Theorem 1   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 satisfies $$ata = a(sas)a = (asa)sa = asa = a.$$ 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 $rb\in\mathfrak{b}$