Unit-regularity and representability for semiartinian *-regular rings

TL;DR

This paper proves that semiartinian subdirectly irreducible *-regular rings can be represented within inner product spaces, linking algebraic properties to geometric representations.

Contribution

It establishes the representability of a class of *-regular rings, specifically semiartinian subdirectly irreducible ones, within inner product spaces, which was previously unknown.

Findings

Semiartinian subdirectly irreducible *-regular rings are representable in inner product spaces.

The result connects algebraic structure with geometric representation.

Provides a new perspective on the structure of *-regular rings.

Abstract

We show that any semiartinian subdirectly irreducible *-regular ring R admits a representation within some inner product space.

Full text

1. Introduction

The motivating examples of $*$-regular rings, due to Murray and von Neumann, were the $*$-rings of unbounded operators affiliated with finite von Neumann algebra factors; to be subsumed, later, as $*$-rings of quotients of finite Rickart $C^{*}$-algebras. All the latter have been shown to be $*$-regular and unit-regular (Handelman [5]). Representations of these as $*$-rings of endomorphisms of suitable inner product spaces have been obtained first, in the von Neumann case, by Luca Giudici (cf. [6]), in general in joint work with Marina Semenova [9]. The existence of such representations implies direct finiteness [7]. In the present note we show that every semiartinian $*$-regular ring is unit-regular and a subdirect product of representables. This might be a contribution to the question, asked by Handelman (cf. [3, Problem 48]), whether all $*$-regular rings are unit-regular. We rely heavily on the result of Baccella and Spinosa [1] that a semiartinian regular ring is unit-regular provided that all its homomorphic images are directly finite. Also, we rely on the theory of representations of $*$-regular rings developed by Florence Micol [12] (cf. [9, 10]).

2. Preliminaries: Regular and $*$-regular rings

We refer to Berberian [2] and Goodearl [3]. Unless stated otherwise, rings will be associative, with unit $1$ as constant. A (von Neumann) regular ring $R$ is such that for each $a\in R$ there is $x\in R$ such that $axa=a$; equivalently, every right (left) principal ideal is generated by an idempotent. The socle $Soc(R)$ is the sum of all minimal right modules. A regular ring $R$ is semiartinian if each of its homomorphic images has non-zero socle; that is, $R$ has Loewy length $\xi+1$ for some ordinal $\xi$. A ring $R$ is directly finite if $xy=1$ implies $yx=1$ for all $x,y\in R$. A ring $R$ is unit-regular if for any $a\in R$ there is a unit $u$ of $R$ such that $aua=a$. The crucial fact to be used, here, is the following result of Baccella and Spinosa [1].

Theorem 1**.**

††margin:

bac

A semiartinian regular ring is unit-regular provided all its homomorphic images are directly finite.

A $*$-ring is a ring $R$ endowed with an involution $r\mapsto r^{*}$. Such $R$ is $*$-regular if it is regular and $rr^{*}=0$ only for $r=0$. A projection is an idempotent $e$ such that $e=e^{*}$; we write $e\in P(I)$ if $e\in I$. A $*$-ring is $*$-regular if and only if for any $a\in R$ there is is a projection $e$ with $aR=eR$; such $e$ is unique and obtained as $aa^{+}$ where $a^{+}$ is the pseudo-inverse of $a$. In particular, for $*$-regular $R$, each ideal $I$ is a $*$-ideal, that is closed under the involution. Thus, $R/I$ is a $*$-ring with involution $a+I\mapsto a^{*}+I$ and a homomorphic image of the $*$-ring $R$. In particular, $R/I$ is regular; and $*$-regular since $aa^{+}+I$ is a projection generating $(a+I)(R/I)$.

If $R$ is a $*$-regular ring and $e\in P(R)$ then the corner $eRe$ is a $*$-regular ring with unit $e$, operations inherited from $R$, otherwise. For a $*$-regular ring, $P(R)$ is a modular lattice, with partial order given by $e\leq f\Leftrightarrow fe=e$, which is isomorphic to the lattice $L(R)$ of principal right ideals of $R$ via $e\mapsto eR$. In particular, $eRe$ is artinian of and only if $e$ is contained in the sum of finitely many minimal right ideals.

