A Bezout ring of stable range 2 which has square stable range 1
aa r X i v : . [ m a t h . R A ] D ec A Bezout ring of stable range 2 whichhas square stable range 1
Bohdan Zabavsky, Oleh Romaniv
Department of Mechanics and MathematicsIvan Franko National University of Lviv, [email protected], [email protected], 2018
Abstract:
In this paper we introduced the concept of a ring of stable range 2 which hassquare stable range 1. We proved that a Hermitian ring R which has (right) square stablerange 1 is an elementary divisor ring if and only if R is a duo ring of neat range 1. Andwe proved that a commutative Hermitian ring R is a Toeplitz ring if and only if R is a ringof (right) square range 1. We proved that if R be a commutative elementary divisor ringof (right) square stable range 1, then for any matrix A ∈ M ( R ) one can find invertibleToeplitz matrices P and Q such that P AQ = (cid:0) e e (cid:1) , where e i is a divisor of e . Key words and phrases:
Hermitian ring, elementary divisor ring, stable range 1, stablerange 2, square stable range 1, Toeplitz matrix, duo ring, quasi-duo ring.
Mathematics Subject Classification : 06F20,13F99.
The notion of a stable range of a ring was introduced by H. Bass, and becameespecially popular because of its various applications to the problem of can-cellation and substitution of modules. Let us say that a module A satisfiesthe power-cancellation property if for all modules B and C , A ⊕ B ∼ = A ⊕ C implies that B n ∼ = C n for some positive integer n (here B n denotes thedirect sum of n copies of B ). Let us say that a right R -module A hasthe power-substitution property if given any right R -module decomposition M = A ⊕ B = A ⊕ B which each A i ∼ = A , there exist a positive integer n and a submodule C ⊆ M n such that M n = C ⊕ B n = C ⊕ B n .Prof. K. Goodearl pointed out that a commutative rind R has the power-substitution property if and only if R is of (right) power stable range 1, i.e.if aR + bR = R than ( a n + bx ) R = R for some x ∈ R and some integer n ≥ depending on a, b ∈ R [1].ecall that a ring R is said to have 1 in the stable range provided thatwhenever ax + b = 1 in R , there exists y ∈ R such that a + by is a unitin R . The following Warfield’s theorem shows that 1 in the stable range isequivalent to a substitution property. Theorem 1. [1] Let A be a right R -module, and set E = End R ( A ) . Then E has 1 in the stable range if and only if for any right R -module decomposition M = A ⊕ B = A ⊕ B with each A i ∼ = A , there exists a submodule C ⊆ M such that M = C ⊕ B = C ⊕ B . A ring R is said to have 2 in the stable range if for any a , . . . , a r ∈ R where r ≥ such that a R + · · · + a r R = R , there exist elements b , . . . , b r − ∈ R such that ( a + a r b ) R + ( a + a r b ) R + · · · + ( a r − + a r b r − ) R = R .K. Goodearl pointed out to us the following result. Proposition 1. [1] Let R be a commutative ring which has 2 in the stablerange. If R satisfies right power-substitution, then so does M n ( R ) , for all n . Our goal this paper is to study certain algebraic versions of the notionof stable range 1. In this paper we study a Bezout ring which has 2 in thestable range and which is a ring square stable range 1.A ring R is said to have (right) square stable range 1 (written ssr ( R ) = 1 )if aR + bR = R for any a, b ∈ R implies that a + bx is an invertible elementof R for some x ∈ R . Considering the problem of factorizing the matrix ( a b ) into a product of two Toeplitz matrices. D. Khurana, T.Y. Lam and ZhouWang were led to ask go units of the form a + bx given that aR + bR = R .Obviously, a commutative ring which has 1 in the stable range is a ringwhich has (right) square stable range 1, but not vice versa in general. Exam-ples of rings which have (right) square stable range 1 are rings of continuousreal-valued functions on topological spaces and real holomorphy rings in for-mally real fields [2]. Proposition 2. [2] For any ring R with ssr ( R ) = 1 , we have that R is rightquasi-duo (i.e. R is a ring in which every maximal right ideal is an ideal). We say that matrices A and B over a ring R are equivalent if there existinvertible matrices P and Q of appropriate sizes such that B = P AQ . If fora matrix A there exists a diagonal matrix D = diag( ε , ε , . . . , ε r , , . . . , such that A and D are equivalent and Rε i +1 R ⊆ ε i R ∩ Rε i for every i thenwe say that the matrix A has a canonical diagonal reduction. A ring R s called an elementary divisor ring if every matrix over R has a canonicaldiagonal reduction. If every (1 × -matrix ( (2 × -matrix) over a ring R has a canonical diagonal reduction then R is called a right (left) Hermitianring. A ring which is both right and left Hermitian is called an Hermitianring. Obviously, a commutative right (left) Hermitian ring is an Hermitianring. We note that a right Hermitian ring is a ring in which every finitelygenerated right ideal is principal. Theorem 2. [3] Let R be a right quasi-duo elementary divisor ring. Thenfor any a ∈ R there exists an element b ∈ R such that RaR = bR = Rb . Ifin addition all zero-divisors of R lie in the Jacobson radical, then R is a duoring. Recall that a right (left) duo ring is a ring in which every right (left) idealis two-sided. A duo ring is a ring which is both left and right duo ring.We have proved the next result.
