Elementary matrix reduction over Bezout duo rings
aa r X i v : . [ m a t h . R A ] F e b Elementary Matrix Reduction Over B´ezout Duo-domains
Huanyin Chen
Department of Mathematics, Hangzhou Normal UniversityHangzhou 310036, [email protected]
Marjan Sheibani
Faculty of Mathematics, Statistics and Computer ScienceSemnan University, Semnan, [email protected]
Abstract:
A ring R is an elementary divisor ring if every matrix over R admitsa diagonal reduction. If R is an elementary divisor domain, we prove that R is aB´ezout duo-domain if and only if for any a ∈ R , RaR = R = ⇒ ∃ s, t ∈ R suchthat sat = 1. We further explore various stable-like conditions on a B´ezoutduo-domain under which it is an elementary divisor domain. Many knownresults are thereby generalized to much wider class of rings, e.g. [3, Theorem3.4.], [5, Theorem 14], [9, Theorem 3.7], [13, Theorem 4.7.1] and [14, Theorem3]. Keywords:
Elementary divisor ring; B´ezout duo-domain; Quasi-duo ring; Lo-cally stable domain; Gelfand range 1.
MR(2010) Subject Classification : 15A21, 16S99, 16L99, 19B10.
1. Introduction
Throughout this paper, all rings are associative with an identity. A matrix A (not necessarilysquare) over a ring R admits diagonal reduction if there exist invertible matrices P and Q such that P AQ is a diagonal matrix ( d ij ), for which d ii is a full divisor of d ( i +1)( i +1) (i.e., Rd ( i +1)( i +1) R ⊆ d ii R T Rd ii ) for each i . A ring R is called an elementary divisor ringprovided that every matrix over R admits a diagonal reduction. A ring R is B´ezout ring ifevery finitely generated right (left) ideal is principal. Clearly, every elementary divisor ringis a B´ezout ring. An attractive problem is to investigate various conditions under which aB´ezout ring is an elementary divisor ring.Commutative elementary divisor domains have been studies by many authors, e.g. [3],[5],[9-10], and [13-14]. But the structure of such rings in the noncommutative case has not been Huanyin Chen and Marjan Sheibani sufficiently studied (cf. [6] and [15]). A ring R is duo if every right (left) ideal of R is atwo-sided ideal. Obviously, every commutative ring is duo, but the converse is not true. Inthis paper, we are concern on when a B´ezout duo-domain is an elementary divisor domain.Here, a ring R is a domain ring if there is no any nonzero zero divisor.In Section 2, We prove that an elementary divisor domain is a B´ezout duo-domain if andonly if for any a ∈ R , RaR = R = ⇒ ∃ s, t ∈ R such that sat = 1. A ring R has stable range1 if, aR + bR = R with a, b ∈ R = ⇒ ∃ y ∈ R such that a + by ∈ R is invertible. This conditionplays an important rule in algebra K-theory. We refer the reader to [1] for the general theoryof stable range 1. Further, we shall introduce certain stable-like conditions on B´ezout duo-domains so that they are elementary divisor domains. In Section 3, we prove that everylocally stable B´ezout duo-domain and every Zabavsky duo-domainis are elementary divisordomains (see Theorem 3.2 and Theorem 3.9). In Section 4, we prove that every B´ezout duo-domain of Gelfand range 1 is an elementary divisor ring (see Theorem 4.3). Finally, in thelast section, we establish related theorems on B´ezout duo-domains by means of countabilityconditions on their maximal ideals. Many known results are thereby generalized to muchwider class of rings, e.g. [3, Theorem 3.4.], [5, Theorem 14], [9, Theorem 3.7], [13, Theorem4.7.1] and [14, Theorem 3].We shall use J ( R ) and U ( R ) to denote the Jacobson radical of R and the set of all unitsin R , respectively. For any a, b ∈ R , a | b means that a ∈ bR .
2. B´ezout Duo-domains
A ring R is right (left) quasi-duo if every right (left) maximal ideal of R is an ideal. Aring R is quasi-duo if it is both right and left quasi-duo. It is open whether there exists aright quasi-duo ring which is not left quasi-duo [8]. We have Theorem 2.1.
