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Homenested ideals in von Neumann regular ring

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# 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 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 $rb\in\mathfrak{b}$. ∎

# References

- 1 David M. Burton: A first course in rings and ideals. Addison-Wesley. Reading, Massachusetts (1970).

## Mathematics Subject Classification

16E50*no label found*

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