σ -Biderivations and σ -commuting maps of triangular algebras
aa r X i v : . [ m a t h . R A ] N ov AUTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTINGMAPS OF TRIANGULAR ALGEBRAS C ´ANDIDO MART´IN GONZ ´ALEZ , JOE REPKA , JUANA S ´ANCHEZ-ORTEGA Abstract.
Let A be an algebra and σ an automorphism of A . A linear map d of A is called a σ -derivation of A if d ( xy ) = d ( x ) y + σ ( x ) d ( y ), for all x, y ∈ A .A bilinear map D : A × A → A is said to be a σ -biderivation of A if it is a σ -derivation in each component. An additive map Θ of A is σ -commuting ifit satisfies Θ( x ) x − σ ( x )Θ( x ) = 0, for all x ∈ A . In this paper, we introducethe notions of inner and extremal σ -biderivations and of proper σ -commutingmaps. One of our main results states that, under certain assumptions, every σ -biderivation of a triangular algebras is the sum of an extremal σ -biderivationand an inner σ -biderivation. Sufficient conditions are provided on a triangularalgebra for all of its σ -biderivations (respectively, σ -commuting maps) to beinner (respectively, proper). A precise description of σ -commuting maps oftriangular algebras is also given. A new class of automorphisms of triangularalgebras is introduced and precisely described. We provide many classes oftriangular algebras whose automorphisms can be precisely described. Introduction
Triangular algebras were introduced by Chase [9] in the early 1960s. He endedup with these structures in the course of his study of the asymmetric behaviorof semi-hereditary rings; he provided an example of a left semi-hereditary ringwhich is not right semi-hereditary. Since their introduction, triangular algebrashave played an important role in the development of ring theory. In 1966, Harada[23] characterized hereditary semi-primary rings by using triangular algebras; inhis paper, triangular algebras are named generalized triangular matrix rings. Lateron, Haghany and Varadarajan [21] studied many properties of triangular algebras;they used the terminology of formal triangular matrix rings.Derivations are a fundamental notion in mathematics. They have been objectsof study by many well-known mathematicians. In the middle/late 1990s, severalauthors undertook the study of derivations and related maps over some particularfamilies of triangular algebras (see for e.g. [12, 14, 20, 25, 37] and references therein).Motivated by those works Cheung [10] initiated, in his thesis, the study of linearmaps of (abstract) triangular algebras. He described automorphisms, derivations,commuting maps and Lie derivations of triangular algebras. Cheung’s work hasinspired several authors to investigate many distinct maps of triangular algebras.The present paper is devoted to the study of automorphisms and the so-called σ -biderivations and σ -commuting maps of triangular algebras. Our motivationarises from recent developments in the theory of maps of triangular algebras; more Mathematics Subject Classification.
Key words and phrases.
Triangular algebras, σ -derivations, σ -biderivations, σ -commutingmaps. concretely, we could say that our interest was sparked by the work of Cheung [11] oncommuting maps of triangular algebras, the paper of Benkoviˇc [3] on biderivationsof triangular algebras, and the study of σ -biderivations and σ -commuting maps ofnest algebras due to Yu and Zhang [36].Because the group of automorphisms is an important tool for understandingthe underlying algebraic structure, it has been most thoroughly investigated. Thegroups of automorphisms of triangular algebras have been investigated by manyauthors; see, for example, [1, 2, 10, 13, 15, 25, 26, 27, 29, 30].The study of biderivations of rings and algebras has a rich history and is stillan active area of research. Also, biderivations have already been shown to be ofutility in connection with distinct areas; for example, Skosyrskii [33] used them toinvestigate noncommutative Jordan algebras, while Farkas and Letzter [19] treatedthem in their study of Poisson algebras.Let A be an algebra over a unital commutative ring of scalars; a bilinear map D : A × A → A is said to be a biderivation of A if it is a derivation in eachargument. The map ∆ λ : ( x, y ) λ [ x, y ] is an example of a biderivation, providedthat λ lies in the center Z( A ) of A . Biderivations of this form are called innerbiderivations ; here, [ x, y ] stands for the commutator xy − yx .Notice that if d is a derivation of a commutative algebra, then the map ( x, y ) d ( x ) d ( y ) is a biderivation. In the noncommutative case, it happens quite oftenthat all biderivations are inner. The classical problem under study is to determinewhether every biderivation of a noncommutative algebra is inner.In 1993, Breˇsar et al. [8] proved that every biderivation of a noncommutativeprime ring R is inner. One year after, Breˇsar [5] extended the result above tosemiprime rings. Those results have been shown to be very useful in the study ofthe so-called commuting maps that, as we will point out below, are closely relatedto biderivations.Recently, motivated by Cheung’s work, Zhang et al. [38, Theorem 2.1] provedthat, under some mild conditions, every biderivation of a nest algebra is inner.Later on, Zhao et al. [39] proved, using the results of [38], that every biderivationof an upper triangular matrix algebra is a sum of an inner biderivation and anelement of a particular class of biderivations. Benkoviˇc [3] was the first author tostudy biderivations of (abstract) triangular algebras. He determined the conditionswhich need to be imposed on a triangular algebra to ensure that all its biderivationare inner.It turns out that biderivations are closely connected to the thoroughly studiedcommuting (additive) maps. A map Θ of an algebra A into itself is said to be commuting if Θ( x ) commutes with x , for every x ∈ A . The usual goal whendealing with a commuting map is to provide a precise description of its form. It isstraightforward to check that the identity map and any central map (a map havingits image in the center of the algebra) are examples of commuting maps. Moreover,the sum and the pointwise product of commuting maps produce commuting maps.For example, it is easy to check that the following mapΘ( x ) = λx + Ω( x ) , ∀ x ∈ A , is commuting, for any λ ∈ Z( A ) and any choice of central map Ω of A . We willrefer to commuting maps of this form as proper commuting maps .The first important result on commuting maps dates back to 1957 and it is dueto Posner [31]. Posner’s theorem states that the existence of a nonzero commuting UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 3 derivation of a prime ring implies the commutativity of the ring. Notice thatPosner’s theorem can be reformulated as follows: Theorem. If d is a commuting derivation of a noncommutative prime ring R ,then d = 0 . osner’s theorem has been generalized by a number of authors in many differentways. We refer the reader to the well-written survey [7], where the development ofthe theory of commuting maps and related maps is presented; see also [7, Section 3]for applications of biderivations and commuting maps to other areas. Let us pointout here that one of the main applications of commuting maps can be found in theirconnection with several conjectures on Lie maps of associative rings, formulated byHerstein [24] in 1961.Let us now come back to the relationship between biderivations and commutingmaps; notice that if Θ is a commuting map of an algebra A , then the map D Θ : A×A → A given by D Θ ( x, y ) = [Θ( x ) , y ] is a biderivation of A . Moreover, suppose thatthere exists λ ∈ Z( A ) such that [Θ( x ) , y ] = λ [ x, y ], for all x, y ∈ A ; that is to saythat D Θ is an inner biderivation. Then it follows that Ω Θ ( x ) := Θ( x ) − λx ∈ Z( A ),for every x ∈ A ; in other words, Θ is a proper commuting map. Therefore, in orderto show that every commuting map is proper, it is enough to prove that everybiderivation is inner.Let σ be an automorphism of an algebra A . Recall that a linear map d of A is called a σ -derivation if d ( xy ) = σ ( x ) d ( y ) + d ( x ) y , for all x, y ∈ A . A σ -biderivation is a bilinear map which is a σ -derivation in each argument.The structure of σ -biderivations of prime rings was investigated by Breˇsar [6].As an application, he characterized additive maps f of a prime ring satisfying f ( x ) x = σ ( x ) f ( x ), for all x in the ring. We will refer to such maps as σ -commutingmaps . In the context of triangular algebras, σ -biderivations and σ -commutingmaps of nest algebras have been studied by Yu and Zhang [36] in 2007.The paper is organized as follows: in Section 2 we gather together basic defini-tions and elementary properties. Section 3 is devoted to describing broad classes oftriangular algebras whose automorphisms are partible in the sense of [10, Definition5.1.6]. We call a triangular algebra partible if all its automorphisms are partible.In Section 4 we introduce and describe precisely a new class of automorphisms oftriangular algebras, the so-called bimodule-preserving . Section 5 deals with thestudy of σ -biderivations of partible triangular algebras, while σ -commuting mapsare investigated in Section 6. 2. Preliminaries
Throughout the paper we consider unital associative algebras, and we assumewithout further mention that all of them are algebras over a fixed commutativeunital ring of scalars R . In the subsequent subsections, we recall some definitionsand basic results, and introduce some notation.2.1. Triangular algebras.
Let A and B be unital associative algebras and M anonzero ( A, B )-bimodule. The following set becomes an associative algebra underthe usual matrix operations: T = Trian ( A, M, B ) = (cid:18)
A MB (cid:19) = (cid:26)(cid:18) a mb (cid:19) : a ∈ A, m ∈ M, b ∈ B (cid:27) . MART´IN GONZ´ALEZ, REPKA, S ´ANCHEZ-ORTEGA
An algebra T is called a triangular algebra if there exist algebras A , B and anonzero ( A, B )-bimodule M such that T is isomorphic to Trian ( A, M, B ).Given T = Trian ( A, M, B ), a triangular algebra, let us denote by π A , π B thetwo natural projections from T onto A , B , respectively defined as follows: π A : (cid:18) a mb (cid:19) a, π B : (cid:18) a mb (cid:19) b, for all (cid:18) a mb (cid:19) ∈ T . The center Z( T ) of T was computed in [10] (see also [11, Proposition 3]); it is thefollowing set:Z( T ) = (cid:26)(cid:18) a b (cid:19) ∈ T : a ∈ Z( A ) , b ∈ Z( B ) and am = mb, for all m ∈ M (cid:27) . At this point, it is worth introducing the following two sets: L := { a ∈ A : aM = 0 } , R := { b ∈ B : M b = 0 } . It is easy to see that L and R are ideals of A and B , respectively. Moreover, LM = M R = 0. Set Z := Z( T ); by an abuse of notation we denote by ¯ a and ¯ b the classes ofthe elements a ∈ π A (Z), b ∈ π B (Z) in π A (Z) / ( π A (Z) ∩ L ) and π B (Z) / ( π B (Z) ∩ R ),respectively. Then the map τ : π A (Z) / ( π A (Z) ∩ L ) → π B (Z) / ( π B (Z) ∩ R ) givenby τ (¯ a ) := ¯ b , where b = π B ( x ) for some x ∈ T with π A ( x ) = a , is well-definedand an algebra isomorphism. In particular, if the bimodule M is faithful as a left A -module and also as a right B -module, we have that L = R = 0 and therefore wecan conclude that there exists a unique algebra isomorphism τ : π A (Z) → π B (Z)satisfying that am = mτ ( a ), for all m ∈ M and a ∈ π A (Z).Upper triangular matrix algebras, block upper triangular matrix algebras, trian-gular Banach algebras and nest algebras constitute the most important examplesof triangular algebras. (See, for example, [10].) Notation 2.1.
Let T = Trian ( A, M, B ) be a triangular algebra. We will write1 A , 1 B to denote the units of the algebras A , B , respectively. The unit of T is theelement: 1 := (cid:18) A B (cid:19) . It is straightforward to check that the following elements are orthogonal idempo-tents of T . p := (cid:18) A (cid:19) , q := 1 − p = (cid:18) B (cid:19) ∈ T . Hence, we can consider the Peirce decomposition of T associated to the idempotent p . In other words, we can write T = p T p ⊕ p T q ⊕ q T q . Note that p T p , q T q aresubalgebras of T isomorphic to A , B , respectively; while p T q is a ( p T p, q T q )-bimodule isomorphic to M . To ease the notation in what follows, we will identify p T p , q T q , p T q with A , B , M , respectively. Thus, any element of T can be expressedas: x = (cid:18) a mb (cid:19) = pap + pmq + qbq = a + m + b, by using the identification above. UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 5 Derivations and related maps.
In this subsection, we collect all the def-initions of the maps that will be used in the paper. Although we have alreadyintroduced some of those maps in the introductory section, we will include theirdefinition here for the sake of completeness.
Definitions 2.2.
