Von Neumann Regular Rings

Introduction

This article is not a rung of either chain of Rings and Fields. It sits after Division Rings, and the ring class it introduces is defined by idempotents rather than by divisibility: it neither strengthens nor weakens its neighbours, and it is written so that it does not. The base is a general ring $A$ with $1 \neq 0$, not assumed commutative; where the article specialises to the commutative case it says so, and the commutative characterisation is its main theorem. Throughout, an idempotent is an element with $e^2 = e$, in the vocabulary of Rings, above, and $Aa$ denotes the principal left ideal generated by $a$.

The definition is von Neumann's: a ring is regular if every principal left ideal is generated by an idempotent, equivalently if every element $a$ satisfies $a = axa$ for some $x$. The first section proves the equivalence of the two forms and the equivalence with the finitely generated left ideals; the second states the commutative characterisation, that a commutative ring is regular exactly when it is reduced of Krull dimension zero, equivalently when its localisation at every maximal ideal is a field; the third treats the strongly regular rings, the reduced ones, and the examples and non-examples. The homological characterisation of the class — that every left module is flat — belongs to the module theory of Linear Spaces, in a later category of this Part, and is not developed here.

The Definition and Its Equivalents

Definition. A ring $A$ with $1 \neq 0$ is von Neumann regular, or regular, if for every $a \in A$ there is $x \in A$ with

$$ a = axa . $$

The element $x$ is a generalised inverse of $a$.

Proposition. Let $A$ be a ring with $1 \neq 0$ and let $a \in A$. The following are equivalent:

$a = axa$ for some $x \in A$;
$Aa = Ae$ for some idempotent $e \in A$;
$aA = fA$ for some idempotent $f \in A$.

Proof. Assume (1) and put $e = xa$. Then

$$ e^2 = xaxa = x(axa) = xa = e , \qquad ae = axa = a , \qquad e = xa \in Aa , $$

so $e$ is an idempotent, $a \in Ae$ because $a = ae$, and $e \in Aa$; hence $Aa = Ae$ and (2) holds. Assume (2), so $a = ae$ for the idempotent $e$ and $e = ya$ for some $y$. Then $a = ae = aya$, which is (1), and the element $y$ is a generalised inverse. Statement (3) is the same argument on the right, with $f = ax$.

Example. In a division ring every nonzero $a$ satisfies $a = aa^{-1}a$, and $0 = 0$, so every division ring is regular. If $a = axa$, then $ax$ and $xa$ are idempotents, since

$$ (ax)^2 = a(xax) = ax , \qquad (xa)^2 = (xax)a = xa , $$

and $xax$ is again a generalised inverse of $a$, because $a(xax)a = (axa)(xa) = axa = a$. So the set of generalised inverses of $a$ is closed under $x \mapsto xax$.

Proposition. A ring $A$ with $1 \neq 0$ is regular if and only if every finitely generated left ideal of $A$ is generated by an idempotent.

Proof. One direction is immediate, a principal left ideal being finitely generated. For the other, let $e$ and $f$ be idempotents and put

$$ g = e + (1 - e)f = e + f - ef . $$

Then $eg = e + ef - ef = e$ and $gf = ef + (1 - e)f^2 = ef + (1 - e)f = f$, so that $e = eg \in Ag$ and $f = gf \in Ag$, whence $Ae + Af \subseteq Ag$; and $g = e + f - ef \in Ae + Af$, whence $Ag \subseteq Ae + Af$. So $Ae + Af = Ag$, and by regularity applied to the element $g$ there is an idempotent $e'$ with $Ag = Ae'$, giving $Ae + Af = Ae'$. Induction on the number of generators gives the statement for every finitely generated left ideal.

Example (the finitely generated ideals are idempotent). If $I = Ae$ with $e$ idempotent, then $I^2 = I$: clearly $I^2 \subseteq I$, and $e = e \cdot 1 \cdot e \in AeAe = I^2$ gives $I = Ae \subseteq I^2$. So in a regular ring every finitely generated left ideal is idempotent-generated, and every left ideal is a sum of principal ideals, hence generated by idempotents, possibly by infinitely many of them.

Proposition. A direct product $\prod_i A_i$ of rings with $1 \neq 0$ is regular if and only if each $A_i$ is regular, and the matrix ring $M_n(A)$ is regular if and only if $A$ is regular. In particular $M_n(D)$ is regular for every division ring $D$ and every $n \geq 1$.

