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 $b r$ and $r b$ 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 $r x b r$ 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).