Let A be an algebra and σ an automorphism of A . Let us denoteby Id A the identity map on A . • A linear map d : A → A is called a derivation of A if it satisfies d ( xy ) = d ( x ) y + xd ( y ) , ∀ x, y ∈ A . • A linear map d : A → A is called a σ -derivation of A if it verifies d ( xy ) = d ( x ) y + σ ( x ) d ( y ) , ∀ x, y ∈ A . Note that every derivation is an Id A -derivation. In the literature (see,for example, [22, 35]), some authors have used the terminology of skew-derivations for σ -derivations. • A bilinear map D : A × A → A is said to be a biderivation of A if it is aderivation in each argument; that is to say that for every y ∈ A , the maps x D ( x, y ) and x D ( y, x ) are derivations of A . In other words: D ( xy, z ) = xD ( y, z ) + D ( x, z ) y, D ( x, yz ) = yD ( x, z ) + D ( x, y ) z ,for all x, y, z ∈ A . The map D is called a σ -biderivation of A providedthat D is a σ -derivation in each argument. • Suppose that A is noncommutative and take λ ∈ Z( A ). The bilinear map∆ λ : A × A → A given by∆ λ ( x, y ) = λ [ x, y ] , ∀ x, y ∈ A is an example of a biderivation. Biderivations of the form above are called inner biderivations . • Given x ∈ A such that x / ∈ Z( A ); suppose that x satisfies [[ x, y ] , x ] = 0for all x, y ∈ A . It was proved in [3, Remark 4.4] that the bilinear map ψ x : A × A → A given by ψ x ( x, y ) = [ x, [ y, x ]] , ∀ x, y ∈ A is a biderivation. Biderivations of this form appear first in [3]; they werenamed extremal biderivations .In this paper, we introduce the notions of inner σ -biderivations and extremal σ -biderivations (see Definitions 5.7 and 5.14 below). Example 2.3.
Let A be an algebra and σ a nontrivial automorphism of A . It iseasy to see that the map d := Id A − σ is a σ -derivation of A , although d is not,in general, a derivation of A . For example, let K be a field of characteristic not 2,and take the algebra of polynomials K [ x ] and σ the automorphism of K [ x ] whichmaps x to − x . Then the σ -derivation d of K [ x ], defined as before, satisfies d ( x ) = x − σ ( x ) = x − σ ( x ) σ ( x ) = x − ( − x )( − x ) = 0 ,d ( x ) x + xd ( x ) = 4 x , since d ( x ) = x − σ ( x ) = 2 x . This shows that d is not a derivation of K [ x ].On the other hand, the map D ( x, y ) = d ( x ) d ( y ) is a σ -biderivation, since K [ x ]is a commutative algebra. But D is not a biderivation, since D ( x , x ) = 0 and D ( x, x ) x + xD ( x, x ) = 8 x . MART´IN GONZ´ALEZ, REPKA, S ´ANCHEZ-ORTEGA
Notice that these examples also work for the finite-dimensional algebra of polyno-mials of degree ≤ N (i.e., the polynomial algebra K [ x ] modulo the ideal generatedby x N +1 ), since σ preserves degrees. Definitions 2.4.
Let A be an algebra, σ an automorphism of A and Θ a linearmap A → A . • The map Θ is a commuting map
A → A if Θ( x ) commutes with x , forevery x ∈ A , i.e., [Θ( x ) , x ] = 0, for all x ∈ A . • Θ is called σ -commuting if it verifies Θ( x ) x = σ ( x )Θ( x ), for all x ∈ A .In particular, in the case where σ = Id A , the notion of a commuting mapis recovered. A this point, it is worth mentioning that the concept of askew-commuting map has a different meaning (see, for example, [4]). Example 2.5.
Let T ( C ) be the algebra of 2 × σ : T ( C ) → T ( C ) given by σ (cid:18) a bc (cid:19) := (cid:18) a − bc (cid:19) , for all (cid:18) a bc (cid:19) ∈ T ( C ) , is an automorphism of T ( C ). It is straightforward to show that the linear map Θof T ( C ) given byΘ (cid:18) a bc (cid:19) := (cid:18) a b − c (cid:19) , for all (cid:18) a bc (cid:19) ∈ T ( C ) , is a σ -commuting map of T ( C ); however, it is not commuting since, for example,for x = (cid:18) (cid:19) , we have that [ x, Θ( x )] = (cid:18) − (cid:19) = 0. Remark 2.6.
When dealing with a triangular algebra T = Trian ( A, M, B ), itis quite common to assume that the bimodule M is faithful as a left A -moduleand also as a right B -module. At this point, it is worth mentioning that we canindeed restrict our attention to triangular algebras having the faithfulness conditionabove. In fact, let T = Trian ( A, M, B ) be a triangular algebra. Notice thatTrian (
A/L, M, B/R ) is a triangular algebra and M is faithful as a left A/L -moduleand as a right
B/R -module, where the annihilators L and R are the ideals of A and B , respectively, defined in Subsection 2.1.It is easy to check that L ⊕ R is an ideal of Trian ( A, M, B ) and thatTrian (
A/L, M, B/R ) ∼ = Trian ( A, M, B ) / ( L ⊕ R ) . Partible automorphisms of triangular algebras
Let T = Trian ( A, M, B ) be a triangular algebra. In what follows, we will assumethat the bimodule M is faithful as a left A -module and also as a right B -module,although these assumptions might not always be necessary.Automorphisms of triangular algebras were studied by Cheung [10, Chapter 5].He introduced and characterized the following class of automorphisms: Definition 3.1. [10, Definition 5.1.6] An automorphism σ of a triangular algebra T = Trian ( A, M, B ) is said to be partible with respect to A , M and B ,if it can be written as σ = φ z ¯ σ , where z ∈ T , φ z is the inner automorphism φ z ( x ) = z − xz , associated to z , and ¯ σ is an automorphism of T satisfying that¯ σ ( A ) = A , ¯ σ ( M ) = M and ¯ σ ( B ) = B . UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 7 As Cheung pointed out inner automorphisms are all partible.
Remark 3.2.
Let T = Trian ( A, M, B ) be a triangular algebra. It is not difficultto prove that every automorphism σ of T satisfying that σ ( A ) = A , σ ( M ) = M and σ ( B ) = B can be written as follows:(1) σ (cid:18) a mb (cid:19) = f σ ( a ) ν σ ( m ) g σ ( b ) , where f σ , g σ are automorphisms of A , B , respectively, and ν σ : M → M is a linearbijective map which satisfies ν σ ( am ) = f σ ( a ) ν σ ( m ) and ν σ ( mb ) = ν σ ( m ) g σ ( b ), forall a ∈ A , b ∈ B , m ∈ M .In this paper we will restrict our attention to the following class of triangularalgebras: Definition 3.3.
A triangular algebra T is said to be partible if every automor-phism of T is partible.See [10, Example 5.2.4] for an example of a triangular algebra which is notpartible. Proposition 3.4.
Upper triangular matrix algebras and nest algebras are partible.Proof.
It follows from the fact that inner automorphisms are partible by takinginto account the fact that automorphisms of upper triangular matrix algebras andnest algebras are all inner. For more details, see [27] for upper triangular matrixalgebras and [29, 30] for nest algebras. (cid:3)
Cheung [10, Theorem 5.3.2] gave sufficient conditions on a triangular algebra T to assert that T is partible. In what follows, we will make use of Cheung’s resultto provide many classes of partible triangular algebras. Let us start by introducingthe following condition based on idempotents. Definition 3.5.
We will say that an algebra A satisfies Condition (I) if anyidempotent e ∈ A such that e A (1 − e ) = 0 must satisfy (1 − e ) A e = 0.The next result collects some examples of algebras satisfying Condition (I). Proposition 3.6.
Let A be an algebra and consider the following properties: (i) A is commutative. (ii) Every idempotent of A is central. (iii) A satisfies Condition (I) .Then (i) ⇒ (ii) ⇒ (iii) .Proof. Straightforward. (cid:3)
Next, we will show that non-degenerate algebras, also called semiprime algebras,satisfy Condition (I). Recall that an algebra A is said to be non-degenerate if,for any a ∈ A , a A a = 0 implies a = 0. Proposition 3.7.
Every non-degenerate algebra satisfies
Condition (I) .Proof.
Let A be a non-degenerate algebra and e ∈ A an idempotent such that e A (1 − e ) = 0. Then we have that ((1 − e ) A e ) A ((1 − e ) A e ) = 0, which implies that(1 − e ) A e = 0, as desired. (cid:3) MART´IN GONZ´ALEZ, REPKA, S ´ANCHEZ-ORTEGA
Theorem 3.8.
Let T = Trian ( A, M, B ) be a triangular algebra such that either A or B satisfies Condition (I) . Then T is partible.Proof. It follows from [10, Theorem 5.3.2 (I) or (II)]. (cid:3)
The following are direct consequences of the theorem above.
Corollary 3.9.
Let T = Trian ( A, M, B ) be a triangular algebra such that either A or B is non-degenerate. Then T is partible. Corollary 3.10.
Let T = Trian ( A, M, B ) be a triangular algebra such that either A or B is commutative. Then T is partible. Corollary 3.11.
Let T = Trian ( A, M, B ) be a triangular algebra such that all theidempotents of A (respectively, B ) are central. Then T is partible. The following result provides a sufficient condition for an automorphism to bepartible.
Theorem 3.12.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an auto-morphism of T such that σ ( M ) = M . Then σ is partible. We first prove a preliminary lemma.
Lemma 3.13.
Let T = Trian ( A, M, B ) be a triangular algebra. We write Lann T ( M ) and Rann T ( M ) for the left annihilator of M in T and the right annihilator of M in T , respectively. Then the following conditions hold: (i) Lann T ( M ) = M ⊕ B , (ii) Rann T ( M ) = A ⊕ M .Proof. (i). It is clear that M ⊕ B ⊆ Lann T ( M ). To show the other containment,take an element x = a + m + b ∈ Lann T ( M ). From here we get that aM = 0, whichyields that a = 0 since we are assuming that M is faithful as a left A -module.Hence x = m + b ∈ M ⊕ B , which concludes the proof.(ii) can be proved in a similar way. (cid:3) Proof of Theorem 3.12.
Let σ be an automorphism of T such that σ ( M ) = M .Then we can find a bijective linear map ν σ : M → M such that σ (cid:18) m (cid:19) = (cid:18) ν σ ( m )0 (cid:19) , for all m ∈ M . Next, keeping in mind that the left and right annihilator of anideal invariant under σ are also invariant under σ , Lemma 3.13 applies to get that σ ( A ⊕ M ) ⊆ A ⊕ M and σ ( M ⊕ B ) ⊆ M ⊕ B . In particular, σ ( A ) ⊆ A ⊕ M and σ ( B ) ⊆ M ⊕ B , which imply that σ ( A ) ∩ B = 0 and A ∩ σ ( B ) = 0. The result nowfollows from [10, Theorem 5.2.3]. (cid:3) K¨othe’s conjecture.
Recall that an ideal of an algebra A is said to be nil ifall its elements are nilpotent. The nil radical Nil ∗ ( A ) of A , also called the K¨otheupper nil radical , is defined as the sum of all nil ideals of A . Note that Nil ∗ ( A )is the largest nil ideal of A and invariant under every automorphism of A . In thissubsection, we will compute the nil radical of a triangular algebra and construct afamily of partible triangular algebras.In 1930, K¨othe [27] conjectured that if a ring has no nonzero nil ideals, then itcan not have any nonzero one-sided nil ideals. Although K¨othe’s conjecture has UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 9 been studied by a significant number of authors in the past few years, it remains anopen problem. Here we show that K¨othe’s conjecture is true for triangular algebras. Lemma 3.14.
Let T = Trian ( A, M, B ) be a triangular algebra. An element x = a + m + b of T is nilpotent if and only if a and b are nilpotent elements of A and B , respectively.Proof. Straightforward. (cid:3)
Proposition 3.15.
The K¨othe radical of a triangular algebra T = Trian ( A, M, B ) is the following triangular algebra Nil ∗ ( T ) = Trian (Nil ∗ ( A ) , M, Nil ∗ ( B )) .Proof. It is easy to see that I := Trian (Nil ∗ ( A ) , M, Nil ∗ ( B )) is an ideal of T .Moreover, I is a nil ideal of T by Lemma 3.14. We claim that I is the largest nilideal of T . In fact, let J be a nil ideal of T . Lemma 3.14 applies to show that π A ( J )and π B ( J ) are nil ideals of A and B , respectively; this implies that π A ( J ) ⊆ Nil ∗ ( A )and π B ( J ) ⊆ Nil ∗ ( B ). Therefore J ⊆ I , as desired. (cid:3) Corollary 3.16.
Let T = Trian ( A, M, B ) be a triangular algebra. Assume that Nil ∗ ( A ) = Nil ∗ ( B ) = 0 . Then the following conditions hold. (i) Nil ∗ ( T ) = M . (ii) Every nilpotent element of T belongs to M . (iii) Every left (or right) nil ideal of T is contained in M . Corollary 3.17.
K¨othe’s conjecture holds for triangular algebras.
Lemma 3.18.
Any unital algebra with zero K¨othe radical is non-degenerate.Proof.
Straightforward. (cid:3)
Theorem 3.19.
Let T = Trian ( A, M, B ) be a triangular algebra such that either Nil ∗ ( A ) = 0 or Nil ∗ ( B ) = 0 . Then T is partible.Proof. It follows from Corollary 3.9 by an application of Lemma 3.18. (cid:3)
Proof.
