Posts Tagged ‘idempotent’

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!).

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.

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

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.

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}.