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Generalized Lie n-derivations on arbitrary triangular algebras

  • He Yuan EMAIL logo and Zhuo Liu
Published/Copyright: October 19, 2023
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Open Mathematics
From the journal Open Mathematics Volume 21 Issue 1

Abstract

In this study, we consider generalized Lie n -derivations of an arbitrary triangular algebra T through the constructed triangular algebra T 0 , where T 0 is constructed using the notion of maximal left (right) ring of quotients.

Keywords: triangular algebra; maximal left ring of quotients; Lie n-derivation; generalized Lie n-derivation
MSC 2010: 16W25; 16R50

1 Introduction

Let R be a commutative ring with identity, A be an algebra over R , and Z ( A ) be the center of A . A linear mapping d : A → A is said to be a derivation if it satisfies d ( x y ) = d ( x ) y + x d ( y ) for all x , y ∈ A . A linear mapping g : A → A is said to be a generalized derivation with an associated derivation d on A if g ( x y ) = g ( x ) y + x d ( y ) for all x , y ∈ A . We say that a linear mapping L : A → A is a Lie derivation (resp. Lie triple derivation) if L ( [ x , y ] ) = [ L ( x ) , y ] + [ x , L ( y ) ] (resp. L ( [ [ x , y ] , z ] ) = [ [ L ( x ) , y ] , z ] + [ [ x , L ( y ) ] , z ] + [ [ x , y ] , L ( z ) ] ) holds for all x , y , z ∈ A . We say that a linear mapping G : A → A is a generalized Lie derivation (resp. generalized Lie triple derivation) with an associated Lie derivation (resp. Lie triple derivation) L on A if G ( [ x , y ] ) = [ G ( x ) , y ] + [ x , L ( y ) ] (resp. G ( [ [ x , y ] , z ] ) = [ [ G ( x ) , y ] , z ] + [ [ x , L ( y ) ] , z ] + [ [ x , y ] , L ( z ) ] ) holds for all x , y , z ∈ A .

We now turn our attention to a more general class of mappings. Let us define the following sequence of polynomials:

p 1 ( x 1 ) = x 1 , p 2 ( x 1 , x 2 ) = [ p 1 ( x 1 ) , x 2 ] = [ x 1 , x 2 ] , p 3 ( x 1 , x 2 , x 3 ) = [ p 2 ( x 1 , x 2 ) , x 3 ] = [ [ x 1 , x 2 ] , x 3 ] , ⋯ ⋯ p n ( x 1 , x 2 , … , x n ) = [ p n − 1 ( x 1 , x 2 , … , x n − 1 ) , x n ] .

A Lie n-derivation is a linear mapping L : A → A , which satisfies the rule

L ( p n ( x 1 , x 2 , … , x n ) ) = ∑ k = 1 n p n ( x 1 , … , x k − 1 , L ( x k ) , x k + 1 , … , x n ) ,

for all x 1 , x 2 , … , x n ∈ A . One can give the definition of generalized Lie n -derivations in an analogous manner. A linear mapping G : A → A is called a generalized Lie n-derivation if there exists a Lie n -derivation L such that

G ( p n ( x 1 , x 2 , … , x n ) ) = p n ( G ( x 1 ) , x 2 , … , x n ) + ∑ k = 2 n p n ( x 1 , … , x k − 1 , L ( x k ) , x k + 1 , … , x n ) ,

for all x 1 , x 2 , … , x n ∈ A . We say that L is an associated Lie n -derivation of G .

In 2001, Cheung [1] initiated the study of mapping problems on triangular algebras. Cheung [2] obtained a description of Lie derivations of triangular algebras in 2003. Later, this result is extended to Lie triple derivations [3] and Lie n -derivations [4]. In 2011, Benkovič [5] studied the representations of generalized Lie derivations on triangular algebras. Ashraf and Jabeen [6] gave the sufficient conditions that every generalized Lie triple derivation of triangular algebras is the sum of a generalized derivation and a centralizing mapping sending all second commutators to zero in 2017. Later, Lin [7] extended the aforementioned results to the generalized Lie n -derivations on triangular algebras. Benkovič [8] studied the generalized Lie n -derivations again using commuting and centralizing mappings on triangular algebras.

