## 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 $M_n(D) \cong \text{End}_D D^n$ 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 Artin-Wedderburn 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.

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”.

• Yaghoub says:

Of course, thank you!

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.