Archive for the ‘von Neumann Regular rings’ Category

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.

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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

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.