Let R be an elementary divisor ring. Then the following are equivalent: (1) R is quasi-duo. (2) R is left (right) quasi-duo ring. (3) For any a ∈ R , RaR = R = ⇒ ∃ s, t ∈ R such that sat = 1 .Proof. (1) ⇒ (2) This is trivial.(2) ⇒ (3) We may assume that R is right quasi-duo. Suppose that RaR = R with a ∈ R .Write 1 = r as + · · · + r n as n . Then r aR + · · · + r n aR = R . In light of [8, Theorem 3.2], Rr a + · · · + Rr n a = R . This shows that sa = 1 for some s ∈ R , as desired.(3) ⇒ (1) Suppose that aR + bR = R with a, b ∈ R . Let A = (cid:18) a b (cid:19) . Since R is anelementary divisor ring, there exist U, V ∈ GL ( R ) such that U AV = (cid:18) w v (cid:19) , where RvR ⊆ wR T Rw . It follows from aR + bR = R that RwR + RvR = R , and so RwR = R . Since R satisfies Zabavsky condition, we can find some s, t ∈ R such that swt = 1. Then we verify that (cid:18) s − wts wt (cid:19) (cid:18) wv (cid:19) (cid:18) sw − tsw t (cid:19) = (cid:18) ∗ ∗ ∗ (cid:19) , lementary Matrix Reduction Over B´ezout Duo-domains (cid:18) s − wts wt (cid:19) U AV (cid:18) (cid:19) (cid:18) sw − tsw t (cid:19) (cid:18) (cid:19) = (cid:18) ∗∗ ∗ (cid:19) . One easily checks that (cid:18) s − wts wt (cid:19) = B ( − wt ) B ( s − B (1) B ( wt − (cid:18) sw − tsw t (cid:19) = B ( − t ) B ( sw − B (1) B ( t − . Here, B ( ∗ ) = (cid:18) ∗ (cid:19) , B ( ∗ ) = (cid:18) ∗ (cid:19) . Thus, (cid:18) s − wts wt (cid:19) , (cid:18) sw − tsw t (cid:19) ∈ GL ( R ) . Therefore we can find some P = ( p ij ) , Q = ( q ij ) ∈ GL ( R ) such that P AQ = (cid:18) u (cid:19) . This implies that ( p a + p b ) q = 1, and then Ra + Rb = R . It follows by [8, Theorem3.2] that R is left quasi-quo ring. Likewise, R is right quasi-duo ring. ✷ Following Zabavsky [13], we say that a ring R satisfies Lam condition if RaR = R = ⇒ a ∈ R is invertible. Corollary 2.2.
Let R be an elementary divisor ring. Then the following are equivalent: (1) R is quasi-duo. (2) R satisfies Lam condition.Proof. (1) ⇒ (2) This is obvious.(2) If RaR = R , by hypothesis, a ∈ U ( R ), which completes the proof by Theorem 2.1. ✷ Recall that a ring R satisfies Dubrovin condition if, for any a ∈ R there exists b ∈ R such that RaR = bR = Rb . We now extend [13, Theorem 4.7.1] as follows: Corollary 2.3.
Let R be an elementary divisor ring. If RaR = R = ⇒ ∃ s, t ∈ R such that sat = 1 , then R satisfies Dubrovin condition.Proof. Let 0 = a ∈ R . Since R is an elementary divisor ring, there exist U = ( u ij ) , V =( v ij ) ∈ GL ( R ) such that (cid:18) a a (cid:19) U = V (cid:18) b c (cid:19) , where RcR ⊆ bR T Rb.
It follows that au = v c and au = v c . Since U ∈ GL ( R ),we see that Ru + Ru = R . In view of Theorem 2.1, R is quasi-duo. By virtue of [8,Theorem 3.2], we have u R + u R = R . Write u x + u y = 1 for some x, y ∈ R. Huanyin Chen and Marjan Sheibani
Then a = au x + au y = v cx + v cy ∈ RcR ;hence,
RaR ⊆ RcR.
But
RcR ⊆ bR \ Rb ⊆ RbR ⊆ RaR.
Thus,
RaR = RcR.
Clearly,
RcR ⊆ bR ⊆ RbR ⊆ RaR = RcR and
RcR ⊆ Rb ⊆ RbR ⊆ RaR = RcR.
Therefore
RaR = RcR = bR = Rb , as desired. ✷ Corollary 2.4 [13, Theorem 4.7.1].
Let R be an elementary divisor ring. If R satisfiesLam condition, then R satisfies Dubrovin condition.Proof. Since R satisfies Lam condition, we see that RaR = R implies a ∈ R is invertible.In view of Corollary 2.3, R satisfies Dubrovin condition, as asserted. ✷ Theorem 2.5.
Let R be an elementary divisor domain. Then the following are equivalent: (1) R is a B´ezout duo-domain. (2) R satisfies Lam condition. (3) For any a ∈ R , RaR = R = ⇒ ∃ s, t ∈ R such that sat = 1 .Proof. (1) ⇒ (2) ⇒ (3) These are obvious.(3) ⇒ (1) In view of Theorem 2.1, R is quasi-duo. In view of Corollary 2.3, R satisfiesDubrovin condition. Therefore R is a duo ring, in terms of [15, Theorem 1]. ✷ An element e ∈ R is infinite if there exist orthogonal idempotents f, g ∈ R such that e = f + g while eR ∼ = f R and g = 0. A simple domain is said to be purely infinite if everynonzero left ideal of R contains an infinite idempotent, As is well known, a ring R is a purelyinfinite simple ring if and only if it is not a division ring and for any nonzero a ∈ R thereexist s, t ∈ R such that sat = 1 (cf. [1]). Corollary 2.6.
Every purely infinite simple domain is not a B´ezout domain.Proof.
