Von Neumann Regular Rings

Q: Why is von Neumann regularity left-right symmetric when the ideal conditions are one-sided?

Because condition (1), a = axa, mentions no side at all, and (4.23) proves it equivalent to each of the one-sided conditions. So the left-hand conditions and the right-hand conditions are both equivalent to the same symmetric statement, hence to each other. This is the same style of argument that makes rad R side-neutral.

Q: Is a von Neumann regular ring the same as a ring with no nilpotent elements?

No. M_2(k) is regular and contains nilpotent matrices. Regular *and reduced* is a stronger condition called strongly regular, characterised by the existence of x with a = a^2x. What regularity does exclude is nonzero nil one-sided ideals.

Q: How does one produce non-semisimple regular rings?

Two reliable recipes. Take End(M_k) for an infinite-dimensional semisimple module M, which is (4.27); or take an infinite direct product, or a direct limit, of semisimple rings. Both break the noetherian condition, which by (4.25) is the only obstruction.

Q: What is the relationship to the Moore–Penrose inverse?

The equation a = axa is the first of Penrose's four conditions. The other three involve an involution and pin the inverse down uniquely. Ring theory keeps only the first condition, so pseudo-inverses exist in abundance but none is canonical; a ring with involution can support the full Moore–Penrose theory.

Executive Summary

Semisimplicity asks that every left ideal be a direct summand, which forces a chain condition. Von Neumann regularity asks the same only of finitely generated left ideals, and that single relaxation removes all finiteness from the theory while keeping most of its module-theoretic comfort.

The condition has a one-line element form — every $a$ admits $x$ with $a = a x a$ — which is visibly left-right symmetric, so the whole theory is side-neutral. Regular rings sit strictly between semisimple and semiprimitive: $rad R = 0$ always, and adding the ascending chain condition on left ideals recovers semisimplicity exactly.

$a = a x a$ The definition

5Equivalent conditions in (4.23)

1936von Neumann

noetherianWhat upgrades regular to semisimple

Overview

John von Neumann introduced these rings around 1935 while building continuous geometry — lattices of subspaces with a continuously varying dimension function. The coordinatising rings of such geometries are exactly the regular rings, and the lattice of principal left ideals is complemented precisely because each of them is generated by an idempotent.

\forall a \in R \exists x \in R : a = a x a

(4.23)(1)

Left-right symmetric as written, which is why every one-sided condition in $(4.23)$ has a valid mirror image.

The element $x$ behaves like a generalised inverse: if $a$ is a unit then $x = a^{- 1}$ is the only solution, and in general $e = x a$ is an idempotent with $R a = R e$ . Turning ideals into idempotents is what makes the class tractable, and it links this page to Idempotents and Peirce Decomposition.

Typical examples are not exotic: every field and every matrix ring over a division ring, every Boolean ring, every direct product of such, and the full endomorphism ring of any vector space, however large.

Learning Objectives

State the conditions $(1)$ , $(2)$ , $(2')$ , $(3)$ , $(3')$ of $(4.23)$ and explain why they are left-right symmetric as a family.
Prove $(1) \Leftrightarrow (2)$ by exhibiting the idempotent $e = x a$ with $R a = R e$ .
Prove $(2) \Rightarrow (3)$ using the idempotent $e + e' - e e'$ .
Derive $rad R = 0$ for regular $R$ from $a (1 - x a) = 0$ .
Prove that left noetherian plus regular is equivalent to semisimple.
Construct a pseudo-inverse for an endomorphism of a semisimple module and apply it to ${End}_{k} (V)$ .

Definitions

Definition(4.23)Von Neumann regular ring

A ring $R$ with identity is von Neumann regular if for every $a \in R$ there exists $x \in R$ with $a = a x a$ . Any of the equivalent conditions of $(4.23)$ may be taken as the definition; since condition $(1)$ mentions no side, the notion is left-right symmetric.

