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nested ideals in von Neumann regular ring


Theorem.

Let π”ž be an ideal of the von Neumann regular ringMathworldPlanetmath R.  Then π”ž itself is a von Neumann regular ring and any ideal π”Ÿ of π”ž is likewise an ideal of R.

Proof.

If  aβˆˆπ”ž, then  asa=a  for some  s∈R.  Setting  t=sas  we see that t belongs to the ideal π”ž and

ata=a(sas)a=(asa)sa=asa=a.

Secondly, we have to show that whenever  bβˆˆπ”ŸβŠ†π”žβ€‰ and  r∈R, then both br and rb lie in π”Ÿ.  Now,  brβˆˆπ”žβ€‰ because π”ž is an ideal of R.  Thus there is an element x in π”ž satisfying  brxbr=br.  Since rxbr belongs to π”ž and π”Ÿ is assumed to be an ideal of π”ž, we conclude that the product  bβ‹…rxbr  must lie in π”Ÿ, i.e.  brβˆˆπ”Ÿ.  Similarly it can be shown that  rbβˆˆπ”Ÿ. ∎

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