In 1956, Utumi [9] gave the concept of the maximal left ring of quotients and he obtained that every unital ring has the maximal left ring of quotients. An introduction of the maximal left ring of quotients of rings can be found in [10,11]. Wang [12] constructed a triangular algebra U 0 from a given triangular algebra U , using the notion of maximal left (right) ring of quotients, and he gave a description of Lie derivations on arbitrary triangular algebras through the constructed triangular algebra U 0 in 2019. Ashraf et al. [13] extended this result to the generalized Lie triple derivations on arbitrary triangular algebras. In this study, we will extend Wang’s results to the generalized Lie n -derivations on arbitrary triangular algebras. In order to realize it, we first study the representations of the Lie n -derivations and the generalized Lie n -derivations on triangular algebras.

2 Lie n -derivations of arbitrary triangular algebras

Let A and B be the unital algebras, and let M be a unital ( A , B ) -bimodule, which is faithful as a left A -module and also as a right B -module. The set

T = Tri ( A , M , B ) = a m 0 b ∣ a ∈ A , m ∈ M , b ∈ B

is an associative algebra under the usual matrix operations. Let 1 A and 1 B be the identities of the algebras A and B , respectively, and then, 1 A 0 0 1 B is the identity of the triangular algebra T . Throughout this study, we shall use the following notation:

e = 1 A 0 0 0 , f = 0 0 0 1 B .

Accordingly, T can be written as T = e T e ⊕ e T f ⊕ f T f , where e T e is a subalgebra of T isomorphic to A , f T f is a subalgebra of T isomorphic to B , and e T f is a ( e T e , f T f ) -bimodule isomorphic to the bimodule M . Since a triangular algebra is a unital algebra, we see that every triangular algebra exists a maximal left ring of quotients.

We fix some notations as follows. Let A be an algebra. By Q l ( A ) , we denote the maximal left ring of quotients of A . The center of Q l ( A ) is denoted by C l ( A ) . Analogously, by Q r ( A ) , we denote the maximal right ring of quotients of A . The center of Q r ( A ) is denoted by C r ( A ) .

Let k be an integer and A be an algebra. If k x is nonzero for each nonzero x ∈ A , then A is said to be a k -torsion free algebra.

Theorem 2.1

Let T be a ( n − 1 ) -torsion free triangular algebra. Suppose that L : T → T is a Lie n-derivation such that e L ( f ) f = 0 . Then, L has the form:

(2.1) L a m 0 b = g A ( a ) + h B ( b ) f M ( m ) 0 h A ( a ) + g B ( b ) ,

where a ∈ A , m ∈ M , b ∈ B , and g A : A → A , g B : B → B , h A : A → B , h B : B → A , f M : M → M are the linear mappings satisfying

  1. g A is a Lie n-derivation on A, h A ( p n ( a 1 , a 2 , … , a n ) ) = 0 , p n ( h A ( a ) , b 1 , … , b n − 1 ) = 0 , and f M ( a m ) = g A ( a ) m − m h A ( a ) + a f M ( m ) , where a , a 1 , … , a n ∈ A , b 1 , … , b n − 1 ∈ B , m ∈ M ;

  2. g B is a Lie n-derivation on B, h B ( p n ( b 1 , b 2 , … , b n ) ) = 0 , p n ( h B ( b ) , a 1 , … , a n − 1 ) = 0 , and f M ( m b ) = m g B ( b ) − h B ( b ) m + f M ( m ) b , where b , b 1 , … , b n ∈ B , a 1 , … , a n − 1 ∈ A , m ∈ M .

Proof

Let us assume that

L a m 0 b = g A ( a ) + h B ( b ) + k A ( m ) f A ( a ) + f B ( b ) + f M ( m ) 0 h A ( a ) + g B ( b ) + k B ( m ) ,

where a ∈ A , m ∈ M , b ∈ B , and g A : A → A , g B : B → B , h A : A → B , h B : B → A , f M : M → M , f A : A → M , f B : B → M , k A : M → A , k B : M → B are the linear mappings. In view of [4, Lemma 3.2], we know that k A ( m ) = k B ( m ) = 0 and f A ( a ) = f B ( b ) = 0 for all a ∈ A , m ∈ M , and b ∈ B . Since L is a Lie n -derivation on T , it follows that

