Clear elements and clear rings
aa r X i v : . [ m a t h . A C ] M a y Clear elements and clear rings
Bohdan Zabavsky, Olha Domsha, Oleh Romaniv
Department of Mechanics and Mathematics, Ivan Franko National University;Lviv Regional Institute for Public Administration of the NationalAcademy for Public Administration under the President of Ukraine;Department of Mechanics and Mathematics, Ivan Franko National University;[email protected], [email protected], [email protected], 2020
Abstract:
An element in a ring R is called clear if it is the sum of unit-regular elementand unit. An associative ring is clear if every its element is clear. In this paper wedefined clear rings and extended many results to wider class. Finally, we proved that acommutative B´ezout domain is an elementary divisor ring if and only if every full matrixorder 2 over it is nontrivial clear. Key words and phrases:
B´ezout domain; clean element; unit-regular element; fullmatrix; elementary divisor ring; clear element; clear ring
Mathematics Subject Classification : 06F20, 13F99.
The work in this paper is prompted by the looking at the two sets U ( R ) and U reg ( R ) in ring R , which denote, respectively the unit group and the set ofunit-regular elements in R . Certainly the units and unit-regular elements arekey elements determining the structure of the ring.The study of rings generated additively by their units started in 1953when K.G. Wolfson [1] and D. Zelinsky [2] proved independently, that everylinear transformation of a vector space V over a division ring D is the sum oftwo nonsingular linear transformations except when dim V = 1 and D = Z .This implies, that the ring of linear transformations End D ( V ) is the sum oftwo units except for one obvious case, when V is a one-dimensional spaceover Z .he ring in which every element is the sum of two units Vamos called2-good rings [3]. The ring R is called von Neumann regular if for any a ∈ R there exists x ∈ R such that axa = a . In 1958 Skornyakov [4, Problem 20,p.167] formulated the question: "Is every element of von Neumann regularring (which does not have Z as a factor-ring) the sum of units?"According to Ehrlich [5] element a ∈ R is unit-regular if a = aua for someunit u ∈ U ( R ) . It is easy to see that a is unit-regular if and only if a is anidempotent times a unit, if and only if a is unit times idempotent. Ring iscalled unit-regular if every its element is such.Note that if a ∈ R is unit-regular element and 2 is unit element in R, then a can be written as a = eu , where e is idempotent and u unit in R .Now, since is unit in R , e is unit with (1 + e ) − = 1 − − e . This givesthat e = ( e + 1) − is the sum of two units and hence a is the sum of twounits.We say that ring R is Henriksen elementary divisor ring if for squarematrix A ∈ R n × n there exist invertible matrices P, Q ∈ R n × n such that P AQ is diagonal matrix [6, p.10].
Theorem 1. [7, Theorem 11] Let R be Henriksen elementary divisor ring.Then R n × n , n > , are 2-good ring. It is important to note that every unit-regular ring is Henriksen elemen-tary divisor ring [8]. Also note that unit and idempotent are unit-regularelements.The next class of rings generated additively by their units and idempo-tents is clean ring. The notion of it was introduced in 1977 by Nicholsonin [9]. Thereafter such rings and their variations were intensively studiedby many authors. Recall that element of ring R is clean if it is the sum ofidempotent and unit of R. The ring R is clean if every element of R is such[9]. Nicholson also showed that in ring where idempotents are central, anyunit-regular element is clean [9, Proposition 1.8], but in noncommutative ringunit-regular element is not necessarily clean. For example, the unit-regularmatrix (
12 50 0 ) ∈ Z × is not clean [10, Example 3.12]. At the same timenote that Camillo and Yu showed that any unit-regular ring is clean [11,Theorem 5].All commutative von Neumann regular rings, local rings, semi-perfectrings and ring R n × n for any clean ring R are the examples of clean rings.Clean rings are closely connected to some important notions of the ring the-ry. Such rings are of interest since they constitute a subclass of the so-calledexchange rings in the theory of noncommutative rings.Module M R has the exchange property if for every module A R and anytwo decompositions A = M ′ ⊕ N = ⊕ i ∈ I A i with M ′ ∼ = M , there exist submodules A ′ i ⊂ A i such that A = M ′ ⊕ ( ⊕ i ∈ I A i ) . The module M R has finite exchange property if the above condition is satis-fied and the index set I is finite. Warfield [12] called ring R an exchange ring if R R has the finite exchange property and he showed that this definition isleft-right symmetric. Independently, Goodearl, Warfield [13] and Nicholson[9] obtained the very useful characterization of R : it is an exchange ring ifand only if for any a ∈ R there exists idempotent e ∈ R such that e ∈ aR and − e ∈ (1 − a ) R . In this case the element a is called an exchange element.In this paper we propose the concept of clear ring based on the conceptof clear element. All, these results are mainly centered on the applicationaround the classical and rather ancient problem of describing all the of anelementary divisor rings. An overview can be found in [6]. In the case ofcommutative rings in [14] the connection of elementary divisor ring with theexistence of clean elements of these rings is proven. Therefore, the problemof studying the matrix rings over elementary divisor rings in this aspect isespecial.We will study matrix rings over elementary divisor ring and reveal con-nection to the theory of full matrices over certain classes of rings.Our main results are the following. Theorem 2.
Let R be a commutative elementary divisor ring and A is a fullnonsingular matrix of R × . Then exist invertible matrices P, Q ∈ GL ( R ) such that P AQ is nontrivial clear element of R × . Theorem 3.
Let R be a commutative elementary divisor ring. Then everyfull nonsingular matrix A ∈ R × is nontrivial clear. Theorem 4.
Let R be a semi-simple commutative B´ezout domain. The nextstatements are equivalent:1. R is an elementary divisor ring;2. any full nonsingular matrix of R × is nontrivial clear. Notations and preliminary results
For ease of explanation, let’s do some notations and recall some definitions.Throughout the paper we suppose R is an associative ring with non-zerounit and U ( R ) its group of units. The vector space of matrices over the ring R of size k × l is denoted by R k × l and group of units of the ring R n × n by GL n ( R ) . The Jacobson radical of R is denoted by J ( R ) . If J ( R ) = 0 , wesay that R is a semi-simple.A ring R is called a right ( left ) B´ezout rin g if each finitely generated right(left) ideal of R is principal. A ring R which is simultaneously right and leftB´ezout ring is called a B´ezout ring .Matrix A over ring R admits diagonal reduction if there exist invertiblematrices P and Q such that P AQ is a diagonal matrix ( d ij ) for which d ii isa total divisor d i +1 ,i +1 (i.e. Rd i +1 ,i +1 R ⊆ d ii R ∩ Rd ii ) for each i . Ring R iscalled an elementary divisor ring provided that every matrix over R admitsa diagonal reduction [6].We can define ranks of a matrix A over R on their rows ρ r ( A ) and theircolumns ρ c ( A ) , respectively (see [15, p. 247]). The smallest m ∈ N such thata matrix A ∈ R k × l is a product of two matrices of size k × m and m × l , iscalled the inner rank ρ ( A ) of A . Note that ρ ( A ) ≤ min { ρ r ( A ) , ρ c ( A ) } andthe number ρ ( A ) does not change under elementary transformations. If R is aright B´ezout domain, then ρ ( A ) = ρ r ( A ) = ρ c ( A ) for any A over R . A matrix A ∈ R n × n is called full if ρ ( A ) = n (see [15, p. 248]), i.e. R n × n AR n × n = R n × n .In the sequel, we use the following result. Proposition 1.
The following statements hold:i) [16, Corollary 3.7] Let R = S × where S is any ring and A = ( a bc d ) ∈ R . If b ∈ U ( S ) or c ∈ U ( S ) , then A is clean element of S × .ii) [17, Theorem 2] A commutative B´ezout domain R is an elementarydivisor ring if and only if for any nonsingular full matrix A ∈ R × right(left) principal ideal AR × ( R × A ) contains nontrivial idempotent.iii) [9, Proposition 1.8] Every clean element of ring is an exchange element. Different classes of rings have generalized clean rings by adding one or moreadjectives such as "almost", "semi", "uniquely" and i.e. to "clean". Weropose another generalization of clean elements namely clear elements. De-scribing full matrices over commutative elementary divisor ring based on thisconcept is able.Element a of ring R is clear if a = r + u where r is unit-regular elementand u ∈ U ( R ) . The ring R is clear if every its element is such. An obviousexample of clear elements are 0 and 1.Let a ∈ R is clear element such that a = r + u , where r is unit-regular and u ∈ U ( R ) . If r = 0 and r / ∈ U ( R ) , we say that clear element a is nontrivial.For example, if R is 2-good ring (any element is the sum of two units), thenany a ∈ R has trivial representation a = u + v , where u, v ∈ U ( R ) , though itdoes not exclude nontrivial representation of elements of R .Since idempotent of ring is obviously unit-regular element, we get thenext result. Proposition 2.
Any clean element is clear. Clean ring is clear.
Moreover, we have the following proposition.
Proposition 3.
Unit-regular element is clear.Proof.
Let aua = a for some u ∈ U ( R ) . Since au is idempotent, so au = 1 − e for some idempotent e ∈ R . Then a = u − − eu − . Since − eu − is unit-regular, a is clear element.We point out that these proposition are not valid in reverse order. Forexample, matrix (
12 50 0 ) ∈ Z × is clear element, but as noted earlier, it isnot clean. Or, to take another example, ring Z is clear ring, but it is notunit-regular.The description of idempotent or unit-regular matrices in the rings ofmatrices over rings is enough actual task. For example: Proposition 4. [18, Proposition 2.1] Let R = S × where S is any ringand let A = ( a b ) ∈ R . A matrix A is unit-regular in R if and only ifexists an idempotent e ∈ S and an unimodular column ( a b ) ∈ S × such that ( ab ) = ( a b ) e . Recall that column ( ab ) is unimodular if Ra + Rb = R . Also note that thiscondition is obviously equivalent to Ra + Rb = Re where e = e , a = a e , b = b e and Ra + Rb = R . Proposition 5.
Every homorphic image of a clear rings is clear.roof.
Every unit-regular element is an idempotent times a unit element andsince multiplication is preserved by every ring homomorphism. The homo-morphic image of an idempotent and unit element are suitable idempotentand unit element of its ring.Since addition is also preserved by every ring homomorphism, the nextresult follows.
Proposition 6.
Every direct product of clear rings is clear.Proof.
It is known that multiplication in direct product of rings is definedcomponentwise, so element in the direct product of rings is unit (resp. idem-potent) of that ring if and only if the entry in each of its components isunit (resp. idempotent) of this ring. Since addition in a direct product ofrings is also defined componentwise, the result follows from a simple compu-tation.
Lemma 1.
Element a ∈ R is clear if and only if ua and au are clean elementsfor some u ∈ U ( R ) .Proof. Let a is clear element, i.e. a = r + v , where r is unit-regular elementand v ∈ U ( R ) . Since rur = r for some u ∈ U ( R ) , then ua = ur + uv where ( ur ) = ur and uv ∈ U ( R ) , i.e. ua is clean element. Similarly au = ru + vu where ( ru ) = ru and vu ∈ U ( R ) .Let for a ∈ R there exists unit element u ∈ U ( R ) such that ua and au are clean elements. Let ua = e + v where e = e and v ∈ U ( R ) . Then a = u − e + u − v . Since ( u − e ) u ( u − e ) = u − e , we have that u − e is an unit-regular element and u − v ∈ U ( R ) , i.e. a is clear. The same is in the case au . In this context for describing clear rings let’s give the following definition.Ring R is said to have unit-regular stable range 1 if for any element a, b ∈ R with aR + bR = R there exists some unit-regular element r of R such that a + br is an unit element of R . Lemma 2.
Ring of unit-regular stable range 1 is clear.Proof.
Let R is ring of unit-regular stable range 1 and a ∈ R . Then aR +( − R = R and we have that a + ( − r = u ∈ U ( R ) for some unit-regularelement r ∈ R , i.e. a = r + u .or the following results recall that element a ∈ R is if a = e + u + v for some idempotent e and unit elements u, v ∈ U ( R ) [19]. Proposition 7.
For any commutative ring clear element is 2-clean. Anytime clear ring is 2-clean ring.Proof.
Let a ∈ R is clear element, i.e. a = r + u for some unit-regular element r and u ∈ U ( R ) . Since every unit-regular element of commutative ring isclean [20] we have that r = e + v for some idempotent e ∈ R and v ∈ U ( R ) .Then a = e + u + v .In addition, we shall notice, that for any ring × and × matricesare 2-clean [20, Lemma 3].Since local ring is clean by Proposition 2 we also have the following result. Proposition 8.