Pseudo-inverse of $a$: Any $x$ with $a x a = a$ . Setting $y = x a x$ produces a reflexive pseudo-inverse satisfying both $a y a = a$ and $y a y = y$ .
Unit-regular: For every $a$ there is a unit $u$ with $a = a u a$ . Strictly stronger than regular; unit-regular rings are Dedekind-finite.
Strongly regular: For every $a$ there is $x$ with $a = a^{2} x$ . Equivalent to being regular and reduced, that is, having no nonzero nilpotent elements.
Semiprimitive: $rad R = 0$ . Every von Neumann regular ring is semiprimitive; the converse fails for $ℤ$ .
Dedekind-finite: $a b = 1 \Rightarrow b a = 1$ . Regular rings need not be Dedekind-finite: ${End}_{k} (V)$ with $\dim V$ infinite is a counterexample.

Regular is heavily overloaded. It does not mean regular local ring, regular sequence, or regular element. When ambiguity is possible, write von Neumann regular in full.

Core Concepts

Idempotents and summands are the same data

A left ideal $I \subseteq R$ is a direct summand of $R R$ if and only if $I = R e$ for some idempotent $e$ . Given $R = I \oplus J$ , write $1 = e + f$ with $e \in I$ , $f \in J$ ; then $e = e^{2}$ and $I = R e$ . Conversely $R = R e \oplus R (1 - e)$ for any idempotent $e$ . This is why $(2) \Leftrightarrow (2')$ and $(3) \Leftrightarrow (3')$ are formalities.

The pseudo-inverse manufactures the idempotent

From $a = a x a$ set $e = x a$ . Then $e^{2} = x a x a = x (a x a) = x a = e$ , so $e$ is idempotent; $e \in R a$ gives $R e \subseteq R a$ , and $a = a e$ gives $R a \subseteq R e$ . Hence $R a = R e$ . Conversely, generation by idempotents produces a pseudo-inverse, so the two viewpoints are interchangeable.

a = a x a

e = x a

idempotent

R a = R e

R a

a direct summand

Why finitely generated, and not all, ideals

Requiring every left ideal to be a summand is semisimplicity. Requiring it of principal ideals turns out to be the same as requiring it of finitely generated ones, because a sum of two idempotent-generated left ideals is again one — the computation with $e + e' - e e'$ below. The gap between finitely generated and all is exactly the gap between regular and semisimple, and it is closed by the noetherian hypothesis.

Key Results

Theorem(4.23)von Neumann

For a ring $R$ with identity the following are equivalent.

For every $a \in R$ there is $x \in R$ with $a = a x a$ .
Every principal left ideal of $R$ is generated by an idempotent.
Every principal left ideal of $R$ is a direct summand of $R R$ .
Every finitely generated left ideal of $R$ is generated by an idempotent.
Every finitely generated left ideal of $R$ is a direct summand of $R R$ .

Because condition $(1)$ is left-right symmetric, the theorem also holds with left replaced by right throughout conditions $(2)$ – $(3')$ .

Proof

** $(2) \Leftrightarrow (2')$ and $(3) \Leftrightarrow (3')$ .** A left ideal is a direct summand of $R R$ exactly when it is generated by an idempotent, as recorded above.

** $(1) \Rightarrow (2)$ .** Given $a$ , pick $x$ with $a x a = a$ and set $e = x a$ . Then $e^{2} = x (a x a) = x a = e$ . Also $e \in R a$ , so $R e \subseteq R a$ ; and $a = a x a = a e$ , so $R a \subseteq R e$ . Hence $R a = R e$ with $e$ idempotent.

** $(2) \Rightarrow (1)$ .** Given $a$ , write $R a = R e$ with $e = e^{2}$ . Then $e = x a$ and $a = y e$ for suitable $x, y \in R$ . Therefore $a x a = a e = (y e) e = y e^{2} = y e = a$ .

