Show that a ring R is von Neumann regular iff IJ = I ∩ J for every right ideal I and every left ideal J in R.
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Expert's answer
2012-10-31T08:59:41-0400
First assume R is von Neumannregular. For I, J as above, it suffices to show that I ∩ J ⊆ IJ. Let a ∈ I ∩ J. There exists x ∈ R such that a = axa. Thus, a ∈(IR)J ⊆ IJ.Conversely, assume that IJ = I ∩ J for any right ideal I andany left ideal J. For any a ∈ R, we have then a ∈(aR) ∩ (Ra) = (aR)(Ra) = aRa.
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