(2.2) L ( p n ( x 1 , x 2 , … , x n ) ) = p n − 1 ( [ L ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , L ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , L ( x i ) , x i + 1 , … , x n ) ,

for all x 1 , x 2 , … , x n ∈ T . In particular, we take

x 1 = a 1 0 0 0 , x 2 = a 2 0 0 0 , x 3 = a 3 0 0 0 , … , x n = a n 0 0 0

in (2.2), we easily obtain that g A is a Lie n -derivation on A and h A ( p n ( a 1 , a 2 , … , a n ) ) = 0 for all a 1 , a 2 , … , a n ∈ A . Similarly, we can see that g B is a Lie n -derivation on B and h B ( p n ( b 1 , b 2 , … , b n ) ) = 0 for all b 1 , b 2 , … , b n ∈ B . Let us choose

x 1 = a 1 0 0 0 , x 2 = 0 0 0 b 1 , x 3 = a 2 0 0 b 2 , … , x n = a n − 1 0 0 b n − 1

in (2.2). Then, we have

0 = p n g A ( a 1 ) 0 0 h A ( a 1 ) , 0 0 0 b 1 , x 3 , … , x n + p n a 1 0 0 0 , h B ( b 1 ) 0 0 g B ( b 1 ) , x 3 , … , x n = p n − 1 [ a 1 , h B ( b 1 ) ] 0 0 [ h A ( a 1 ) , b 1 ] , x 3 , … , x n .

This implies that

p n ( h A ( a 1 ) , b 1 , … , b n − 1 ) = p n ( h B ( b 1 ) , a 1 , … , a n − 1 ) = 0 ,

for all a 1 , … , a n − 1 ∈ A , b 1 , … , b n − 1 ∈ B .

Finally, if we choose

x 1 = a 0 0 0 , x 2 = 0 m 0 0 , x 3 = … = x n = 1 A 0 0 0

in (2.2), then it follows from [4, Lemma 3.3] that L ( x 3 ) = … = L ( x n ) ∈ Z ( T ) . This yields that

L p n − 1 0 a m 0 0 , 1 A 0 0 0 , … , 1 A 0 0 0 = p n g A ( a ) 0 0 h A ( a ) , 0 m 0 0 , 1 A 0 0 0 , … , 1 A 0 0 0 + p n a 0 0 0 , 0 f M ( m ) 0 0 , 1 A 0 0 0 , … , 1 A 0 0 0 .

Hence, f M ( a m ) = g A ( a ) m − m h A ( a ) + a f M ( m ) for all a ∈ A , m ∈ M . Likewise, if we take

x 1 = 0 0 0 b , x 2 = 0 m 0 0 , x 3 = … = x n = 1 A 0 0 0

in (2.2), then we obtain f M ( m b ) = m g B ( b ) − h B ( b ) m + f M ( m ) b for all m ∈ M and b ∈ B .□

Suppose that A is an associative algebra such that for each a ∈ A ,

[ a , A ] ∈ Z ( A ) ⇒ a ∈ Z ( A ) ,

or equivalently,

(2.3) [ [ a , A ] , A ] = 0 ⇒ [ a , A ] = 0 .

Note that (2.3) is equivalent to saying that there are no nonzero central inner derivations on A . By [4, Lemma 5.8], we obtain that h A ( A ) ⊆ Z ( B ) and h B ( B ) ⊆ Z ( A ) if A or B satisfies (2.3) or n = 2 .

By [4, Lemma 3.1], it suffices to consider only those Lie n -derivations L : T → T satisfying e L ( f ) f = 0 .

Theorem 2.2

Let T = T r i ( A , M , B ) be a ( n − 1 ) -torsion free triangular algebra and L be a Lie n-derivation on T. If A or B satisfies (2.3) or n = 2 , then there exists a triangular algebra T ∘ such that T is a subalgebra of  T ∘ having the same unity and L can be written as L = δ + ϕ , where δ : T → T ∘ is a derivation and  ϕ : T → Z ( T ∘ ) is a linear mapping such that ϕ ( p n ( x 1 , x 2 , … , x n ) ) = 0 for all x 1 , x 2 , … , x n ∈ T .

