it would be trivial to show

$$ax(1+a)=a(1+x)$$

which is as close as I could get to the suggested equality.

And if you meant for $$xax=x$$, that turns into

$$ax(1+x)=ax+x=a(1+x)$$ also, of course.

• Thanks rschwieb and others also for their valuable comments. – Explanation Maths Jun 1 at 15:46