** $(3) \Rightarrow (2)$ ** is trivial.

** $(2) \Rightarrow (3)$ .** By induction it suffices to treat $I = R e + R f$ with $e, f$ idempotent. Since $f = f e + f (1 - e)$ and $f e \in R e$ , we have $I = R e + R f (1 - e)$ . By $(2)$ there is an idempotent $e'$ with $R f (1 - e) = R e'$ . Then $e' \in R f (1 - e)$ gives $e' e \in R f (1 - e) e = 0$ , so $e' e = 0$ . Put

e'' = e + e' - e e' .

Using $e^{2} = e$ , $e'^{2} = e'$ and $e' e = 0$ one computes $e''^{2} = e + e' - e e' = e''$ , so $e''$ is idempotent. Moreover $e e'' = e + e e' - e e' = e$ and $e' e'' = e' - (e' e) e' = e'$ , so both $e$ and $e'$ lie in $R e''$ ; conversely $e'' \in R e + R e'$ . Hence $I = R e + R e' = R e''$ , which is $(3)$ .

Corollary(4.24)Position in the hierarchy

Every semisimple ring is von Neumann regular, and every von Neumann regular ring is semiprimitive: $rad R = 0$ .

Proof

If $R$ is semisimple then every left ideal is a direct summand of $R R$ , so in particular $(2')$ holds and $R$ is regular. For the second implication let $a \in rad R$ and choose $x$ with $a = a x a$ . Then $a (1 - x a) = 0$ ; but $x a \in rad R$ , so $1 - x a \in U (R)$ and multiplying on the right by its inverse gives $a = 0$ .

Theorem(4.25)Noetherian regular equals semisimple

A ring $R$ is semisimple if and only if it is left noetherian and von Neumann regular. The same statement holds with right noetherian in place of left noetherian.

Proof

A semisimple ring is left noetherian and, by $(4.24)$ , regular. Conversely, if $R$ is left noetherian then every left ideal is finitely generated, hence by condition $(3')$ of $(4.23)$ every left ideal is a direct summand of $R R$ . A ring all of whose left ideals are direct summands is semisimple.

Corollary(4.26)One chain condition gives all of them

If a von Neumann regular ring is left noetherian, then it is semisimple and therefore left and right noetherian and left and right artinian.

Proposition(4.27)Endomorphism rings of semisimple modules

Let $k$ be a ring and $M$ a semisimple right $k$ -module. Then $R = End (M_{k})$ is von Neumann regular.

Proof

Let $f \in R$ and put $K = \ker f$ . Because $M$ is semisimple, $K$ has a complement: $M = K \oplus N$ for some submodule $N$ . Then $f$ restricts to an isomorphism $f |_{N} : N \to N' : = f (N) = f (M)$ . Again by semisimplicity choose a complement $M = K' \oplus N'$ . Define $g \in R$ by $g |_{K'} = 0$ and $g |_{N'} = (f |_{N})^{- 1}$ .

For $m \in M$ write $m = u + v$ with $u \in K$ , $v \in N$ . Then $f (m) = f (v) \in N'$ , so $g (f (m)) = v$ and $f (g (f (m))) = f (v) = f (m)$ . Hence $f g f = f$ .

Remark—Closure properties

Direct products and quotient rings of von Neumann regular rings are again von Neumann regular, as is $M_{n} (R)$ for regular $R$ . Subrings are not: $ℤ \subseteq ℚ$ shows this at once. Neither is a polynomial extension — for $R \neq 0$ the ring $R [t]$ is never regular, since $t = t g t = g t^{2}$ would compare a degree-one element with something of degree at least two.

Proof Techniques and Method

How these proofs work, and the reusable move.

Move 1

Turn an equation into an idempotent

From $a x a = a$ the element $x a$ is idempotent, and $a x$ is too. Every structural statement about regular rings is obtained by producing idempotents and using them to split modules.

Move 2

Orthogonalise before adding

To combine $R e$ and $R f$ , replace $f$ by $f (1 - e)$ so that the new idempotent $e'$ satisfies $e' e = 0$ . Then $e + e' - e e'$ is idempotent and generates the sum. Orthogonalisation is the standard first step in idempotent arithmetic.