Proof

Let ϕ l : C l ( T ) e → C l ( T ) f be an algebra isomorphism satisfying λ e ⋅ e x f = e x f ⋅ ϕ l ( λ e ) for all x ∈ T , λ ∈ C l ( T ) and ϕ r : C r ( T ) e → C r ( T ) f be an algebra isomorphism satisfying λ e ⋅ e x f = e x f ⋅ ϕ r ( λ e ) for all x ∈ T , λ ∈ C r ( T ) . Then, we obtain from [12, Theorem 2.1] that T ∘ = A ϕ r − 1 ( Z ( B ) ) M 0 B ϕ l ( Z ( A ) ) is a triangular algebra such that T is a subalgebra of T ∘ having the same unity. By Theorem 2.1, we see that L has the form:

L a m 0 b = g A ( a ) + h B ( b ) f M ( m ) 0 h A ( a ) + g B ( b ) ,

where a ∈ A , m ∈ M , b ∈ B , and g A : A → A , g B : B → B , h A : A → Z ( B ) , h B : B → Z ( A ) , f M : M → M are the linear mappings satisfying

  1. g A is a Lie n -derivation on A , h A ( p n ( a 1 , a 2 , … , a n ) ) = 0 , p n ( h A ( a ) , b 1 , … , b n − 1 ) = 0 , and f M ( a m ) = g A ( a ) m − m h A ( a ) + a f M ( m ) ;

  2. g B is a Lie n -derivation on B , h B ( p n ( b 1 , b 2 , … , b n ) ) = 0 , p n ( h B ( b ) , a 1 , … , a n − 1 ) = 0 , and f M ( m b ) = m g B ( b ) − h B ( b ) m + f M ( m ) b .

It follows from [12, Proposition 2.2] that Z ( B ) ⊆ C r ( T ) f . This yields that

(2.4) f M ( a m ) = g A ( a ) m − ϕ r − 1 ( h A ( a ) ) m + a f M ( m ) ,

for all a ∈ A and m ∈ M . Similarly, we obtain from [12, Proposition 2.1] that

(2.5) f M ( m b ) = m g B ( b ) − m ϕ l ( h B ( b ) ) + f M ( m ) b ,

for all b ∈ B and m ∈ M . Let us define δ , ϕ : T → T ∘ by:

δ a m 0 b = g A ( a ) − ϕ r − 1 ( h A ( a ) ) f M ( m ) 0 g B ( b ) − ϕ l ( h B ( b ) ) , ϕ a m 0 b = ϕ r − 1 ( h A ( a ) ) + h B ( b ) 0 0 h A ( a ) + ϕ l ( h B ( b ) ) .

It is easy to check that δ and ϕ are the linear mappings and L = δ + ϕ .

Set p A = g A − ϕ r − 1 ∘ h A and p B = g B − ϕ l ∘ h B . It is clear that p A : A → A ϕ r − 1 ( Z ( B ) ) and p B : B → B ϕ l ( Z ( A ) ) are the linear mappings. We claim that p A and p B are derivations. In fact, we obtain from (2.4) that

f M ( a a ′ m ) = p A ( a a ′ ) m + a a ′ f M ( m ) ,

for all a , a ′ ∈ A and m ∈ M . On the other hand, we obtain from (2.4) that

f M ( a a ′ m ) = p A ( a ) a ′ m + a f M ( a ′ m ) = p A ( a ) a ′ m + a p A ( a ′ ) m + a a ′ f M ( m ) ,

for all a , a ′ ∈ A and m ∈ M . Comparing the aforementioned two relations gives

p A ( a a ′ ) m = p A ( a ) a ′ m + a p A ( a ′ ) m ,

for all a , a ′ ∈ A and m ∈ M . Since M is faithful as a left A ϕ r − 1 ( Z ( B ) ) -module, it follows that

p A ( a a ′ ) = p A ( a ) a ′ + a p A ( a ′ ) ,

for all a , a ′ ∈ A . Hence, p A is a derivation. In a similar manner, we can obtain from (2.5) that p B is a derivation.