Proof. In a direct product the operations are componentwise, so $a = axa$ holds in the product exactly when it holds in every component. For the matrix ring, regularity of $M_n(A)$ is equivalent to regularity of $A$, the generalised inverse of a matrix over a regular ring being computed in the standard way; this is cited from the literature.

The Commutative Characterisation

Theorem. Let $R$ be a commutative ring with $1 \neq 0$. The following are equivalent:

$R$ is regular;
$R$ is reduced of Krull dimension zero;
$R_{\mathrm{M}}$ is a field for every maximal ideal $\mathrm{M}$;
$R_{\mathrm{P}}$ is a field for every prime ideal $\mathrm{P}$;
every prime ideal of $R$ is maximal and $R$ is reduced.

Proof. (1) $\Rightarrow$ (2). Let $a^2 = 0$. Regularity at $a$ gives $a = axa = a^2x = 0$, so $R$ is reduced. Let $\mathrm{P}$ be prime and let $\mathrm{M}$ be a maximal ideal containing it, with $\mathrm{P} \neq \mathrm{M}$; choose $a \in \mathrm{M} \setminus \mathrm{P}$ and $x$ with $a = a^2x$. Then $a(1 - ax) = 0 \in \mathrm{P}$, and $a \notin \mathrm{P}$, so $1 - ax \in \mathrm{P} \subseteq \mathrm{M}$, while $ax \in \mathrm{M}$; hence $1 \in \mathrm{M}$, a contradiction. So $\mathrm{P} = \mathrm{M}$ and the dimension is zero.

(2) $\Rightarrow$ (3). Let $\mathrm{M}$ be maximal. The localisation $R_{\mathrm{M}}$ is reduced, the localisation of a reduced ring being reduced, and it has a unique prime ideal, namely $\mathrm{M}R_{\mathrm{M}}$, since the primes of the localisation correspond to the primes of $R$ contained in $\mathrm{M}$, of which there is only $\mathrm{M}$. Hence $\mathrm{M}R_{\mathrm{M}}$ is the nilradical of the reduced ring $R_{\mathrm{M}}$, by Reduced Rings and the Nilradical, above, so $\mathrm{M}R_{\mathrm{M}} = 0$ and $R_{\mathrm{M}}$ is a field.

(3) $\Rightarrow$ (4). A prime ideal is contained in a maximal ideal, so it suffices to note that $R_{\mathrm{P}} = (R_{\mathrm{M}})_{\mathrm{P}R_{\mathrm{M}}}$ is a localisation of a field, hence a field.

(4) $\Rightarrow$ (5). If $\mathrm{P}$ is prime, then $\mathrm{P}R_{\mathrm{P}}$ is the maximal ideal of the field $R_{\mathrm{P}}$, hence is zero; so $\mathrm{P}$ is maximal, since for $a \notin \mathrm{P}$ the image of $a$ in $R_{\mathrm{P}}$ is nonzero and therefore a unit, which exhibits $1 - ab \in \mathrm{P}$ for some $b$. If $a^2 = 0$ then the image of $a$ in every $R_{\mathrm{P}}$ is zero, so $a = 0$; hence $R$ is reduced.

(5) $\Rightarrow$ (1). This is the substantive direction and is cited from the literature: a reduced ring of Krull dimension zero is regular. The content is that every prime ideal is maximal, so that distinct maximal ideals are comaximal and the maximal ideals are the only primes; the localisation at each of them is a field, by the argument of the previous paragraphs read backwards, and an element of $R$ is regular because the set of maximal ideals containing a given element is both open and closed in the prime spectrum, so that the element which inverts $a$ at those maximal ideals not containing it is the localisation of an element of $R$. The reader is referred to the literature for the details of that last step.

Corollary. A commutative regular ring is a subdirect product of fields, and a reduced ring of Krull dimension zero is a subdirect product of fields. Conversely a subdirect product of fields is reduced; it is regular when the product is a direct product.

Proof. The localisations at the maximal ideals are fields by the theorem, and the intersection of the kernels of the localisation maps is the nilradical, which is zero when $R$ is reduced; hence the product map into the product of the fields is injective.

Corollary. A commutative regular ring in which every idempotent is $0$ or $1$ is a field. In particular a local ring is regular exactly when it is a field, and a domain is regular exactly when it is a field.

Proof. In a local ring the idempotents are $0$ and $1$, so a principal ideal generated by an idempotent is $0$ or the whole ring and every element is zero or a unit. A domain is regular exactly when every prime ideal is maximal and the ring is reduced with no nonzero nilpotents, which for a domain means that the zero ideal is maximal.

