We saw in part (2) that von Neumann regular rings live somewhere between semisimple and semiprimitive rings. The goal in this post is to prove a theorem of Armendariz and others which gives a necessary and sufficient condition for a ring to be both regular and reduced. This result extends Kaplansky’s result for commutative rings (see the corollary at the end of this post). We remark that a commutative von Neumann regular ring is necessarily reduced. That is because if for some then choosing with we will get
Definition . A von Neumann regular ring is called strongly regular if is reduced.
Theorem 1. (Armendariz, 1974) A ring with 1 is strongly regular if and only if is a division ring for all maximal ideals of
Proof. Suppose first that is strongly regular and let be a maximal ideal of Let So for all Since is regular, there exists some such that Then is an idempotent and thus because in a reduced ring every idempotent is central. Since we have and hence Thus is a right inverse of Similarly and is a left inverse of Therefore is invertible and hence is a division ring. Conversely, suppose that is a division ring for all maximal ideals of If is not reduced, then there exists such that Let Clearly is a proper ideal of and hence for some maximal ideal of But then in which is a division ring. Thus i.e. there exists some such that which is absurd. To prove that is von Neumann regular, we will assume, to the contrary, that is not regular. So there exists such that for all Let Clearly is a proper ideal of and so for some maximal ideal of It is also clear that if for some then because we may choose Thus in and hence there exists some and such that Therefore for some But then and so which is nonsense. This contradiction proves that must be regular.
Corollary. (Kaplansky) A commutative ring is regular if and only if is a field for all maximal ideals of
At the end let me mention a nice property of strongly regular rings.
Theorem 2. (Pere Ara, 1996) If is strongly regular and for some then is a unit for some