Posts Tagged ‘strongly regular’

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 R is necessarily reduced. That is because if x^2=0  for some x \in R, then choosing y \in R with x=xyx we will get x=yx^2=0.

Definition . A von Neumann regular ring R is called strongly regular if R is reduced.

Theorem 1. (Armendariz, 1974) A ring  R with 1 is strongly regular if and only if R_M is a division ring for all maximal ideals M of Z(R).

Proof. Suppose first that R is strongly regular and let M be a maximal ideal of Z(R). Let 0 \neq s^{-1}x \in R_M. So tx \neq 0 for all t \in Z(R) \setminus M. Since R is regular, there exists some y \in R such that xyx = x. Then xy=e is an idempotent and thus e \in Z(R) because in a reduced ring every idempotent is central.  Since (1-e)x=0 we have 1-e \in M and hence e \in Z(R) \setminus M. Thus e^{-1}sy is a right inverse of s^{-1}x. Similarly f=yx \in Z(R) \setminus M and f^{-1}sy is a left inverse of s^{-1}x. Therefore s^{-1}x is invertible and hence R_M is a division ring. Conversely, suppose that R_M is a division ring for all maximal ideals M of Z(R). If R is not reduced, then there exists 0 \neq x \in R such that x^2=0. Let I=\{s \in Z(R): \ sx = 0 \}. Clearly I is a proper ideal of Z(R) and hence I \subseteq M for some maximal ideal M of Z(R). But then (1^{-1}x)^2=0 in R_M, which is a division ring. Thus 1^{-1}x=0, i.e. there exists some s \in Z(R) \setminus M such that sx = 0, which is absurd. To prove that R is von Neumann regular, we will assume, to the contrary, that R is not regular. So there exists x \in R such that xzx \neq x for all z \in R. Let J= \{s \in Z(R): \ xzx=sx \ \text{for some} \ z \in R \}. Clearly J is a proper ideal of Z(R) and so J \subseteq M for some maximal ideal M of Z(R). It is also clear that if sx = 0 for some s \in Z(R), then s \in J because we may choose z = 0. Thus 1^{-1}x \neq 0 in R_M and hence there exists some y \in R and t \in Z(R) \setminus M such that 1^{-1}x t^{-1}y = 1. Therefore u(xy-t)=0 for some u \in Z(R) \setminus M. But then x(uy)x=utx and so ut \in J, which is nonsense. This contradiction proves that R must be regular. \Box

Corollary. (Kaplansky) A commutative ring R is regular if and only if R_M is a field for all maximal ideals M of R. \ \Box

At the end let me mention a nice property of strongly regular rings.

Theorem 2. (Pere Ara, 1996) If R is strongly regular and Ra+Rb=R, for some a, b \in R, then a+rb is a unit for some r \in R.

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