Strongly Regular Rings and Examples

Definition. A ring $A$ with $1 \neq 0$ is strongly regular if for every $a \in A$ there is $x \in A$ with

$$ a = a^2x . $$

Theorem. A ring is strongly regular if and only if it is regular and reduced.

Proof. If $a^2 = 0$ then $a = a^2x = 0$, so a strongly regular ring is reduced; that it is also regular is the theorem of Arens and Kaplansky, cited from the literature. Conversely let $A$ be regular and reduced, and let $a \in A$ with $a = axa$; put $e = ax$, so that $e$ is idempotent and $ea = axa = a$, whence $(1 - e)a = 0$. Then

$$ (a - a^2x)^2 = (a - ae)^2 = \big(a(1 - e)\big)^2 = a(1 - e)a(1 - e) = 0 , $$

the last step using $(1 - e)a = 0$. Reducedness gives $a - a^2x = 0$, that is $a = a^2x$. Hence $A$ is strongly regular.

Remark. The equivalence of regularity with strong regularity plus reducedness is the theorem of Arens and Kaplansky, and the argument above for the converse is the standard one. For commutative rings it says that a commutative regular ring is always strongly regular, reducedness being automatic by the first implication.

Example (Boolean rings). A ring in which every element satisfies $a^2 = a$ is called a Boolean ring; such a ring is commutative of characteristic two, and every element is its own generalised inverse, so it is regular and reduced. The field $\mathbb{F}_2$ and the power set ring of a set, with symmetric difference and intersection, are the standard examples. Every Boolean ring is commutative, regular and of characteristic two, but a commutative regular ring of characteristic two need not be Boolean: the field $\mathbb{F}_4$ has characteristic two and is regular, and $x^2 = x$ fails for the two elements of $\mathbb{F}_4 \setminus \mathbb{F}_2$.

Example (products of division rings). Let $\{D_i\}$ be a family of division rings and let $A = \prod_i D_i$. Then $A$ is regular by the proposition above: an element is regular in a component either because it is a unit or because it is zero. For $A = \prod_{n \geq 1} M_2(F)$ over a field $F$, the ring is regular by the proposition above, and it is not reduced.

Example (the non-examples). $\mathbb{Z}$ is not regular: the equation $2 = 2^2x$ has no solution in $\mathbb{Z}$. The polynomial ring $k[x]$ over a field is not regular, for the same reason applied to $x$. The ring $\mathbb{Z}/4\mathbb{Z}$ is not regular, since $2^2 = 0$ and $2 \neq 0$; it is not reduced, so no power of it lying in the reduced classes of this article can repair the failure. A local ring that is not a field is not regular by the corollary above, and a domain that is not a field is not regular. In particular the class is strictly wider than the class of division rings and it is incomparable with the class of reduced rings: $\mathbb{Z}$ is reduced and not regular, while $M_2(F)$ is regular and not reduced.

Theorem (closure under quotients). Every quotient of a regular ring is regular.

Proof. Let $A$ be regular, let $I \trianglelefteq A$ and let $\pi : A \to A/I$ be the quotient map. If $a = axa$ in $A$ then

$$ \pi(a) = \pi(axa) = \pi(a)\pi(x)\pi(a) , $$

so the equation defining regularity is satisfied in $A/I$ by the image $\pi(a)$ and the element $\pi(x)$; since $\pi$ is surjective, every element of $A/I$ is such an image.

Remark. The class is therefore closed under quotients, and the property of being regular, being expressed by the equation $a = axa$, is inherited by every homomorphic image; the parallel class of reduced rings, in contrast, is not closed under quotients, as the quotient $k[x]/(x^2)$ of Reduced Rings and the Nilradical, above, records.

Proposition. Every regular ring is semiprime.

Proof. Let $A$ be regular and let $I$ be a two-sided ideal. For $a \in I$ choose $x$ with $a = axa$; then $a = axa \in IAI \subseteq I^2$, so $I \subseteq I^2$, while $I^2 \subseteq I$ always holds. Hence $I = I^2$ and no nonzero ideal of $A$ is nilpotent, which is semiprimeness as in Semiprime Rings, above.

Theorem. The centre $Z(A)$ of a regular ring is a commutative regular ring, hence strongly regular and reduced.

Proof. This is the standard theorem of the theory, cited from the literature: for $z \in Z(A)$ the equation $z = zxz$ holds with $x \in A$, and the idempotent $zx$ can be replaced by a central idempotent $e$ with $Z(A)z = Z(A)e$, which is the assertion of regularity for $Z(A)$.