Let σ be an automorphism of T . By Corollary 3.16 (i) we have thatNil ∗ ( T ) = M . In particular, we have that σ ( M ) = M . The result now followsfrom Theorem 3.12. (cid:3) Remark 3.20.
Notice that for example the algebra of matrices M n ( R ) satisfiesthat Nil ∗ (M n ( R )) = 0, and M n ( R ) has many nontrivial idempotents.4. Bimodule preserving automorphisms
Let T = Trian ( A, M, B ) be a triangular algebra. Along this section, M needsnot to be either faithful as a left A -module or as a right B -module. Inspired byTheorem 3.12, we introduce the following definition: Definition 4.1.
Let T = Trian ( A, M, B ) be a triangular algebra. We say that anendomorphism σ of T is M -preserving or bimodule-preserving if σ ( M ) = M .Our goal in this section is to provide a precise description of bimodule-preservingautomorphisms of triangular algebras. Our results contribute to the development ofthe study of automorphism of triangular algebras initiated by Cheung [10, Chapter5]. We first provide a class of triangular algebras such that all its automorphismsare bimodule-preserving. Proposition 4.2.
Let T = Trian ( A, M, B ) be a triangular algebra such that thealgebras A and B satisfy Condition (I) . Then every automorphism of T is M -preserving.Proof. Let σ be an automorphism of T and denote e = σ ( p ) and f = σ ( q ) = 1 − e .Then applying σ to the identities pT q = M and qT p = 0 gives eT f = σ ( M )and f T e = 0, respectively. If e = (cid:18) a m b (cid:19) then a and b are idempotents of A and B , respectively. On the other hand, from 0 = f T e = (1 − e ) T e we get that(1 − a ) Aa = 0 and (1 − b ) Bb = 0. Applying that A and B satisfy Condition (I) wehave that aA (1 − a ) = 0 and bB (1 − b ) = 0. But then σ ( M ) = eT f ⊆ M and σ is M -preserving, as desired. (cid:3) Notation 4.3.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an M -preserving endomorphism of T . We write σ as follows: σ (cid:18) a m b (cid:19) = χ ( a ) + γ ( b ) χ ( a ) + h ( m ) + γ ( b )0 χ ( a ) + γ ( b ) , where χ : A → A , χ : A → M , χ : A → B , γ : B → B , γ : B → M , γ : B → A are linear maps and h : M → M is an endomorphism of the bimodule M .Our goal is to determine the necessary and sufficient condition on x i , y i and h to σ be M -preserving. Theorem 4.4.
The following conditions hold for a bimodule-preserving endomor-phism σ of a triangular algebra T = Trian ( A, M, B ) . (i) χ : A → A , γ : B → A are algebra homomorphisms, and Im( χ )Im( γ ) =Im( γ )Im( χ ) = 0 . (ii) Im( χ ) and Im( γ ) are ideals of A . (iii) A = Im( χ ) ⊕ Im( γ ) . (iv) χ : A → B , γ : B → B are algebra homomorphisms, and Im( χ )Im( γ ) =Im( γ )Im( χ ) = 0 . (v) Im( χ ) and Im( γ ) are ideals of B . (vi) B = Im( χ ) ⊕ Im( γ ) .Proof. (i). Given x = (cid:18) a m b (cid:19) and x ′ = (cid:18) a ′ m ′ b ′ (cid:19) in T , from σ ( xx ′ ) = σ ( x ) σ ( x ′ ) we get that χ ( aa ′ ) + γ ( bb ′ ) = (cid:0) χ ( a ) + γ ( b ) (cid:1)(cid:0) χ ( a ′ ) + γ ( b ′ ) (cid:1) == χ ( a ) χ ( a ′ ) + χ ( a ) γ ( b ′ ) + γ ( b ) χ ( a ′ ) + γ ( b ) γ ( b ′ ) , for all a , a ′ ∈ A and b , b ′ ∈ B . It implies that χ and γ are algebra homomorphismsand Im( χ )Im( γ ) = Im( γ )Im( χ ) = 0, proving (i).(ii). Since σ is A = Im( χ ) + Im( γ ). Then given a ∈ A and χ ( a ), γ ( b ) for a ∈ A and b ∈ B . Writing a = χ ( a ′ ) + γ ( b ′ ) for a ′ ∈ A and b ′ ∈ B , we have that χ ( a ) a = χ ( a ) χ ( a ′ ) + χ ( a ) γ ( b ′ ) = χ ( aa ′ ) ∈ Im( χ ) ,γ ( b ) a = γ ( b ) χ ( a ′ ) + γ ( b ) γ ( b ′ ) = γ ( bb ′ ) ∈ Im( γ ) , by (i). Similarly, a χ ( a ) ∈ Im( χ ) and a γ ( b ) ∈ Im( γ ), which concludes theproof of (ii). UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 11 (iii). It remains to show that Im( χ ) ∩ Im( γ ) = 0. From A = Im( χ ) + Im( γ )we have that (cid:0) Im( χ ) ∩ Im( γ ) (cid:1) A ⊆ (cid:0) Im( χ ) ∩ Im( γ ) (cid:1) Im( χ ) + (cid:0) Im( χ ) ∩ Im( γ ) (cid:1) Im( γ ) ⊆⊆ Im( γ )Im( χ ) + Im( χ )Im( γ ) = 0 . Thus (cid:0)
Im( χ ) ∩ Im( γ ) (cid:1) A = 0, which yields that Im( χ ) ∩ Im( γ ) = 0.(iv), (v) and (vi) can be proved in a similar way. (cid:3) Theorem 4.5.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an M -preserving endomorphism of T such that its restriction to M is an automorphismof M . Then the following conditions hold. (i) Im( γ ) ⊆ Lann A ( M ) and Im( χ ) ⊆ Rann B ( M ) . (ii) ker( χ ) ⊆ Lann A ( M ) and ker( γ ) ⊆ Rann B ( M ) . (iii) χ ( aa ′ ) = χ ( a ) χ ( a ′ ) , for all a, a ′ ∈ A . (iv) γ ( bb ′ ) = γ ( b ) γ ( b ′ ) , for all b, b ′ ∈ B . (v) ker( χ ) ⊆ ker( χ ) and ker( γ ) ⊆ ker( γ ) .Proof. Let x = (cid:18) a m b (cid:19) and x ′ = (cid:18) a ′ m ′ b ′ (cid:19) in T . From σ ( xx ′ ) = σ ( x ) σ ( x ′ )we have that χ ( aa ′ ) + h ( am ′ + mb ′ ) + γ ( bb ′ ) = (cid:0) χ ( a ) + γ ( b ) (cid:1)(cid:0) χ ( a ′ ) + h ( m ′ ) + γ ( b ′ ) (cid:1) ++ (cid:0) χ ( a ) + h ( m ) + γ ( b ) (cid:1)(cid:0) χ ( a ′ ) + γ ( b ′ ) (cid:1) . It implies that χ ( aa ′ ) = χ ( a ) χ ( a ′ ) + χ ( a ) χ ( a ′ ) , (2) γ ( bb ′ ) = γ ( b ) γ ( b ′ ) + γ ( b ) γ ( b ′ ) , (3) h ( am ′ ) = χ ( a ) h ( m ′ ) , (4) h ( mb ′ ) = h ( m ) γ ( b ′ ) , (5) 0 = χ ( a ) γ ( b ′ ) + χ ( a ) γ ( b ′ ) , (6) 0 = γ ( b ) χ ( a ′ ) + γ ( b ) χ ( a ′ ) , (7) 0 = h ( m ) χ ( a ′ ) , (8) 0 = γ ( b ) h ( m ′ ) . (9)(i) follows from (8) and (9), since h is an automorphism of M .(ii). The equality ker( χ ) ⊆ Lann A ( M ) can be obtained from (4):0 = ah ( m ′ ) = h ( am ′ ) , ∀ m ′ ∈ M, since h is an automorphism of M . An application of (5) proves that ker( γ ) ⊆ Rann B ( M ).(iii) (respectively, (iv)) follows from (2) (respectively, (3)) by an application ofIm( χ ) ⊆ Rann B ( M ) (respectively, Im( γ ) ⊆ Lann A ( M )).(v). Making a ′ = 1 in (3) gives that χ ( a ) = χ ( a ) χ (1), for all a ∈ A , whichimplies that ker( χ ) ⊆ ker( χ ). Similarly, ker( γ ) ⊆ ker( γ ). (cid:3) Remark 4.6.
Notice that condition (iii) in theorem above is equivalent to thefollowing one:(iii)’ χ ( a ) = χ ( a ) χ (1) , for all a ∈ A. It is clear that (iii) implies (iii)’ by making a ′ = 1. Conversely, assume that (iii)’holds and take a, a ′ ∈ A , from χ ( aa ′ ) = χ ( aa ′ ) χ (1) applying that χ is analgebra homomorphism we have that χ ( aa ′ ) = χ ( aa ′ ) χ (1) = χ ( a ) χ ( a ′ ) χ (1) (iii) ′ = χ ( a ) χ ( a ′ ) . In a similar way, one can show that condition (iv) is equivalent to the followingone:(iv)’ γ ( b ) = γ (1) γ ( b ) , for all b ∈ B. Theorem 4.7.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an M -preserving endomorphism of T . Then σ is a monomorphism of T if and only if thefollowing three conditions hold. (M1) ker( h ) = 0 . (M2) ker( χ ) ∩ ker( χ ) = 0 . (M3) ker( γ ) ∩ ker( γ ) = 0 .Proof. Assume first that σ is an M -preserving monomorphism of T . Then (M1)clearly follows. If a ∈ ker( χ ) ∩ ker( χ ) then a ∈ ker( χ ) by Theorem 4.5 (v). Butalso σ (cid:18) a
00 0 (cid:19) = (cid:18) χ ( a ) χ ( a )0 χ ( a ) (cid:19) = (cid:18) (cid:19) , which implies that a = 0. It shows (M2). Similarly, (M3) can be proved.Conversely, assume that (M1), (M2) and (M3) hold. If x = (cid:18) a m b (cid:19) ∈ ker( σ )then px = (cid:18) a m (cid:19) ∈ ker( σ ). But σ ( px ) = (cid:18) χ ( a ) χ ( a ) + h ( m )0 χ ( a ) (cid:19) . Thus a ∈ ker( χ ) ∩ ker( χ ) (M2) = 0 and so a = 0, which yields that h ( m ) = 0, and by (M1)we get that m = 0. Apply that x ∈ ker( σ ) to get that b ∈ ker( γ ) ∩ ker( γ ) (M3) = 0,which finishes the proof. (cid:3) Theorem 4.8.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an M -preserving endomorphism of T . Then σ is an epimorphism of T if and only if thefollowing two conditions hold. (E1) h is an epimorphism of M . (E2) A = χ (cid:0) ker( χ ) (cid:1) ∩ γ (cid:0) ker( γ ) (cid:1) . (E3) B = γ (cid:0) ker( γ ) (cid:1) ∩ χ (cid:0) ker( χ ) (cid:1) .Proof. Assume first that σ is an M -preserving epimorphism of T . It is clear that(E1) follows. Given (cid:18) a
00 0 (cid:19) ∈ T we find an element x = (cid:18) a m b (cid:19) ∈ T suchthat σ ( x ) = χ ( a ) + γ ( b ) χ ( a ) + h ( m ) + γ ( b )0 χ ( a ) + γ ( b ) = (cid:18) a
00 0 (cid:19) , which implies that χ ( a ) + γ ( b ) = a ,χ ( a ) + h ( m ) + γ ( b ) = 0 ,χ ( a ) + γ ( b ) = 0 . UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 13 Theorem (vi) allows us to conclude that a ∈ ker( χ ) and b ∈ ker( γ ). It shows(E2). Similarly, (E3) follows.Conversely, assume that (E1), (E2) and (E3) hold. Then given a ∈ A , apply(E2) to write it as a = χ ( a ) + γ ( b ), where χ ( a ) = 0 and γ ( b ) = 0. Thus: (cid:18) a
00 0 (cid:19) = σ (cid:18) a m b (cid:19) ∈ Im( σ ) , where m ∈ M is such that h ( m ) = − χ ( a ) − γ ( b ). It proves that (cid:18) A
00 0 (cid:19) ⊆ Im( σ ). Similarly, (cid:18) B (cid:19) ⊆ Im( σ ). To finish, apply (E1). (cid:3) Theorem 4.9.