Let R be a purely infinite ring. Suppose that R is a B´ezout domain. In view of [13,Theorem 1.2.6], R is an Hermite ring. Thus, every 1 × × R admit adiagonal reduction. Let A = ( a ij ) ∈ M ( R ). There exists Q ∈ GL ( R ) such that AQ = (cid:18) a b c (cid:19) for some a, b, c ∈ R . If a = b = c = 0, then AQ = 0 is a diagonal matrix. Otherwise, wemay assume that b = 0. Thus, we can find s, t ∈ R such that sbt = 1. Then we check that (cid:18) sb − tsb t (cid:19) (cid:18) a b c (cid:19) (cid:18) s − bts bt (cid:19) = (cid:18) ∗ ∗ ∗ (cid:19) , lementary Matrix Reduction Over B´ezout Duo-domains B ( ∗ ) (cid:18) (cid:19) (cid:18) sb − tsb t (cid:19) (cid:18) a b c (cid:19) (cid:18) s − bts bt (cid:19) B ( ∗ ) = (cid:18) ∗ (cid:19) . As in the proof of Theorem 2.1, we see that (cid:18) s − bts bt (cid:19) , (cid:18) sb − tsb t (cid:19) ∈ GL ( R ) . Therefore, A is equivalent to a diagonal matrix diag (1 , ∗ ). Hence, R is an elementary divisorring. If RaR = R , then a = 0, and so sat = 1 for some s, t ∈ R . In light of Theorem 2.1, R is a quasi-duo ring. Thus, R is Dedekind-finite, i.e., uv = 1 in R = ⇒ vu = 1. Let 0 = x ∈ R .Then there exists c, d ∈ R such that cxd = 1. Hence, xdc = dcx = 1. This implies that x ∈ U ( R ), and thus R is a division ring, a contradiction. Therefore we conclude that R isnot a B´ezout domain. ✷ Lemma 2.7.
Let R be a B´ezout duo-domain. Then the following are equivalent: (1) R is an elementary divisor ring. (2) aR + bR + cR = R with a, b, c ∈ R = ⇒ ∃ p, q ∈ R such that ( pa + qb ) R + qcR = R .Proof. This is clear, by [15, Lemma 2 and Lemma 3]. ✷ Theorem 2.8.
Let R be a B´ezout duo-domain. Then the following are equivalent: (1) R is an elementary divisor ring. (2) Ra + Rb + Rc = R with a, b, c ∈ R = ⇒ ∃ r, s, t ∈ R such that s | rt and ra + sb + tc = 1 .Proof. (1) ⇒ (2) Suppose Ra + Rb + Rc = R with a, b, c ∈ R . Then aR + bR + cR = R , as R is duo. In view of Lemma 2.7, there exist p, q ∈ R such that ( pa + qb ) R + qcR = R . Hence, R ( pa + qb ) + R ( qc ) = R . Write 1 = x ( pa + qb ) + y ( qc ). Then ( xp ) a + ( xq ) b + ( yq ) c = 1. As R is a duo-domain, we see that ( xp )( yq ) ∈ xRq = Rxq , as desired.(2) ⇒ (1) Let a, b, c ∈ R be such that aR + bR + cR = R . Then Ra + Rb + Rc = R . Byhypothesis, there exist r, s, t ∈ R such that tr ∈ Rs and ta + sb + rc = 1. Since R is a B´ e zoutring, we can find a d ∈ R such that Rd = dR = Rs + Rt = sR + tR , as R is a duo-domain.Case I. d = 0. Then c ∈ U ( R ), and so ( a + c − · b ) R + ( c − · c ) R = R .Case II. d = 0. Write s = dq and t = dp for some q, p ∈ R . Then Rdr = Rsr + Rtr ⊆ Rs = sR , and so dRr = Rdr ⊇ sR = dqR . This shows that r ∈ qR = Rq . Therefore1 = ta + sb + rc = d ( pa + qb ) + kqc ∈ R ( pa + qb ) + R ( qc ) = ( pa + qb ) R + ( qc ) R ; hence,( pa + qb ) R + qcR = R . In light of Lemma 2.7, R is an elementary divisor ring. ✷ Example 2.9.
Let R be the ring of all quaternions. Then R is a noncommutative divisionring. Thus, R is an elementary divisor duo-domain, but it is not commutative. Theorem 2.10.
Let R be a B´ezout duo-domain. Then R is an elementary divisor ring ifand only if R/J ( R ) is an elementary divisor ring.Proof. = ⇒ This is obvious. ⇐ = Suppose that aR + bR + cR = R with a, b, c ∈ R . Then (cid:18) a b c (cid:19) ∈ M ( R/J ( R ))admits a diagonal reduction. Thus, we can find some ( p ij ) , ( q ij ) ∈ GL ( R/J ( R )) such that( p ij ) (cid:18) a b c (cid:19) ( q ij ) = diag ( u, v ) , Huanyin Chen and Marjan Sheibani where
RvR ⊆ uR T Ru. As R is a duo-domain, we see that uR + vR = R . Hence, u ∈ R/J ( R ) is right invertible. This implies that u ∈ R is invertible, as R is quasi-duo. Therefore( p a + p b ) q + p cq = u , and so ( p a + p b ) R + p cR = R. In light of Lemma 2.7, R is an elementary divisor domain. ✷ Corollary 2.11.
Let R be a B´ezout duo-domain. Then R is an elementary divisor ring ifand only if R [[ x , · · · , x n ]] is an elementary divisor ring.Proof. ⇐ = This is obvious, as R is a homomorphic image of R [[ x , · · · , x n ]].= ⇒ Let φ : R [[ x , · · · , x n ]] → R is defined by φ ( f ( x , · · · , x n )) = f (0 , · · · , I := Ker ( φ ) ⊆ J ( R [[ x , · · · , x n ]]) and R [[ x , · · · , x n ]] /I ∼ = R . Since R/I/J ( R ) /I ∼ = R/J ( R ), R/I/J ( R ) /I is an elementary divisor ring if and only if so is R/J ( R ). Since R/I ∼ = R , the result follows by Theorem 2.10. ✷
3. Locally Stable domains
We say that an element a of a duo-ring R is stable if, R/aR has stable range 1. A duoring R is locally stable if, aR + bR = R with a, b ∈ R = ⇒ ∃ y ∈ R such that a + by ∈ R isstable. For instance, every duo-ring of stable range 1 is locally stable. The purpose of thissection is to investigate matrix diagonal reduction over locally stable B´ezout duo-domain. Lemma 3.1.
