WHITEHEAD GROUPS OF EXCHANGE RINGS WITH PRIMITIVE FACTORS ARTINIAN

We show that if R is an exchange ring with primitive factors artinian then K1(R) U(R)/V(R), where U(R) is the group of units of R and V(R) is the subgroup generated by {(1+ab)(1+ba)−1 | a,b ∈ R with 1+ab ∈ U(R)}. As a corollary, K1(R) is the abelianized group of units of R if 1/2∈ R. 2000 Mathematics Subject Classification. 16E50, 19B10. Very recently, Ara et al. [2] showed that the natural homomorphismGL1(R)→K1(R) is surjective provided that R is a separative exchange ring. A natural problem is the description of the kernel of the epimorphism GL1(R) → K1(R). In [9, Theorems 1.2 and 1.6], Menal and Moncasi showed that if R satisfies unit 1-stable range or is unitregular, then K1(R) U(R)/V(R). Here U(R) is the group of units of R while V(R) is a subgroup described later. In [7], Goodearl and Menal remarked that if for each x,y ∈ R there exists a unit u ∈ R such that x−u and y −u−1 are both units, then K1(R) U(R)ab. In this paper, we investigate the above kernel for exchange rings with primitive factors artinian. Recall that R is called an exchange ring if for every right R-module A and two decompositions A=M⊕N =⊕i∈IAi, where MR R and the index set I is finite, there exist submodules Ai ⊆ Ai such that A = M ⊕ (⊕i∈IAi). It is well known that regular rings, π -regular rings, semiperfect rings, left or right continuous rings, clean rings and unit C∗-algebras of real rank zero [1] are all exchange rings. Many authors have studied exchange rings with primitive factors artinian. Fisher and Snider proved that every regular ring with primitive factors artinian is unitregular (see [6, Theorem 6.10]). Moreover, Menal [8, Thereom B] proved that every π -regular ring with primitive factors artinian has stable range one. Recently, Yu [13, Thereom 1] extended these results to exchange rings and showed that every exchange ring with primitive factors artinian has stable range one. On the other hand, Pardo [10] investigated the Grothendieck group K0 of exchange rings. In this paper, we show that the Whitehead group K1(R) U(R)/V(R) for an exchange ring R with primitive factors artinian. We refer the reader to [11] for the general theory of Whitehead groups. Throughout this paper, all rings are associative with identity. The set U(R) denotes the set of all units of R, V(R) denotes the subgroup generated by {p(a,b)p(b,a)−1 | p(a,b)∈ U(R), a,b ∈ R}, and J(R) denotes the Jacobson radical of R (see below for the definition of p(a,b) and other continuant polynomials). If G is a group and G′ its 394 H. CHEN AND F. LI commutator subgroup, then Gab denotes G/G′. Let GLn(R) be the group of units of Mn(R), the ring of all n×n matrices over R. We start with the following new element-wise property of exchange rings with primitive factors artinian. Lemma 1. Let R be an exchange ring with primitive factors artinian. Then, for any x,y ∈ R, there exists a unit-regular w ∈ R such that 1+xy−xw ∈U(R). Proof. Assume that there are some x,y ∈ R such that 1+xy −xw ∈ U(R) for any unit-regular w ∈ R. Let Ω be the set of all two-sided ideals A of R such that 1+xy−xw is not a unit modulo A for any unit-regular w+A∈ R/A. Clearly, Ω ≠∅. Given any ascending chain A1 ⊆A2 ⊆ ··· ⊆An ⊆ ··· in Ω, set M =∪1≤i≤∞Ai. Then M is a two-sided ideal of R. Assume that M is not in Ω. Then there exists a unitregular w+M ∈ R/M such that (1+xy−xw)+M is a unit of R/M . Since w+M is unit-regular in R/M , we have e+M = (e+M)2, u+M ∈ U(R/M) such that w+M = (e+M)(u+M). As R/M is also exchange, we may assume that e= e2 ∈ R. Thus we can find positive integers ni (1 ≤ i ≤ 5) such that (1+xy−xw)s−1 ∈ An1 , s(1+xy− xw)−1 ∈ An2 , w−eu ∈ An3 , ut−1 ∈ An4 , and tu−1 ∈ An5 for some s,t ∈ R. Let n=max{n1,n2,n3,n4,n5}. Then 1+xy−xw+An is a unit of R/An for a unit-regular w+An ∈ R/An. This contradicts the choice of An. So M ∈ Ω. That is, Ω is inductive. By using Zorn’s lemma, we have a two-sided ideal Q of R such that it is maximal in Ω. Let S = R/Q. If J(R/Q)≠ 0, then J(R/Q)=K/Q for some K ⊃Q. Clearly, S/J(S) R/K. By the maximality of Q, we can find a unit-regular (v +Q)+ J(S) such that ((1+xy−xv)+Q)+J(S) is a unit of S/J(S). Since idempotents and units of S/J(S) can be lifted modulo J(S), we may assume that v+Q is unit-regular in S. On the other hand, (1+xy −xv)+Q = (m+Q)+ (r +Q) for some m+Q ∈ U(S), r +Q ∈ J(S). Thus, (1+xy−xv)+Q is a unit of S. This gives a contradiction, whence J(R/Q)= 0. By the maximality of Q, one easily checks that R/Q is an indecomposable ring. According to [14, Lemma 3.7], R/Q is a simple artinian ring. Clearly, (1+xy−xy)+ Q= 1+Q is a unit of R/Q. Since R/Q is unit-regular, y+Q is a unit-regular element of R/Q. This contradicts the choice of Q. Therefore the proof is complete. We note that the following conditions for a ring R are equivalent. (1) For any x,y ∈ R, there exists unit-regularw ∈ R such that 1+x(y−w)∈U(R). (2) Given aR+bR = R, then a+bw ∈U(R) for some unit-regular w ∈ R. (3) Given ax + b = 1, then aw + b ∈ U(R) for some unit-regular w ∈ R (cf. [5, Theorem 2.1]). Clearly, the conditions above are stronger than the stable range one condition and can be viewed as a generalization of rings with many idempotents [4]. Call R a ring with many unit-regular elements if the equivalent conditions above hold. In [5, Theorem 3.1], the authors showed that such rings are all GE2. Furthermore, Chen [3] proved that if R has many unit-regular elements then so doesMn(R). Now we include the fact that every exchange ring with primitive factors artinian has many unit-regular elements to make our paper self-contained. Recall that p(a)= a, p(a,b)= 1+ab, and p(a,b,c)= a+c+abc for any a,b,c ∈ R. It is easy to verify that p(a,b,c)=p(a,b)c+p(a), p(a,b,c)p(b,a)=p(a,b)p(c,b,a). WHITEHEAD GROUPS OF EXCHANGE RINGS . . . 395 Lemma 2. Let a,b,c ∈ R with p(a,b,c) ∈ U(R). If a is unit-regular, then p(a,b,c) ≡ p(c,b,a) (modV(R)). Proof. Since a is unit-regular, there is an idempotent e and a unit u such that a= eu. So we have p(a,b,c) = eu+ c + eubc, and then p(a,b,c)u−1 = e+ cu−1 + eubcu−1. Obviously, p(e,−ub(1−e))= p(e,−ub(1−e))(p(−ub(1−e),e))−1 ∈ V(R). Thus we see that p(a,b,c)u−1 ≡ pe,−ub(1−e)p(a,b,c)u−1 modV(R) = 1−eub(1−e)e+cu−1+eubcu−1 = e+cu−1+eubecu−1 modV(R). (1) On the other hand, we can verify that p(c,b,a)u−1=e+cu−1+cbe≡ e+cu−1+cu−1ubep(1−e,−ube) modV(R) =e+cu−1+cu−1eube= 1+cu−1−(1−e)(1+eube) ≡1+(1+eube)cu−1−(1−e) modV(R) ≡p(a,b,c)u−1 modV(R). (2) Consequently, p(a,b,c)≡ p(c,b,a) (modV(R)), as asserted. In [4, Theorem 16], Chen showed that K1(R) U(R)/V(R) provided that R has idempotent 1-stable range. Now we extend this fact to exchange rings with primitive

Very recently, Ara et al. [2] showed that the natural homomorphism GL 1 (R) → K 1 (R) is surjective provided that R is a separative exchange ring.A natural problem is the description of the kernel of the epimorphism GL 1 (R) → K 1 (R).In [9, Theorems 1.2 and 1.6], Menal and Moncasi showed that if R satisfies unit 1-stable range or is unitregular, then K 1 (R) U (R)/V (R).Here U(R) is the group of units of R while V (R) is a subgroup described later.In [7], Goodearl and Menal remarked that if for each x, y ∈ R there exists a unit u ∈ R such that x − u and y − u −1 are both units, then K 1 (R) U(R) ab .In this paper, we investigate the above kernel for exchange rings with primitive factors artinian.