Let T = Trian ( A, M, B ) be a triangular algebra. An endomorphism σ : T → T is an M -preserving automorphism of T if and only if the associated maps χ i , γ j and h satisfy the following conditions: (i) A = Im( χ ) ⊕ Im( γ ) , B = Im( χ ) ⊕ Im( γ ) . (ii) h is an automorphism of M . (iii) χ , γ , χ and γ are algebra homomorphisms, and A = Im( χ ) ⊕ Im( γ ) = χ (cid:0) ker( χ ) (cid:1) ∩ γ (cid:0) ker( γ ) (cid:1) ,B = Im( χ ) ⊕ Im( γ ) = γ (cid:0) ker( γ ) (cid:1) ∩ χ (cid:0) ker( χ ) (cid:1) , where Im( χ ) , Im( γ ) are ideals of A , and Im( γ ) , Im( χ ) are ideals of B . (iv) Im( γ ) ⊆ Lann A ( M ) and Im( χ ) ⊆ Rann B ( M ) . (v) ker( χ ) ⊆ Lann A ( M ) and ker( γ ) ⊆ Rann B ( M ) . (vi) χ ( a ) = χ ( a ) χ (1) , for all a ∈ A. (vii) γ ( b ) = γ (1) γ ( b ) , for all b ∈ B. Remark 4.10.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an auto-morphism of T . There are two extreme situations: Case 1. ker( χ ) = ker( γ ) = 0. Case 2. ker( χ ) = ker( γ ) = 0.In Case 1, we have that A = ker( χ ) and B = ker( γ ), and so χ = γ = 0 and σ is indeed partible. In Case 2, we get that A = ker( χ ) and B = ker( γ ). Then χ = γ = 0, which imply that χ = γ = 0. Then (4) and (5) (see the proofof Theorem 4.5) yield that h = 0. Notice that this situation can only happensprovided M = 0. In such a case, we will refer to T as a diagonal triangularalgebra , while σ will be called an “anti-partible” automorphism . Theorem 4.11.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an M -preserving automorphism of T . Then there exist ideals I and J of T such that (i) σ ( I ) = I and σ ( J ) = J . (ii) T = I ⊕ J . (iii) σ | I is a partible automorphism of the triangular algebra I . (iv) σ | J is an anti-partible automorphism of the diagonal triangular algebra J .Proof. Take I = Trian (ker( χ ) , M, ker( γ )) and J = Trian (ker( χ ) , , ker( γ )). (cid:3) Corollary 4.12.
Let T = Trian ( A, M, B ) be a triangular algebra such that thealgebras A and B are indecomposable. Then any M -preserving automorphism σ of T is either partible or anti-partible. Moreover, σ is anti-partible if and only if T isdiagonal. σ -biderivations of partible triangular algebras This section is devoted to the study of σ -biderivations of partible triangularalgebras. We start by proving a preliminary result. Lemma 5.1.
Let T = Trian ( A, M, B ) be a triangular algebra and σ a partibleautomorphism of T . Let σ = φ z ¯ σ , where φ z ( x ) = z − xz , for all x ∈ T ; and ¯ σ isan automorphism of T such that ¯ σ ( A ) = A , ¯ σ ( M ) = M , ¯ σ ( B ) = B . Then (i) a linear map d : T → T is a σ -derivation of T if and only if ¯ d ( x ) = zd ( x ) ,for all x ∈ T , is a ¯ σ -derivation. (ii) a bilinear map D : T × T → T is a σ -biderivation of T if and only if ¯ D ( x, y ) = zD ( x, y ) , for all x, y ∈ T , is a ¯ σ -biderivation.Proof. (i) Take x, y, z ∈ T , and let us assume first that d is a σ -derivation. Thenwe have that¯ d ( xy ) = zd ( xy ) = z ( d ( x ) y + σ ( x ) d ( y )) = ¯ d ( x ) y + zz − ¯ σ ( x ) zd ( y ) == ¯ d ( x ) y + ¯ σ ( x ) ¯ d ( y ) , which shows that ¯ d is a ¯ σ -derivation. Conversely, suppose that ¯ d is a ¯ σ -derivation.Then: d ( xy ) = z − ¯ d ( xy ) = z − (cid:0) ¯ d ( x ) y + ¯ σ ( x ) ¯ d ( y ) (cid:1) = z − ¯ d ( x ) y + z − ¯ σ ( x ) zz − ¯ d ( y ) == d ( x ) y + σ ( x ) d ( y ) , as desired.(ii) follows from (i). (cid:3) Since our objects of study are σ -biderivations of partible triangular algebras, theresult above allows us to deal with a nicer class of automorphisms. More precisely,we can assume that σ satisfies that σ ( A ) = A , σ ( M ) = M , σ ( B ) = B , and thereforeit can be written as in (1). In what follows, we will make this assumption withoutfurther mention.The following notion will play an important role for our purposes. Definition 5.2. [35, Definition 2.3] Let A be an algebra and σ an automorphismof A . The σ -center of A is the set Z σ ( A ) given byZ σ ( A ) = { λ ∈ A : σ ( x ) λ = λx, for all x ∈ A} . A linear map from A to Z σ ( A ) will be called σ -central . Lemma 5.3. [35, Lemma 2.5]
Let A be an algebra and σ an automorphism of A .The σ -center of A is a subspace of A which is invariant under σ ; in other words: σ ( λ ) ∈ Z σ ( A ) for every λ ∈ Z σ ( A ) . Lemma 5.4.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an automor-phism of T such that σ ( A ) = A , σ ( M ) = M , σ ( B ) = B , written as in (1) . Then: Z σ ( T ) = (cid:26)(cid:18) a b (cid:19) ∈ T : a ∈ Z f σ ( A ) , b ∈ Z g σ ( B ) , am = ν σ ( m ) b, ∀ m ∈ M (cid:27) . Moreover, there exists a unique algebra isomorphism η : π B (Z σ ( T )) → π A (Z σ ( T )) such that η ( b ) m = ν σ ( m ) b , for all b ∈ Z σ ( T ) , m ∈ M .Proof. Straightforward. (cid:3)
UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 15 Let A be an algebra and σ an automorphism of A . We introduce a new bilinearoperation on A :(10) [ x, y ] σ = σ ( x ) y − yx, for all x, y ∈ A . We will call (10) the σ -commutator of A . Note thatZ σ ( A ) = { λ ∈ A : [ x, λ ] σ = 0 for all x ∈ A} . The next result collects some elementary properties of the σ -commutator. We omitits proof since it consists of elementary calculations. Lemma 5.5.
Let A be an algebra and σ an automorphism of A . For x, y, z ∈ A itfollows that (i) [ xy, z ] σ = [ x, z ] σ y + σ ( x )[ y, z ] σ , (ii) [ x, [ y, z ] σ ] σ = [[ x, y ] , z ] σ + [ y, [ x, z ] σ ] σ . Set x ∈ A . Notice that Lemma 5.5 (i) says that the map δ x : A → A given by(11) δ x ( x ) = [ x, x ] σ , ∀ x ∈ A is a σ -derivation of A . We will refer to (11) as an inner σ -derivation .We continue by investigating what should be the suitable generalization of aninner biderivation in the σ -maps setting. Proposition 5.6.
Let A be a noncommutative algebra, σ an automorphism of A and λ ∈ Z σ ( A ) . Then the map ∆ λ : A × A → A given by (12) ∆ λ ( x, y ) = λ [ x, y ] , ∀ x, y ∈ A is a σ -biderivation of A .Proof. Take x, y, z ∈ A .∆ λ ( x, z ) y + σ ( x )∆ λ ( y, z ) = λxzy − λzxy + σ ( x ) λyz − σ ( x ) λzy == λxzy − λzxy + λxyz − λxzy = λxyz − λzxy, since λ ∈ Z σ ( A ). On the other hand, we have that∆ λ ( xy, z ) = λxyz − λzxy. Therefore, ∆ λ ( xy, z ) = ∆ λ ( x, z ) y + σ ( x )∆ λ ( y, z ). Similarly, one can show that∆ λ ( x, yz ) = ∆ λ ( x, y ) z + σ ( y )∆ λ ( x, z ), concluding the proof. (cid:3) The result above makes it possible to introduce the following concept.
Definition 5.7.
Let A be a noncommutative algebra and σ an automorphism of A . Maps of the form (12) will be called inner σ -biderivations .One of our main results in this section is the following theorem. Of course, anyinner σ -biderivation satisfies ∆ λ ( x, x ) = 0, for every x ∈ A . For a triangular algebraand the idempotent p defined in Notation (2.1), the theorem provides sufficientconditions to ensure that the σ -biderivations vanishing at ( p, p ) are indeed inner.At this point, a natural question arises: Question 5.8.
What can be said about the σ -biderivations of T which do notvanish at ( p, p )?This question will be treated in Theorem 5.15 below. Theorem 5.9.
Let T = Trian ( A, M, B ) be a partible triangular algebra and σ anautomorphism of T . Suppose that T satisfies the following conditions: (i) π A (Z σ ( T )) = Z f σ ( A ) and π B (Z σ ( T )) = Z g σ ( B ) . (ii) Either A or B is noncommutative. (iii) If λ ∈ Z σ ( T ) satisfies λx = 0 , for some nonzero element x in T , then λ = 0 . (iv) Every linear map ξ : M → M satisfying ξ ( amb ) = f σ ( a ) ξ ( m ) b , for all a ∈ A , m ∈ M , b ∈ B , can be expressed as ξ ( m ) = λ m + ν σ ( m ) µ , for all m ∈ M and certain λ ∈ Z f σ ( A ) and µ ∈ Z g σ ( B ) .Then every σ -biderivation D of T such that D ( p, p ) = 0 is an inner σ -biderivation. Remark 5.10.
The results of the present section generalize the results of [3, Section4] (see Section 7 for details). Theorem 5.9 is the analogue of [3, Theorem 4.11] for σ -maps. (See also [3, Remark 4.12].) Note that Condition (iv) could be replacedby the following condition:(iv) ′ Every σ -derivation of T is inner.The σ -derivations of triangular algebras have been already studied by Han and Wei[22]; however, the innerness of σ -derivations has not been treated yet. It is notdifficult to prove that a σ -derivation of T is inner if and only if its associated linearmap ξ : M → M (see [22, Theorem 3.12] for a precise description of σ -derivations)is of the following form: ξ ( m ) = λ m + ν σ ( m ) µ , ∀ m ∈ M, for certain λ ∈ Z f σ ( A ) and µ ∈ Z g σ ( B ). From this, one can easily derive thatCondition (iv) ′ above implies Condition (iv) in Theorem 5.9.In what follows, we will collect the appropriate results and develop the machineryneeded to prove Theorem 5.9. Lemma 5.11. (See [6, Lemma 2.3] and [3, Lemma 4.2])
Let A be an algebra, σ anautomorphism of A , and D a σ -biderivation of A . Then (i) D ( x, y )[ u, v ] = σ ([ x, y ]) D ( u, v ) = [ σ ( x ) , σ ( y )] D ( u, v ) , for all x, y, u, v ∈ A . (ii) D ( x,
1) = D (1 , x ) = D ( x,
0) = D (0 , x ) = 0 , for all x ∈ A . (iii) For an idempotent e of A , it follows that D ( e, e ) = − D ( e, − e ) = − D (1 − e, e ) = D (1 − e, − e ) . Lemma 5.12.
Let T = Trian ( A, M, B ) be a partible triangular algebra, σ anautomorphism of T , and D a σ -biderivation of T . If x, y ∈ T are such that [ x, y ] =0 , then D ( x, y ) = pD ( x, y ) q , where p , q are as in Notation 2.1.Proof. Take x, y ∈ T with [ x, y ] = 0. Considering the Peirce decomposition of T associated to the idempotent p of T , we can write D ( x, y ) as follows:(13) D ( x, y ) = pD ( x, y ) p + pD ( x, y ) q + qD ( x, y ) q. For m ∈ M , applying Lemma 5.11 (i) we get that0 = D ( p, pmq )[ x, y ] = σ ([ p, pmq ]) D ( x, y ) = σ ( pmq ) D ( x, y ) = pν σ ( m ) qD ( x, y ) , which implies that pM qD ( x, y ) = 0, since ν σ is a bijection. From this, we canconclude that M qD ( x, y ) q = 0, which yields qD ( x, y ) q = 0, since M is a faithfulright B -module. Given m ∈ M , a second application of Lemma 5.11 (i) produces: D ( x, y ) pmq = D ( x, y )[ p, pmq ] = σ ([ x, y ]) D ( p, pmq ) = 0 . UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 17 Thus, pD ( x, y ) pM = 0, which yields pD ( x, y ) p = 0. The result now follows from(13). (cid:3) Motivated by the fact that extremal biderivations played an important role inthe study of biderivations of triangular algebras; our next task will be to investigatewhat should be called an extremal σ -biderivation. Proposition 5.13.
