The cancellable range of rings

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1 Arch. Math. 85 (2005) X/05/ DOI /s Birkhäuser Verlag, Basel, 2005 Archiv der Mathematik The cancellable range of rings By Hongbo Zhang and Wenting Tong Abstract. We unify the cancellation property of rings with stable range one and the principal ideal domain by introducing a new notion which is called cancellable range. It is proved that if a ring R has cancellable range n for some positive integer n, then for any n-generated module B and any module C, R B = R C implies B = C; ifr is a Noetherian ring and R has cancellable range n for any n 1, then R has the cancellation property. 1. Introduction. All rings in this paper are associative with identity, modules are unital right modules. The cancellation problem has been studied in a wide variety of categories: Does A B = A C imply B = C? The most famous results in this area are: (1) The integer domain Z has the cancellation property (i.e., Z B = Z C implies B = C); (2) (Evans [6]) Let A be an R-module, if E = End(A) has stable range one, then A has the cancellation property, where a ring E is said to have stable range one (denoted by E Sr1) provided that whenever ax + b = 1inE, there exists y E such that a + by is invertible. The stable range one plays a key role in the research of the cancellation problem (see also Ara [1], Camillo and Yu [3], Camps and Dicks [4], Herbera and Shamsuddin [9], Vaserstein [10], Warfield [11]). Many authors have generalized the stable range one condition in different ways, cf. Ara et al. [2]; Chen [5]; Goodearl [7]; Wu and Tong [12]. There are many advantages and applications of these generalizations. The rings with these generalized stable range conditions have partial cancellation properties, that is, generally speaking, for a ring R with these generalized stable ranges, R B = R C does not imply B = C (see [2, Lemma 8.5]; [5, Theorem 9]; [7, Corollary 2.2]; [12, Theorem 4.1(1)]). Meanwhile, it is well known that the integer ring Z does not have stable range one, although Z has the cancellation property. A problem arises naturally: is there a common Mathematics Subject Classification (2000): 16D70, 16D40, 16E50. This research was partially supported by the Doctorate Foundation of China Education Ministry (Grant No ).

2 328 H. Zhang and W. Tong arch. math. property, possessed by both the rings with Sr1 and the integer ring Z, which ensures the cancellation property of such rings? We solve the problem in this paper. The conception which is called cancellable range of rings is introduced in Section 2. It is a common generalization of rings with stable range one and the principal ideal domain (PID). The properties of cancellable range are studied, and it is proved that if a ring R has cancellable range n, then for any n-generated module B, R B = R C implies B = C. If R is a Noetherian ring (e.g. the integer ring Z), and if R has cancellable range n for any n 1, then R has the cancellation property. Given an index set N (or a positive integer n), a module M is called N-generated (resp. n-generated)ifm can be generated by some set {x i x i M} (resp. {x i x i M} i {1,...,n} ). We use C(R) to denote the center of a ring R; U(R) the set of all units of R. A B means that A is a submodule of B. R n represents direct sums of n copies of a ring R. Every element x R will be written as a row; while x R will be written as a column, i I i I and x T is the transpose of x. 2. Main results. Definition 2.1. Let R be a ring. An integer n > 0 is said to be in the cancellable range of R (denoted by R Cr n) provided that whenever a 0 b 0 + (a 1,...,a n ) (b 1,...,b n ) T a 0 b 0 + a 1 b 1 + +a n b n = 1 R, where a i,b i R(i= 0,...,n), there is a non zero-divisor a C(R) such that (i) (a 1,...,a n ) = a(a 1,...,a n ) for some a i R, 1 i n; (ii) a 0 + (a 1,...,a n )(y 1,...,y n ) T = u is left invertible for some (y 1,...,y n ) T R n ; (iii) If a 0 x 0 = ay with x 0,y R, then x 0 = ax 0 for some x 0 R. A ring R is called FG-cancellable provided that R Cr n for any n 1. Proposition 2.2. If R Cr n for some n 2, then R Cr(n 1). Proof. Wecangetthis result by setting the a n = 0 in Definition 2.1. Proposition 2.3. Let R be a ring with R Cr n for some n 1. Then R is directly finite, i.e., any right invertible element of R is left invertible. Proof. Assume that v is right invertible, i.e., vu = 1 R for some u R. Then vu + 0 = 1. By Proposition 2.2 we have R Cr 1, so there is a non zero-divisor a C(R) such that 0 = aa 1 for some a 1 R and v + a 1 y = u 1 is left invertible. So a 1 = 0 and v = u 1 is left invertible. Proposition 2.4. Let R be a commutative ring. Then R Cr n if and only if whenever a 0 b 0 + a 1 b a n b n = 1 R, where a i,b i R(i = 0,...,n), there is a non zero-divisor a R, such that

