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.

Advertisements
Comments
  1. Leon Lampret says:

    In the last line of the proof of the theorem, we should have “e=e_1+e_3” instead of “e=e_1+e_2”.

  2. seventhson says:

    Thank you Yaghoub:)

  3. seventhson says:

    Why every direct product of division rings is regular ? Help please.

    • Yaghoub says:

      In general, every direct product of regular rings is regular. This is easy to see:
      let R_i, \ i \in I, be a collection of regular rings and R=\prod_{i \in I} R_i. Let r=(r_i) \in R. Now, for every i \in I there exists some x_i \in R_i such that r_i=r_ix_ir_i, because R_i is regular. Let x=(x_i) \in R. Then rxr=(r_i)(x_i)(r_i)=(r_ix_ir_i)=(r_i)=r and so R is regular.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s