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 $1+x.$

  • $\begingroup$ 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 $ax(1 + x) = 1 + x.$ $\endgroup$ – Carlo Jun 1 at 3:02
  • $\begingroup$ @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. $\endgroup$ – Arturo Magidin Jun 1 at 3:33
  • $\begingroup$ 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. $\endgroup$ – 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



$1+x=(0,1)$ 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


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.

| cite | improve this answer | |
  • $\begingroup$ Thanks rschwieb and others also for their valuable comments. $\endgroup$ – Explanation Maths Jun 1 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.