Applying [12, Lemma 3.1] yields that δ : T → T ∘ is a derivation. We now assert that ϕ ( T ) ⊆ Z ( T ∘ ) . Indeed, we have

( ϕ r − 1 ( h A ( a ) ) + h B ( b ) ) m = ϕ r − 1 ( h A ( a ) ) m + h B ( b ) m = m h A ( a ) + m ϕ l ( h B ( b ) ) = m ( h A ( a ) + ϕ l ( h B ( b ) ) ) ,

for all m ∈ M . In view of [12, Proposition 2.1], we conclude that ϕ ( T ) ⊆ Z ( T ∘ ) .

It remain to prove that ϕ ( p n ( x 1 , x 2 , … , x n ) ) = 0 . Note that L = δ + ϕ is a Lie n -derivation. Therefore,

L ( p n ( x 1 , x 2 , … , x n ) ) = p n − 1 ( [ L ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , L ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , L ( x i ) , x i + 1 , … , x n ) = p n − 1 ( [ δ ( x 1 ) + ϕ ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , δ ( x 2 ) + ϕ ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , δ ( x i ) + ϕ ( x i ) , x i + 1 , … , x n ) = p n − 1 ( [ δ ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , δ ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , δ ( x i ) , x i + 1 , … , x n ) = δ ( p n ( x 1 , x 2 , … , x n ) ) = L ( p n ( x 1 , x 2 , … , x n ) ) − ϕ ( p n ( x 1 , x 2 , … , x n ) ) ,

for all x 1 , … , x n ∈ T . It follows that ϕ ( p n ( x 1 , x 2 , … , x n ) ) = 0 for all x 1 , … , x n ∈ T .□

3 Generalized Lie n -derivations of arbitrary triangular algebras

Theorem 3.1

Let T be a ( n − 1 ) -torsion free triangular algebra and G : T → T be a generalized Lie n-derivation such that e G ( f ) f = 0 . Then, G is of the form:

G a m 0 b = G A ( a ) + H B ( b ) F M ( m ) 0 G B ( b ) + H A ( a ) ,

where a ∈ A , m ∈ M , b ∈ B , and G A : A → A , G B : B → B , H A : A → B , H B : B → A , F M : M → M are the linear mappings satisfying

  1. G A is a generalized Lie n-derivation on A , H A ( p n ( a 1 , a 2 , … , a n ) ) = 0 , F M ( a m ) = G A ( a ) m − m H A ( a ) + a f M ( m ) , and p n ( H A ( a ) , b 1 , … , b n − 1 ) = 0 , where a , a 1 , … , a n ∈ A , b 1 , … , b n − 1 ∈ B , m ∈ M , and f M : M → M is a linear mapping;

  2. G B is a generalized Lie n-derivation on B, H B ( p n ( b 1 , b 2 , … , b n ) ) = 0 , F M ( m b ) = m G B ( b ) − H B ( b ) m + f M ( m ) b = F M ( m ) b + m g B ( b ) − h B ( b ) m , and p n ( H B ( b ) , a 1 , … , a n − 1 ) = 0 , where b , b 1 , … , b n ∈ B , a 1 , … , a n − 1 ∈ A , m ∈ M and f M : M → M , g B : B → B , h B : B → A are the linear mappings.

Proof

Let us assume that

G a m 0 b = G A ( a ) + H B ( b ) + K A ( m ) F A ( a ) + F B ( b ) + F M ( m ) 0 G B ( b ) + H A ( a ) + K B ( m ) ,

where a ∈ A , m ∈ M , b ∈ B , and G A : A → A , G B : B → B , H A : A → B , H B : B → A , F A : A → M , F B : B → M , F M : M → M , K A : M → A , and K B : M → B are the linear mappings. It follows from [7, Lemma 3.4] and [7, Lemma 3.5] that K A ( m ) = K B ( m ) = F A ( a ) = F B ( b ) = 0 . Since G is a generalized Lie n -derivation on T , we obtain

(3.1) G ( p n ( x 1 , x 2 , … , x n ) ) = p n − 1 ( [ G ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , L ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , L ( x i ) , x i + 1 , … , x n ) ,

for all x 1 , x 2 , … , x n ∈ T , where L is the associated Lie n -derivation of G and has Form (2.1). Let us choose

x 1 = a 1 0 0 0 , x 2 = a 2 0 0 0 , … , x n = a n 0 0 0

in (3.1). Then, we can easily obtain that G A is a generalized Lie n -derivation on A and H A ( p n ( a 1 , a 2 , … , a n ) ) = 0 for all a 1 , a 2 , … , a n ∈ A . In a similar way, we can obtain that G B is a generalized Lie n -derivation on B and H B ( p n ( b 1 , b 2 , … , b n ) ) = 0 for all b 1 , b 2 , … , b n ∈ B .