3 Vol. 85, 2005 The cancellable range of rings 329 (i) a i = a i a, 1 i n; (ii) a 0 + a 1 y 1 + +a n y n = u U(R) for some y i R. Proof. We need only to prove that R satisfies the condition (iii) of Definition 2.1. Suppose a 0 x 0 = ay for some x 0,y R. Since a 0 b 0 + a(a 1 b 1 + +a n b n) = 1 R, x 0 = (a 0 b 0 + a(a 1 b 1 + +a n b n))x 0 (a 0 x 0 ay)b 0 = a[(a 1 b 1 + +a n b n)x 0 + yb 0 ], as asserted. The following corollary shows that Cr1 is a large class of rings. Corollary 2.5. If R is a commutative domain, then R Cr1. Proof. Suppose a 0 b 0 + a 1 b 1 = 1 R. In the case of a 1 = 0. Let a = a 1, a 1 = 1 and y 1 = 1 a 0, then a 0 + a 1 y 1 = 1 U(R). In the case of a 1 = 0, then a 0 b 0 = 1. Let a = 1, a 1 = 0 and y 1 R, then a 0 + a 1 y 1 = a 0 U(R). We see that R Cr1 from Proposition 2.4. Example 2.6. If R is a commutative principal ideal domain, then R is FG-cancellable. In fact, given a 0 b 0 + a 1 b 1 + +a n b n = 1 R for some n 1. Set a be the greatest common divisor of a 1,...,a n and let a i = a i/a. Since R is a PID, a 1 y 1 + +a n y n = 1 for some y i R. Whence a 0 + a 1 y 1 (1 a 0) + +a n y n (1 a 0) = 1 U(R). Example 2.7. R Sr1 implies that R is FG-cancellable. In fact, suppose a 0 b 0 + a 1 b 1 + +a n b n = 1 R for some integer n 1. Since R Sr1, there are y R and u U(R) such that a 0 + (a 1 b 1 + +a n b n )y = u. Set a = 1 and y i = b i y. Then R is FG-cancellable. Proposition 2.8. If R is the direct product of a family {R α } of rings, then R Cr n if and only if so is each R α. Proof. Straightforward. Proposition 2.9. If R Cr n and e = e 2 R, then ere Cr n. Proof. Suppose that a 0 b 0 + a 1 b a n b n = e ere with a i,b i ere(i = 0,...,n). Then (a e)(b e) + a 1 b a n b n = 1inR. Since R Cr n, there exists a non zero-divisor a C(R) such that (i) a i = a i a for some a i R,1 i n; (ii) (a e) + a 1 y 1 + +a n y n = u U(R) for some y i R; (iii) If (a e)x 0 = ay for some x 0,y R, then x 0 = ax 0 for some x 0 R. First note that a i ere, multiplying the equation (i) on the left by (1 e), we obtain 0 = (1 e)a i a,so(1 e)a i = 0, i.e., a i = ea i ; Note that a C(R), multiplying (i) on the right by (1 e) gives a i = a i e,soa i = ea ie ere. Thus we have (1) a i = (eae)a i,(i = 1,...,n),

4 330 H. Zhang and W. Tong arch. math. where eae is a non zero-divisor in C(eRe). Next, by (ii) we have (2) [(a e) + a 1 y 1 + +a n y n]u 1 = 1. Multiplying this equation on the left by (1 e), wehave(1 e)u 1 = 1 e, so (1 e)u 1 e = 0, i. e., u 1 e = eu 1 e. Multiplying (2) on the left and right by e, we obtain (a 0 + a 1 y 1 + +a n y n)u 1 e = e. Then (3) [a 0 + a 1 ey 1e + +a n ey ne]eu 1 e = [a 0 + a 1 y 1 + +a n y n]u 1 e = e. On the other hand, we have So u 1 [(a e) + a 1 y 1 + +a n y n] = 1. (4) eu 1 e[a 0 + a 1 ey 1e + +a n ey ne] = eu 1 [(a e) + a 1 y 1 + +a n y n]e = e. Combining the equations (3) and (4), we have (5) a 0 + a 1 (ey 1e) + +a n (ey ne) = (eu 1 e) 1 U(eRe). Now assume that a 0 x 0 = eaey with x 0,y ere. Then (a 0 +1 e)x 0 = a 0 x 0 = ay R. By the condition (iii), there is a x 0 R such that x 0 = ax 0. Since x 0 ere, wehave (6) x 0 = eae(ex 0 e). Thus we see that ere Cr n from the equations (1), (5) and (6). Theorem Let R be a ring. If R Cr n, then for any R-module C and any n-generated R-module B, R B = R C implies B = C. In particular, if R is FG-cancellable, then R B = R C implies B = C for any finitely generated module B. Proof. LetB be a n generated module. We have an exact sequence ( ) 0 K R n q B 0 with K = ker q. The condition R B = R C implies that there is a split exact sequence 0 C R B π R 0. Let πφ = 1 R for some φ Hom(R, R B). Set π = (a 0,p) and φ = (s 0,ϕ) for some a 0,s 0 R, p Hom(B, R) and ϕ Hom(R, B). Since R is projective, there is a σ Hom(R, R n ) such that ϕ = qσ. Then there are {b i R} {i=1,...,n} such that σ = (b 1,...,b n ) T : R R n with σ(r) = (b 1 r,...,b n r) T for any r R. Set