Theorem 3.
Let R be an elementary divisor ring which has (right) squarestable range 1 and which all zero-divisors of R lie in Jacobson radical of R ,then R is a duo ring.Proof. By Proposition 2 we have that R is a right quasi-duo ring. By Theo-rem 2 we have that R is a duo ring. Proposition 3.
Let R be a Hermitian duo ring. For every a, b, c ∈ R suchthat aR + bR + cR = R the following conditions are equivalent:1) there exist elements p, q ∈ R such that paR + ( pb + qc ) R = R ;2) there exist elements λ, u, v ∈ R such that b + λc = vu , where uR + aR = R and vR + cR = R .Proof. ⇒
2) Since paR + ( pb + qc ) R = R we have pR + qcR = R and since R is a duo ring we have pR + cR = R . Than Rp + Rc = R , i.e. vp + jc = 1 for some elements v, j ∈ R . Then vpb + jcb = b and b − vpb = jcb = cj ′ b = ct where t = j ′ b and jc = cj ′ . Element j ′ exist, since R is a duo ring. Then v ( pb + qc ) = vpb + vqc = b + ct + vqc = b + ct + ck , where vqc = ck for someelement k ∈ R . That is, we have v ( pb + qc ) − b = cλ for some element λ ∈ R .We have b + cλ = v ( pb + qc ) . Let u = pb + qc . We have b + cλ = vu , where vR + cR = R , since vp + cj ′ = 1 and uR + aR = R , since paR +( pb + qc ) R = R .) ⇒
1) Since vR + cR = R then Rv + Rc = R . Let pv + jc = 1 forsome elements p, j ∈ R . Then pR + cR = R . Since b + λc = vu , we have pb = p ( vu − λc ) = ( pv ) u − pλc = (1 − jc ) u + pλc = u − ju ′ c + pλc = u + qc forsome element q = pλ − ju ′ , where cu = u ′ c for some element u ′ ∈ R . Since u = pb + qc , therefore ( pb + qc ) R + aR = R . Since R is an Hermitian duo ringthen we have pR + qR = dR where p = dp , q = dq and p R + q R = R . Then p R + ( p b + q c ) R = R since pR ⊂ p R and pR + cR = R , p R + q R = R ,i.e. we have p R + ( p b + q c ) R = R . Hence, aR + ( p b + q c ) R we have p aR + ( p b + q c ) R = R . Remark 1.
In Proposition 3 we can choose the elements u and v such that uR + vR = R . Proposition 4.