Move 3

Split the kernel, invert on a complement

For a map out of a semisimple module, choose complements of the kernel and of the image and define the pseudo-inverse to be zero on one and inverse on the other. This is precisely the construction of a generalised inverse in linear algebra.

Move 3 shows where the name pseudo-inverse comes from. Over a field the same recipe applied to a matrix gives a reflexive generalised inverse; adding orthogonality requirements for a chosen inner product would give the Moore–Penrose inverse, which is unique. Ring-theoretically nothing selects a preferred $x$ , and non-uniqueness is the normal state of affairs.

Given $a$ Look for any

x

with

a x a = a

; symmetry means you may work on whichever side is convenient.

NormaliseReplace

x

y = x a x

to get

a y a = a

and

y a y = y

; the pair

(a, y)

is then reflexive.

Extract idempotents

e = y a

and

e^{'} = a y

are idempotents with

R a = R e

and

a R = e^{'} R

SplitUse

R = R e \oplus R (1 - e)

to reduce the problem to the two corners of a Peirce decomposition.

Worked Example

Linear operators on an infinite-dimensional space

Let $k$ be a field, $V$ a vector space with countable basis $e_{1}, e_{2}, e_{3}, \dots$ , and $R = {End}_{k} (V)$ . Since $V$ is a semisimple $k$ -module, $(4.27)$ says $R$ is von Neumann regular. Take the shift

f (e_{i}) = e_{i + 1} (i \geq 1), g (e_{1}) = 0, g (e_{i + 1}) = e_{i} (i \geq 1) .

(E.1)

Then $g f = 1$ while $f g$ is the projection onto the span of $e_{2}, e_{3}, \dots$ , so $f g \neq 1$ . Nevertheless $g$ is a pseudo-inverse of $f$ :

(f g f) (e_{i}) = (f g) (e_{i + 1}) = f (e_{i}) = e_{i + 1} = f (e_{i}) for all i .

(E.2)

$f g f = f$ , so $f$ is regular even though it is not invertible.

$R$ is not semisimple: it is not left noetherian, so $(4.25)$ forbids it. Concretely, $I_{n} = {h : h (e_{i}) = 0 for all i > n}$ is a strictly ascending chain of left ideals, and $J_{n} = {h : h (e_{i}) = 0 for all i \leq n}$ a strictly descending one.
$R$ is not Dedekind-finite: $g f = 1 \neq f g$ . So regularity does not imply Dedekind-finiteness, and $R$ is not unit-regular.
$R$ is semiprimitive, as $(4.24)$ requires: any $a$ with $a = a x a$ and $a \in rad R$ must vanish.

An infinite product of matrix rings

Let $R = \prod_{n \geq 1} M_{n} (k)$ . Each factor is semisimple, hence regular, and a product of regular rings is regular — solve $a = a x a$ coordinatewise. So $R$ is regular and $rad R = 0$ . It is not semisimple: the ideal of sequences with only finitely many nonzero entries is not a direct summand, and $R$ is not left noetherian.

A commutative example and a commutative non-example

Boolean rings. If $a^{2} = a$ for all $a$ , then $a = a \cdot 1 \cdot a$ , so $R$ is regular. The power set of a set $X$ with symmetric difference and intersection is the standard model. More generally any ring in which every $a$ satisfies $a^{n (a)} = a$ with $n (a) \geq 2$ is regular, via $x = a^{n (a) - 2}$ .
** $ℤ$ is not regular.** The equation $2 = 2 x \cdot 2 = 4 x$ has no solution in $ℤ$ . Yet $rad ℤ = 0$ , so semiprimitive does not imply regular and the second implication of $(4.24)$ is strict.
A criterion. A commutative ring is von Neumann regular exactly when it is reduced and every prime ideal is maximal; equivalently, when every localisation at a maximal ideal is a field. So $ℤ$ fails on the dimension count, and $k [x] / (x^{2})$ fails on reducedness.