5 Vol. 85, 2005 The cancellable range of rings 331 g = pq Hom(R n,r) = n R, then g = (a 1,...,a n ) for some a i R. and i=1 a 0 s 0 + gσ = a 0 s 0 + p(qσ) = a 0 s 0 + pϕ = πφ = 1 R. By the hypothesis, there are a non zero-divisor a C(R), u U(R), {a i R} {i=1,...,n} and {y i R} {i=1,...,n} satisfying the conditions of Definition 2.1. Let g = (a 1,...,a n ) and y = (y 1,...,y n ) T. Set t 2 = q(yu 1 g 1 R n) Hom(R n,b). Then for any k K = ker q ker g, t 2 (k) = q( k) = 0. By (*) there is an t 2 Hom(B, B) such that t 2 = t 2 q. Set t = (u 1 p, t 2 ) Hom(B, R B). Now we prove that the following sequence is exact 0 B t R B π R 0. (1) t(b) ker π. For any b B = q(r n ), let b = q(r) for some r R n. Then πt(b) = πt(q(r)) = a 0 u 1 pq(r) + pt 2 q(r) = a 0 u 1 g(r) + pq(yu 1 g 1)(r) = (a 0 u 1 g + gyu 1 g g)(r) = [(a 0 u 1 1)g + g yu 1 (a 1,...,a n )](r) = [(a 0 u 1 1)g + (u a 0 )u 1 g](r) = 0. (2) ker π t(b).if(x 0, q(r)) ker π, then a 0 x 0 + pq(r) = a 0 x 0 + g(r) = 0, that is, a 0 x 0 = ag (r). By condition (iii) of Definition 2.1, x 0 = ax 0 for some x 0 R. Set b = q(yx 0 r) B. Then u 1 p(b) = u 1 g(yx 0 r) = u 1 ag (yx 0 ) u 1 g(r) = u 1 a(u a 0 )x 0 u 1 g(r) = u 1 a(u a 0 )x 0 u 1 ( a 0 x 0 ) = x 0. Whence u 1 g(yx 0 r) = x 0, i.e., u 1 g (yx 0 r) = x 0,so t 2 (b) = t 2 q(yx 0 r) = q(yu 1 g 1)(yx 0 r) = qy[u 1 g (yx 0 r)] q(yx 0 r) = qy(x 0 ) q(yx 0 r) = q(r). Thus we have t(b) = (u 1 p(b), t 2 (b)) = (x 0, q(r)). Therefore ker π t(b). (3) t is monic. For any q(r) ker t B with r R n.nowt(q(r)) = 0 shows that u 1 g(r) = 0 and q(yu 1 g 1)(r) = t 2 q(r) = 0, u 1 g(r) = 0givesg (r) = 0. So q(r) = q(yu 1 g 1)(r) = 0, as required. Therefore B = t(b) = ker π = C. Remark From Theorem 2.10, we see that the cancellable range can be regarded as a criterion to evaluate how much cancellation property a ring has. Corollary Let R be a FG-cancellable ring. If P is a finite generated projective R-module, then for any finite generated module B,P B = P C implies B = C.

6 332 H. Zhang and W. Tong arch. math. Corollary If R is a FG-cancellable ring, then R is stably finite (i.e., every left invertible matrix over R is right invertible). We know that the class of rings over which every stably free module is free is important in the solution of the Serre conjecture and the research of the K 1 -group. We have Corollary If R is an FG-cancellable ring, then every stably free module is free. Recall that a ring R is called an exchange ring if for every right R-module A and two decompositions A = M M = A i, where M R = R and N is an index set, there exist submodules A i A i such that A = M ( 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 are all exchange rings. Corollary Let R be an exchange ring. Then the following conditions are equivalent: (1) R has the cancellation property; (2) R Sr1; (3) R Cr n for some n 1. Proof. (1) (2) is proved in Yu [13, Theorem 10]. (2) (3) follows from Example 2.7. It only remains to prove (3) (2). Since R is an exchange ring, by Yu [13, Theorem 10], R Sr1 if and only if R has the internal cancellation property, i.e., R = M 1 N 1 = M 2 N 2 with M 1 = M2 implies N 1 = N2.IfR = M 1 N 1 = M 2 N 2 with M 1 = M2, then N 1 is 1-generated. Observe that R N 1 = N 1 (M 2 N 2 ) = N 1 M 1 N 2 = R N 2. By Theorem 2.10 we have N 1 = N2, as desired. Proposition Let R be a Noetherian ring. If R is FG-cancellable, then R has the cancellation property. Proof. Given any index set N. Suppose that a 0 b 0 + gσ = 1 R, where g = (a 1,..., a i,...) R, σ = (b 1,...,b i,...) T R and a 0,b 0,a i,b i R(). We prove first that there is a non zero-divisor a C(R) such that (i) g = ag for some g = (a 1,...,a i,...) R; (ii) There exists Y R such that a 0 + g Y = u U(R); (iii) If a 0 x 0 = ay with x 0,y R, then x 0 = ax 0 for some x 0 R. Since σ = (b 1,...,b i,...) R, there are only finite b i = 0. Let {b ij = 0 j {1,...,n 1 }} be the subset of all the nonzero elements of {b i }, where n 1 is a positive integer. Then a 0 b 0 + gσ = 1 R is equivalent to the equation a 0 b 0 + (a 1,...,a n1 ) (b 1,...,b n1 ) T = 1 R.