Recall that R is called an exchange ring if for every right R-module A and two decompositions A = M ⊕ N = ⊕ i∈I A i , where M R R and the index set I is finite, there exist submodules A i ⊆ A i such that A = M ⊕ (⊕ i∈I A i ).It is well known that regular rings, π -regular rings, semiperfect rings, left or right continuous rings, clean rings and unit C * -algebras of real rank zero [1] are all exchange rings.
Many authors have studied exchange rings with primitive factors artinian.Fisher and Snider proved that every regular ring with primitive factors artinian is unitregular (see [6,Theorem 6.10]).Moreover, Menal [8,Thereom B] proved that every π -regular ring with primitive factors artinian has stable range one.Recently, Yu [13, Thereom 1] extended these results to exchange rings and showed that every exchange ring with primitive factors artinian has stable range one.On the other hand, Pardo [10] investigated the Grothendieck group K 0 of exchange rings.In this paper, we show that the Whitehead group K 1 (R) U (R)/V (R) for an exchange ring R with primitive factors artinian.We refer the reader to [11] for the general theory of Whitehead groups.
Throughout this paper, all rings are associative with identity.The set U(R) denotes the set of all units of R, V (R) denotes the subgroup generated by {p(a, b)p(b, a) −1 | p(a, b) ∈ U(R), a,b ∈ R}, and J(R) denotes the Jacobson radical of R (see below for the definition of p(a, b) and other continuant polynomials).If G is a group and G its commutator subgroup, then G ab denotes G/G .Let GL n (R) be the group of units of M n (R), the ring of all n × n matrices over R.
We start with the following new element-wise property of exchange rings with primitive factors artinian.Lemma 1.Let R be an exchange ring with primitive factors artinian.Then, for any x, y ∈ R, there exists a unit-regular w ∈ R such that 1 + xy − xw ∈ U(R).
Proof.Assume that there are some x, y ∈ R such that 1 + xy − xw ∈ U(R) for any unit-regular w ∈ R. Let Ω be the set of all two-sided ideals A of R such that 1 + xy − xw is not a unit modulo A for any unit-regular w + A ∈ R/A.Clearly, Ω ≠ ∅.
Given any ascending chain As R/M is also exchange, we may assume that e = e 2 ∈ R. Thus we can find positive integers n i (1 is a unit of S/J(S).Since idempotents and units of S/J(S) can be lifted modulo J(S), we may assume that v +Q is unit-regular in S. On the other hand, (1 Thus, (1+xy −xv)+Q is a unit of S. This gives a contradiction, whence J(R/Q) = 0.
By the maximality of Q, one easily checks that R/Q is an indecomposable ring.According to [14,Lemma 3.7], R/Q is a simple artinian ring.Clearly, (1 We note that the following conditions for a ring R are equivalent. (1) For any x, y ∈ R, there exists unit-regular Clearly, the conditions above are stronger than the stable range one condition and can be viewed as a generalization of rings with many idempotents [4].Call R a ring with many unit-regular elements if the equivalent conditions above hold.In [5, Theorem 3.1], the authors showed that such rings are all GE 2 .Furthermore, Chen [3] proved that if R has many unit-regular elements then so does M n (R).Now we include the fact that every exchange ring with primitive factors artinian has many unit-regular elements to make our paper self-contained.
