## Rings satisfying x^3 = x are commutative

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Throughout $R$ is a ring.

Theorem (Jacobson). If for every $x \in R$ there exists some $n > 1$ such that $x^n=x,$ then $R$ is commutative.

The proof of Jacobson’s theorem can be found in standard ring theory textbooks. Here we only discuss a very special case of the theorem, i.e. when $x^3=x$ for all $x \in R.$

Definitions. An element $x \in R$ is called idempotent if $x^2=x.$ The center of $R$ is

$Z(R)=\{x \in R: \ xy=yx \ \text{for all} \ y \in R \}.$

It is easy to see that $Z(R)$ is a subring of $R.$ An element $x \in R$ is called central if $x \in Z(R).$ Obviously $R$ is commutative iff $Z(R)=R,$ i.e. every element of $R$ is central.

Problem. Prove that if $x^3=x$ for all $x \in R,$ then $R$ is commutative.

Solution.  Clearly $R$ is reduced, i.e. $R$ has no nonzero nilpotent element.  For every $x \in R$ we have $(x^2)^2=x^4 = x^2$ and so $x^2$ is idempotent for all $x \in R.$ Hence, by Remark 3 in this post, $x^2$ is central for all $x \in R.$ Now, since

$(x^2+x)^2=x^4+2x^3+x^2=2x^2+2x$

we have $2x=(x^2+x)^2-2x^2$ and thus $2x$ is central. Also, since

$x^2+x=(x^2+x)^3=x^6+3x^5+3x^4+x^3=4x^2+4x,$

we have $3x=-3x^2$ and hence $3x$ is central. Therefore $x = 3x-2x$ is central. $\Box$

A similar argument shows that if $x^4=x$ for all $x \in R,$ then $R$ is commutative (see here!).

## 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.

## Clean rings

Posted: September 15, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Definition. A ring $R$ with 1 is called clean if for every $a \in R$ there exist a unit $b \in R$ and an idempotent $c \in R$ such that $a = b+c.$

Examples. Every commutative local ring $R$ is clean. The reason is that for every $a \in R,$ either $a$ or $a-1$ is a unit. It is also obvious that a direct sum or product of clean rings is clean.

Problem. Let $R$ be a commutative clean ring. Prove that if the number of idempotent elements of $R$ is finite, then $R$ is a finite direct product of local rings.

Solution. The proof is by induction on $n,$ the number of idempotent elements of $R.$ Suppose that $n = 2,$ i.e. the only idempotent elements of $R$ are $0$ and $1$. Then for any $x \in R,$ either $x$ or $1 - x$ is a unit because $R$ is clean. Now let $M$ be a maximal ideal of $R$ and $x \notin M$. Then $Rx + M = R$ and hence $rx + y = 1 - x,$ for some $y \in M, \ r \in R$. Thus $1-y=(r+1)x$ and so, since $y \in M$ and hence $y$ cannot be a unit, $1-y$ is a unit. Therefore $x$ is a unit and so $R$ is local. Now suppose $n > 2.$ Choose an idempotent $e \in R \setminus \{0,1\}.$ Clearly $R = Re \oplus R(1-e)$ and the number of idempotent elements in both $Re$ and $R(1-e)$ is at most $n-1$ because $1 \notin Re$ and $1 \notin R(1-e)$. So to apply induction hypotheis on $Re$ and $R(1-e)$ and finish the proof, we only need to prove that both $Re$ and $R(1-e)$ are clean. First note that the identity element of $Re$ is $e$. Now let $x = re \in Re.$ Since $R$ is clean, we have $r=u+f$ for some unit $u \in R$ and an idempotent $f \in R.$ Then  $x=re=ue + fe.$ Clearly  $ue$ is a unit in $Re$ and $fe$ is an idempotent in $Re.$ So $Re$ is clean. A similar argument shows that $R(1-e)$ is clean too. $\Box$