Let R be a B´ezout domain. If pR + qR = R with p, q ∈ R , then there existsa matrix (cid:18) p q ∗ ∗ (cid:19) ∈ GL ( R ) .Proof. As every B´ezout domain is a Hermite domain, then there exists some Q ∈ GL ( R )such that ( p, q ) Q = ( d,
0) for some d ∈ R . We have p, q ∈ dR and also R = pR + qR ⊆ dR this implies that d is right invertible so there exists some x ∈ R such that dx = 1, now xdx = x then ( xd − x = 0. Since R is a domain, then xd − xd = 1 andso d is invertible. Now ( p, q ) Q (cid:18) d − d − (cid:19) = (1 , U − = Q (cid:18) d − d − (cid:19) . Then( p, q ) = (1 , U , this means that ( p, q ) is the first row of U So U = (cid:18) p q ∗ ∗ (cid:19) ∈ GL ( R ) asrequired. ✷ Let R be a B´ezout domain. If Rx + Ry = R with x, y ∈ R , similarly, we can find amatrix (cid:18) x ∗ y ∗ (cid:19) ∈ GL ( R ). Theorem 3.2.
Every locally stable B´ezout duo-domain is an elementary divisor ring.Proof.
Let R be a locally stable B´ezout duo-domain. Suppose that aR + bR + cR = R with a, b, c ∈ R . Then ax + by + cz = 1 for some x, y, z ∈ R . Since R is locally stable, there existssome s ∈ R such that R/ ( b + ( ax + cz ) s ) R has stable range 1. Set w = b + ( ax + cz ) s . Then aR + cR = 1 in R/wR . Thus, we have (cid:18) (cid:19) (cid:18) xz (cid:19) (cid:18) a b c (cid:19) (cid:18) zs (cid:19) = (cid:18) w ca (cid:19) . lementary Matrix Reduction Over B´ezout Duo-domains R is a Hermite ring. Thus, we have a Q ∈ GL ( R ) such that( w, c ) Q = ( v, (cid:18) w ca (cid:19) W = (cid:18) v k l (cid:19) . Clearly, w ∈ vR ; hence, R/vR ∼ = R/wRvR/wR has stable range 1.One easily checks that vR + kR + lR = R , and so kR + lR = R . Thus, we have a t ∈ R such that vR + ( k + lt ) R = R . Since R is duo, we see that Rv + ( k + lt ) R = R . Write pv + p ′ ( k + lt ) = 1. Then (cid:18) v k l (cid:19) (cid:18) t (cid:19) = (cid:18) v k + lt (cid:19) . By virtue of Lemma 3.1, we can find some
P, Q ∈ GL ( R ) such that P (cid:18) a b c (cid:19) Q = (cid:18) d (cid:19) . Write P = (cid:18) p qp ′ q ′ (cid:19) and Q − = (cid:18) m nm ′ n ′ (cid:19) . Then pa + qb = m and qc = n . As mR + nR = R , we see that ( pa + qb ) R + qcR = R . In light of Lemma 2.7, we complete theproof. ✷ As an immediate consequence, we extend [9, Theorem 3.7] from commutative case toduo rings.
Corollary 3.3.
Let R be a B´ezout duo-domain. If R/aR has stable range 1 for all nonzero a ∈ R , then R is an elementary divisor domain.Proof. Given aR + bR = R with a, b ∈ R , then a = 0 or a + b = 0. Hence, a + by = 0 forsome y ∈ R . By hypothesis, R ( a + by ) R has stable ring 1, i.e., R is locally stable. Therefore R is an elementary divisor domain, by Theorem 3.2. ✷ Since every B´ezout duo-domain of stable range 1 satisfies the condition in Corollary 3.3,we derive
Lemma 3.4.
Every B´ezout duo-domain of stable range 1 is an elementary divisor domain.
Theorem 3.5.
Let R be a B´ezout duo-domain. If aR + bR = R = ⇒ ∃ x, y ∈ R such that xR + yR = R, xy ∈ J ( R ) , a | x, b | y , then R is an elementary divisor ring.Proof. Suppose that aR + bR = R with a, b ∈ R . Then there exist x, y ∈ R such that xR + yR = R, xy ∈ J ( R ) , a | x, b | y . Write xr + ys = 1 for some r, s ∈ R . Since R is aduo-domain, we see that xr ∈ Rx , and then ( xr ) ≡ xr ( mod J ( R )).Set g = xr and e = g + gxr (1 − g ). Then e ∈ xR ⊆ aR . One easily checks that1 − e = (cid:0) − gxr (cid:1) (1 − g ) ≡ − g ( mod J ( R )). Thus, we have some d ∈ J ( R ) such that1 − e − d = 1 − g = ys ∈ yR ⊆ bR . Set f = 1 − e − d . Then e + f = 1 − d ∈ U ( R ), and so e (1 − d ) − + f (1 − d ) − = 1. This implies that eR + f R = R . One easily checks that a | e and b | f . Moreover ef = ( e − e ) − ed ∈ J ( R ). Therefore R/J ( R ) is a duo exchange ring. Inview of [1, Corollary 1.3.5], R/J ( R ) has stable range 1, and then so does R . According toLemma 3.4, we complete the proof. ✷ Huanyin Chen and Marjan Sheibani
Recall that a ring R is feckly clean if for any a ∈ R , there exists e ∈ R such that a − e ∈ U ( R ) and eR (1 − e ) ⊆ J ( R ) [2]. We now derive Corollary 3.6.
Every feckly clean B´ezout duo-domain is an elementary divisor ring.Proof.
Let R be a feckly clean B´ezout duo-ring. Suppose that aR + bR = R with a, b ∈ R .Then ax + by = 1 for some x, y ∈ R . Set c = ax . Then there exist f ∈ R and u ∈ U ( R )such that c = f + u and f R (1 − f ) ∈ J ( R ). Let g = (1 − f ) + ( f − f ) u − . Then g − g ≡ f − f ≡ (cid:0) mod J ( R ) (cid:1) . Hence, g − g ∈ J ( R ). As f u (1 − f ) ∈ J ( R ), we see that f u − f uf ∈ J ( R ). On the other hand, ( R (1 − f ) uf R ) = R (1 − f ) u ( f R (1 − f )) uf R ⊆ J ( R ),and then (1 − f ) uf ∈ J ( R ). This implies that uf = f uf . Therefore, f u − uf ∈ J ( R ).Write uf = f u + r for some r ∈ J ( R ). Then there exists some r ′ ∈ J ( R ) such that a − g = f + u − f − ( f − f ) u − = ( u − f u − u + f − f )( − u − ) = ( c − c )( − u − ) + r ′ .Write e = g − r ′ . Then a − e ∈ ( c − c ) R . Write c − e = ( c − c ) s . Then e = c (cid:0) − (1 − c ) cs (cid:1) ∈ cR ⊆ aR , and that 1 − e = (1 − c )(1 + cs ) ∈ (1 − c ) R ⊆ bR . We easily check that eR + (1 − e ) R = R, e (1 − e ) ∈ J ( R ) and a | e, b | − e . According to Theorem 3.5, R is anelementary divisor ring. ✷ Example 3.7.
Let R := Z (2) T Z (3) = { mn | m, n ∈ Z , , ∤ n } . Then R is a feckly cleanB´ezout duo-domain, and then it is an elementary divisor domain.We say that c ∈ R is feckly adequate if for any a ∈ R there exist some r, s ∈ R such that(1) c ≡ rs ( mod J ( R )); (2) rR + aR = R ; (3) s ′ R + aR = R for each non-invertible divisor s ′ of s . The following is an generalization of [3, Lemma 3.1] in the duo case. Lemma 3.8.
Every feckly adequate element in B´ezout duo-domains is stable.Proof.
Let R be a B´ezout duo-domain, and let a ∈ R be feckly adequate. Set S = R/aR . Let b ∈ S . Then there exist r, s ∈ R such that a ≡ rs ( mod J ( R )) , rR + bR = R and s ′ R + bR = R for any noninvertible divisor s ′ of s . Hence, a ≡ rs (cid:0) mod J ( S ) (cid:1) , i.e., rs ∈ J ( S ). Clearly, rS + bS = S . Let t ∈ S be a noninvertible divisor of s . As in the proof of [3, Lemma 3.1],we see that sR + tR = R . Since R is a B´ e zout ring, we have a noninvertible u ∈ R such that sR + tR = uR . We infer that u is a noninvertible divisor of s . Hence, uR + bR = R . Thisproves that uS + bS = S ; otherwise, there exist x, y, z ∈ R such that ux + by = 1 + az . Thisimplies that ux + by = 1 + w ′ z + rsz = 1 + w ′ z + ucrz for c ∈ R and w ′ ∈ J ( R ), because a − rs ∈ J ( R ). Hence, u ( x − crz )(1 + w ′ z ) − + by (1 + w ′ z ) − = 1, a contradiction. Thus tS + bS = S . Therefore, 0 ∈ S is feckly adequate.Let x ∈ S . by the preceding discussion, we have some r, s ∈ R such that rs ∈ J ( S ),where rS + xS = S and s ′ S + xS = S for any noninvertible divisor s ′ of s . Similarly,we have rS + sS = S . Since rS + xS = S , we get rS + sxS = S , as S is duo. Write rc + sxd = 1 in S . Set e = rc . Then e − e = ( rc ) − rc = − ( rc )( sxd ) ∈ rsS ⊆ J ( S ). Let u be an arbitrary invertible element of S . As in the proof of [3, Lemma 3.1], we verify that( u − ex ) S + rsS = S , as sS = Ss . Since rs ∈ J ( S ), we get u − ex ∈ U ( S ). This impliesthat (cid:0) x − (1 − e ) x (cid:1) − u = ex − u ∈ U ( S ). Furthermore, S has stable range 1. We infer that x − (1 − e ) x ∈ J ( S ). Clearly, 1 − e = sxd ∈ xR . Therefore x = x ( sd ) x in S/J ( S ), and so S/J ( S ) is regular.As R is duo, so is S/J ( S ). Hence, S has stable range 1, by [1, Corollary 1.3.15]. Therefore a ∈ R is stable, as required. ✷ A B´ezout domain R is a Zabavsky domain if, aR + bR = R with a, b ∈ R = ⇒ ∃ y ∈ R lementary Matrix Reduction Over B´ezout Duo-domains a + by ∈ R is feckly adequate. For instance, every domain of stable range 1 is aZabavsky domain. We can derive Theorem 3.9.
Every Zabavsky duo-domain is an elementary divisor ring.Proof.
Suppose that aR + bR = R with a, b ∈ R . Then there exists a y ∈ R such that a + by ∈ R is adequate. In view of Lemma 3.8, a + by ∈ R is stable. Therefore R is locallystable. This completes the proof, by Theorem 3.2. ✷ A ring R is called adequate if every nonzero element in R is adequate. We can extend[13, Theorem 1.2.13] as follows: Corollary 3.10.
Every adequate B´ezout duo-domain is an elementary divisor domain.Proof.
Let R be a adequate B´ezout duo-domain. If aR + bR = R with a, b ∈ R , then a or a + b = 0. Hence, a + by ∈ R is adequate, where y = 0 or 1. This implies that R is aZabavsky duo-domain. The result follows by Theorem 3.9. ✷ Theorem 3.11.
Let R be a B´ezout duo-domain. If for any a ∈ R either a or − a isadequate, then R is an elementary divisor domain.Proof. Given aR + bR = R with a, b ∈ R . Write ax + by = 1 for some x, y ∈ R . Then a ( x − y ) + ( a + b ) y = 1. By hypothesis, either a (1 − x ) or 1 − a (1 − x ) is adequate. In viewof Lemma 3.8, a (1 − x ) or 1 − a (1 − x ) is stable. If a (1 − x ) ∈ R is stable, then R/a (1 − x ) R has stable range 1. As R/aR ∼ = R/a (1 − x ) aR/a (1 − x ) R , we easily check that R/aR has stable range 1.If ( a + b ) y ∈ R is stable, similarly, R/ ( a + b ) R has stable range 1. Therefore R/ ( a + by ) hasstable range 1 where y = 0 or 1. Hence, R is locally stable, whence the result by Theorem3.2. ✷ ✷ Corollary 3.12.
Let R be a B´ezout duo-domain. If for any a ∈ R either J ( RaR ) = 0 or J ( R (1 − a ) R ) = 0 , then R is an elementary divisor domain.Proof. Let a ∈ R . If a = 0, then 1 − a ∈ R is adequate. If a = 1, then a ∈ R is adequate.We now assume that a = 0 ,
1. By hypothesis, J ( RaR ) = 0 or J ( R (1 − a ) R ) = 0. In view of [11,Theorem 17], a or 1 − a is sqaure-free. It follows by [11, Proposition 16] that a or 1 − a isadequate in R . Therefore we complete the proof, by Theorem 3.11. ✷ A ring R is called homomorphically semiprimitive if every ring homomorphic image(including R ) of R has zero Jacobson radical. Von Neumann regular rings are clearly ho-momorphically semiprimitive. But the converse is not true in general, for example, let R = W [ F ] be the first Weyl algebra over a field F of characteristic zero. As an immediateconsequence of Corollary 3.12, we have Corollary 3.13.
Let R be a B´ezout duo-domain. If R is homomorphically semiprimitive,then R is an elementary divisor domain.
4. Rings of Gelfand Range 1 Huanyin Chen and Marjan Sheibani
A ring R is a PM ring if every prime ideal of R contains in only one maximal ideal.An element a of a B´ezout duo-domain R is PM if, RaR is a PM ring. We easily extend [4,Theorem 4.1] to duo-rings.
Lemma 4.1.
Let R be a duo-ring. Then the following are equivalent:(1) R is a PM ring. (2) a + b = 1 in R implies that (1 + ar )(1 + bs ) = 0 for some r, s ∈ R . Lemma 4.2.
Let R be a B´ezout duo-domain, and let a ∈ R . Then the following areequivalent: (1) a ∈ R is PM. (2) If bR + cR = R , then there exist r, s ∈ R such that a = rs, rR + bR = sR + cR = R . (3) If aR + bR + cR = R , then there exist r, s ∈ R such that a = rs, rR + bR = sR + cR = R .Proof. (1) ⇒ (3) Suppose aR + bR + cR = R . Then b ( R/aR ) + c ( R/aR ) =
R/aR . Byvirtue of Lemma 4.1, we can find some p, q ∈ R such that (1 + bp )(1 + cq ) = 0 in R/aR . Set k = 1 + bp and l = 1 + cq . Then kR + bR = lR + cR = R and that kl ∈ aR .Write kl = ap for some p ∈ R . Since R is a B´ezout duo-domain, we have an r ∈ R suchthat kR + aR = rR . Thus, there are some s, t ∈ R such that a = rs, k = rt and sR + tR = R .Clearly, rR + bR = R . It follows that rtl = rsp , and so tl = sp . Write lu + cv = 1. Then tlu + ctv = spu + ctv = t . But sR + tR = R , we get sR + cR = R , as required.(3) ⇒ (2) This is obvious.(2) ⇒ (1) Suppose b + c = 1 in R/aR . Then b + c + ax = 1 for some x ∈ R . Byhypothesis, there exist r, s ∈ R such that a = rs, rR + bR = sR + ( c + ax ) R = R . Write rk + bl = sp + ( c + ax ) q = 1. Then (1 − bk )(1 − cq ) = rskp = 0 in R/aR . Therefore
R/aR is a PM ring, by Lemma 4.1. That is, a ∈ R is PM, as asserted. ✷ A B´ezout duo-ring has Gelfand range 1 if, aR + bR = R with a, b ∈ R implies that thereexists a y ∈ R such that a + by ∈ R is PM. We come now to the main result of this section: Theorem 4.3.
Every B´ezout duo-domain of Gelfand range 1 is an elementary divisor ring.Proof.
Let R be a B´ezout duo-domain of Gelfand range 1. Suppose that aR + bR + cR = R with a, b, c ∈ R . Write ax + by + cz = 1 with x, y, z ∈ R . Then we can find some k ∈ R suchthat w := b + ( ax + cz ) k ∈ R is PM. Hence, (cid:18) xk (cid:19) A (cid:18) zk (cid:19) = (cid:18) a w c (cid:19) . Clearly, wR + aR + cR = R . In view of Lemma 4.2, there exist r, s ∈ R such that w = rs, rR + aR = sR + cR = R . Write sp ′ + cl = 1. Then rsp ′ + rcl = r . Set q ′ = rl . Then wp ′ + cq ′ = r . Write p ′ R + q ′ R = dR . Then p ′ = dp, q ′ = dq and pR + qR = R . Hence,( wp + cq ) R + aR = R . Clearly, cR + pR = qR + pR = R , we get cqR + pR = R . This impliesthat ( wp + cq ) R + pR = R . Thus, we have apR +( wp + cq ) R = R . Write aps +( wp + cq ) t = 1.Then p ( as + wt ) + q ( ct ) = 1. Hence, (cid:18) s twp + cq − ap (cid:19) (cid:18) a w c (cid:19) (cid:18) p − ctq as + wt (cid:19) = (cid:18) ∗ (cid:19) , lementary Matrix Reduction Over B´ezout Duo-domains (cid:12)(cid:12)(cid:12)(cid:12) s twp + cq − ap (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) p − ctq as + wt (cid:12)(cid:12)(cid:12)(cid:12) = 1. As in the proof of Theorem 3.2, we have p, q ∈ R such that ( pa + qb ) R + qcR = R . This completes the proof, by Lemma 2.7. ✷ Corollary 4.4 [14, Theorem 3].
Every commutative B´ezout domain of Gelfand range 1is an elementary divisor ring.We note that Theorem 4.3 can not be extended to any rings with zero divisor as thefollowing shows.
Example 4.5.
Let R + be the subset { ( x,
0) : x > } of the plane S + = { ( x, sinπx ) : x ≥ } and let X = S + S R + . If R = C ( X − β ( X )) where β ( X ) the largest compact Hausdorffspace generated by X in the sense that any map from X to a compact Hausdorff spacefactors through β ( X ), and C ( X − β ( X ) is the ring of all continuous functions on X − β ( X )In view of [7], C ( X − β ( X )) is a B´ezout duo-ring. On the other hand, C ( X − β ( x )) is aPM ring (cf. [4]). Thus C ( X − β ( x )) is a B´ezout duo-ring of Gelfand range 1. But it is notHermite ring (cf. [7]), and therefore R is not an elementary divisor ring. Lemma 4.6.
Let I be an ideal of a PM ring R . Then R/I is a PM ring.Proof.
Let P be a prime ideal of R . Then P = Q/I where Q is a prime ideal of R . Since R is a PM ring, there exists a unique maximal M of R such that Q ⊆ M . Hence, P ⊆ M/I where
M/I is a maximal of
R/I . The uniqueness is easily checked, and therefore
R/I isPM. ✷ Theorem 4.7.
Let R be B´ezout duo-domain. Suppose for any a ∈ R , either a or − a isPM. Then R is an elementary divisor ring.Proof. Given aR + bR = R with a, b ∈ R . Write ax + by = 1 for some x, y ∈ R . As in theproof of Theorem 3.11, we see that a (1 − x ) or 1 − a (1 − x ) is PM. If a (1 − x ) ∈ R is PM,then R/a (1 − x ) R is PM. Since R/aR ∼ = R/a (1 − x ) aR/a (1 − x ) R , it follows by Lemma 4.6 that R/aR isPM. If ( a + b ) y ∈ R is PM, R/ ( a + b ) R is PM. Hence, R/ ( a + by ) is PM, and therefore R isan elementary divisor ring, by Theorem 4.3. ✷ Corollary 4.8.
Every B´ezout PM duo-domain is an elementary divisor ring.
5. Countability Conditions
Let R be a ring and a ∈ R . We use mspec ( a ) to denote the set of all maximal ideals whichcontain a . We now determine elementary divisor domains by means of the countability ofthe sets of related elements. Lemma 5.1.