The Place of the Class among the Rings of the Category

Remark. The class of regular rings is not a rung of either chain of this category, and the comparison with the rungs shows why. A regular ring that is a domain is a field, by the corollary above, so the intersection of the class with the commutative chain is the class of fields; and it contains, besides the division rings, the matrix rings $M_n(D)$, which are prime and not domains, and the products $\prod_i D_i$ of division rings, which are semiprime and not domains. So the class neither strengthens nor weakens its neighbours: $M_2(F)$ is regular and prime but not a domain, and $\mathbb{Z}$ is a domain and not regular.

class	regular?	reduced?	a domain?
$\mathbb{F}_2$, a field	yes	yes	yes
$M_2(F)$, $F$ a field	yes	no	no
$\prod_{n} D_n$, $D_n$ division rings	yes	yes	no
$\mathbb{Z}$	no	yes	yes
$k[x]$	no	yes	yes
$\mathbb{Z}/4\mathbb{Z}$	no	no	no
$\mathbb{Z}_{(2)}$, a local ring that is not a field	no	yes	yes

Proposition. The regular rings are closed under direct products, under matrix rings, under corner rings and under quotients, and they are not closed under subrings or under polynomial extensions; the reduced rings are closed under subrings, products and localisation, and the two classes are incomparable.

Proof. The closure under products and matrix rings is the proposition above, the closure under quotients is the theorem above, and that a corner ring of a regular ring is regular is standard, cited from the literature. For the failures, $\mathbb{Q}$ is regular and its subring $\mathbb{Z}$ is not, and for a field $k$ the polynomial ring $k[x]$ is not regular by the example above. The incomparability is shown by $M_2(F)$, regular and not reduced, and by $\mathbb{Z}$, reduced and not regular.

Remark. Idempotents are the vocabulary of Rings, above, reducedness and the nilradical are Reduced Rings and the Nilradical', above, and Krull dimension is Integral Extensions and Krull Dimension'. The module-theoretic characterisation of the class — that a ring is regular exactly when every left module over it is flat, equivalently when it has weak global dimension zero — is the module theory of Linear Spaces, in a later category of this Part, and is not developed or used in this article.

Summary

A ring $A$ with $1 \neq 0$ is von Neumann regular when every principal left ideal is generated by an idempotent, equivalently when every element satisfies $a = axa$, equivalently when every finitely generated left ideal is generated by an idempotent, and every finitely generated left ideal of a regular ring is idempotent-generated. Division rings are regular; products of regular rings and matrix rings over regular rings are regular, so $M_n(D)$ is regular for every division ring $D$; and a commutative ring is regular exactly when it is reduced of Krull dimension zero, equivalently when its localisation at every maximal ideal is a field, equivalently when every prime ideal is maximal and the ring is reduced. The strongly regular rings are the regular reduced rings, by the theorem of Arens and Kaplansky; Boolean rings are the commutative regular rings of characteristic two. The non-examples are $\mathbb{Z}$, $k[x]$, $\mathbb{Z}/4\mathbb{Z}$, the local rings that are not fields and the domains that are not fields. The class is closed under direct products, matrix rings, corner rings and quotients, and is not closed under subrings or polynomial extensions.

Summary of Notation

Symbol	Meaning
$A$	A ring with $1 \neq 0$, not assumed commutative; the base of this article
$a = axa$	The regularity condition; $x$ is a generalised inverse of $a$
$Aa$	The principal left ideal generated by $a$
$e$, $f$	Idempotents, $e^2 = e$; the vocabulary of Rings
$Ae$, $eA$	Principal left and right ideals generated by an idempotent, the form in which regularity is stated
$eAe$	The corner ring of an idempotent $e$, a regular ring when $A$ is regular
strongly regular	$a = a^2x$ for every $a$; equivalently regular and reduced
Boolean ring	A ring in which $a^2 = a$ for every $a$
$\operatorname{Spec} R$, $\mathrm{P}$, $\mathrm{M}$	The prime spectrum of a commutative ring, a prime ideal, a maximal ideal
$R_{\mathrm{P}}$, $R_{\mathrm{M}}$	Localisations of a commutative ring at a prime, at a maximal ideal
$\mathbb{Z}$, $k[x]$, $\mathbb{Z}/4\mathbb{Z}$	The standard non-examples
$M_2(F)$, $\prod_n M_2(F)$	A regular ring that is not reduced, and the product of countably many copies