Let R be an Hermitian duo ring. Then the following condi-tions are equivalent:1) R is an elementary divisor duo ring;2) for every x, y, z, t ∈ R such that xR + yR = R and zR + tR = R thereexists an element λ ∈ R such that x + λy = vu , where vR + zR = R and uR + tR = R .Proof. ⇒
2) Let R be an elementary divisor ring. By [4] for any a , b , c such that aR + bR + cR = R there exist elements p, q ∈ R such that paR + ( pb + qc ) R = R .Since xR + yR = R , zR + tR = R and the fact that R is a Hermitian duoring we have zR + xR + ytR = R . By Proposition 3 we have x + λyt = uv where uR + zR = R , vR + ytR = R . Since x + ( λt ) y = x + µy = uv where µ = λt , we have uR + zR = R , vR + yR = R .2) ⇒
1) Let aR + bR + cR = R and Rb + Rc = Rd and b = b d , c = c d ,where Rb = Rc = R . Since R is a duo ring then b R + c R = R . So now dR = Rd and aR + bR + cR = R , Rb + Rc = Rd we have aR + dR = R , i.e. dd + ax = 1 for some elements d , x ∈ R . Then − dd ∈ aR .Since b R + c R = R , by Conditions 2 of Proposition 3 there exists anelement λ ∈ R such that b + c λ = vu where u R + (1 − dd ) R = R and vR + dd R = R . Since (1 − dd ) ∈ aR and u R + (1 − dd ) R = R . We have uR + aR = R . Let u = u d . Since u R + aR = R and dR + aR = R we have uR + aR = R . Since b + c λ = vu , we have b + cµ + vu , where λd = dµ .ecall that vR + dd R = R then vR + dR = R . Since vR + cR = vR + c dR = vR + c R . So b + c λ = vu and b R + c R = R then vR + c R = R .Therefore, vR + cR = R . This means that the Condition 2 of Proposition 3is true. By Proposition 3 we conclude that for every a, b, c ∈ R with aR + bR + cR = R there exist elements p, q ∈ R such that paR + ( pb + qc ) R = R ,i.e. according to [4], R is an elementary divisor ring. Definition 1.
Let R be a duo ring. We say that an element a ∈ R \{ } isneat if for any elements b, c ∈ R such that bR + cR = R there exist elements r, s ∈ R such that a = rs , where rR + bR = R , sR + cR = R , rR + sR = R . Definition 2.
We say that a duo ring R has neat range 1 if for every a, b ∈ R such that aR + bR = R there exists an element t ∈ R such that a + bt is aneat element.According to Propositions 3, 4 and Remark 1 we have the following result. Theorem 4.
A Hermitian duo ring R is an elementary divisor ring if andonly if R has neat range 1. The term "neat range 1" substantiates the following theorem.
Theorem 5.
Let R be a Hermitian duo ring. If c is a neat element of R then R/cR is a clean ring.Proof.
Let c = rs , where rR + aR = R , sR + (1 − a ) R = R for any element a ∈ R . Let ¯ r = r + cR , ¯ s = s + aR . From the equality rR + sR = R wehave ru + sv = 1 for some elements u, v ∈ R . Hence r u + srv = r and rsu + s v = s we have ¯ r ¯ u = ¯ r , ¯ s ¯ v = ¯ s . Let ¯ s ¯ v = ¯ e . It is obvious that ¯ e = ¯ e and ¯1 − ¯ e = ¯ u ¯ r . Since rR + aR = R , we have rx + ay = 1 for elements x, y ∈ R . Hence rxsv + aysv = sv we have rsx ′ v + aysv = sv where xs = sx ′ for some element x ′ ∈ R . Then ¯ a ¯ y ¯ e = ¯ e , i.e. ¯ e ∈ ¯ aR . Similarly from theequality sR + (1 − a ) R = R , it follows ¯1 − ¯ e ∈ (¯1 − ¯ a ) R . According to [5] R/cR is an exchange ring. Since R is a duo ring, R/cR is a clean ring.Taking into account the Theorem 3 and Theorem 4 we have the followingresult.
Theorem 6.
A Hermitian ring R which has (right) square stable range 1 isan elementary divisor ring if and only if R is a duo ring of neat range 1. et R be a commutative Bezout ring. The matrix A of order 2 over R issaid to be a Toeplitz matrix if it is of the form (cid:18) a bc a (cid:19) where a, b, c ∈ R .Notice that if A is an invertible Toeplitz matrix, then A − is also aninvertible Toeplitz matrix. Definition 3.
A commutative Hermitian ring R is called a Toeplitz ringif for any a, b ∈ R there exist an invertible Toeplitz matrix T such that ( a, b ) T = ( d, for some element d ∈ R . Theorem 7.
