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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 $1+x.$

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  • $\begingroup$ 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
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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.

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

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