Comparison and Classification

Properties across the hierarchy
	Semisimple	von Neumann regular	Semiprimitive	Arbitrary
$rad R = 0$	yes	yes	yes	no
every f.g. left ideal is a summand	yes	yes	no	no
every left ideal is a summand	yes	no	no	no
left noetherian	yes	no	no	no
every module is projective	yes	no	no	no
every f.p. module is projective	yes	yes	no	no
closed under arbitrary direct products	no	yes	yes	yes
closed under subrings	no	no	no	yes

Properties across the hierarchy

Reference examples
Ring	Regular?	Semisimple?	Why
$M_{n} (D)$ , $D$ a division ring	yes	yes	artinian and semiprimitive
${End}_{k} (V)$ , $\dim_{k} V$ infinite	yes	no	$(4.27)$ ; not noetherian, not Dedekind-finite
$\prod_{n \geq 1} M_{n} (k)$	yes	no	products of regular rings are regular
Boolean rings	yes	no in general	$a = a \cdot 1 \cdot a$ ; commutative of Krull dimension $0$
$ℤ$	no	no	$2 = 4 x$ is unsolvable, though $rad ℤ = 0$
$k [t]$	no	no	$t = g t^{2}$ fails on degrees
$k [t] / (t^{2})$	no	no	$rad R = (t) \neq 0$
$T_{2} (k)$ upper triangular	no	no	nonzero radical

Relationship Map

Semisimplevon Neumann regularSemiprimitive

Both arrows are strict, and each becomes an equivalence when a chain condition is added: regular plus left noetherian is semisimple by $(4.25)$ , and semiprimitive plus left artinian is semisimple by Lam's $(4.14)$ .

Von Neumann regular rings — $a = a x a$ for all $a$
- Strengthenings
  - unit-regular: $x$ may be taken to be a unit; implies Dedekind-finite
  - strongly regular: $a = a^{2} x$ ; equivalent to regular and reduced
  - semisimple: regular plus left noetherian
- Consequences
  - $rad R = 0$
  - no nonzero nil one-sided ideals, hence $R$ is semiprime
  - every finitely presented module is projective; $R$ is coherent
  - the lattice of principal left ideals is complemented
- Stability
  - closed under direct products, quotients and $M_{n} (-)$
  - not closed under subrings or polynomial extension

That there are no nonzero nil one-sided ideals follows quickly: if $a$ lies in a nil one-sided ideal and $a = a x a$ , then $a x$ is an idempotent lying in that ideal, hence nilpotent, hence $a x = 0$ and $a = a x a = 0$ .

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Operator algebras

Continuous geometry and finite von Neumann algebras

Von Neumann introduced regular rings to coordinatise lattices with a continuous dimension function. The ring of affiliated operators of a finite von Neumann algebra is regular, and this is the algebraic shadow of the type II $1$ dimension theory.

Numerical linear algebra

Generalised inverses

$a = a x a$ is the first Penrose condition. Least-squares solvers compute a Moore–Penrose inverse, which is the unique reflexive pseudo-inverse compatible with the adjoint. The ring-theoretic statement is what survives when no inner product is available.

K-theory

Ultramatricial algebras and dimension groups

Direct limits of finite products of matrix rings are regular, and their ordered $K_{0}$ groups classify them — the algebraic counterpart of Elliott's classification of AF algebras.

Module theory

Coherence and finite presentation

Over a regular ring every finitely presented module is projective. This makes regular rings a standard test class for homological conjectures where noetherian hypotheses would trivialise the question.

