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Question #17361
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.
First assume R is von Neumannregular. For I, J as above, it suffices to show that I &cap; J &sube; IJ. Let a &isin; I &cap; J. There exists x &isin; R such that a = axa. Thus, a &isin; (IR)J &sube; IJ.Conversely, assume that IJ = I &cap; J for any right ideal I andany left ideal J. For any a &isin; R, we have then
a &isin; (aR) &cap; (Ra) = (aR)(Ra) = aRa.

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