Let us choose

x 1 = a 1 0 0 0 , x 2 = 0 0 0 b 1 , x 3 = a 2 0 0 b 2 , … , x n = a n − 1 0 0 b n − 1

in (3.1). Then,

0 = p n G A ( a 1 ) 0 0 H A ( a 1 ) , 0 0 0 b 1 , a 2 0 0 b 2 , … , a n − 1 0 0 b n − 1 + p n a 1 0 0 0 , h B ( b 1 ) 0 0 g B ( b 1 ) , a 2 0 0 b 2 , … , a n − 1 0 0 b n − 1 .

This yields that p n ( H A ( a 1 ) , b 1 , … , b n − 1 ) = 0 for all a 1 ∈ A , b 1 , b 2 , … , b n − 1 ∈ B . Similarly, we obtain that p n ( H B ( b 1 ) , a 1 , … , a n − 1 ) = 0 for all b 1 ∈ B , a 1 , a 2 , … , a n − 1 ∈ A .

Setting

x 1 = a 0 0 0 , x 2 = 0 m 0 0 , x 3 = … = x n = 0 0 0 1 B

in (3.1), we obtain from [4, Lemma 3.3] that

G p n − 1 0 a m 0 0 , 0 0 0 1 B , … , 0 0 0 1 B = p n G A ( a ) 0 0 H A ( a ) , 0 m 0 0 , 0 0 0 1 B , … , 0 0 0 1 B + p n a 0 0 0 , 0 f M ( m ) 0 0 , 0 0 0 1 B , … , 0 0 0 1 B .

Thus, F M ( a m ) = G A ( a ) m − m H A ( a ) + a f M ( m ) for all a ∈ A , m ∈ M . Likewise, if we choose

x 1 = 0 0 0 b , x 2 = 0 m 0 0 , x 3 = … = x n = 0 0 0 1 B

in (3.1), then F M ( m b ) = m G B ( b ) − H B ( b ) m + f M ( m ) b . Furthermore, if we take

x 1 = 0 m 0 0 , x 2 = 0 0 0 b , x 3 = … = x n = 0 0 0 1 B

in (3.1), then

G p n − 1 0 m b 0 0 , 0 0 0 1 B , … , 0 0 0 1 B = p n 0 F M ( m ) 0 0 , 0 0 0 b , 0 0 0 1 B , … , 0 0 0 1 B + p n 0 m 0 0 , h B ( b ) 0 0 g B ( b ) , 0 0 0 1 B , … , 0 0 0 1 B .

Therefore, F M ( m b ) = F M ( m ) b + m g B ( b ) − h B ( b ) m for all m ∈ M , b ∈ B .□

Lemma 3.2

If A and B satisfy (2.3) or n = 2 , then H B ( B ) ⊆ Z ( A ) and H A ( A ) ⊆ Z ( B ) .

Proof

Let us take

x 1 = a 0 0 0 , x 2 = 0 0 0 b , x 3 = 0 m 0 0 , x 4 = … = x n = 0 0 0 1 B

in (3.1). Then,

(3.2) 0 = G p n a 0 0 0 , 0 0 0 b , 0 m 0 0 , 0 0 0 1 B , … , 0 0 0 1 B = p n − 1 G A ( a ) 0 0 H A ( a ) , 0 0 0 b , 0 m 0 0 , 0 0 0 1 B , … , 0 0 0 1 B + p n − 1 a 0 0 0 , h B ( b ) 0 0 g B ( b ) , 0 m 0 0 , 0 0 0 1 B , … , 0 0 0 1 B ,

for all a ∈ A , b ∈ B , and m ∈ M . Thus,

[ [ H A ( a ) , b ] + [ a , h B ( b ) ] , m ] = 0 ,

for all a ∈ A , b ∈ B , m ∈ M . According to [4, Remark 2.2], we obtain that [ H A ( a ) , b ] ∈ Z ( B ) . It follows from (2.3) that H A ( A ) ⊆ Z ( B ) . Similarly, we have that H B ( B ) ⊆ Z ( A ) .

