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.

Expert's answer

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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