nested ideals in von Neumann regular ring
Theorem.
Let π be an ideal of the von Neumann regular ring R. βThen π itself is a von Neumann regular ring and any ideal π of π is likewise an ideal of R.
Proof.
If βaβπ, then βasa=aβ for some βsβR. βSetting βt=sasβ we see that t belongs to the ideal π and
ata=a(sas)a=(asa)sa=asa=a. |
Secondly, we have to show that whenever βbβπβπβ and βrβR, then both br and rb lie in π. βNow, βbrβπβ because π is an ideal of R. βThus there is an element x in π satisfying βbrxbr=br. βSince rxbr belongs to π and π is assumed to be an ideal of π, we conclude that the product βbβ rxbrβ must lie in π, i.e. βbrβπ. βSimilarly it can be shown that βrbβπ. β
References
- 1 David M. Burton: A first course in rings and ideals. βAddison-Wesley. βReading, 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 |