## Left ideals generated by idempotents

Posted: June 10, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem 1. (Richard Brauer) Let $I$ be a minimal left ideal of a ring $R$. Then either $I^2= \{0\}$ or $I=Re,$ for some non-zero idempotent $e \in I.$

Solution. Suppose that $I^2 \neq \{0\}.$ Then there exists some $x \in I$ such that $Ix \neq \{0\}.$ Thus $Ix=I,$ because $Ix \subseteq I$ is a non-zero left ideal of $R$ and $I$ is a minimal left ideal of $R.$ Hence there exists $0 \neq e \in I$ such that $ex=x$ and so

$(e^2-e)x=0. \ \ \ \ (1)$

On the other hand, $J= \{r \in I: \ rx = 0 \}$ is obviously a left ideal of $R$ which is contained in $I.$ Since $Ix \neq \{0\},$ we have $J \neq I$ and thus, by the minimality of $I,$ we must have $J=\{0\}$. Therefore $e^2-e=0$, by $(1).$ So $e \in I$ is a non-zero idempotent. Now $Re$ is a left ideal of $R$ which is contained in $I.$ Also $0 \neq e=e^2 \in Re$ and so $Re \neq \{0\}.$ Therefore $Re=I,$ by the minimality of $I. \ \Box$

Note that we did not need $R$ to have $1.$ Also, a similar result holds for minimal right ideals of $R$, i.e. if $I$ is a minimal right ideal of $R,$ then either $I^2 = \{0\}$ or $I=eR,$ for some non-zero idempotent $e \in I.$  If $R$ is a semisimple ring (with $1$), for example full matrix rings over division algebras, then every left (resp. right) ideal of $R$ is generated by some idempotent, as the next problem shows.

Problem 2. Let $R$ be a semisimple ring and $I$ any left (resp. right) ideal of $R$. Then there exists some idempotent $e \in I$ such that $I=Re$ (resp. $I=eR$).

Solution. We’ll only prove the claim for left ideals of $R.$ The proof for right ideals is similar. Since $R$ is semisimple, there exists a left ideal $J$ of $R$ such that $R=I \oplus J.$ So $1=e+f,$ for some $e \in I$ and $f \in J.$ Hence $e=e^2+ef$ and so $e^2=e$ and $ef=0$, because $e, e^2 \in I$ and $ef \in J$ and the sum is direct. So $e$ is an idempotent. It is clear that $Re \subseteq I$. Now if $x \in I$, then $x=xe+xf$ and therefore $x=xe.$ Thus $I \subseteq Re$ and we’re done.

## “Almost Boolean” rings are commutative

Posted: March 12, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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It is easy to prove that if every element of a ring is idempotent, then the ring is commutative. This fact can be generalized as follows.

Problem. 1) Let $R$ be a ring with identity and suppose that every element of $R$ is a product of idempotent  elements. Prove that $R$ is commutative.
2)  Give an example of a noncommutative ring with identity $R$ such that every element of $R$ is a product of some elements of the set $\{r \in R: \ r^n=r, \ \text{for some} \ n \geq 2 \}.$

Solution. 1) Obviously we only need to prove that every idempotent is central. Suppose first that $ab = 1,$ for some $a,b \in R.$ We claim that $a = b = 1.$ So suppose the claim is false. Then $a = e_1e_2 \cdots e_k,$ where $e_j$ are idempotents and $e_1 \neq 1.$ Let $e = e_2 \cdots e_kb.$ Then $e_1e = 1$ and hence $1 - e_1 = (1 - e_1)e_1e = 0.$ Thus $e_1 = 1.$ Contradiction! Now suppose that $x^2 = 0,$ for some $x \in R.$ Then $(1 - x)(1 + x) = 1$ and therefore $x = 0$, by what we just proved. Finally, since $(ey-eye)^2=(ye-eye)^2=0$ for any idempotent $e \in R$ and any $y \in R,$ we have $ey = ye$ and so $e$ is central.
2) One example is the ring of $2 \times 2$ upper triangular matrices with entries from $\mathbb{Z}/2\mathbb{Z}.$