The honest statement is that regularity is a structural hypothesis rather than an applied one: its value is that it supports splitting arguments without any finiteness, which is exactly what infinite-dimensional analysis and $K$ -theory need.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Preferred termvon Neumann regular, written in full to avoid collision with regular local rings

Common abbreviationregular ring — safe only in a context where commutative regular local rings do not appear

Element condition

a = a x a

; Penrose's first condition in the numerical literature

Variantsunit-regular, strongly regular,

π

-regular, right weakly regular

Standard monographGoodearl, von Neumann Regular Rings

ImplementationsGAP and Magma expose idempotent and radical computations; regularity is tested indirectly through the radical

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

In $M_{n} (k)$ a pseudo-inverse of $A$ is computed from a rank factorisation $A = B C$ with $B$ of full column rank and $C$ of full row rank: any $X$ with $C X B = I_{r}$ works, and Gaussian elimination supplies one in $O (n^{3})$ field operations.
The reflexive normalisation $y = x a x$ costs two extra multiplications and is worth doing: it makes $a y$ and $y a$ idempotent and stabilises subsequent Peirce decompositions.
Deciding whether a finitely presented ring is von Neumann regular is not algorithmic in general, since the word problem for finitely presented rings is undecidable. Practical tests assume a finite ring or a finite-dimensional algebra.
For a finite-dimensional algebra $A$ over a field, regularity is equivalent to semisimplicity by $(4.25)$ , so the test reduces to computing $rad A$ and checking that it vanishes — one nullspace computation in characteristic $0$ .
Boolean rings admit the cheapest test of all: regularity is automatic, and computations reduce to set operations, which is why they underpin decision procedures based on Stone duality.

Failure Modes and Common Mistakes

Do not assume a regular ring is Dedekind-finite. The shift operator gives $g f = 1 \neq f g$ inside a regular ring.
Do not confuse von Neumann regular with regular local, with a regular element, or with a regular sequence.
Do not expect every module over a regular ring to be projective; only the finitely presented ones are.
Do not conclude that a regular ring has no nilpotent elements. $M_{2} (k)$ is regular and full of them; the correct statement is that it has no nonzero nil one-sided ideals.

Historical Notes and Lessons Learned

1936von Neumann defines regular ringsIn a short note in the Proceedings of the National Academy of Sciences, von Neumann isolates the condition a equals axa and proves the equivalence with complementation of principal ideals.
1936–37Continuous geometryThe lecture notes that became Continuous Geometry coordinatise complemented modular lattices with a continuous dimension function; regular rings are the coordinate rings, and the theorem reproduced here is Part II, Chapter 2 of that work.
1950sAlgebraic developmentRegular rings enter general ring theory as the natural chain-condition-free weakening of semisimplicity, and the relation to the Jacobson radical is clarified.
1979Goodearl's monographGoodearl's book organises the theory: unit-regularity, directly finite rings, ideal lattices and the K-theoretic classification of ultramatricial algebras.
1980sOperator algebra connectionsRings of affiliated operators and the classification of AF algebras by ordered K-theory make the regular-ring dimension theory a working tool in analysis.

The lesson is about how to weaken a definition. Von Neumann did not weaken semisimplicity by relaxing the conclusion; he restricted the class of ideals to which it applies. Restricting the scope of a universally quantified condition, rather than diluting the condition itself, is what produced a class large enough to contain infinite-dimensional endomorphism rings and still small enough to have a structure theory.

Quick Reference

Definition

\forall a \exists x : a = a x a

Ideal formevery f.g. left ideal is

R e

with

e = e^{2}

, hence a direct summand

Symmetryleft and right versions agree, since

a = a x a

mentions no side

Radical

rad R = 0

always

Hierarchysemisimple

\Rightarrow

regular

\Rightarrow

semiprimitive, both strict

Upgraderegular

+

left noetherian

=

semisimple

Main source

End (M_{k})

for

M_{k}

semisimple

Closureproducts, quotients,

M_{n} (-)

yes; subrings and

R [t]

The five conditions of (4.23)
Label	Condition	Role
(1)	$a = a x a$ for all $a$	the definition; manifestly side-neutral
(2)	principal left ideals are $R e$ , $e = e^{2}$	the idempotent form
(2 $'$ )	principal left ideals are direct summands	the module form
(3)	f.g. left ideals are $R e$ , $e = e^{2}$	obtained from (2) by orthogonalising
(3 $'$ )	f.g. left ideals are direct summands	the form used to prove (4.25)

Frequently Asked Questions

Why is von Neumann regularity left-right symmetric when the ideal conditions are one-sided?