Let R be a B´ezout domain, and let aR + bR + cR = R with a, b, c ∈ R . If mspec ( a ) is at most countable, then there exist p, q ∈ R such that ( pa + qb ) R + qcR = R .Proof. In view of [13, Theorem 1.2.6], R is an Hermite ring.2 Huanyin Chen and Marjan Sheibani
Case I. mspec ( a ) is empty. Then a ∈ U ( R ). Hence, (cid:18) a b c (cid:19) (cid:18) − a − b (cid:19) = (cid:18) a c (cid:19) . Case II. mspec ( a ) = ∅ . Set mspec ( a ) = { M , M , M , · · · } . If b ∈ M , then c M ;hence, b + c M . By an elementary transformation of column, b can be replaced by theelement b + c . Thus, we may assume that b M . As R is a Hermite ring, there exists P ∈ GL ( R ) such that P A = (cid:18) a b c (cid:19) . This implies that aR + bR = a R . If a ∈ U ( R ),then P A (cid:18) − a − b (cid:19) = diag ( a , c )is a diagonal matrix.Assume that a U ( R ). Then mspec ( a ) = ∅ . As mspec ( a ) ⊆ mspec ( a ), we mayassume that a ∈ M . If b ∈ M , then c M since a R + b R + c R = R . Hence, b + c M . By an elementary transformation of row, b can be replaced by the element b + c . Thus, we may assume that b M . Since R is a Hermite ring, there exists Q ∈ GL ( R ) such that P AQ = (cid:18) a b c (cid:19) . Clearly, a R + b R = a R . Moreover, wesee that a ∈ M , a ∈ M \ M , a ∈ M \ (cid:0) M S M (cid:1) . By iteration of this process, there is acollection of matrices of the form P k AQ k = (cid:18) a i ∗ ∗ (cid:19) . If there exists some a i ∈ U ( R ), then A admits a diagonal reduction. Otherwise, we get aninfinite chain of ideals aR $ a R $ a R $ · · · . Further, each a n M n ( n ∈ N . Let I = ∞ S i =1 a i R . Then I = R ; hence, there exists a maximalideal J of R such that I ⊆ J ( R . This implies that a ∈ J , and so J ∈ mspec ( a ). Thus, J = M k for some k ∈ N . As a k M k , we deduce that a k J , a contradiction. Therefore A admits a diagonal reduction. As in the proof of Theorem 3.2, there exist p, q ∈ R such that( pa + qb ) R + qcR = R , as required. ✷ We say that an element a of a ring R is a non-P element if, RaR do not satisfy the property P , where P is some ring theoretical property. We now derive Theorem 5.2.
Let R be a B´ezout duo-domain. Then R is an elementary divisor ring if (1) the set of nonstable elements in R is at most countable, or (2) the set of non-PM elements in R is at most countable.Proof. (1) If R has stable range 1, then it is an elementary divisor ring, by Lemma 3.4. Wenow assume that R has no stable range 1. Suppose that aR + bR + cR = R with a, b, c ∈ R .If a ∈ R is stable, as in the proof of Theorem 3.2, we can find some p, q ∈ R such that( pa + qb ) R + qcR = R . If a ∈ R is nonstable, then a = 0; otherwise, R has stable range1. Since R is a domain, we see that a m = a n for any distinct m, n ∈ N . If mspec ( a ) isuncountable, then { a, a , · · · , a k , · · · } is uncountable as a subset of mspec ( a ). This gives acontradiction. This implies that mspec ( a ) is at most countable. By virtue of Lemma 5.1, lementary Matrix Reduction Over B´ezout Duo-domains p, q ∈ R such that ( pa + qb ) R + qcR = R . Therefore proving (1), in terms ofLemma 2.7.(2) Suppose that aR + bR + cR = R with a, b, c ∈ R . If a ∈ R is stable, similarly toTheorem 3.2, we have some p, q ∈ R such that ( pa + qb ) R + qcR = R . If a ∈ R is a non-PMelement, by hypothesis, mspec ( a ) is at most countable. In light of Lemma 5.1, we can find p, q ∈ R such that ( pa + qb ) R + qcR = R . Thus, we obtain the result by Lemma 2.7. ✷ Recall that an ideal of a duo ring R is a maximally nonprincipal ideal if it is maximal inthe set of nonprincipal ideals of R with respect to the ordering by set inclusion. Lemma 5.3.
Let R be a B´ezout domain, and let aR + bR + cR = R with a, b, c ∈ R . If a does not belong to any maximally nonprincipal ideal, then there exist p, q ∈ R such that ( pa + qb ) R + qcR = R .Proof. Set A = (cid:18) a b c (cid:19) ∈ M ( R ). Since a does not belong to any maximally nonprincipalideal, it follows by [7, Theorem 1] that A admits a diagonal reduction. Therefore we completethe proof, as in the proof of Theorem 3.2. ✷ Theorem 5.4.
Let R be a B´ezout duo-domain. Then R is an elementary divisor ring if (1) every nonstable element in R is contained in a maximal nonprincipal ideal, or (2) every non-PM element in R is contained in a maximal nonprincipal ideal.Proof. (1) Suppose that aR + bR + cR = R with a, b, c ∈ R . If a ∈ R is stable, similarly toTheorem 3.2, there are some p, q ∈ R such that ( pa + qb ) R + qcR = R . If a ∈ R is nonstable,by hypothesis, a ∈ R is contained in a maximal nonprincipal ideal. In light of Lemma 5.3,( pa + qb ) R + qcR = R for some p, q ∈ R . It follows by Lemma 2.7 that R is an elementarydivisor ring.(2) Analogously to Theorem 4.3 and the preceding discussion, we complete the proof, byLemma 5.3 and Lemma 2.7. ✷ Corollary 5.5.
Let R be a B´ezout duo-domain. Then R is an elementary divisor ring if (1) every nonlocal element in R is contained in a maximal nonprincipal ideal, or (2) every nonregular element in R is contained in a maximal nonprincipal ideal.Proof. This is obvious by Theorem 5.4, as every local duo-ring and every regular duo-ringare both PM rings. ✷ References [1] H. Chen,
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