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).

Title	nested ideals in von Neumann regular ring
Canonical name	NestedIdealsInVonNeumannRegularRing
Date of creation	2013-03-22 14:48:24
Last modified on	2013-03-22 14:48:24
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	12
Author	CWoo (3771)
Entry type	Theorem
Classification	msc 16E50