## von Neumann regular rings (3)

Posted: October 31, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
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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 $r \in R.$

## von Neumann regular rings (2)

Posted: October 22, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
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Theorem 1. (von Neumann, 1936) The center of a regular ring $R$ is regular.

Proof.  Let $a$ be a central element of $R$ and let $x \in R$ be such that $a=axa=a^2x.$ So $a^2x$ is central. Let $z \in R.$ Then $a^2xz=za^2x$ and hence $xa^2z=a^2zx,$ i.e. $a^2z$ commutes with $x$ and so it commutes with $x^3.$ Therefore $a^2x^3z=za^2x^3,$ i.e. $y=a^2x^3$ is central. But, since $a^2x^2=ax,$ we have $y=ax^2$ and clearly $aya=a^2x^2a=axa=a.$ $\Box$

Remark. In a commutative regular ring $R$ every prime ideal $P$ is maximal. To see this, let $a \in R \setminus P$ and $r \in R$ be such that $a=ara.$ Then $a(1-ra)=0 \in P$ and so $1-ra \in P.$ Therefore $P+Ra=R$ and hence $P$ is maximal.

Notation. Let $M$ be a maximal ideal of $Z(R),$ the center of $R.$ The localization of $R$ at $M$ is denoted by $R_M.$

Lemma. If $R$ is a commutative regular ring and $M$ is a maximal ideal of $R,$ then $R_M$ is a field.

Proof. The unique maximal ideal of $R_M$ is $M_M.$ So, to prove that $R_M$ is a field, we only need to show that $M_M=\{0\},$ i.e. for every $a \in M,$ there exists some $s \notin M$ such that $sa=0.$ This is easy to see: we have $a=xa^2,$ for some $x \in R,$ because $R$ is a commutative regular ring, and thus $(1-xa)a=0$ and clearly $s=1-xa \notin M$ because $a \in M. \ \Box$

The converse of the lemma is also true and we will prove it in a more general setting in part (3). This result that a commutative ring $R$ is regular if and only if $R_M$ is a field for any maximal ideal $M$ of $R$ is due to Kaplansky.

Theorem 2. (Armendariz, 1974) Let $R$ be a ring with the center $Z(R).$ If $Z(R)$ is regular, then $R/MR \cong R_M$ for any maximal ideal $M$ of $Z(R).$

Proof. Let $f: R \longrightarrow R_M$ be the natural homomorphism defined by $f(x)=1^{-1}x,$ for all $x \in R.$ Let $S=Z(R) \setminus M.$

1) $f$ is surjective. To see this, let $s^{-1}x \in R_M.$ Since $Z(R)$ is regular, there exists some $c \in Z(R)$ such that $s=cs^2.$ Hence $s(1-cs)=0$ and therefore $f(cx)=1^{-1}cx = s^{-1}x.$

2) $\ker f = MR.$ To see this, let $f(x)=1^{-1}x=0.$ That means $sx = 0$ for some $s \in S.$ Since $Z(R)$ is regular, there exists some $c \in Z(R)$ such that $s(1-cs)=0 \in M.$ Thus $1-cs \in M$ because $s \notin M.$ Therefore $x=(1-cs)x \in MR.$ This proves $\ker f \subseteq MR.$ For the other side of the inclusion, we first apply the above lemma to $Z(R)$ to get $1^{-1}a=0$ for all $a \in M.$ Thus for every $a \in M$ and $x \in R$ we have $f(ax)=1^{-1}ax = 1^{-1}a 1^{-1}x = 0.$ So $ax \in \ker f$ and therefore $MR \subseteq \ker f. \ \Box$

## von Neumann Regular rings (1)

Posted: October 22, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
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Definition. A ring $R$ is called von Nemann regular, or just regular, if for every $a \in R$ there exists $x \in R$ such that $a=axa.$

Remark 1. Regular rings are semiprimitive. To see this, let $R$ be a regular ring. Let $a \in J(R),$ the Jacobson radical of $R,$ and choose $x \in R$ such that $a=axa.$ Then $a(1-xa)=0$ and, since $1-xa$ is invertible because $a$ is in the Jacobson radical of $R,$ we get $a=0.$

Examples 1. Every division ring is obviously regular because if $a = 0,$ then $a=axa$ for all $x$ and if $a \neq 0,$ then $a=axa$ for $x = a^{-1}.$

Example 2. Every direct product of regular rings is clearly a regular ring.

Example 3. If $V$ is a vector space over a division ring $D,$ then ${\rm End}_D V$ is regular.

Proof. Let $R={\rm End}_D V$ and $f \in R.$ There exist vector subspaces $V_1, V_2$ of $V$ such that $\ker f \oplus V_1 = {\rm im}(f) \oplus V_2 = V.$  So if $u \in V,$ then $u=u_1+u_2$ for some unique elements $u_1 \in {\rm im}(f)$ and $u_2 \in V_2.$ We also have $u_1 = v_1 + v$ for some unique elements $v_1 \in \ker f$ and $v \in V_1.$ Now define $g: V \longrightarrow V$ by $g(u)=v.$ It is obvious that $g$ is well-defined and easy to see that $g \in R$ and $fgf=f. \ \Box$

Example 4. Every semisimple ring is regular.

Proof. For a division ring $D$ the ring $M_n(D) \cong End_D D^n$ is regular by Example 3. Now apply Example 2 and the Wedderburn-Artin theorem.

Theorem. A ring $R$ is regular if and only if every finitely generated left ideal of $R$ is generated by an idempotent.

Proof. Suppose first that every finitely generated left ideal of $R$ can be generated by an idempotent. Let $x \in R.$ Then $I=Rx = Re$ for some idempotent $e.$ That is $x = re$ and $e=sx$ for some $r,s \in R.$ But then $xsx=xe=re^2=re=x.$ Conversely, suppose that $R$ is regular. We first show that every cyclic left ideal $I=Rx$ can be generated by an idempotent. This is quite easy to see: let $y \in R$ be such that $xyx=x$ and let $yx=e.$ Clearly $e$ is an idempotent and $xe=x.$ Thus $x \in Re$ and so $I \subseteq Re.$ Also $e=yx \in I$ and hence $Re \subseteq I.$ So $I=Re$ and we’re done for this part. To complete the proof of the theorem we only need to show that if $J=Rx_1 + Rx_2,$ then there exists some idempotent $e \in R$ such that $J=Re.$ To see this, choose an idempotent $e_1$ such that $Rx_1=Re_1.$ Thus $J=Re_1 + Rx_2(1-e_1).$  Now choose an idempotent $e_2$ such that $Rx_2(1-e_1)=Re_2$ and put $e_3=(1-e_1)e_2.$ See that $e_3$ is an idempotent, $e_1e_3=e_3e_1=0$ and $Re_2=Re_3.$ Thus $J=Re_1 + Re_3.$ Let $e=e_1+e_3.$ Then $e$ is an idempotent and $J=Re. \Box$

Corollary. If the number of idempotents of a regular ring $R$ is finite, then $R$ is semisimple.

Proof. By the theorem, $R$ has only a finite number of left principal ideals. Since every left ideal is a sum of left principal ideals, it follows that $R$ has only a finite number of left ideals and hence it is left Artinian. Thus $R$ is semisimple because $R$ is semiprimitive by Remark 1. $\Box$

Remark 2. The theorem is also true for finitely generated right ideals. The proof is similar.

Remark 3. Since, by the Wedderburn-Artin theorem, a commutative ring is semisimple if and only if it is a finite direct product of fields, it follows from the Corollary that if the number of idempotents of a commutative von Neumann regular ring $R$ is finite, then $R$ is a finite direct product of fields.