# nested ideals in von Neumann regular ring

###### Theorem.

Let $\mathfrak{a}$ be an ideal of the von Neumann regular ring $R$.  Then $\mathfrak{a}$ itself is a von Neumann regular ring and any ideal $\mathfrak{b}$ of $\mathfrak{a}$ is likewise an ideal of $R$.

###### Proof.

If  $a\in\mathfrak{a}$, then  $asa=a$  for some  $s\in R$.  Setting  $t=sas$  we see that $t$ belongs to the ideal $\mathfrak{a}$ and

 $ata=a(sas)a=(asa)sa=asa=a.$

Secondly, we have to show that whenever  $b\in\mathfrak{b}\subseteq\mathfrak{a}$  and  $r\in R$, then both $br$ and $rb$ lie in $\mathfrak{b}$.  Now,  $br\in\mathfrak{a}$  because $\mathfrak{a}$ is an ideal of $R$.  Thus there is an element $x$ in $\mathfrak{a}$ satisfying  $brxbr=br$.  Since $rxbr$ belongs to $\mathfrak{a}$ and $\mathfrak{b}$ is assumed to be an ideal of $\mathfrak{a}$, we conclude that the product  $b\cdot rxbr$  must lie in $\mathfrak{b}$, i.e.  $br\in\mathfrak{b}$.  Similarly it can be shown that  $rb\in\mathfrak{b}$. ∎

## References

• 1 David M. Burton: A first course in rings and ideals.  Addison-Wesley.  Reading, Massachusetts (1970).
Title nested ideals in von Neumann regular ring NestedIdealsInVonNeumannRegularRing 2013-03-22 14:48:24 2013-03-22 14:48:24 CWoo (3771) CWoo (3771) 12 CWoo (3771) Theorem msc 16E50