## von Neumann regular rings (3)

Posted: October 31, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
Tags: , , , , ,

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 $r \in R.$

2. To add some more context to Theorem 2, that a ring with that property (that if $Ra+Rb=R$, then there exists $r$ such that $a+rb$ is a unit) is said to have “stable range 1.” It is indeed an interesting proprety.
Well, strongly regular rings are strongly $\pi$-regular and so the theorem in my post is just a special case of Pere’s theorem.