Because condition $(1)$ , $a = a x a$ , mentions no side at all, and $(4.23)$ proves it equivalent to each of the one-sided conditions. So the left-hand conditions and the right-hand conditions are both equivalent to the same symmetric statement, hence to each other. This is the same style of argument that makes $rad R$ side-neutral.

Is a von Neumann regular ring the same as a ring with no nilpotent elements?

No. $M_{2} (k)$ is regular and contains nilpotent matrices. Regular and reduced is a stronger condition called strongly regular, characterised by the existence of $x$ with $a = a^{2} x$ . What regularity does exclude is nonzero nil one-sided ideals.

Does regularity imply that every module is projective?

Only for finitely presented modules. Over a regular ring every finitely presented module is projective, which makes the ring coherent, but there are plenty of non-projective modules — otherwise the ring would be semisimple by a standard characterisation.

How does one produce non-semisimple regular rings?

Two reliable recipes. Take $End (M_{k})$ for an infinite-dimensional semisimple module $M$ , which is $(4.27)$ ; or take an infinite direct product, or a direct limit, of semisimple rings. Both break the noetherian condition, which by $(4.25)$ is the only obstruction.

What is the relationship to the Moore–Penrose inverse?

The equation $a = a x a$ is the first of Penrose's four conditions. The other three involve an involution and pin the inverse down uniquely. Ring theory keeps only the first condition, so pseudo-inverses exist in abundance but none is canonical; a ring with involution can support the full Moore–Penrose theory.

Are von Neumann regular rings semiprime?

Yes, and more: they have no nonzero nil one-sided ideals. If $a$ lies in a nil one-sided ideal and $a = a x a$ , then $a x$ is an idempotent in that ideal and so nilpotent, forcing $a x = 0$ and hence $a = 0$ . Semiprimeness and $rad R = 0$ both follow.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §4, statements (4.23)–(4.27).
J. von Neumann, “On regular rings”, Proceedings of the National Academy of Sciences of the USA 22 (1936), 707–713.
J. von Neumann, Continuous Geometry, edited by I. Halperin, Princeton Mathematical Series 25, Princeton University Press, 1960.
K. R. Goodearl, von Neumann Regular Rings, Pitman, 1979; second edition, Krieger, 1991.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988.

AI Suggested Questions

Prove that $M_{n} (R)$ is von Neumann regular whenever $R$ is, and identify where the pseudo-inverse comes from.
Give an example of a von Neumann regular ring that is not unit-regular, and explain the role of Dedekind-finiteness.
Characterise the commutative von Neumann regular rings in terms of their prime spectrum and Stone duality.
How does the ordered group $K_{0}$ classify ultramatricial algebras, and what does it say about AF algebras?
What is a $π$ -regular ring, and how does it relate to strongly $π$ -regular rings and to the artinian condition?
Show that the ring of affiliated operators of a finite von Neumann algebra is von Neumann regular.
Does von Neumann regularity pass to fixed rings under a finite group action?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Idempotents and summands are the same data

The pseudo-inverse manufactures the idempotent

Why finitely generated, and not all, ideals

06Key Results

07Proof Techniques and Method

Turn an equation into an idempotent

Orthogonalise before adding

Split the kernel, invert on a complement

08Worked Example

Linear operators on an infinite-dimensional space

An infinite product of matrix rings

A commutative example and a commutative non-example

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Continuous geometry and finite von Neumann algebras

Generalised inverses

Ultramatricial algebras and dimension groups

Coherence and finite presentation

12Standards and Notation

13Computational Notes

14Failure Modes and Common Mistakes

15Historical Notes and Lessons Learned

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions