Posts Tagged ‘von Neumann regular’

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.

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

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.