Let A be an algebra and σ an automorphism of A . The sym-metric bilinear map ψ x : A × A → A given by (14) ψ x ( x, y ) = [ x, [ y, x ] σ ] σ , ∀ x, y ∈ A is a σ -biderivation, provided that the element x ∈ A satisfies x / ∈ Z σ ( A ) and [[ A , A ] , x ] σ = 0 .Proof. Let x ∈ A such that x / ∈ Z σ ( A ) and [[ A , A ] , x ] σ = 0. Clearly, ψ x is abilinear map. To show its symmetry, apply Lemma 5.5 (ii) to get that ψ x ( x, y ) = [ x, [ y, x ] σ ] σ = [[ x, y ] , x ] σ + [ y, [ x, x ] σ ] σ = ψ x ( y, x ) , as desired. In order to prove that ψ x is a biderivation, it is enough to check thatit is a derivation in its first argument. Given x, y, z ∈ A , applying Lemma 5.5 (i)we obtain that ψ x ( xy, z ) = [ xy, [ z, x ] σ ] σ = [ x, [ z, x ] σ ] σ y + σ ( x )[ y, [ z, x ] σ ] σ == ψ x ( x, z ) y + σ ( x ) ψ x ( y, z ) , which concludes the proof. (cid:3) In view of the previous result, we introduce the following terminology:
Definition 5.14.
Let A be an algebra and σ an automorphism of A . An extremal σ -biderivation is a bilinear map of A of the form (14).Now, we are in a position to answer Question 5.8. Theorem 5.15.
Let T = Trian ( A, M, B ) be a partible triangular algebra, σ anautomorphism of T , and D a σ -biderivation of T . Then D can be written as a sumof an extremal σ -biderivation and a biderivation vanishing at ( p, p ) . Moreover, D = ψ D ( p,p ) + D , where ψ D ( p,p ) is the extremal σ -biderivation associated to theelement x = D ( p, p ) , and D is a biderivation of T satisfying D ( p, p ) = 0 .Proof. Let D be a σ -biderivation of T such that D ( p, p ) = 0. Note that from[ p, p ] = 0, we get that D ( p, p ) = pD ( p, p ) q , by an application of Lemma 5.12.Lemma 5.4 allows us to conclude that D ( p, p ) / ∈ Z σ ( T ). Next, we claim that[[ x, y ] , D ( p, p )] σ = 0, for every x, y ∈ T . In fact, given x, y ∈ T , we have that[[ x, y ] , D ( p, p )] σ = σ ([ x, y ]) D ( p, p ) − D ( p, p )[ x, y ]= [ σ ( x ) , σ ( y )] D ( p, p ) − D ( p, p )[ x, y ] = 0 , by Lemma 5.11 (i). Therefore, it makes sense to consider the extremal σ -biderivation ψ D ( p,p ) . Let D = D − ψ D ( p,p ) ; it remains to check that D ( p, p ) = 0. We will showthat ψ D ( p,p ) ( p, p ) = D ( p, p ). ψ D ( p,p ) ( p, p ) = [ p, [ p, D ( p, p )] σ ] σ = [ p, σ ( p ) D ( p, p ) − D ( p, p ) p ] σ = σ ( p ) σ ( p ) D ( p, p ) − σ ( p ) D ( p, p ) p. Taking into account that D ( p, p ) = pD ( p, p ) q and that σ ( p ) = (cid:18) A (cid:19) , the righthand side of the equality above becomes D ( p, p ); that is, ψ D ( p,p ) ( p, p ) = D ( p, p ),which finishes the proof. (cid:3) Proof of Theorem 5.9.
Let D be a σ -biderivation of T such that D ( p, p ) = 0.Take x = (cid:18) a x m x b x (cid:19) = a x + m x + b x , y = (cid:18) a y m y b y (cid:19) = a y + m y + b y , two arbitrary elements of T . The bilinearity of D implies that D ( x, y ) = D ( a x , a y ) + D ( a x , m y ) + D ( a x , b y ) + D ( m x , a y ) + D ( m x , m y )(15) + D ( m x , b y ) + D ( m x , a y ) + D ( m x , m y ) + D ( m x , b y ) + D ( b x , a y )+ D ( b x , m y ) + D ( b x , b y ) . Note that an application of Lemma 5.11 yields that D ( p, q ) = D ( q, p ) = D ( q, q ) = 0.In what follows, we will distinguish several cases: ⋄ Case 1. D ( a, b ) = D ( b, a ) = 0, for all a ∈ A and b ∈ B .Applying Lemma 5.12, taking into account that [ a, b ] = 0, we get that D ( a, b ) = pD ( a, b ) q and D ( b, a ) = pD ( b, a ) q . Thus: D ( a, b ) = pD ( a, b ) q = pD ( ap, qb ) q = pD ( a, qb ) pq + pσ ( a ) D ( p, qb ) q = σ ( a ) D ( p, qb ) q = σ ( a ) D ( p, q ) bq + σ ( a ) σ ( q ) D ( p, b ) q = σ ( aq ) D ( p, b ) q = 0 . It remains to check that D ( b, a ) = 0. Similar calculations give: D ( b, a ) = pD ( b, a ) q = pD ( qb, ap ) q = pD ( q, ap ) bq + pσ ( q ) D ( b, ap ) q = pD ( q, ae ) b = pD ( q, a ) pb + pσ ( a ) D ( q, p ) b = 0 , which finishes the proof of Case 1 . ⋄ Case 2. D ( p, b ) = D ( b, p ) = D ( p, a ) = D ( a, p ) = 0, for all a ∈ A and b ∈ B .By Case 1 we have that D ( p, b ) = D ( b, p ), for all b ∈ B . Given a ∈ A , anapplication of Lemma 5.11 (ii) gives:0 = D (1 , a ) = D ( p, a ) + D ( q, a ) Case 1 = D ( p, a ) , D ( a,
1) = D ( a, p ) + D ( a, q ) Case 1 = D ( a, p ) , concluding the proof of Case 2 . ⋄ Case 3. D ( q, b ) = D ( b, q ) = D ( q, a ) = D ( a, q ) = 0, for all a ∈ A and b ∈ B .The proof is analogous to that of Case 2. ⋄ Case 4. D ( a, m ) p = D ( m, a ) p = 0, for all a ∈ A and m ∈ M .Given a ∈ A and m ∈ M , we have that D ( a, m ) p = D ( a, mq ) p = σ ( m ) D ( a, q ) p + D ( a, m ) qp Case 3 = 0 ,D ( m, a ) p = D ( mq, a ) p = σ ( m ) D ( q, a ) p + D ( m, a ) qp Case 3 = 0 , finishing the proof of Case 4 .The next case can be proved in a similar way. ⋄ Case 5. σ ( q ) D ( b, m ) = σ ( q ) D ( m, b ) = 0, for all m ∈ M and b ∈ B . UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 19 ⋄ Case 6.
There exists ˜ λ ∈ Z f σ ( A ) such that D ( a, m ) = − D ( m, a ) = ˜ λ am, D ( m, b ) = − D ( b, m ) = ˜ λ mb, for all a ∈ A , m ∈ M and b ∈ B .Consider the map ξ : M → M given by ξ ( m ) = D ( p, m ), for all m ∈ M . Noticethat ξ is well-defined, i.e., ξ ( m ) ∈ M , for all m ∈ M . In fact, given m ∈ M , wehave that D ( p, m ) = D ( p, pmq ) = D ( p, pm ) q + σ ( pm ) D ( p, q ) = σ ( p ) D ( p, m ) q + D ( p, p ) mq = σ ( p ) D ( p, m ) q ∈ M, since σ ( p ) = (cid:18) A (cid:19) . Trivially, ξ is an additive map. Moreover, ξ satisfies ξ ( amb ) = f σ ( a ) ξ ( m ) b , for all a ∈ A , m ∈ M , b ∈ B . In fact, given a ∈ A , m ∈ M , b ∈ B , we have that ξ ( amb ) = D ( p, amb ) = σ ( a ) D ( p, mb ) + D ( p, a ) mb Case 2 = σ ( a ) σ ( m ) D ( p, b )+ σ ( a ) D ( p, m ) b Case 2 = σ ( a ) D ( p, m ) b = f σ ( a ) ξ ( m ) b, since σ ( a ) = (cid:18) f σ ( a ) 00 (cid:19) , and D ( p, m ) ∈ M . Thus Condition (iv) implies thatthere exist λ ∈ Z f σ ( A ) and µ ∈ Z g σ ( B ) such that ξ ( m ) = λ m + ν σ ( m ) µ ,for all m ∈ M . Lemma 5.4 and Condition (i) allow us to consider the element˜ λ := λ + η ( µ ) ∈ Z f σ ( A ). The calculations above jointly with a second use ofLemma 5.4 imply that D ( p, m ) = ξ ( m ) = λ m + ν σ ( m ) µ = ( λ + η ( µ )) m = ˜ λ m, ∀ m ∈ M. Similarly, one can find ˜ µ ∈ Z f σ ( A ) such that D ( m, p ) = ˜ µ m , for all m ∈ M .Next, we claim that ˜ λ + ˜ µ = 0. From Condition (ii) we have that either A or B is noncommutative. Let us assume, for example, that A is noncommutative. Then,there exist a, a ′ ∈ A such that [ a, a ′ ] = 0. Applying Lemma 5.11 (i), we get that D ( a, a ′ )[ p, m ] = σ ([ a, a ′ ]) D ( p, m ) = f σ ([ a, a ′ ])˜ λ m = ˜ λ [ a, a ′ ] m,D ( a, a ′ )[ m, p ] = σ ([ a, a ′ ]) D ( m, p ) = f σ ([ a, a ′ ])˜ µ m = ˜ µ [ a, a ′ ] m, for all m ∈ M . Thus: (˜ λ + ˜ µ )[ a, a ′ ] M = 0, which implies that (˜ λ + ˜ µ )[ a, a ′ ] = 0,since M is a faithful left A -module. From this, Condition (iii) applies to show that˜ λ + ˜ µ = 0, which finishes the proof of our claim. Notice that we have just provedthat D ( p, m ) = − D ( m, p ) = ˜ λ m, ∀ m ∈ M. From here, an application of Lemma 5.11 (ii) gives that D ( m, q ) = − D ( q, m ) = ˜ λ m, ∀ m ∈ M. For a ∈ A , it follows that D ( a, m ) = D ( ap, m ) = σ ( a ) D ( p, m ) + D ( a, m ) p Case 4 = f σ ( a )˜ λ m = ˜ λ am,D ( m, a ) = D ( m, ap ) = σ ( a ) D ( m, p ) + D ( m, a ) p Case 4 = − f σ ( a )˜ λ m = − ˜ λ am, for all m ∈ M . Proceeding as above, applying Case 5 one can prove that D ( m, b ) = − D ( b, m ) = ˜ λ mb, for all m ∈ M and b ∈ B , as desired. ⋄ Case 7. D ( a, a ′ ) = ˜ λ [ a, a ′ ] for every a, a ′ ∈ A .Given a, a ′ ∈ A , we start by showing that D ( a, a ′ ) ∈ A . D ( a, a ′ ) = D ( pap, a ′ ) = σ ( p ) D ( ap, a ′ ) + D ( p, a ′ ) ap Case 2 = σ ( p ) σ ( a ) D ( p, a ′ ) + σ ( p ) D ( a, a ′ ) p Case 2 = σ ( p ) D ( a, a ′ ) p ∈ A. For m ∈ M , apply Lemma 5.11 (i) to obtain that D ( a, a ′ ) m = D ( a, a ′ )[ p, m ] = σ ([ a, a ′ ]) D ( p, m ) = f σ ([ a, a ′ ])˜ λ m = ˜ λ [ a, a ′ ] m, ∀ m ∈ M. Thus: ( D ( a, a ′ ) − ˜ λ [ a, a ′ ]) M = 0, which yields that D ( a, a ′ ) = ˜ λ [ a, a ′ ]. ⋄ Case 8. D ( b, b ′ ) = η − (˜ λ )[ b, b ′ ] , for all b, b ′ ∈ B .Take b, b ′ ∈ B and write D ( b, b ′ ) = (cid:18) a ′′ m ′′ b ′′ (cid:19) . On the other hand, we havethat D ( b, b ′ ) = D ( qbq, b ′ ) = σ ( q ) D ( bq, b ′ ) + D ( q, b ′ ) bq = σ ( q ) σ ( b ) D ( q, b ′ ) + σ ( q ) D ( b, b ′ ) q Case 3 = σ ( q ) D ( b, b ′ ) q = (cid:18) b ′′ (cid:19) , which implies that a ′′ = 0 and m ′′ = 0. Given m ∈ M , from Lemma 5.11 (i) weobtain that σ ( m ) D ( b, b ′ ) = σ ([ p, m ]) D ( b, b ′ ) = D ( p, m )[ b, b ′ ] = ˜ λ m [ b, b ′ ] = ν σ ( m ) η − (˜ λ )[ b, b ′ ] . Taking into account the fact that σ ( m ) D ( b, b ′ ) = ν σ ( m ) b ′′ , we have that ν ( m )( b ′′ − η − (˜ λ )[ b, b ′ ]) = 0 , ∀ m ∈ M. Combining the bijectivity of ν with the fact that M is faithful as a B -module, weobtain that b ′′ = η − (˜ λ ), finishing the proof of Case 8 . ⋄ Case 9. D ( m, m ′ ) = 0, for all m, m ′ ∈ M .Given m, m ′ ∈ M , since [ m, m ′ ] = 0, Lemma 5.12 can be applied to get that D ( m, m ′ ) = pD ( m, m ′ ) q ∈ M . Next, fix m ∈ M and define a map ξ : M → M by ξ ( m ) = D ( m, m ), for all m ∈ M . The above calculation yields that ξ is well-defined. Moreover, ξ is an additive map satisfying ξ ( amb ) = f ( a ) ξ ( m ) b , for all a ∈ A , m ∈ M and b ∈ B . In fact, let a ∈ A , m ∈ M , and b ∈ B . Then ξ ( amb ) = D ( amb, m ) = σ ( a ) D ( mb, m ) + D ( a, m ) mb Case 6 = σ ( a ) σ ( m ) D ( b, m )+ σ ( a ) D ( m, m ) b + ˜ λ am mb Case 6 = − σ ( a ) σ ( m )˜ λ m b + f ( a ) D ( m, m ) b = f ( a ) ξ ( m ) b, since ˜ λ m = ν σ ( m ) η − (˜ λ ) and ν σ ( m ) ν σ ( m ) = 0. Now apply Condition (iv) tofind λ ′′ ∈ Z f σ ( A ) and µ ′′ ∈ Z g σ ( B ) such that ξ ( m ) = λ ′′ m + ν σ ( m ) µ ′′ , for all m ∈ M .Set λ m := λ ′′ + η ( µ ′′ ) ∈ Z f σ ( A ). Notice that ξ ( m ) = λ ′′ m + ν σ ( m ) µ ′′ = λ m m , forall m ∈ M . From here, proceeding as in the proof of Case 6 , we get that λ m = 0,which implies that ξ ( m ) = D ( m, m ) = 0, for all m ∈ M and m ∈ M .Taking into account what has already been proved, (15) can be rewritten as: D ( x, y ) = ˜ λ [ a x , a y ] + ˜ λ a x m y − ˜ λ a y m x + ˜ λ m x b y − ˜ λ m y b x + η − (˜ λ )[ b x , b y ] . UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 21 To finish, we will show that D = ∆ λ , where ∆ λ is the inner σ -biderivationassociated to the element λ = ˜ λ η − (˜ λ ) . Notice that it makes sense toconsider ∆ λ since λ ∈ Z σ ( T ). On the other hand, from[ x, y ] = [ a x , a y ] a x m y + m x b y − a y m x − m y b x [ b x , b y ] , it is easy to check that ˜ λ [ x, y ] = D ( x, y ), which concludes the proof of the theorem. (cid:3) σ -commuting maps of triangular algebras Let A be an algebra and σ an automorphism of A . Notice that in terms of the σ -commutator, a linear map Θ of A is σ -commuting if it satisfies( σ -c1) [ x, Θ( x )] σ = 0 , ∀ x ∈ A . A linearization of ( σ -c1) gives the equivalent condition( σ -c2) [ x, Θ( y )] σ = − [ y, Θ( x )] σ , ∀ x, y ∈ A . Given λ ∈ Z σ ( A ) and a σ -central map Ω of A , i.e., a linear map Ω : A → Z σ ( A ), itis straightforward to check that the map Θ : A → A defined by(16) Θ( x ) = λx + Ω( x ) , ∀ x ∈ A is σ -commuting. Motivated by this fact and keeping in mind the concept of propercommuting maps, we introduce the following notion: Definition 6.1.
