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 any standard ring theory textbooks. Note that n, in Jacobson’s theorem, doesn’t have to be fixed, i.e. it could depend on x. See this post for the proof of the theorem when n is fixed. Here we only discuss a very special case of the theorem, i.e. when n=3.

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.  First note that 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


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


we have 3x=-3x^2 and so 3x is central. Thus x = 3x-2x is central and so R is commutative.  \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}.