7 Vol. 85, 2005 The cancellable range of rings 333 Set H = g( R) {gr r = (r 1,...,r i,...) T R} = {a i } R R. Since R is Noetherian, and {a i } is a generating set of H, we see that H can be generated by a finite set {a ij } j {n2,...,n m } {a i }, where m is a positive integer. Set N ={i 1,...,i n1 } {i n2,...,i nm } N. For simplicity, write N ={j 1,...,j n } for some positive integer n. Thus, we see that a 0 b 0 + (a 1,...,a n1 )(b 1,...,b n1 ) T = 1 R is equivalent to a 0 b 0 + (a j1,...,a jn )(b j1,...,b jn ) T = 1 R. Since R is FG-cancellable, there is a non zero-divisor a C(R) such that (i ) (a j1,...,a jn ) = a(a j 1,...,a j n ) for some a j i R, i {1,...,n}; (ii ) There exists (y j1,...,y jn ) T R n such that a 0 + (a j 1,...,a j n )(y j1,...,y jn ) T = u U(R); (iii) If a 0 x 0 = ay with x 0,y R, then x 0 = ax 0 for some x 0 R. Since H = {a i } can be generated by {a j1,...,a jn }, by (i ) there exists a i R such that a i = aa i for any i N. So (i) is proved. Set Y = (0,...,0,y j1, 0,...,0,y j2, 0,...) T R. Then (ii) follows from (ii ). By (i), (ii) and (iii), similar to the proof of Theorem 2.10, we can get that for any N-generated module B and any module C, R B = R C implies B = C. SoR has the cancellation property. Acknowledgements. The authors express their sincere thanks to the referee for his/her careful reading and valuable suggestions to the earlier version of the present paper. References [1] P. Ara, Strongly π-regular rings have stable range one. Proc. Amer. Math. Soc. 124(11), (1996). [2] P. Ara, G. K. Pedersen and F. Perera, An infinite analogue of rings with stable rank one. J. Algebra 230, (2000). [3] V. P. Camillo and H. P. Yu, Stable range one for rings with many idempotents. Trans. Amer. Math. Soc. 347(8), (1995). [4] R. Camps and W. Dicks, On semilocal rings. Israel J. Math. 81, (1993). [5] H. Chen, On generalized stable rings. Comm. Algebra 28(4), (2000). [6] E. G. Jr. Evans, Krull-Schmidt and cancellation over local rings. Pacific J. Math. 46(1), (1973). [7] K. R. Goodearl, Power-cancellation of groups and modules. Pacific J. Math. 94(2), (1976). [8] K. R. Goodearl and P. Menal, Stable range one for rings with many units. J. Pure Appl. Algebra 54, (1988). [9] D. Herbera and A. Shamsuddin, Modules with semilocal endomorphism rings. Proc. Amer. Math. Soc. 123, (1995). [10] L. N. Vaserstein, Bass s first stable range condition. J. Pure Appl. Algebra 34, (1984). [11] R. B. Warfield, JR., Cancellation of modules and stable range of endomorphism rings. Pacific J. Math. 91(2), (1980).

8 334 H. Zhang and W. Tong arch. math. [12] T. Wu and W. Tong, Stable range condition and cancellation of modules. In: Pitman Res. Notes Math. Pitman. London Ser. 346, (1996). [13] H. P. Yu, Stable range one for exchange rings. J. Pure Appl. Algebra 98, (1995). Received: 16 November 2004 Hongbo Zhang Wenting Tong Department of Mathematics Department of Mathematics Nanjing University Nanjing University Nanjing Nanjing P.R. China P.R. China and Department of Information Science Jiangsu Polytechnic University Changzhou , Jiangsu P.R. China