Let A be an algebra and σ an automorphism of A . Maps of theform (16) will be called proper σ -commuting maps .In this section, we will investigate when a σ -commuting map of a partible trian-gular algebra is proper. Sufficient conditions will be given on a partible triangularalgebra for all its σ -commuting maps to be proper.Let us start with the following elementary result: Lemma 6.2.
Let T = Trian ( A, M, B ) be a triangular algebra and σ a partibleautomorphism of T . Let σ = φ z ¯ σ , where φ z ( x ) = z − xz , for all x ∈ T ; and ¯ σ isan automorphism of T such that ¯ σ ( A ) = A , ¯ σ ( M ) = M , ¯ σ ( B ) = B . Then a linearmap Θ :
T → T is σ -commuting if and only if ¯Θ( x ) = z Θ( x ) , for all x ∈ T , is ¯ σ -commuting. Given a partible triangular algebra T = Trian ( A, M, B ), the result above allowsus to assume that the automorphisms σ of T satisfying that σ ( A ) = A , σ ( M ) = M , σ ( B ) = B , and therefore they are written as in (1). Theorem 6.3.
Let T = Trian ( A, M, B ) be a partible triangular algebra and σ anautomorphism of T . Then every σ -commuting map Θ of T is of the following form: ( σ -cm)Θ (cid:18) a mb (cid:19) = δ ( a ) + δ ( m ) + δ ( b ) δ (1 A ) m − ν σ ( m ) µ (1 A ) µ ( a ) + µ ( m ) + µ ( b ) , where δ : A → A, δ : M → Z f σ ( A ) , δ : B → Z f σ ( A ) ,µ : A → Z g σ ( B ) , µ : M → Z g σ ( B ) , µ : B → B, are linear maps such that (i) δ is an f σ -commuting map of A , (ii) µ is a g σ -commuting map of B , (iii) δ ( a ) m − ν σ ( m ) µ ( a ) = f σ ( a )( δ (1 A ) m − ν σ ( m ) µ (1 A )) , (iv) ν σ ( m ) µ ( b ) − δ ( b ) m = ( ν σ ( m ) µ (1 B ) − δ (1 B ) m ) b , (v) δ ( m ) m = ν σ ( m ) µ ( m ) , (vi) δ (1 A ) m − ν σ ( m ) µ (1 A ) = ν σ ( m ) µ (1 B ) − δ (1 B ) m ,for all a ∈ A , m ∈ M and b ∈ B .Proof. Let Θ be a σ -commuting map of T ; write:Θ (cid:18) a mb (cid:19) = δ ( a ) + δ ( m ) + δ ( b ) τ ( a ) + τ ( m ) + τ ( b )) µ ( a ) + µ ( m ) + µ ( b ) , for all (cid:18) a mb (cid:19) ∈ T . In what follows, we will apply equations ( σ -c1) and ( σ -c2)many times with appropriate elements of T . Apply ( σ -c1) with x = (cid:18) a (cid:19) toget that(17) [ a, δ ( a )] f σ = 0 , f σ ( a ) τ ( a ) = 0 , for all a ∈ A . This shows (i) and implies that(18) τ (1 A ) = 0 , by letting a = 1 A in the second equation of (17). A second application of ( σ -c1)with x = (cid:18) b (cid:19) proves (ii), and also that(19) τ (1 B ) = 0 . Next, take x = (cid:18) a b (cid:19) and y = q in ( σ -c2) to obtain (using (19)) that(20) [ a, δ (1 B )] f σ = 0 , [ b, µ (1 B )] g σ = 0 , (21) − τ (1 B ) b − τ ( a ) − τ ( b ) = 0for all a ∈ A and b ∈ B . By (19) equation (21) becomes(22) τ ( a ) + τ ( b ) = 0 , for all a ∈ A and b ∈ B . Next, let a = 1 A (respectively, b = 1 B ) in (22) and apply(18) (respectively, (19)) to obtain that(23) τ ( a ) = 0 , τ ( b ) = 0 , for all a ∈ A and b ∈ B . On the other hand, (v) follows from ( σ -c1) by considering x = (cid:18) m (cid:19) . Now letting x = (cid:18) a (cid:19) and y = (cid:18) m (cid:19) in ( σ -c2) produces UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 23 the following equations:[ a, δ ( m )] f σ = 0 , (24) f σ ( a ) τ ( m ) = δ ( a ) m − ν σ ( m ) µ ( a ) , (25)for all a ∈ A and m ∈ M . Note that (24) says that δ ( M ) ⊆ Z f σ ( A ). Let a = 1 A in (25) to obtain that(26) τ ( m ) = δ (1 A ) m − ν σ ( m ) µ (1 A ) , for all m ∈ M . Notice that (iii) follows from (25) and (26).Now taking x = (cid:18) b (cid:19) and y = (cid:18) m (cid:19) in ( σ -c2), we get that[ b, µ ( m )] g σ = 0 , (27) − τ ( m ) b + ν σ ( m ) µ ( b ) − δ ( b ) m = 0 , (28)for all m ∈ M and b ∈ B . Equation (27) gives µ ( M ) ⊆ Z g σ ( B ). An application of(27) in (28) allows us to write that(29) τ ( m ) b = ν σ ( m ) µ ( b ) − δ ( b ) m, for all m ∈ M . Letting b = 1 B in (29) we obtain that(30) τ ( m ) = ν σ ( m ) µ (1 B ) − δ (1 B ) m, for all m ∈ M . From (29) and (30) we prove (iv). On the other hand, (vi) followsfrom (26) and (30). Next notice that an application of (23) and (26) gives: τ ( a ) + τ ( m ) + τ ( b ) = δ (1 A ) m − ν σ ( m ) µ (1 A ) , as desired. It remains to show that µ ( A ) ⊆ Z g σ ( B ) and δ ( B ) ⊆ Z f σ ( A ), whichfollows from an application of ( σ -c2) with x = (cid:18) a (cid:19) and y = (cid:18) b (cid:19) ,finishing the proof. (cid:3) In what follows, we will use this theorem without further mention. In otherwords, whenever we are given a σ -commuting map Θ of a partible triangular al-gebra T , we will assume that Θ is of the form ( σ -cm). Our next result providesa characterization of the properness of a σ -commuting map of T in terms of itsbehavior with the σ -center of T . Theorem 6.4.
Let T = Trian ( A, M, B ) be a partible triangular algebra and σ an automorphism of T . Then for every σ -commuting map Θ of T , the followingconditions are equivalent: (i) Θ is a proper σ -commuting map. (ii) µ ( A ) ⊆ π B (Z σ ( T )) , δ ( B ) ⊆ π A (Z σ ( T )) and δ ( m ) 0 µ ( m ) ∈ Z σ ( T ) , ∀ m ∈ M. (iii) δ (1 A ) ∈ π A (Z σ ( T )) , µ (1 A ) ∈ π B (Z σ ( T )) and δ ( m ) 0 µ ( m ) ∈ Z σ ( T ) , ∀ m ∈ M. In order to prove the theorem above, we need a preliminary result.
Lemma 6.5.
Let T = Trian ( A, M, B ) be a triangular algebra and σ an automor-phism of T such that σ ( A ) = A , σ ( M ) = M and σ ( B ) = B . Then for every σ -commuting map Θ of T , it follows that [ A, A ] ⊆ µ − ( π B (Z σ ( T ))) ✁ A, [ B, B ] ⊆ δ − ( π A (Z σ ( T ))) ✁ B. Proof.
Let us show, for example, that [
A, A ] ⊆ µ − ( π B (Z σ ( T ))) ✁ A . The othercondition can be proved analogously. Set a, a ′ ∈ A and m ∈ M . Applications ofTheorem 6.3 (iii) give the following: δ ( a ′ a ) m − ν σ ( m ) µ ( a ′ a ) = f σ ( a ′ a )( δ (1 A ) m − ν σ ( m ) µ (1 A ))(31) = f σ ( a ′ )( δ ( a ) m − ν σ ( m ) µ ( a )) ,δ ( aa ′ ) m − ν σ ( m ) µ ( aa ′ ) = f σ ( aa ′ )( δ (1 A ) m − ν σ ( m ) µ (1 A ))(32) = f σ ( a )( f σ ( a ′ ) δ (1 A ) m − f σ ( a ′ ) ν σ ( m ) µ (1 A ))= f σ ( a )( δ (1 A ) a ′ m − ν σ ( a ′ m ) µ (1 A ))= δ ( a ) a ′ m − ν σ ( a ′ m ) µ ( a ) . From (31) and (32) we get that δ ([ a, a ′ ]) m − ν σ ( m ) µ ([ a, a ′ ]) + f σ ( a ′ ) δ ( a ) m − δ ( a ) a ′ m + ν σ ( a ′ m ) µ ( a ) − f σ ( a ′ ) ν σ ( m ) µ ( a ) = 0 . Taking into account the facts that ν σ ( a ′ m ) = f σ ( a ′ ) ν σ ( m ) and δ ( A ) ⊆ Z f σ ( A ),the identity above becomes: δ ([ a, a ′ ]) m = ν σ ( m ) µ ([ a, a ′ ]) , which allows us to conclude that δ ([ a, a ′ ]) 0 µ ([ a, a ′ ]) ∈ Z σ ( T ) , for all a, a ′ ∈ A . This implies that µ ([ A, A ]) ⊆ π B (Z σ ( T )). Therefore [ A, A ] ⊆ µ − ( π B (Z σ ( T ))).Let us show now that µ − ( π B (Z σ ( T ))) is an ideal of A . To this end, take a ∈ µ − ( π B (Z σ ( T ))) and a ′ ∈ A , m ∈ M . Applying (31), taking into account thefact that ν σ ( m ) µ ( a ) = η ( µ ( a )) m , which makes sense since µ ( a ) ∈ π B (Z σ ( T )),we get that ( δ ( a ′ a ) − f σ ( a ′ ) δ ( a ) + f σ ( a ′ ) η ( µ ( a ))) m = ν σ ( m ) µ ( a ′ a ) . This yields δ ( a ′ a ) − f σ ( a ′ ) δ ( a ) + f σ ( a ′ ) η ( µ ( a )) 0 µ ( a ′ a ) ∈ Z σ ( T ) , which says that µ ( a ′ a ) ∈ π B (Z σ ( T )), that is, a ′ a ∈ µ − ( π B (Z σ ( T ))). To show that aa ′ ∈ µ − ( π B (Z σ ( T ))), proceed as before, this time applying (32). This proves that µ − ( π B (Z σ ( T ))) is an ideal of A , as desired. (cid:3) UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 25 Proof of Theorem 6.4.
We are going to show that (i) ⇔ (iii) and (ii) ⇔ (iii).Let Θ be a σ -commuting map of T .(i) ⇒ (iii). Suppose that Θ is proper. Then there exist λ ∈ Z σ ( T ) and a linearmap Ω : T → Z σ ( T ) such that Θ( x ) = λx + Ω( x ), for all x ∈ T . We can express λ as follows: λ = a λ η − ( a λ ) , for some a λ ∈ π A (Z σ ( T )). Take m ∈ M and compute Θ( x ) for x = (cid:18) m (cid:19) ; letus assume that Ω( x ) = a m η − ( a m ) , where a m ∈ π A (Z σ ( T )). Then, wehave that(33) Θ( x ) = a m a λ mη − ( a m ) . On the other hand, we can write(34) Θ( x ) = δ ( m ) δ (1 A ) m − ν σ ( m ) µ (1 A ) µ ( m ) . Therefore, (33) and (34) allow us to conclude that a m = δ ( m ) , µ ( m ) = η − ( a m ) , (35) a λ m = δ (1 A ) m − ν σ ( m ) µ (1 A ) . (36)From (35) we get that δ ( m ) 0 µ ( m ) ∈ Z σ ( T ) , for all m ∈ M . Note that (36) becomes(37) a λ m = δ (1 A ) m − ν σ ( m ) µ (1 A ) , ∀ m ∈ M, by an application of (35). Taking into account the fact that a λ ∈ Z σ ( T ), (37) canbe rewritten as δ (1 A ) m = ν σ ( m )( µ (1 A ) + η − ( a λ )) , ∀ m ∈ M, which implies that δ (1 A ) 0 µ (1 A ) + η − ( a λ ) ∈ Z σ ( T ) , for all m ∈ M . In particular, we have that δ (1 A ) ∈ π A (Z σ ( T )). On the otherhand, writing (37) as( δ (1 A ) − a λ ) m = ν σ ( m ) µ (1 A ) , ∀ m ∈ M, we obtain that δ (1 A ) − a λ µ (1 A ) ∈ Z σ ( T ) , for all m ∈ M . Thus: µ (1 B ) ∈ π B (Z σ ( T )), concluding the proof of (iii).(iii) ⇒ (i).Let us start by noticing that δ (1 A ) ∈ π A (Z σ ( T )) and µ (1 A ) ∈ π B (Z σ ( T ))allow us to write η ( µ (1 A )) and η − ( δ (1 A )), respectively. Thus, it makes sense toconsider the element λ := δ (1 A ) − η ( µ (1 A )) 0 η − ( δ (1 A )) − µ (1 A ) ∈ T . Note that λ ∈ Z σ ( T ) since( δ (1 A ) − η ( µ (1 A ))) m = ν σ ( m )( η − ( δ (1 A )) − µ (1 A )) , ∀ m ∈ M. Next, we claim that Ω( x ) := Θ( x ) − λx ∈ Z σ ( T ), for all x ∈ T . Given x = (cid:18) a x m x b x (cid:19) ∈ T , we have that λx = ( δ (1 A ) − η ( µ (1 A ))) a x ( δ (1 A ) − η ( µ (1 A ))) m x ( η − ( δ (1 A )) − µ (1 A )) b x , which implies thatΩ( x ) = δ ( a x ) − δ (1 A ) a x + η ( µ (1 A )) a x µ ( a x ) (38) + δ ( b x ) − η − ( δ (1 A )) b x + µ (1 A ) b x ) µ ( b x ) − η − ( δ (1 A )) b x + µ (1 A ) b x + δ ( m x ) 0 µ ( m x ) . To finish, we will show that the three terms in (38) are indeed in Z σ ( T ). Noticethat the last term belongs to Z σ ( T ), by hypothesis. Given m ∈ M , it is enough toshow that ( δ ( a x ) − δ (1 A ) a x + η ( µ (1 A )) a x ) m − ν σ ( m ) µ ( a x ) = 0 ,δ ( b x ) m − ν σ ( m )( µ ( b x ) − η − ( δ (1 A )) b x + µ (1 A ) b x )= 0 . Let us start by proving the first identity above. Apply Theorem 6.3 (iii) to get that( δ ( a x ) − δ (1 A ) a x + η ( µ (1 A )) a x ) m − ν σ ( m ) µ ( a x ) = f σ ( a x ) δ (1 A ) m − f σ ( a x ) ν σ ( m ) µ (1 A ) − δ (1 A ) a x m + η ( µ (1 A )) a x m =[ a x , δ (1 A )] f σ m + η ( µ (1 A )) a x m − ν σ ( a x m ) µ (1 A ) = 0 , since δ (1 A ) ∈ π A (Z σ ( T )) ⊆ Z f σ ( A ) and µ (1 A ) ∈ π B (Z σ ( T )), by (iii). UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 27 It remains to prove the second of the two displayed identities above. ApplyTheorem 6.3 (iv) and (vi) to obtain that δ ( b x ) m − ν σ ( m )( µ ( b x ) − η − ( δ (1 A )) b x + µ (1 A ) b x ) =( δ (1 B ) m − ν σ ( m ) µ (1 B )) b x + ν σ ( m ) η − ( δ (1 A )) b x − ν σ ( m ) µ (1 A ) b x =( ν σ ( m ) µ (1 A ) − δ (1 A ) m ) b x + ν σ ( m ) η − ( δ (1 A )) b x − ν σ ( m ) µ (1 A ) b x =( ν σ ( m ) η − ( δ (1 A )) − δ (1 A ) m ) b x = 0 , since δ (1 A ) ∈ π A (Z σ ( T )).(ii) ⇒ (iii).It is only necessary to show that δ (1 A ) ∈ π A (Z σ ( T )). Given m ∈ M , noticethat δ (1 B ) ∈ π A (Z σ ( T )) allows us to write δ (1) m = ν σ ( m ) η − ( δ (1 B )). Keepingthis fact in mind, an application of Theorem 6.3 (vi) gives the following: δ (1 A ) m = ν σ ( m )( µ (1 A ) + µ (1 B ) − η − ( δ (1 B ))) , which implies that δ (1 A ) 0 µ (1 A ) + µ (1 B ) − η − ( δ (1 B )) ∈ Z σ ( T ) . Thus: δ (1 A ) ∈ π A (Z σ ( T )), proving (iii).(iii) ⇒ (ii).From µ (1 A ) ∈ π B (Z σ ( T )), we get that 1 A ∈ µ − ( π B (Z σ ( T ))), which is anideal of A by Lemma 6.5. Hence: µ − ( π B (Z σ ( T ))) = A , and therefore µ ( A ) ⊆ π B (Z σ ( T )). In order to show that δ ( B ) ⊆ π A (Z σ ( T )), we first need to prove that δ (1 B ) ∈ π A (Z σ ( T )). To this end, apply Theorem 6.3 (vi), taking into account thefact that δ (1 A ) ∈ π A (Z σ ( T )), to obtain that δ (1 B ) m = ν σ ( m )( µ (1 B ) + µ (1 A )) − δ (1 A ) m = ν σ ( m )( µ (1 B ) + µ (1 A ) − η − ( δ (1 A ))) , ∀ m ∈ M. This implies that δ (1 B ) 0 µ (1 A ) + µ (1 B ) − η − ( δ (1 A )) ∈ Z σ ( T ) , and therefore δ (1 B ) ∈ π A (Z σ ( T )). Now take b ∈ B , m ∈ M and apply Theorem6.3 (iv), taking into account the fact that δ (1 B ) m = ν σ ( m ) η − ( δ (1 B )), to getthat δ ( b ) m = ν σ ( m ) µ ( b ) − δ (1 B ) mb + ν σ ( m ) µ (1 B ) b = ν σ ( m )( µ ( b ) − η − ( δ (1 B )) b + µ (1 B ) b ) , which says that δ ( b ) 0 µ ( b ) + µ (1 B ) b − η − ( δ (1 B )) b ∈ Z σ ( T ) . Thus: δ ( b ) ∈ π A (Z σ ( T ), concluding the proof of Theorem 6.4. (cid:3) Our last result states sufficient conditions on a triangular algebra T to guaranteethat all its σ -commuting maps are proper. Theorem 6.6.
Let T = Trian ( A, M, B ) be a partible triangular algebra and σ anautomorphism of T . Suppose that T satisfies the following conditions: (i) Either Z f σ ( A ) = π A (Z σ ( T )) or B = [ B, B ] . (ii) Either Z g σ ( B ) = π B (Z σ ( T )) or A = [ A, A ] . (iii) There exists m ∈ M such that Z σ ( T ) = (cid:26)(cid:18) a b (cid:19) ∈ T , such that am = ν σ ( m ) b (cid:27) . Then every σ -commuting map of T is proper.Proof. Let Θ be a σ -commuting map of T . By Theorem 6.3 we have that µ ( A ) ⊆ Z g σ ( B ). If Z g σ ( B ) = π B (Z σ ( T )), we can conclude that µ ( A ) ⊆ π B (Z σ ( T )). Onthe other hand, if A = [ A, A ], apply Lemma 6.5 to get that A ⊆ µ − ( π B (Z σ ( T ))),that is, µ ( A ) ⊆ π B (Z σ ( T )). Reasoning as above, now using the fact that δ ( B ) ⊆ Z f σ ( A ) and applying (i), we get that δ ( B ) ⊆ π A (Z σ ( T )).In view of Theorem 6.4, to prove that Θ is proper, it remains to show that(39) δ ( m ) 0 µ ( m ) ∈ Z σ ( T ) , ∀ m ∈ M. From Theorem 6.3 we have that δ ( M ) ⊆ Z f σ ( A ) and µ ( M ) ⊆ Z g σ ( B ), whichimply that(40) δ ( m ) m = ν σ ( m ) µ ( m ) , ∀ m ∈ M. In particular, for m = m we get that δ ( m ) m = ν σ ( m ) µ ( m ), which gives δ ( m ) 0 µ ( m ) ∈ Z σ ( T ) , by an application of (iii). Therefore: m , δ ( m ) 0 µ ( m ) σ = 0 , ∀ m ∈ M. Expanding the above product, we get that(41) δ ( m ) m = ν σ ( m ) µ ( m ) , ∀ m ∈ M. Given m ∈ M , applying (40) with m + m and making use of (41), we get that δ ( m ) m = ν σ ( m ) µ ( m ) , which, by an application of (iii), proves (39), concluding the proof of the theorem. (cid:3) Remark 6.7.
Let σ be an automorphism of an algebra A . Notice that, as hap-pened with biderivations and commuting maps, the study of the properness of σ -commuting maps of A can be reduced to the study of the innerness of its σ -biderivations. Let Θ be any σ -commuting map of A ; applying Lemma 5.5 (i)and ( σ -c2), it is not difficult to prove that the map D Θ : A × A → A given by D Θ ( x, y ) = [ x, Θ( y )] σ , for all x, y ∈ A , is a σ -biderivation of A . Let us assume now UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 29 that D Θ is inner, i.e., D Θ ( x, y ) = λ [ x, y ], for some λ ∈ Z σ ( A ). This implies thatthe map Ω Θ ( y ) := Θ( y ) − λy is σ -central; in fact:[ x, Ω Θ ( y )] σ = [ x, Θ( y ) − λy ] σ = σ ( x )Θ( y ) − σ ( x ) λy − Θ( y ) x + λyx = [ x, Θ( y )] σ − λxy + λyx = [ x, Θ( y )] σ − λ [ x, y ]= D Θ ( x, y ) − λ [ x, y ] = 0 , for all x ∈ A . Therefore, we can conclude that Θ is a proper σ -commuting map.Suppose now that A is a triangular algebra and notice that D Θ ( p, p ) = 0. Inother words, D Θ is one of the σ -biderivations studied in Theorem 5.9. Taking intoaccount the calculations above, under the assumptions of Theorem 5.9, we canguarantee that every σ -commuting map of A is proper. Nevertheless, Theorem 6.6above gives us more precise information when dealing directly with σ -commutingmaps; it is a generalization of [11, Theorem 2].The next result shows that, under some mild conditions, the identity is the onlycommuting automorphism of a triangular algebra. Theorem 6.8.