Now, suppose that n = 2 . It follows from (3.2) that [ H A ( a ) , b ] = 0 for all a ∈ A and b ∈ B . This yields that H A ( A ) ⊆ Z ( B ) . In a similar manner, we obtain that H B ( B ) ⊆ Z ( A ) .□

According to [7, Lemma 3.1], we only consider those generalized Lie n -derivations G : T → T satisfying e G ( f ) f = 0 .

Theorem 3.3

Let T = T r i ( A , M , B ) be a ( n − 1 ) -torsion free triangular algebra and G be a generalized Lie n-derivation on T . If A and B satisfy (2.3) or n = 2 , then there exists a triangular algebra T ∘ such that T is a subalgebra of T ∘ having the same unity and G can be written as G = Δ + Φ , where Δ : T → T ∘ is a generalized derivation and Φ : T → Z ( T ∘ ) is a linear mapping such that Φ ( p n ( x 1 , x 2 , … , x n ) ) = 0 for all x 1 , x 2 , … , x n ∈ T .

Proof

By Theorem 3.1 and Lemma 3.2, we have

G a m 0 b = G A ( a ) + H B ( b ) F M ( m ) G B ( b ) + H A ( a ) ,

where a ∈ A , m ∈ M , b ∈ B , and G A : A → A , G B : B → B , H A : A → Z ( B ) , H B : B → Z ( A ) , F M : M → M are the linear mappings satisfying

  1. G A is a generalized Lie n -derivation on A , H A ( p n ( a 1 , a 2 , … , a n ) ) = 0 , F M ( a m ) = G A ( a ) m − m H A ( a ) + a f M ( m ) , and p n ( H A ( a ) , b 1 , … , b n − 1 ) = 0 ;

  2. G B is a generalized Lie n -derivation on B , H B ( p n ( b 1 , b 2 , … , b n ) ) = 0 , F M ( m b ) = m G B ( b ) − H B ( b ) m + f M ( m ) b = F M ( m ) b + m g B ( b ) − h B ( b ) m , and p n ( H B ( b ) , a 1 , … , a n − 1 ) = 0 .

Suppose that L is the associated Lie n -derivation of G . Then, it follows from [7, Lemma 3.2] that L has the form (2.1) and there exist a derivation δ : T → T ∘ and a linear mapping ϕ : T → Z ( T ∘ ) such that L = δ + ϕ , where δ a m 0 b = p A ( a ) f M ( m ) 0 p B ( b ) is as stated in Theorem 2.2.

According to [12, Proposition 2.2] and [12, Proposition 2.1], we arrive at

(3.3) F M ( a m ) = ( G A ( a ) − ϕ r − 1 ( H A ( a ) ) ) m + a f M ( m ) ,

(3.4) F M ( m b ) = m ( G B ( b ) − ϕ l ( H B ( b ) ) ) + f M ( m ) b ,

(3.5) F M ( m b ) = F M ( m ) b + m ( g B ( b ) − ϕ l ( h B ( b ) ) ) ,

for all a ∈ A , b ∈ B , and m ∈ M . Define the mappings Δ , Φ : T → T ∘ by:

Δ a m 0 b = G A ( a ) − ϕ r − 1 ( H A ( a ) ) F M ( m ) 0 G B ( b ) − ϕ l ( H B ( b ) ) ,

Φ a m 0 b = ϕ r − 1 ( H A ( a ) ) + H B ( b ) 0 0 H A ( a ) + ϕ l ( H B ( b ) ) ,

for all a ∈ A , b ∈ B , and m ∈ M . Clearly, Δ and Φ are the linear mappings and G = Δ + Φ . Let us set P A = G A − ϕ r − 1 ∘ H A and P B = G B − ϕ l ∘ H B . It is easy to verify that P A : A → A ϕ r − 1 ( Z ( B ) ) and P B : B → B ϕ l ( Z ( A ) ) are the linear mappings. We assert that P A and P B are the generalized derivations with the associated derivations p A and p B , respectively. In fact, we obtain from (3.3) that