Local ring is clear.
There are not nontrivial idempotents in the local ring. Let’s describe clearring R which has not nontrivial idempotent (domain, for example). Recallthat element a ∈ R is if a is the sum of two unit elements. Proposition 9.
The following are equivalent for any ring R :1. R is a clear and has no nontrivial idempotents;2. for any element a ∈ R : (a) a ∈ U ( R ) or(b) a is 2-good element.Proof. Let R be a clear and has not nontrivial idempotents.Since 0 and 1 are the only ones idempotents in R , then 0 and u ∈ U ( R ) are unit-regulars, i.e. all elements of R are trivial clear elements, i.e. for any a ∈ R we have that a ∈ U ( R ) or a is 2-good element.The implication (2) → (1) is obviously. We start our proofs with the following lemma. emma 3.
Let R be a commutative elementary divisor ring, then for anyfull matrix A ∈ R × there exist invertible matrices P, Q ∈ GL ( R ) such that P AQ = (cid:18) d (cid:19) for some element d ∈ R. Proof.
As noted earlier, if R is commutative elementary divisor ring, thenfor any full matrix A ∈ R × there exist invertible matrices P, T ∈ GL ( R ) such that P AT = (cid:18) d d (cid:19) = D, where d is a total divisor of d . Since R is commutative ring, then obviously d is divisor of d . From R = R × A R × = R × DR × and from d is divisor d follows d ∈ U ( R ) , i.e. we can assume that d =1 . Proof of Theorem 2.
Before starting Proof, note that if A = ( d ) ∈ S × forany ring S, then (cid:18) d (cid:19) (cid:18) (cid:19) = (cid:18) d (cid:19) and (cid:18) d (cid:19) = (cid:18) d + 1 1 (cid:19) + (cid:18) − − (cid:19) . So (by Proposition 1), we have that ( d ) is clean matrix. Also note that ( ) ∈ GL ( R ) .According to our comments and Lemma 3 we are finishing the proving ofTheorem 2. Proof of Theorem 3.
By Theorem 2 we have that
P AQ = E + U where E = E and invertible matrices P, Q, U ∈ GL ( R ) . So A = P − EQ − + P − U Q − and P − EQ − = ( P − EQ − ) QP ( P − EQ − ) . Obviously QP ∈ GL ( R ) and P − EQ − is a unit-regular matrix and P − U Q − ∈ GL ( R ) . That is A isclear matrix.As well known, every commutative principal ideal domain is elementarydivisor domain [18]. Consequently of Theorem 3 we have the following result. Corollary 1.
Let R be a commutative principal ideal domain. Then any fullmatrix A ∈ R × is clear. ow let R be a commutative B´ezout domain in which any nonzero primeideal contained in a unique maximal ideal, i.e. R is P M ∗ ring. By [21,Theorem 1] R is an elementary divisor ring. So, in the same way fromTheorem 3 we get the following result. Corollary 2.
Let R be a commutative P M ∗ B´ezout domain. Then any fullmatrix A ∈ R × is clear. Note that elementary divisor ring is B´ezout ring [6, Theorem 1.2.7]. Con-sequently of Theorem 3 and Proposition 1(ii) we can proving Theorem 4.
Proof of Theorem 4.
By Theorem 3 we have implication → . → . Let full matrix A ∈ R × is clear. By Lemma 1 there existsinvertible matrix U ∈ GL ( R ) such that U A and AU are clean matricies.By Proposition 1 U A ( AU ) is an exchange element. By [9, Proposition 1.1]a right (left) ideal AU R × ( R × U A ) contains idempotent E ( F ). Since R is semi-simple and J ( R × ) = ( J ( R )) × by proving of Proposition 1.9 [9]idempotent E ( F ) is not trivial. By Proposition 1(ii) we obtain that R is anelementary divisor domain. Is the commutative clear ring a ring of unit-regular stable range 1? Is the notion of a ring of unit-regular stable range 1 a left-right symmetric?
Disclosure statement
No potential conflict of interest was reported by the author(s).
ORCID
Bohdan Zabavsky http://orcid.org/0000-0002-2327-5379
Olha Domsha http://orcid.org/0000-0003-1803-6055
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