Recall  mod V (R) ( On the other hand, we can verify that ( Consequently, p(a, b, c) ≡ p(c, b, a) (mod V (R)), as asserted.
In [4, Theorem 16], Chen showed that K 1 (R) U(R)/V (R) provided that R has idempotent 1-stable range.Now we extend this fact to exchange rings with primitive factors artinian.

Theorem 3. Let R be an exchange ring with primitive factors artinian. Then K 1 (R) U (R)/V (R).
Proof.Given a, b, c ∈ R with p(a, b, c) ∈ U(R), we have p(c, b, a) ∈ U(R).From Lemma 1, we can find some unit-regular w ∈ R such that 1 In view of Lemma 2, we know that On the other hand, ( Recall that an exchange ring is said to be of bounded index if there exists some positive integer n such that x n = 0 for all nilpotent x ∈ R. We can derive the following corollary.

Corollary 4. Let R be an exchange ring of bounded index. Then
Proof.It is easy to obtain the first isomorphism by an argument of Menal.In view of [13,Theorem 3], R is an exchange ring with primitive factors artinian.Therefore we complete the proof by Theorem 3.
Write L 1 (R) for the subgroup generated by the elements in U(R) ∪L, where L is the subgroup generated by all 1+er (1−e) with e = e 2 , r ∈ R.
Proof.Since a is unit-regular, there is an idempotent e and a unit u such that a = ue.So we have Lemma 6.Let R be an exchange ring with primitive factors artinian.If R does not have Z/2Z as a homomorphic image, then, for any x, y ∈ R, there exists a w ∈ U(R) such that 1 + x(y − w) ∈ U(R) and y − w is unit-regular.
Proof.Assume that there exist some x, y ∈ R such that 1 + x(y − w) ∈ U(R) or y − w is not unit-regular for any unit w ∈ R. Let Ω be the set of all two-sided ideals A of R such that 1 + x(y − w) is not a unit or y − w is not unit-regular modulo A for any unit w + A ∈ R/A.Obviously, Ω ≠ ∅.
Analogously to the discussion in Lemma 1, we can find a two-sided ideal Q of R such that it is maximal in Ω. Set S = R/Q.If J(S) ≠ 0, then J(S) = K/Q for some K ⊃ Q. Obviously, S/J(S) R/K.By the maximality of Q, we have a unit (v + Q) + J(S) such that ((1 + x(y − v)) + Q) + J(S) is a unit of S/J(S) and ((y − v) + Q) + J(S) is unitregular in S/J(S).Since units of S/J(S) can be lifted modulo J(S) because S is an exchange ring, we may assume that v + Q is unit of S. Similarly to Lemma 1, we see that (1+x(y −v))+Q = (m+Q)+(r +Q) for some m+Q ∈ U(S), r +Q ∈ J(S).Thus, (1+x(y −v))+Q is a unit of S. On the other hand, idempotents of S/J(S) can be lifted modulo J(S) because S is an exchange ring.So we may assume that ((y This contradicts the choice of Q. Therefore J(S) = 0.The maximality of Q implies that S is indecomposable as a ring.By virtue of [14,Lemma 3.7], S is a simple artinian ring.Assume that R/Q M n (D), where n is a positive integer and D is a division ring.Since R does not have Z/2Z as a homomorphic image, we claim that n ≥ 2, or n = 1, and D Z/2Z.Thus R/Q satisfies unit 1-stable range.From ((1 On the other hand, 2 ∈ U(R) implies that R does not have Z/2Z as a homomorphic image.Using Theorem 7, we conclude that K 1 (R) U(R)/L 1 (R) U(R)/U(R) U(R) ab , as desired.
that p(a) = a, p(a, b) = 1+ab, and p(a, b, c) = a+c +abc for any a, b, c ∈ R. It is easy to verify that p(a, b, c) = p(a, b)c +p(a), p(a, b, c)p(b, a) = p(a, b)p(c, b, a).

Corollary 9 .
Let R be an exchange ring of bounded index of nilpotence.If 2 ∈ U(R), then K 1 (R) U(R) ab .Proof.By [12, Proposition 2.1], we see that the primitive factors of R are artinian.Thus the result follows from Corollary 8.