A commutative Hermitian ring R is a Toeplitz ring if and onlyif R is a ring of (right) square range 1.Proof. Let R be a commutative Hermitian ring of (right) square stable range1 and aR + bR = R for some elements a, b ∈ R . Then a + bt = u , where u is an invertible element of R .Let S = (cid:18) a − bt a (cid:19) , K = (cid:18) u − u − (cid:19) . Then ( a, b ) S = ( u, , ( u, K = (1 , , i.e. we have ( a, b ) SK = (1 , . Since (cid:18) a − bt a (cid:19) (cid:18) u − u − (cid:19) = (cid:18) au − − bu − − tu − au − (cid:19) = T we have that T = SK is a Toeplitz matrix. So ( a, b ) T = (1 , . If a, b ∈ R and aR + bR = dR then by a = da , b = db and a R + b R = R [4]. Thenthere exists an element t ∈ R such that a + b t = u , where a is an invertibleelement of R .Let (cid:18) a − b t a (cid:19) (cid:18) u − u − (cid:19) . Note that T is an invertible Toeplitz matrix. Then ( a, b ) T = ( d, , i.e. R isa Toeplitz ring.et R be a Toeplitz ring and aR + bR = R . The exists an invertibleToeplitz matrix T such that ( a, b ) T = (1 , . Let S = T − = (cid:18) x ty x (cid:19) , where x, y, t ∈ R . So det T − = z + ty = u is an invertible element of R . Since ( a, b ) = (1 , T − , we have a = x , b = t . By equality x + ty = u we have a + by = u , i.e. R is a ring of (right) square stable range 1. Theorem 8.
Let R be a commutative ring of square stable range 1. Thenfor any row ( a, b ) , where aR + bR = R , there exists an invertible Toeplitzmatrix T = (cid:18) a bx a (cid:19) , where x ∈ R .Proof. By Theorem 7 we have ( a, b ) = (1 , T for some invertible Toeplitzmatrix T . Let T = (cid:18) x ty x (cid:19) . Then a = x , b = t and T = (cid:18) a by a (cid:19) is aninvertible Toeplitz matrix.Recall that GE n ( R ) denotes a group of n × n elementary matrices overring R . The following theorem demonstrated that it is sufficient to consideronly the case of matrices of order 2 in Theorem 7. Theorem 9. [4] Let R be a commutative elementary divisor ring. Then forany n × m matrix A ( n > , m > ) one can find matrices P ∈ GE n ( R ) and Q ∈ GE m ( R ) such that P AQ = e . . . e . . . . . . . . . . . . . . . . . . . . . e s
00 0 . . . A where e i is a divisor of e i +1 , ≤ i ≤ s − , and A is a × k or k × matrixfor some k ∈ N . Theorem 10.
Let R be a commutative elementary divisor ring of (right)square stable range 1. Then for any × matrix A one can find invertibleToeplitz matrices P and Q such that P AQ = (cid:18) e e (cid:19) , here e i is a divisor of e .Proof. Since R is a Toeplitz ring it is enough to consider matrices of the form A = (cid:18) a b c (cid:19) , where aR + bR + cR = R . Since R is an elementary divisor ring by [4] thereexist elements p, q ∈ R such that paR +( pb + qc ) R = R , i.e. par +( pb + qc ) s = 1 for some elements r, s ∈ R . Since pR + qR = R and rR + sR = R , byTheorem 8 we have the invertible Toeplitz matrices P = (cid:18) p q ∗ ∗ (cid:19) , Q = (cid:18) r ∗ s ∗ (cid:19) such that P AQ = (cid:18) xy z (cid:19) = A . Then (cid:18) − y (cid:19) A (cid:18) − x (cid:19) = (cid:18) ac (cid:19) , where S = (cid:18) − y (cid:19) and T = (cid:18) − x (cid:19) are invertible Toeplitz matrices. So SP AQT = (cid:18) ac (cid:19) . Theorem is proved.
Open Question . Is it true that every commutative Bezout domain ofstable range 2 which has (right) square stable range 1 is an elementary divisorring?
References [1] K. R. Goodearl, Power-cancellation of groups and modules,
Pacific J.Math . (1976) 487–411.2] D. Khurana, T. Y. Lam and Zh. Wang, Rings of square stable rangeone, J. Algebra (2011) 122–143.[3] B. V. Zabavsky and M. Ya. Komarnytskii, Distributive elementary di-visor domains,
Ukr. Math. J. (1990) 890–892.[4] B. V. Zabavsky Diagonal reduction of matrices over rings (Mathemat-ical Studies, Monograph Series, volume XVI, VNTL Publishers, Lviv,2012).[5] W. K. Nicholson, Lifting idempotents and exchange rings,
Trans. Amer.Math. Soc.229