A $*$-ring is subdirectly irreducible if it has an unique minimal ideal, denoted by $M(R)$. For the following see Lemma 2 and Theorem 3 in [8].

Fact 2**.**

††margin:

simp

If $R$ is a subdirectly irreducible $*$-regular ring then $eRe$ is simple for all $e\in P(M(R))$ and $R$ a homomorphic image of a $*$-regular sub-$*$-ring of some ultraproduct of the $eRe$, $e\in P(M(R))$.

3. Preliminaries: Representations

We refer to Gross [4] and Sections 1 of [9], 2–4 of [10]. By an inner product space $V_{F}$ we will mean a right vector space (also denoted by $V_{F}$) over a division $*$-ring $F$, endowed with a sesqui-linear form $\langle.\mid.\rangle$ which is anisotropic ($\langle v\mid v\rangle=0$ only for $v=0$) and orthosymmetric ( $\langle v\mid w\rangle=0$ if and only if $\langle w\mid v\rangle=0$). Let $\operatorname{{\sf End}}^{*}(V_{F})$ denote the $*$-ring consisting of those endomorphisms $\varphi$ of the vector space $V_{F}$ which have an adjoint $\varphi^{*}$ w.r.t. $\langle.\mid.\rangle$.

A representation of a $*$-ring $R$ within $V_{F}$ is an embedding of $R$ into $\operatorname{{\sf End}}^{*}(V)$. $R$ is representable if such exists. The following is well known, cf. [11, Chapter IV.12]

Fact 3**.**

††margin:

art

Each simple artinian $*$-regular ring is representable.

The following two facts are consequences of Propositions 13 and 25 in [9] (cf. Micol [12, Corollary 3.9]) and, respectively, [7, Theorem 3.1] (cf. [8, Theorem 4]).

Fact 4**.**

††margin:

F2

A $*$-regular ring is representable provided it is a homomorphic image of a $*$-regular sub-$*$-ring of an ultraproduct of representable $*$-regular rings.

Fact 5**.**

††margin:

fr

Every representable $*$-regular ring is directly finite.

4. Main results

Theorem 6**.**

††margin:

thm

If $R$ is a subdirectly irreducible $*$-regular ring such that $Soc(R)\neq 0$, then $Soc(R)=M(R)$, each $eRe$ with $e\in P(M(R)$ is artinian, and $R$ is representable.

Proof.

Consider a minimal right ideal $aR$. As $R$ is subdirectly irreducible, $M(R)$ is contained in the ideal generated by $a$; that is, for any $0\neq e\in P(M(R))$ one has $e=\sum_{i}r_{i}as_{i}$ for suitable $r_{i},s_{i}\in R$, $r_{i}as_{i}\neq 0$. By minimality of $aR$, one has $as_{i}R=aR$ and $r_{i}as_{i}R=r_{i}aR$ is minimal, too. Thus, $e\in\sum_{i}r_{i}aR$ means that $eRe$ is artinian. By Facts 3, 2, and 4, $R$ is representable.

It remains to show that $Soc(R)\subseteq M(R)$. Recall that the congruence lattice of $L(R)$ is isomorphic to the ideal lattice of $R$ ([13, Theorem 4.3] with an isomorphism $\theta\mapsto I$ such that $aR/0\in\theta$ if and only if $a\in I$. In particular, since $R$ is subdirectly irreducible so is $L(R)$. Choose $e\in M(R)$ with $eR$ minimal. Then for each minimal $aR$ one has $eR/0$ in the lattice congruence $\theta$ generated by $aR/0$. Since both quotients are prime, by modularity this means that they are projective to each other. Thus, $aR\in\theta$ and $a$ in the ideal generated by $e$, that is in $M(R)$. ∎

Theorem 7**.**

Every semiartinian $*$-regular ring $R$ is unit-regular and a subdirect product of representable homomorphic images.

Proof.