Let T = Trian ( A, M, B ) be a partible triangular algebra. If σ is acommuting automorphism of T , then σ = Id T .Proof. Let σ be a commuting automorphism of T and consider the elements x = (cid:18) a (cid:19) and y = (cid:18) m (cid:19) ; since [ x, σ ( y )] + [ y, σ ( x )] = 0, we get that aν σ ( m ) − f σ ( a ) m = 0, for all a ∈ A and m ∈ M . In particular, for a = 1 A we get that ν σ = Id M , which implies that( a − f σ ( a )) M = 0 , M ( b − g σ ( b )) = 0 , for all a ∈ A , b ∈ B . Therefore: f σ = Id A and g σ = Id B , since M is a faithful( A, B )-bimodule. (cid:3)
We close the section by showing that Posner’s theorem also holds for partibletriangular algebras.
Theorem 6.9.
Let T = Trian ( A, M, B ) be a partible triangular algebra and σ anautomorphism of T . If a σ -derivation d of T is σ -commuting, then d = 0 .Proof. Let d be a σ -derivation of T which is σ -commuting. By [22, Theorem 3.12],we know that d is of the following form: d (cid:18) a mb (cid:19) = d A ( a ) f σ ( a ) m d − m d b + ξ d ( m ) d B ( b ) , where d A is an f σ -derivation of A , d B is a g σ -derivation of B , m d is a fixed elementof M , and ξ d : M → M is a linear map which satisfies(42) ξ d ( am ) = d A ( a ) m + f σ ( a ) ξ ( m ) , ξ ( mb ) = ξ ( m ) b + ν σ ( m ) d B ( b ) , for all a ∈ A , b ∈ B , m ∈ M . An application of ( σ -c2) with x = (cid:18) a (cid:19) and y = (cid:18) m (cid:19) gives that f σ ( a ) ξ ( m ) − d A ( a ) m = 0, for all a ∈ A and m ∈ M . Inparticular, substituting a = 1 A we get that ξ = 0. Then from (42) we obtain that d A ( a ) M = 0 , ν σ ( M ) d B ( b ) = 0 , for all a ∈ A and b ∈ B . Since ν σ is bijective and M is a faithful ( A, B )-bimodule,we see that d A = d B = 0. Applying ( σ -c1) with x = p , we get that m d = 0,finishing the proof. (cid:3) Corollary 6.10. If d is a commuting derivation of a partible triangular algebra,then d = 0 . Some facts about ( α, β ) -biderivations, ( α, β ) -commuting maps andgeneralized matrix algebras. In 2011, Xiao and Wei [34] introduced a generalization of σ -biderivations and σ -commuting maps that they named ( α, β )-biderivations and ( α, β )-commuting maps.In their paper, they studied ( α, β )-biderivations and ( α, β )-commuting maps ofnest algebras. In the present paper, we have investigated σ -biderivations and σ -commuting maps since, as will be shown below, the study of ( α, β )-biderivations(respectively, ( α, β )-commuting maps) of any algebra can be reduced to the studyof its σ -biderivations (respectively, σ -commuting maps).Let A be an algebra and α, β, σ automorphisms of A . Recall that a linear map d : A → A is called an ( α, β ) -derivation of A if it satisfies d ( xy ) = d ( x ) β ( y ) + α ( x ) d ( y ) , ∀ x, y ∈ A . A bilinear map D : A × A → A is said to be an ( α, β ) -biderivation of A if itis an ( α, β )-derivation in each argument. An ( α, β ) -commuting map of A is alinear map Θ satisfying that Θ( x ) α ( x ) = β ( x )Θ( x ), for all x ∈ A . Clearly, every σ -derivation can be seen as an ( α, β ) -derivation with α = Id A and β = σ . Thesame can be said for σ -biderivations and σ -commuting maps.It is straightforward to check that every ( α, β )-biderivation D (respectively,( α, β )-commuting map Θ) of A gives rise to an α − β -biderivation (respectively, α − β -commuting map) by considering α − D (respectively, α − Θ). Accordingly,it is sufficient to restrict attention to σ -biderivations (respectively, σ -commutingmaps).In the last few years, many results on maps of triangular algebras have beenextended to the setting of generalized matrix algebras (GMAs); see, for example,[17, 18, 28, 34] and references therein. GMAs were introduced by Sands [32] in theearly 1970s. He ended up with these structures during his study of radicals of ringsin Morita contexts. Specifically, a Morita context ( A, B, M, N, Φ MN , Ψ NM ) , consists of two R -algebras A and B , two bimodules A M B and B N A , and two bi-module homomorphisms Φ MN : M ⊗ B N → A and Ψ NM : N ⊗ A M → B such thatthe following diagrams commute. M ⊗ B N ⊗ A M Φ MN ⊗ I M −−−−−−→ A ⊗ A M y ∼ = y M ⊗ B B ∼ = −−−−→ M N ⊗ A M ⊗ B N Ψ NM ⊗ I N −−−−−−→ B ⊗ B N y ∼ = y N ⊗ A A ∼ = −−−−→ N UTOMORPHISMS, σ -BIDERIVATIONS AND σ -COMMUTING MAPS OF TA’S 31 Let (
A, B, M, N, Φ MN , Ψ NM ) be a Morita context, where at least one of the twobimodules is nonzero. The set G = (cid:18) A MN B (cid:19) = (cid:26)(cid:18) a mn b (cid:19) : a ∈ A, m ∈ M, n ∈ N, b ∈ B (cid:27) , can be endowed with an R -algebra structure under the usual matrix operations.We will call G a generalized matrix algebra . Note that any triangular algebracan be seen as a generalized matrix algebra with N = 0.The extension of our results proceeds through the study of automorphisms ofGMAs. However, the description of automorphisms of GMAs is still an open prob-lem. (See [28, Question 4.4]) Acknowledgements
The second and third authors were supported by a grant from the Natural Sci-ences and Engineering Research Council (Canada). The first and third authorswere supported by the Spanish MEC and Fondos FEDER jointly through projectMTM2010-15223, and by the Junta de Andaluc´ıa (projects FQM-336, FQM2467and FQM7156). The third author thanks Professor Nantel Berger´on, the Depart-ment of Mathematics at the University of Toronto and the Fields Institute for hervisit from August 2012 to September 2013.
References [1]
P. N. ´Anh, L. van Wyk : Automorphism groups of generalized triangular matrix rings.
LinearAlgebra Appl. (2011), 1018–1026.[2]
G. P. Barker, T. P. Kezlan : Automorphisms of algebras of upper triangular matrices.
Arch. Math. (1990), 38–43.[3] D. Benkoviˇc : Biderivations of triangular algebras.
Linear Algebra Appl. (2009), 1587–1602.[4]
M. Breˇsar : On skew-commuting mappings of rings,
Bull. Austral. Math. Soc. (1993),291–296.[5] M. Breˇsar : On certain pairs of functions of semiprime rings.
Proc. Amer. Math. Soc. (1994), 709–713.[6]
M. Breˇsar : On Generalized Biderivations and Related Maps.
J. Algebra (1995), 764–786.[7]
M. Breˇsar : Commuting maps: a survey.
Taiwanese J. Math. (2004) 361–397.[8] M. Breˇsar, W. S. Martindale 3rd, C. R. Miers : Centralizing maps in prime rings withinvolution.
J. Algebra (1993), 342–357.[9]
S. U. Chase : A generalization of the ring of triangular matrices.
Nagoya Math. J. (1961),13–25.[10] W. S. Cheung : Mappings on triangular algebras . Ph.D. Dissertation, University of Victoria,2000.[11]
W. S. Cheung : Commuting maps of triangular algebras.
J. London Math. Soc. (2001),117–127.[12] E. Christensen : Derivations of nest algebras.
Math. Ann. (1977), 155–161.[13]
S. P. Coelho : Automorphism groups of certain algebras of triangular matrices.
Arch.Math. (1993), no. 2, 119–123.[14] S. P. Coelho, C. P. Milies : Derivations of upper triangular matrix rings.
Linear AlgebraAppl. (1993), 263–267.[15]
C. Cao, Z. Xian : Multiplicative semigroup automorphisms of upper triangular matrices overrings.
Linear Algebra Appl. (1998), 85–90.[16]
K. R. Davidson : Nest algebras in: Pitman Research Notes in Mathematical Series, vol. 191,Longman, London/ New York, 1988.[17]
Y. Du, Y. Wang : Lie derivations of generalized matrix algebras.
Linear Algebra Appl. (2012), 2719–2726. [18]
Y. Du, Y. Wang : Biderivations of generalized matrix algebras.
Linear Algebra Appl. (2013), 4483–4499.[19]
D. R. Farkas, G. Letzter : Ring theory from symplectic geometry.
J. Pure Applied Algebra (1998), 401–416.[20]
B. E. Forrest, L. W. Marcoux : Derivations of triangular Banach algebras.
Indiana Univ.Math. J. (1996), 441–462.[21] A. Haghany, K. Varadarajan : Study of formal triangular matrix rings.
Comm. Algebra (1999), 5507–5525.[22] D. Han, F. Wei : Jordan ( α, β )-derivations on triangular algebras and related mappings.
Linear Algebra Appl. (2011), 259–284.[23]
M. Harada : Hereditary semi-primary rings and triangular matrix rings.
Nagoya Math. J. (1966), 463–484.[24] I. N. Herstein : Lie and Jordan structures in simple, associative rings.
Bull. Amer. Math.Soc. (1961), 517–531.[25] S. Jøndrup : Automorphisms and derivations of upper triangular matrix rings.
Linear AlgebraAppl. (1995), 205–218.[26]
R. Khazal, S. Dˇascˇalescu, L. V. Wyk : Isomorphism of generalized triangular matrix ringsand recovery of titles.
Int. J. Math. Sci. (2003), 553–538.[27] T. P. Kezlan : A note on Algebra Automorphisms of Triangular Matrices over Commutativerings.
Linear Algebra Appl. (1990), 181–184.[28]
Y. B. Li, F. Wei : Semi-centralizing maps of generalized matrix algebras.
Linear AlgebraAppl. (2012), 1122–1153.[29]
F. Lu : Isomorphisms of subalgebras of nest algebras.
Proc. Amer. Soc.
No. 12 (2003),3883–3892.[30]
F. Lu : Multiplicative mappings of operator algebras.
Linear Algebra Appl. (2002), 283–291.[31]
E. C. Posner : Derivations in prime rings.
Proc. Amer. Math. Soc. (1957) 1093–1100.[32] A. D. Sands : Radicals and Morita contexts.
J. Algebra (1973), 335–345.[33] Skosyrskii : Strongly prime noncommutative Jordan algebras.
Trudy Inst. Mat. (Novosibirsk) (1989), 131–164 (in Russian).[34] Z. Xiao, F. Wei : Commuting mappings of generalized matrix algebras.
Linear Algebra Appl. (2010) 2178–2197.[35]
W. Yang, J. Zhu : Characterizations of additive (generalized) ξ -Lie ( α, β )-derivations ontriangular algebras. Linear and Multilinear A. (2013) no. 6, 811–830.[36] W.-Y. Yu, J.-H. Zhang : σ -biderivations and σ -commuting maps on nest algebras. ActaMath. Sin. (Chin. Ser.) (2007) 1391–1396.[37] J.-H. Zhang : Jordan derivations of nest algebras.
Acta Math. Sin. (Chin. Ser.) (1998)205–212.[38] J.-H. Zhang, S. Feng, H.-X. Li, R.-H. Wu : Generalized biderivations of nest algebras.
Linear Algebra Appl. (2006) 225–233.[39]
Y. Zhao, D. Wang, R. Yao : Biderivations of upper triangular matrix algebras over commu-tative rings.
Int. J. Math. Game Theory Algebra (2009) no. 6, 473–778.(1) Departamento de ´Algebra, Geometr´ıa y Topolog´ıa, Universidad de M´alaga, M´alaga,Spain (2)
University of Toronto, Toronto, ON, Canada (3)
Department of Mathematics and Applied Mathematics. University of Cape Town,Cape Town, South Africa
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