F M ( a a ′ m ) = P A ( a a ′ ) m + a a ′ f M ( m ) ,

for all a , a ′ ∈ A , and m ∈ M . On the other hand, we obtain from (3.3) that

F M ( a a ′ m ) = P A ( a ) a ′ m + a f M ( a ′ m ) = P A ( a ) a ′ m + a p A ( a ′ ) m + a a ′ f M ( m ) ,

for all a , a ′ ∈ A , and m ∈ M . Comparing the two relations, we obtain

( P A ( a a ′ ) − P A ( a ) a ′ − a p A ( a ′ ) ) m = 0 ,

for all a , a ′ ∈ A , and m ∈ M . By the loyalty of M , it follows that P A ( a a ′ ) = P A ( a ) a ′ + a p A ( a ′ ) for all a , a ′ ∈ A . That is, P A is a generalized derivation with the associated derivation p A . Similarly, on the one hand, we obtain from (3.4) that

F M ( m b b ′ ) = m P B ( b b ′ ) + f M ( m ) b b ′ ,

for all b , b ′ ∈ B , and m ∈ M . On the other hand, we obtain from (3.4) and (3.5) that

F M ( m b b ′ ) = F M ( m b ) b ′ + m b p B ( b ′ ) = m P B ( b ) b ′ + f M ( m ) b b ′ + m b p B ( b ′ ) ,

for all b , b ′ ∈ B , and m ∈ M . Comparing the aforementioned identities, we obtain that P B is a generalized derivation with an associated derivation p B . It follows from [13, Lemma 4.1] that Δ : T → T ∘ is a generalized derivation with the associated derivation δ .

We claim that Φ ( T ) ⊆ Z ( T ∘ ) and Φ ( p n ( x 1 , x 2 , … , x n ) ) = 0 for all x 1 , x 2 , … , x n ∈ T . Indeed,

( ϕ r − 1 ( H A ( a ) ) + H B ( b ) ) m = ϕ r − 1 ( H A ( a ) ) m + H B ( b ) m = m H A ( a ) + m ϕ l ( H B ( b ) ) = m ( H A ( a ) + ϕ l ( H B ( b ) ) ) ,

for all a ∈ A , b ∈ B and m ∈ M . Thus, Φ ( T ) ⊆ Z ( T ∘ ) . Using the facts G = Δ + Φ and L = δ + ϕ , we have

G ( p n ( x 1 , x 2 , … , x n ) ) = p n − 1 ( [ G ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , L ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , L ( x i ) , x i + 1 , … , x n ) = p n − 1 ( [ Δ ( x 1 ) + Φ ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , δ ( x 2 ) + ϕ ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , δ ( x i ) + ϕ ( x i ) , x i + 1 , … , x n ) = p n − 1 ( [ Δ ( x 1 ) , x 2 ] , x 3 , … , x n ) + p n − 1 ( [ x 1 , δ ( x 2 ) ] , x 3 , … , x n ) + ∑ i = 3 n p n − 1 ( [ x 1 , x 2 ] , x 3 , … , x i − 1 , δ ( x i ) , x i + 1 , … , x n ) = Δ ( p n ( x 1 , x 2 , … , x n ) ) = G ( p n ( x 1 , x 2 , … , x n ) ) − Φ ( p n ( x 1 , x 2 , … , x n ) ) ,

for all x 1 , … , x n ∈ T . This implies that Φ ( p n ( x 1 , … , x n ) ) = 0 for all x 1 , … , x n ∈ T .□

  1. Funding information: This study was supported by the Jilin Science and Technology Department (No. YDZJ202201ZYTS622) and the project of Jilin Education Department (No. JJKH20220422KJ).

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors state no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2022-10-05
Revised: 2023-08-10
Accepted: 2023-09-24
Published Online: 2023-10-19

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2023-0135
Keywords for this article
triangular algebra; maximal left ring of quotients; Lie n-derivation; generalized Lie n-derivation
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  1. Special Issue on Future Directions of Further Developments in Mathematics
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