Consider an ideal $I$ of $R$. Then $I=\bigcap_{x\in X}I_{x}$ with completely meet irreducible $I_{x}$, that is subdirectly irreducible $R/I_{x}$. Since $R$ is semiartinian one has $Soc(R/I_{x})\neq 0$, whence $R/I_{x}$ is representable by Theorem 6 and directly finite by Fact 5. Then $R/I$ is directly finite, too, being a subdirect product of the $R/I_{x}$. By Theorem 1 it follows that $R$ is unit-regular. ∎

5. Examples

It appears that semiartinian $*$-regular rings form a very special subclass of the class of unit-regular $*$-regular rings, even within the class of those which are subdirect products of representables. E.g. the $*$-ring of unbounded operators affiliated to the hyperfinite von Neumann algebra factor is representable, unit-regular, and $*$-regular with zero socle. On the other hand, due to the following, for every simple artinian $*$-regular ring $R$ and any natural number $n>0$ there is a semiartinian $*$-regular ring having ideal lattice an $n$-element chain and $R$ as a homomorphic image.

Proposition 8**.**

††margin:

pro

Every representable $*$-regular ring $R$ embeds into some subdirectly irreducible representable $*$-regular ring $\hat{R}$ such that $R\cong\hat{R}/M(\hat{R})$. In particular, $\hat{R}$ is semiartinian if and only if so is $R$.

The proof needs some preparation. Call a representation $\iota:R\to\operatorname{{\sf End}}^{*}(V_{F})$ large if for all $a,b\in R$ with $\operatorname{{\sf im}}\iota(b)\subseteq\operatorname{{\sf im}}\iota(a)$ and finite $\dim(\operatorname{{\sf im}}\iota(a)/\operatorname{{\sf im}}\iota(b))_{F}$ one has $\operatorname{{\sf im}}\iota(a)=\operatorname{{\sf im}}\iota(b)$.

Lemma 9**.**

††margin:

large

Any representable $*$-regular ring admits some large representation.

Proof.

Inner product spaces can be considered as $2$-sorted structures $V_{F}$ with sorts $V$ and $F$. In particular, the class of inner product spaces is closed under formation of ultraproducts. Representations of $*$-rings $R$ can be viewed as $R$-$F$-bimodules ${}_{R}V_{F}$, that is as $3$-sorted structures, with $R$ acting faithfully on $V$. It is easily verified that the class of representations of $*$-rings is closed under ultraproducts cf. [9, Proposition 13].

Now, given a representation $\eta$ of $R$ in $W_{F}$, form an ultrapower $\iota$, that is ${}_{S}V_{F^{\prime}}$, such that $\dim F^{\prime}_{F}$ is infinite (recall that $F^{\prime}$ is an ultrapower of $F$). Observe that $\operatorname{{\sf End}}^{*}(V_{F^{\prime}})$ is a sub-$*$-ring of $\operatorname{{\sf End}}^{*}(V_{F})$ and $\dim(U/W)_{F}$ is infinite for any subspaces $U\supseteq W$ of $V_{F^{\prime}}$. Also, $S$ is an ultrapower of $R$ with canonical embedding $\varepsilon:R\to S$. Thus, $\varepsilon\circ\iota$ is a large representation of $R$ in $V_{F}$. ∎

Proof.

of Proposition 8. In view of Lemma 9 we may assume a large representation $\iota$ of $R$ in $V_{F}$. Identifying $R$ via $\iota$ with its image, we have $R$ a $*$-regular sub-$*$-ring of $\operatorname{{\sf End}}^{*}(V_{F})$. Let $I$ denote the set of all $\varphi\in\operatorname{{\sf End}}(V_{F})$ such that $\dim(\operatorname{{\sf im}}\varphi)_{F}$ is finite. According to Micol [12, Proposition 3.12] (cf. Propositions 4.4 (i),(iii) and 4.5 in [10]) $R+I$ is a $*$-regular sub-$*$-ring of $\operatorname{{\sf End}}^{*}(V_{F})$, with unique minimal ideal $I$. By Theorem 6 one has $I=Soc(R+I)$. Moreover, $R\cap I=\{0\}$ since the representation $\iota$ of $R$ in $V_{F}$ is large. Hence, $R\cong(R+I)/I$. ∎